Your SlideShare is downloading.
×

×

Saving this for later?
Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.

Text the download link to your phone

Standard text messaging rates apply

Like this document? Why not share!

- Chapter5 Jeopardy Ac by kwest 3820 views
- Day 4 writing linear equations by Erik Tjersland 171 views
- Algebra 1 Warm-ups by Yvonne Mafunga 2265 views
- Naveed a. sherwani algorithms for v... by Souvik Maity 7323 views
- Factory Integration by JohnsonDon 2375 views
- MEMC (WFR) Initiation by zacsadow 4289 views
- Vlsi by soumya968 6158 views
- 2013 Guide to a Contamination Free ... by m0rtsl4k 16252 views
- Resources for algebra goals and obj... by The Math Magazine 167 views
- Sapphire Applications and Market fr... by Yole Developpement 1493 views
- RFID's by PeterSam67 951 views
- High Volume, Low Cost Production of... by PeterSam67 1548 views

Like this? Share it with your network
Share

441

views

views

Published on

No Downloads

Total Views

441

On Slideshare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

5

Comments

0

Likes

1

No embeds

No notes for slide

- 1. These lessons are designed to be copied for the classroom, with both a question and an answer page. If you have questions or would like to submit lessons for future publication, contact Micron’s K-12 Programs. Algebra Analysis and Probability Geometry Measurements Numbers and Operations Accountant Accountant/CPA Architect Area Sales Manager Assembly Operator Bank Teller Benefits Manager BLM Operations Coordinator Bookkeeper Business Manager Cardiovascular Specialist Carpenter Ceramic Tile Setter Chemist Construction Contractor Construction Supervisor Container Supply Dental Assistant Director International Finance Disk Drive Engineer Electrical Contractor Math in the Workplace
- 2. Electrician Engineer - Semiconductor Equipment Engineer Estimator Excavator Facility Representative - Utilities Family Physician Farm Owner/Manager Financial Analyst Financial Manager Financial Services - Vice President Food Broker Forest Mensurationist Geologist Geologist (Mining) GIS Analyst Glass Installer Graphic Designer Home Improvement: Concrete Home Improvement: Remodeling Homemaker: Redecorator Homemaker: Seamstress Lawn Mower Manufacturer Lead Estimator Licensed Practical Nurse
- 3. Lumber Sales Rep Machine Operator Manager - Government Agency Manager - Manufacturing Manager - Produce Department Manager - Property Development Manager - R&D Program Manager - Retail Store Manager - Telemarketing Manufacturing Engineer Manufacturing Technician Medical Technologist Mortgage Loan Officer Nutrition Specialist Office Manager/Bookkeeper Operating Room Nurse Painter Pastor Payroll Clerk Process Engineer Process Engineer - Cost Process Engineer - Design Physician Plumber Police Officer
- 4. Production Planner Production Supervisor Property Manager Purchasing Agent Quality Control Railroad Conductor Real Estate Planner Realtor Registered Dietician Registered Nurse Road Management Technician Sales / Adminstrative Assistant Salesperson - Shoe Department Software Specialist Tax Analyst Tax Specialist Teacher - Elementary Technical Writer Test Engineer Textile Screen Printing Treasury Analyst Veterinarian - Equine Veterinarian - Small Animal
- 5. Measurements: Road Management Technician Ada County Highway District Job Description: Inspects pavement, calculates asphalt estimates and reports inspection data to the ACHD maintenance department. Problem: The Ada County Highway District Pavement Management Technician needs to know how many tons of asphalt will be needed on a section of road. The section of road measures 5,280 feet (1 mile) in length and 26 feet wide. The asphalt needs to be 3 inches in depth. Asphalt weighs 144 tons per 2,000 ft³.
- 6. Measurements: Road Management Technician Ada County Highway District Job Description: Inspects pavement, calculates asphalt estimates and reports inspection data to the ACHD maintenance department. Problem: The Ada County Highway District Pavement Management Technician needs to know how many tons of asphalt will be needed on a section of road. The section of road measures 5,280 feet (1 mile) in length and 26 feet wide. The asphalt needs to be 3 inches in depth. Asphalt weighs 144 tons per 2,000 ft³. Solution: 3 inches x 12 inches = .25 ft or ¼ ft 144 tons 2000 ft³ = .072 tons of asphalt per cubic foot length x width x depth x weight per ft³ = tons of asphalt 5,280 ft x 26 ft x .25 ft x .072 tons = 2,471 tons of asphalt needed
- 7. Numbers and Operations: Assistant Produce Manager Albertsons, Inc. Job Description: Oversee the daily operation of the produce department for Albertsons grocery store. Problem: Find the percent of gross profit on sales of $100,000. The cost of the goods sold is $75,000.
- 8. Numbers and Operations: Assistant Produce Manager Albertsons, Inc. Job Description: Oversee the daily operation of the produce department for Albertsons grocery store. Problem: Find the percent of gross profit on sales of $100,000. The cost of the goods sold is $75,000. Solution: Subtract the cost of the goods sold from the sales and divide the sum by the sales. Sales 100,000 Cost - 75,000 25,000 25,000 ÷ 100,000 = .25 Your percent of gross profit would be 25% on the items you sold.
- 9. Numbers and Operations: Food Broker Albertsons, Inc. Job Description: Brings the seller and the buyer of food together. Problem: What is the retail price at Albertson's grocery store of a package of Gallo Pepperoni? The case cost of the pepperoni is $24 per 12 (8 oz.) packages. The warehouse markup will be 7%. The store markup will be 38%. Store markup is on top of warehouse markup.
- 10. Numbers and Operations: Food Broker Albertsons, Inc. Job Description: Brings the seller and the buyer of food together. Problem: What is the retail price at Albertson's grocery store of a package of Gallo Pepperoni? The case cost of the pepperoni is $24 per 12 (8 oz.) packages. The warehouse markup will be 7%. The store markup will be 38%. Store markup is on top of warehouse markup. Solution: Package cost = $24.00 ÷12 packages = $2.00/package Warehouse markup =$2.00 x .07 = $.14 Albertsons markup =$2.14 (door cost) x .38 = $.81 Albertsons retail price =$2.00 + .14 + .81 = $2.95/package
- 11. Numbers and Operations: Real Estate Planner Albertsons, Inc. Job Description: Responsible for day-to-day operations of commercial properties, provides lease administration for the commercial portfolio, and participates in budgeting and financial reporting to maximize financial returns. Problem: A real estate manager has selected a location for a new grocery store. How would Management determine the total cost of project? The building will be 55,922 square feet at a building cost of $50/sq. ft. The land purchase is 250,000 square feet at a cost of $4.25/sq. ft. However, the company will only use 240,000 square feet of the land (retained land) and the remaining amount will be sold as surplus land at the same rate as purchased. Improvements to the site (site costs) will be $1.25/sq. ft. 250,000 sq. ft. 55,922 sq. ft.
- 12. Numbers and Operations: Real Estate Planner Albertsons, Inc. Job Description: Responsible for day-to-day operations of commercial properties, provides lease administration for the commercial portfolio, and participates in budgeting and financial reporting to maximize financial returns. Problem: A real estate manager has selected a location for a new grocery store. How would Management determine the total cost of project? The building will be 55,922 square feet at a building cost of $50/sq.ft. The land purchase is 250,000 square feet at a cost of $4.25/sq. ft. However, the company will only use 240,000 square feet of the land (retained land) and the remaining amount will be sold as surplus land at the same rate as purchased. Improvements to the site (site costs) will be $1.25/sq. ft. Solution: Store Occupancy Costs Total Cost of Retained Land (ft² ): 240,000 x $4.25 =$1,020,000 Site Costs: 240,000 x $1.25 = $300,000 Building Costs: 55,922 x $50.00 = $2,796,100 Total Building and Site Improvement Costs: 300,000 + 2,796,100 = $3,096,100 Total Cost of Project: 3,096,100 + 1,020,000 = $4,116,100 250,000 sq. ft. 55,922 sq. ft.
- 13. Analysis & Probability/Algebra: Forest Mensurationist Boise Cascade Corporation Job Description: Use mathematics and statistics to inventory forest resources and predict the state of those resources in the future. Problem: In forestry, we can describe the relationship between the age of stand and the yield (timber volume) in that stand with the following "yield equation": y = ke-b⁄A , y = yield, e is the base of natural logarithms, A = stand age, and k and b are constants that depend on the type of stand (species, geography, etc.) The yield equation can be used to draw a yield curve, which looks like this: If the yield at any age is divided by the age, we can derive the Mean Annual Increment (MAI), a measure of the rate of growth for that stand up to that age. The age at which MAI is maximized is known as the "rotation age", the age at which the stand should be harvested, and then begin a new stand (a new rotation). Here is the problem, which can be solved with calculus: Find the age at which MAI is maximized or use the definition of MAI to develop an expression for the rotation age. [Hint: the yield equation shown above is popular because the rotation age = b; so the rotation age can be read directly from the yield equation]
- 14. Analysis & Probability/Algebra: Forest Mensurationist Boise Cascade Corporation Job Description: Use mathematics and statistics to inventory forest resources and predict the state of those resources in the future. See problem for details. Solution: Use calculus to show the age at which MAI is maximized is equal to b.
- 15. Algebra: Treasury Analyst Boise Cascade Forest Products Job Description: Has direct responsibility for managing corporate short-term borrowings and investments. Problem: The company has a debt payment due on May 1st in the amount of $5,000,000. It currently has $5,000,000 available to meet the obligation; but is not allowed to make the payment until the due date, May 1st. The company would like to purchase an investment, which at maturity would have a value equal to the amount of the debt payment-thus allowing the excess cash to be used elsewhere. Using the following assumptions, determine the price of the required investment and the amount of excess cash they are holding. Assumptions Debt payment required: $5,000,000 (P) Cash on hand: $5,000,000 Interest rate (or discount rate) on investment: 6% (IR) Days until debt payment date: 63 (DM) Dollar Price: amount required to invest today to meet debt obligation in 63 days (DP) A year consist of 360 days for interest accrual formula: P x (1 - (IR· x DM ÷ 360)) = DP
- 16. Algebra: Treasury Analyst Boise Cascade Forest Products Job Description: Has direct responsibility for managing corporate short-term borrowings and investments. Problem: The company has a debt payment due on May 1st in the amount of $5,000,000. It currently has $5,000,000 available to meet the obligation; but is not allowed to make the payment until the due date, May 1st. The company would like to purchase an investment, which at maturity would have a value equal to the amount of the debt payment-thus allowing the excess cash to be used elsewhere. Using the following assumptions, determine the price of the required investment and the amount of excess cash they are holding. Assumptions Debt payment required: $5,000,000 (P) Cash on hand: $5,000,000 Interest rate (or discount rate) on investment: 6% (IR) Days until debt payment date: 63 (DM) Dollar Price: amount required to invest today to meet debt obligation in 63 days (DP) A year consists of 360 days for interest accrual formula: P x (1 - (IR x DM ÷ 360)) = DP Solution: P x (1-(IR ÷ DM360)) = DP $5,000,000 x (1-(.06 x 63 ÷ 360)) = DP $5,000,000 x (1-(.0105)) = DP $5,000,000 x (.9895) = DP $4,947,500 = DP Cash available for other purposes = $52,500.00 ($5,000,000 - $4,947,500) Financial instruments such as this one are referred to as "discounted" and include Banker's Acceptances, Commercial Paper, Treasury Bills, and secondary Discount Notes.
- 17. Numbers and Operations: Full Charge Bookkeeper Boise Hardware, Inc. Job Description: Responsible for accounts receivable, accounts payable, payroll, general ledger, inventory maintenance. Problem: All employees have to have Federal Social Security (FICA) and Medicare taxes taken from their gross wages. Figure the FICA and Medicare taxes for the following employee's gross wages and for the total gross wages. Gross wages is the total earned before any deductions. FICA is 6.2% of the gross wages. Medicare is 1.45% of the gross wages. employee a $ 540.00 employee b $ 870.00 employee c $ 250.00 employee d $ 125.00 employee e $1,600.00
- 18. Numbers and Operations: Full Charge Bookkeeper Boise Hardware, Inc. Job Description: Responsible for accounts receivable, accounts payable, payroll, general ledger, inventory maintenance. Problem: All employees have to have Federal Social Security (FICA) and Medicare taxes taken from their gross wages. Figure the FICA and Medicare taxes for the following employee's gross wages and for the total gross wages. Gross wages is the total earned before any deductions. FICA is 6.2% of the gross wages. Medicare is 1.45% of the gross wages. employee a $ 540.00 employee b $ 870.00 employee c $ 250.00 employee d $ 125.00 employee e $1,600.00 Solution: employee a $ 540.00 employee b $ 870.00 employee c $ 250.00 employee d $ 125.00 employee e +$1,600.00 Total Gross Wages $3,385.00 FICA = 6.2% of $3,385 = $209.87 Medicare = 1.45% of $3,385 = $49.08 Employee FICA Medicare a 6.2% of $540 = $33.48 1.45% of $540 = $7.83 b 6.2% of $870 = $53.94 1.45% of $870 = $12.62 c 6.2% of $250 = $15.50 1.45% of $250 = $3.63 d 6.2% of $125 = $ 7.75 1.45% of $125 = $1.81 e 6.2% of $1600 = $99.20 1.45% of $1600 = $23.20 Total $209.87 $49.09
- 19. Analysis & Probability/Algebra: Cardiovascular Specialist Boise Heart Clinic Job Description: Active cath lab and cardiovascular surgeon at the hospitals and clinical physician of primary or secondary heart disease at Boise Heart Clinic, P.A. Problem: #1 A man comes to the hospital and needs a medication based on his body size. He weighs 170 lbs. The Doctor prescribes Dobutamine, which stimulates the heart to increase the number of contractions and increases strength of contraction of heart muscle to pump blood more effectively. He needs to be given 10 mg/kg every minute of this medication intravenously. If we put 500 mg of Dobutamine in one liter of normal saline, how much solution do we need to give per hour? #2 A 59-year old man comes to the Boise Heart Clinic short of breath. Dr. Lee hears a heart murmur and suspects the patient's aortic valve is tight. He orders an echocardiogram (ultrasound of the heart) to find the valve area of the aortic valve. If it is too tight, he will have surgery to replace the valve. You must calculate the aortic valve area: CSA = cross sectional area CSA=πr² AVA = aortic valve area LVOT x CSA/AO LVOT = minimal flow (before blood goes through the valve) AO = maximum flow (after blood goes through the valve) π = 3.14 LVOT = 1.0 m/sec AO = 5.8 m/sec LVOT Diameter = 2.5 cm
- 20. Analysis & Probability/Algebra: Cardiovascular Specialist Boise Heart Clinic Job Description: Active cath lab and cardiovascular surgeon at the hospitals and clinical physician of primary or secondary heart disease at Boise Heart Clinic, P.A. See problem for details. Solution: #1 (Include diagrams as needed) How much Dobutamine does this man need per min? 2.2 lbs = 1 kg 170 lbs ÷ 22 lb/kg = 77.3 kg 77.3 kg x (10 mg/kg)/min = 773 mg/min How many liters per min? (1,000 ml ÷ 500 mg) x 773 mg/min = 155 ml/min = 1.55 liters/min How many liters per hour? 1.55 l/min x 60 min/hr = 93 l/hr #2 (Include diagrams as needed) CSA = πr² = 3.14(2.5cm ÷ 2)² = 4.9cm² r = LVOT DIAM ÷ 2 AVA = LVOT x CSA ÷ AO AVA = (1.0m/sec)(4.9cm) ÷ 5.8m/sec = .85 cm² Aortic valve area ranges: Normal - 2.6 cm² to 1.6 cm² Mild - 1.6 cm² to 1.2 cm² Moderate - 1.2 cm² to 1.0 cm² Severe - 1.0 cm² to 0.8 cm² Critical - less than 0.75 cm² This patient has now been diagnosed with severe aortic stenosis. He will have echocardiograms to monitor the tightness of the valve. He will probably have a valve replacement in the near future.
- 21. Numbers and Operations: Registered Dietitian Boise Heart Clinic Job Description: Provide Medical Nutrition Therapy for hyperlipidemia (high cholesterol and high triglycerides), hypertension, diabetes and obesity to lower individual's risk of primary or secondary heart disease. Problem: #1 Reading Food Labels This label came from a can of soup (see label information at bottom of page). What are percentages of calories from fat, saturated fat, total carbohydrate, sugars, and protein? Saturated fats have the same calories per gram as total fat. Sugars have the same calories per gram as total carbohydrates. #2 Reading Food Labels If Americans are encouraged to eat no more than 40% of their calories from total fat, and no more than 10% of their calories from saturated fat, and about 50% of their calories from total carbohydrates (not too high, nor too low), how would you judge this product?
- 22. Numbers and Operations: Registered Dietitian Boise Heart Clinic Job Description: Provide Medical Nutrition Therapy for hyperlipidemia (high cholesterol and high triglycerides), hypertension, diabetes and obesity to lower individual's risk of primary or secondary heart disease. See problem for details. Solution: #1 13 grams fat x 9 cal/gram = 117 fat calories 5 grams saturated fat x 9 cal/gram = 45 saturated fat calories 31 grams carbohydrates x 4 cal/gram = 124 carbohydrate calories 5 grams sugar x 4 cal/gram = 20 sugar calories 5 grams protein x 4 cal/gram = 20 protein calories Percent (%) of Calories from: 117 ÷ 260 = 45% fat 45 ÷ 260 = 17% saturated fat 124 ÷ 260 = 48% carbohydrates 20 ÷ 260 = 8% sugar 20 ÷ 260 = 8% protein #2 Goal Actual Rate =< 40% 45% total fat: very high "not at goal" =< 10% 17% saturated fat: very high "not at goal" 50% 48% total carbohydrates: good "at goal"
- 23. Analysis & Probability/Algebra: Registered Dietitian Boise Heart Clinic Job Description: Provide Medical Nutrition Therapy for hyperlipidemia (highcholesterol and high triglycerides), hypertension, diabetes and obesity to lower individual's risk of primary or secondary heart disease. Problem: Leona is a 75-year-old woman with high blood pressure, coronary artery disease, and adult-onset diabetes. Lab values on 11-11-97 showed the following: LDL - 136 HDL - 28 Triglycerides - 326 Blood sugar - 239 Total cholesterol equals HDL (high density lipids), plus LDL (low density lipids), plus VLDL (very low density lipids), and VLDL equals 1/5 of Triglycerides. Total Cholesterol = HDL + LDL + VLDL. 1. What are Leona's total cholesterol level, VLDL levels and total cholesterol/HDL ratio? (Round to the nearest tenth) The risk ratio for heart disease is a ratio between the total cholesterol and HDL Levels. What is Leona's risk ration? (Round to the nearest tenth) Circle her results as desirable or high risk on the supplemental form. How likely is Leona to have a heart attack?
- 24. Supplemental Form National Cholesterol Education Program's Guidelines Desirable Borderline High Total Cholesterol "DIET: low fat/high fiber" Less than 200 mg/dl 200-239 mg/dl 240 mg/dl or higher HDL Cholesterol (good cholesterol) "DIET: weight loss/exercise" 45 mg/dl or higher 35-44 mg/dl Less than 35 mg/dl LDL Cholesterol (bad cholesterol) "DIET: low fat/high fiber" No Coronary Heart Disease Less than 130 mg/dl 130-159 mg/dl 160 mg/dl or higher With Coronary Heart Disease Less than 100 mg/dl 100-129 mg/dl 130 mg/dl or higher Triglycerides "DIET: weight loss/exercise" Low refined sugars, low alcohol intake Less than 165 mg/dl 165-249 mg/dl 250 mg/dl or higher Risk Ratio (Total Cholesterol divided by HDL) Less than 4.5 4.5 More than 4.5 Fasting Glucose (Blood Sugar) "DIET: Weight control, exercise, low refined sugars, balanced meal times" Less than 120 mg/dl 120-139 mg/dl 140 mg/dl or higher HDL, cholesterol, and your heart attack risk Instructions: To help you track your progress and compare your risk with others, plot your lipid levels on this graph after each blood test. With a pencil, mark each axis at the points corresponding to your high-density lipoprotein (HDL) and total cholesterol levels. Then draw an imaginary line from each mark, perpendicular to the axis, and mark the point on the chart where both lines intersect. Write the date beneath the mark.
- 25. Your triglyceride level is: 326 Chart your risk reduction over time DATE HDL TOTAL TRIGS VISIT 1 11-11-97 28 229 326 VISIT 2 VISIT 3 VISIT 4
- 26. Analysis & Probability/Algebra: Registered Dietitian Boise Heart Clinic Job Description: Provide Medical Nutrition Therapy for hyperlipidemia (highcholesterol and high triglycerides), hypertension, diabetes and obesity to lower individual's risk of primary or secondary heart disease. See problem for details. Solution: #1 (Use diagrams on supplemental form as needed) Triglycerides VLDL = 326 ÷ 5 = 65.2 VLDL VLDL + HDL + LDL = Total Cholesterol 65.2 + 28 + 136 = 229 Formula: Total cholesterol ÷ HDL = Risk ratio 229 ÷ 28 = 8.2 ration #2 Leona has the lab values of someone who just had a heart attack. Therefore, if she keeps these values, Leona is VERY likely to have a heart attack.
- 27. Algebra: Police Officer Boise Police Department Job Description: Traffic investigation Problem: Background: A car travels 1.467 feet per second for every 1 mph. The formula for speed at the start of skid (in miles per hour) is S = f = coefficient of friction; d = skid distance Total feet per second (fps) = miles per hour (mph) x 1.467 Total Stopping Distance = (reaction time) x (fps) + skid length Time to impact = d ÷ v Questions: 1. Assuming a reaction time of .75 seconds, how fast was car A traveling at the beginning of its skid? The coefficient of friction (f) on the road is .80. The coefficient of friction is given for different circumstances, such as dry pavement, snow floor, or black ice. 2. What was the total stopping distance of car A? 3. How long did it take car B to turn if driver A reacted immediately when car B began its turn?
- 28. Algebra: Police Officer Boise Police Department Job Description: Traffic investigation See problem page for details. Solution: 1. Speed at the start of skid = Speed = 35 mph at the start of the skid 2. Total stopping distance = reaction time x (speed at start of skid x 1.467) + skid length speed at start of skid = fps = 35 x 1.467 = 51.345 Total stopping distance = .75 x 51.345 ft + 50 foot skid = 88.5 ft Car A was 88–89 feet from point of impact when car B started the left turn. 3. Time = d ÷ v = Total stopping distance ÷ fps Time = 88.5 ft ÷ 51.345 = 1.7236 Time = 1.72 seconds for car B to begin turning and get hit
- 29. Numbers and Operations: Operations Coordinator Bureau of Land Management Job Description: Coordinates logistical support for natural disasters and movement of incident support resources. Problem: The company has been asked to help fight a fire in the Phoenix, Arizona area. How much it will cost to fly a Boeing 787 from the Boise Airport to Phoenix, Arizona to assist in this fire fighting effort? Flight time to Phoenix is 1 hour and 47 minutes. Cost to fly one hour is $3,833.
- 30. Numbers and Operations: Operations Coordinator Bureau of Land Management Job Description: Coordinates logistical support for natural disasters and movement of incident support resources. Problem: The company has been asked to help fight a fire in the Phoenix, Arizona area. How much it will cost to fly a Boeing 787 from the Boise Airport to Phoenix, Arizona to assist in this fire fighting effort? Flight time to Phoenix is 1 hour and 47 minutes. Cost to fly one hour is $3,833. Solution: 1. Convert flight time in minutes to hundredths of an hour. 47 minutes x 1 hour divided by 60 minutes = 0.78 hours + 1 hour = 1.78 hours or... 60 minutes + 47 minutes = 107 minutes 107 minutes x 1.6667 (convert minutes to hundredths) = 1.78 hours 2. Total flight time = 1.78 hr 1.78 hr x $3,833/hour = $6,822.74 (cost to fly)
- 31. Analysis & Probability/Geometry: Geologist (Mining) Bureau of Land Management Job Description: Assessment of mineral resources, oversight of active mining operations, geologic and mineral resource investigations. Problem: Acme Mining Company is considering developing a small phosphate mine in southeastern Idaho. The area has been explored, geology has been mapped, and two drill holes (DH1 and DH2) were drilled. A vertical, east-west oriented cross section of the area has been drawn, as shown below. Note that the phosphate bed (a sedimentary layer) is not flat, but is inclined. The angle of inclination from a horizontal plane is known as the "dip" (d on the cross section). The drill holes were drilled vertically and rock samples were collected. The geologist on the drill rig recognized the change in rock types as the hole was drilled downward. The following drill data is given: Drill hole depth (feet) Rock type DH1 DH2 Surface, alluvium 0 - 3' 0' - 3' Shale 3' - 75' 3' - 15' Phosphate 75' - 105' 15' - 42' Limestone 105' - 135' 42' - 80' Since mining is expensive, it is important to the company to obtain an accurate estimate of phosphate resources in this area to assess the economic feasibility of developing a mine. If the resource is poor, the costs of mining could well exceed the profits generated, thus the company would go broke (and the geologist lose his or her job!). As a first step in estimation of the volume and tonnage of resource, the geologist needs to determine the average true thickness of the phosphate bed. True thickness (t, on the cross section) is measured perpendicular to the phosphate bed. Since the drill holes were drilled vertically (at an angle to the inclined bed), the thickness observed from the drill hole data is not true thickness, but is "apparent" thickness. Assuming the ground level of DH1 is the same level as the highest phosphate level of DH2, determine the dip angle (d) and the average true thickness (t) of the phosphate bed between the two drill holes.
- 32. Analysis & Probability/Geometry: Geologist (Mining) Bureau of Land Management Job Description: Assessment of mineral resources, oversight of active mining operations, geologic and mineral resource investigations. See problem for details. Solution: 1. Determine the dip angle of bed (d) d = angle of true dip (inclination of bed from horizontal) A right triangle is formed as shown, thus trigonometry can be used to solve the problem. tan d = opposite/adjacent = 75 ft./ 150 ft. = .500 Using a calculator to determine angle: tan-1 ≈26.6° 2. Determine the true thickness (t) of phosphate bed (average of "t" from PH1 + PH2). True thickness can also be calculated using trigonometry. A right triangle is formed as shown. For DH1: "Apparent" thickness from drill-data = 105' - 75' = 30' sin 63.4° = opposite / hypotenuse = t / 30' t = 30sin 63.4° = 26.8 feet for t1 For DH2: "Apparent" thickness from drill-data = 42' - 15' = 27' sin 63.4° = t/27' t = 27 sin 63.4° = 24.1 feet for t2 3. Average thickness = (26.8' + 24.1' ) / 2 ≈ 25.5 feet Note: This problem could also be solved graphically as an alternative. The result of 25.5 feet average thickness is significantly different from an average of (30 +27)2 = 28.5 feet that would have been determined from averaging the apparent thickness from drill-hole data. A difference of 3 feet is significant when the volume of the entire resource block is calculated.
- 33. Numbers and Operations: Office Manager/Bookkeeper Computer Arts, Inc. Job Description: General office operations including telephones, ordering supplies, supervising employees, Accounts Receivable bookkeeping, Accounts Payable, and payroll. Problem: Our company has a 401(K) Profit Sharing Plan for employees. Employees may defer up to 15% of their salary to the profit sharing plan. This is a great benefit to employees. Not only is the deferral "pre-tax", the company offers a "match" on the deferral. The 401(K) amount is deducted from the gross payroll before Federal and State taxes are deducted. The company matches $.50 on the dollar up to 3% of salary. Payment for the deferrals and company match must be sent to the Trust each pay period. The names of six employees, their monthly salary, and their chosen deferral percentage are listed below. What is the total check amount to the Trust each pay period? Complete the following table: Employee Salary % Deferral Amount Match* Total Judy $1,000 15% Tom 1,200 10% Joe 2,000 3% Sue 900 5% Sally 1,500 3% John 1,750 7%
- 34. Numbers and Operations: Office Manager/Bookkeeper Computer Arts, Inc. Job Description: General office operations including telephones, ordering supplies, supervising employees, Accounts Receivable bookkeeping, Accounts Payable, and payroll. See problem page for details. Solution: Deferral Amount = Salary x % Deferral Match* =Salary x 3% = Salary x 0.03 Example: Judy invests 15%, which is $1000 x .15 = $150/month. The company matches $.50 on the dollar up to 3%, which is $1000 x 0.03 = $30 x .50 = $15. The company will add $15/month to Judy's investment. Total = Deferral amount plus match amount. For Judy that is $150 + $15 = $165. Employee Payroll % Deferral Amount Match* Total Judy $1,000 15% $150.00 $15.00 $165.00 Tom 1,200 10% 120.00 18.00 138.00 Joe 2,000 3% 60.00 30.00 90.00 Sue 900 5% 45.00 13.50 58.50 Sally 1,500 3% 45.00 22.50 67.50 John 1,750 7% 122.50 26.25 148.75 Total check to trust: $667.75
- 35. Numbers and Operations: Container Supply Container & Packaging Supply, Inc. Job Description: Figures costs, selling prices, manage employees, purchase equipment, arrange for long-term financing and make decisions for the direction of the company. Problem: I buy a product that costs me $10.00. I need to make a 20% gross profit (selling price – cost of goods sold). What should my selling price be? What will my gross profit be?
- 36. Numbers and Operations: Container Supply Container & Packaging Supply, Inc. Job Description: Figures costs, selling prices, manage employees, purchase equipment, arrange for long-term financing and make decisions for the direction of the company. Problem: I buy a product that costs me $10.00. I need to make a 20% gross profit (selling price – cost of goods sold). What should my selling price be? What will my gross profit be? Solution: Selling price (100%) – Cost of Goods (80%) = Gross Profit (20%) To calculate my selling price (x): x - $10 = (.20) x - .2x = $10 .8x = $10 x = $10 ÷ .8 = $12.50 $12.50 - $10.00 = $2.50 gross profit
- 37. Numbers and Operations: Assembly Operator Coretronics, Inc. Job Description: Order and maintain sufficient parts to complete customer orders. Assemble and test print heads. Problem: I buy a product that costs me $10.00. I need to make a 20% gross profit (selling price – cost of goods sold). What should my selling price be? What will my gross profit be?
- 38. Numbers and Operations: Assembly Operator Coretronics, Inc. Job Description: Order and maintain sufficient parts to complete customer orders. Assemble and test print heads. Problem: I buy a product that costs me $10.00. I need to make a 20% gross profit (selling price – cost of goods sold). What should my selling price be? What will my gross profit be? Solution: 1. 173 hammers in supply ÷ 6 hammers per print head = 28.83 print heads Therefore, 28 print heads can be built with the hammers currently stocked. 2. 45 print heads x 6 hammers per print head = 270 total hammers needed for entire order. 270 total hammers needed - 173 hammers in supply = 97 hammers needed to complete the order. This is an actual interview question. About 40% of applicants get it right.
- 39. Numbers and Operations: Machine Operator Coretronics, Inc. Job Description: Grind print heads using surface grinders. Must know Imperial (U.S.) system to Metric system conversion. Problem: Upon measuring the part you are machining, your inspection tool says that an additional 0.45 mm needs to be removed. The machine you are using only has controls calibrated in inches. (25.4 mm = 1 inch) How many inches do you need to remove from the part? Do not use a calculator. Answer is to be given with 4 decimal places.
- 40. Numbers and Operations: Machine Operator Coretronics, Inc. Job Description: Grind print heads using surface grinders. . Must know Imperial (U.S.) system to Metric system conversion. Problem: Upon measuring the part you are machining, your inspection tool says that an additional 0.45 mm needs to be removed. The machine you are using only has controls calibrated in inches. (25.4 mm = 1 inch) How many inches do you need to remove from the part? Do not use a calculator. Answer is to be given with 4 decimal places. Solution: 0.45 mm x (1 inch ÷ 25.4 mm) = 0.45 inches / mm ÷ 25.4 mm = 0.0177 inches This is an actual interview question used. About 40% of applicants get it right.
- 41. Algebra: Production Planner Darigold Inc. Job Description: Plan ice cream production and supervise employees Problem: How much ice cream mix and vanilla flavor will it take to make 1000 gallons of vanilla ice cream at 90% overrun with the vanilla flavor usage rate at 1 oz. per 10 gallon mix? (90% overrun means that enough air is put into the frozen mix to increase its volume by 90%). Note: 1 gallon = 128 fluid oz.
- 42. Algebra: Production Planner Darigold Inc. Job Description: Plan ice cream production and supervise employees Problem: How much ice cream mix and vanilla flavor will it take to make 1000 gallons of vanilla ice cream at 90% overrun with the vanilla flavor usage rate at 1 oz. per 10 gallon mix? (90% overrun means that enough air is put into the frozen mix to increase its volume by 90%). Note: 1 gallon = 128 fluid oz. Solution: Total volume required = 1000 gal ÷ (1 + 0.9) = 526.3 gal Now let x = volume of ice cream mix y = volume of vanilla flavor So x + y = 526.3 but y = (0.00078)(x) - which is negligible So x = 526.3 But since we need 1 oz. of flavoring for 10 gal of mix We get y oz ÷ 526.3 gal = 1 oz ÷ 10 gal Y = 52.6 oz of flavor
- 43. Measurements: Ceramic Tile Setter Davis Tile Job Description: Installs ceramic tile. Problem: Find the total square footage in the countertop illustrated. A = 2' B = 11' C = 3' D = 6'
- 44. Measurements: Ceramic Tile Setter Davis Tile Job Description: Installs ceramic tile. Problem: Find the total square footage in the countertop illustrated. A = 2' B = 11' C = 3' D = 6' Solution: A x (B + A) = E 2 x 13 = 26 ft2 A x A = F 2 x 2 = 4 ft2 A x C = G 2 x 3 = 6 ft2 C x D = H 3 x 6 = 18 ft2 E + F + G + H = Total area of countertop = 54 ft2
- 45. Geometry: Computer Manufacturer – Electrician Electrical Company Problem: When an electrician wires a house, the switch boxes must be properly sized. Each insulated wire (wire size called #14) requires 2.25 cubic inches of space. Any number of bare wires all together only needs another 2.25 cubic inches of space. Each cable contains two (2) insulated and one (1) bare wire. How many cables can be put into a box that is 3 inches deep?
- 46. Geometry: Computer Manufacturer – Electrician Electrical Company Problem: When an electrician wires a house, the switch boxes must be properly sized. Each insulated wire (wire size called #14) requires 2.25 cubic inches of space. Any number of bare wires all together only needs another 2.25 cubic inches of space. Each cable contains two (2) insulated and one (1) bare wire. How many cables can be put into a box that is 3 inches deep? Solution: Box volume is 3" x 3.5" x 2" or 21 in3 Space needed for insulated wires: 21 in3 - 2.25 in3 = 18.75 in3 (all bare wires need only 2.25 in3 ). Space needed for each cable that contains two insulated wires: 2.25 in3 x 2 = 4.50 in3 Maximum number of cables allowed: 18.75 in3 ÷ 4.50 in3 = 4.166… Therefore, 4 cables can be put into the box.
- 47. Measurements: Homemaker – Redecorate Elizabeth Crandlemire Job Description: Handyman, carpenter, plumber, seamstress, cook, maintenance, engineer, mother, etc. Problem: I want to wallpaper a 10' x 12' room that has 8' ceilings. The wallpaper has 44 sq. ft. per double roll. How many double rolls of wallpaper do I need to order? (Wallpaper is not available in single rolls.)
- 48. Measurements: Homemaker – Redecorate Elizabeth Crandlemire Job Description: Handyman, carpenter, plumber, seamstress, cook, maintenance, engineer, mother, etc. Problem: I want to wallpaper a 10' x 12' room that has 8' ceilings. The wallpaper has 44 sq. ft. per double roll. How many double rolls of wallpaper do I need to order? (Wallpaper is not available in single rolls.) Solution: Width x Height ÷ square footage = # of double rolls needed 44 ft x 8 feet ÷ 44 sq ft/roll = 8 rolls ...OR add the area of each wall [2 (10 x 8) + 2 (12 x 8) ] = 352 square feet 352 sq ft ÷ 44 sq ft/roll = 8 double rolls
- 49. Numbers and Operations: Veterinarian Equine Hospital and Lameness Center Job Description: Veterinary medicine and surgery. Problem: A horse weighs 1,200 pounds. He is sick and has been diagnosed with a certain disease. This disease is treated with Drug X. Instructions are to give 3 mg/kg orally twice a day for 5 days. The medicine is provided in 200 mg tablets. 2.2 pounds = 1 kg How many tablets need to be dispensed each day? How many tablets need to be dispensed for the 5 days of treatment?
- 50. Numbers and Operations: Veterinarian Equine Hospital and Lameness Center Job Description: Veterinary medicine and surgery. Problem: A horse weighs 1,200 pounds. He is sick and has been diagnosed with a certain disease. This disease is treated with Drug X. Instructions are to give 3 mg/kg orally twice a day for 5 days. The medicine is provided in 200 mg tablets. 2.2 pounds = 1 kg How many tablets need to be dispensed each day? How many tablets need to be dispensed for the 5 days of treatment? Solution: 1,200 lbs ÷ 2.2 lbs/kg = 545 kgs (weight of horse) 545 kgs x 3 mg/kg = 1,635 mg 1,635 mg x 2 times/day = 3,270 mg per day 3,270 mg/day ÷ 200 mg/pill = 16.35 or 16 pills per day 16.35 pills/day x 5 days = 81.75 pills to be dispensed Open-ended discussion: What would you do about the partial pill?
- 51. Numbers and Operations: Retail Store Manager Fred Meyer Job Description: Retail store operations including pricing control, inventory control, employee supervision, customer service and loss prevention. Budgeting and scheduling are also common duties. Problem: 1. In a retail store, the profit on a product is calculated as a percent of the retail price to the customer. This is called margin. If the store pays $8.00 for a music CD and wants to make a 20% margin, what would the retail sales price need to be? 2. Margin differs from markup. Markup is the percentage of profit based on the cost of the item while margin is based on the sales price of the item. If the store pays $8.00 for a music CD and wishes a 20% markup, what would the sale price be? Generally speaking, businesses use the margin method to calculate the sales prices of their merchandise.
- 52. Numbers and Operations: Retail Store Manager Fred Meyer Job Description: Retail store operations including pricing control, inventory control, employee supervision, customer service and loss prevention. Budgeting and scheduling are also common duties. Problem: 1. In a retail store, the profit on a product is calculated as a percent of the retail price to the customer. This is called margin. If the store pays $8.00 for a music CD and wants to make a 20% margin, what would the retail sales price need to be? 2. Margin differs from markup. Markup is the percentage of profit based on the cost of the item while margin is based on the sales price of the item. If the store pays $8.00 for a music CD and wishes a 20% markup, what would the sale price be? Generally speaking, businesses use the margin method to calculate the sales prices of their merchandise. Solution: 1. Retail price with a 20% margin: Divide the cost of the CD by 100% less the margin (20%) $8.00 /(100% - 20%) = $8.00/.80 = $10.00 retail sales price 2. Retail price with a 20% markup: Multiply the cost of the CD by 100% plus the markup (20%) $8.00 x (100% + 20%) = $8.00 x 1.2 = $9.60 retail sales price
- 53. Numbers and Operations: Church Pastor Grace Bible Church of Boise Job Description: Preach and teach the Bible; care for people's spiritual needs through training and counseling; perform weddings, funerals and other church services; train younger men to become pastors. Problem: 1. In 1995, Grace Bible Church had 400 people attending. Forty of those people left to help start Nampa Bible Church. Two years later, Grace Bible Church had 520 people, and Nampa Bible Church had 200. a. What is the percentage of growth for Grace Bible Church in the two years 1995 - 1997? b. What is the percentage of growth for Nampa Bible Church in the two years 1995 - 1997? c. What is the combined growth for the two churches in the two years 1995 - 1997? 2. If Grace Bible Church wants to build a building large enough for its congregation in the year 2000, how many seats should it have?
- 54. Numbers and Operations: Church Pastor Grace Bible Church of Boise Job Description: Preach and teach the Bible; care for people's spiritual needs through training and counseling; perform weddings, funerals and other church services; train younger men to become pastors. Problem: 1. In 1995, Grace Bible Church had 400 people attending. Forty of those people left to help start Nampa Bible Church. Two years later, Grace Bible Church had 520 people, and Nampa Bible Church had 200. a. What is the percentage of growth for Grace Bible Church in the two years 1995 - 1997? b. What is the percentage of growth for Nampa Bible Church in the two years 1995 - 1997? c. What is the combined growth for the two churches in the two years 1995 - 1997? 2. If Grace Bible Church wants to build a building large enough for its congregation in the year 2000, how many seats should it have? Solution: 1: The percent of growth a. 520 - 400 / 400 = 120/400 = .3 = 30% growth for Grace Bible Church. b. 200 - 40/40 = 160/40 = 4.0 = 400% growth for Nampa Bible Church. c. 720 - 400/400 = 320/400 = .8 = 80% combined growth 2: Seating anticipated in the year 2000: Two years growth = 30% What is 3 years growth 1997-2000? Estimate 1 year growth = 30% / 2 = 15% Three years growth = 15% x 3 = 45% 520 + 520(.45) = 520 + 234 = 754
- 55. Numbers and Operations: HP Business Manager Hewlett Packard Job Description: Responsible for business operations planning. Problem: We often travel from the United States to Japan for business. As such, we need to convert US Dollars to Japanese Yen for any business transaction. If a certain item cost 150,000 yen (¥), how much would it be in US Dollars? Assume an exchange rate of 115 ¥ to $1.
- 56. Numbers and Operations: HP Business Manager Hewlett Packard Job Description: Responsible for business operations planning. Problem: We often travel from the United States to Japan for business. As such, we need to convert US Dollars to Japanese Yen for any business transaction. If a certain item cost 150,000 yen (¥), how much would it be in US Dollars? Assume an exchange rate of 115 ¥ to $1. Solution: 150,000 ¥ ÷ 115 ¥ = about $1,304
- 57. Analysis & Probability/Algebra: Disk Drive Engineer Hewlett-Packard Company Job Description: Design digital magnetic recording systems for use in hard disk drives. Problem: A magnetic recording engineer wants to determine the minimum write current (Iw) he can use to drive a head to write magnetic transitions (bits of information) onto a disk in a disk drive. The head cannot get any closer than 25 nanometers (1 nanometer = 1 x 10-9 meters) to the surface of the disk. The magnetic field at the head gap (Hg) is roughly equal to product of the number of windings on the head (N) times the write current (expressed in amps), divided by the distance of the gap (G). Hg = (N x Iw) ÷ G The magnetic field is expressed in a couple of terms, Oersteds (Oe) and Amp•Turns/Meters. One Orested is equal to 80•πAmp•Turns/Meter. The head has 48 turns (or windings,N) and the gap distance,G, is 1 micrometer (1 x 10-6 meters or 1 µm). From previous modeling, he knows that the magnetic field decreases with distance away from the gap, such that it is roughly one half of Hg when the head is 25 nanometers away from the disk. He knows there is a demagnetizing field (Hd) that occurs inside the disk during the write process. This field opposes the field from the head. Previous calculations showed that Hd is 500 Oersteds. The net field for writing (Hx) must be greater than the coercivity (Hc) of the media plus the demagnetizing field. The coercivity of the media has been measured at 2000 Oersteds. What is the minimum write current?
- 58. Analysis & Probability/Algebra: Disk Drive Engineer Hewlett-Packard Company Job Description: Design digital magnetic recording systems for use in hard disk drives. See problem for details. Solution: Hx > Hc + Hd Hx = net field for writing Hc = coercivity of media Hd = Demagnetizing field Hc + Hd = 2000 Oe + 500 Oe = 2500 Oe Hx = 0.5 Hg Therefore, Hg = 2 • Hx > 5000 Oe 1 Oe = 80 • π Amp • Turns / Meter, abbreviated (A • T) / m Therefore, Hg > 5000 Oe (80 • π) (A • T)/m Hg > 1.26 • 106 A • T/m Hg = (Iw • N)/G Iw = Hg • G/N = (1.26 • 106 A • T/m)(1 • 10-6 m) / 48 T Iw > 0.026 Amps or 26 • 10-3 Amps (26 milliamps)
- 59. Analysis & Probability/Algebra: R&D Program Manager Hewlett-Packard Company Job Description: Develop full business models for development programs and work with all functional areas (R&D, marketing, finance, manufacturing, etc.) to bring new products to market. Problem: A new technology computer storage tape format has been designed by a computer company. Other companies "license" the right to produce products that use this format. They pay an initial fee of $10,000 and then pay $10 for every tape cartridge sold. The original company has expenses to develop, document and test this new tape format. It wants to know if the new tape license format will make money for the company. Given the following information . . . Original company expenses: $1,000,000 to develop format. 1st year expenses to set up documents and tests - $200,000. Each year after 1st - $50,000 / year to maintain. Assume 5 companies license the format: 2 during the 1st year and 3 more during the 2nd year. Assume each new company will sell 5,000 tape cartridges the 1st year and 10,000 tape cartridges every year after the 1st. Questions: 1. How many years before the original development company begins to earn money on its $1,000,000 R&D investment? 2. How much money does the original company make per year in the year three after the "market is mature?" 3. Do you think this investment was a good investment? 4. Which would affect the decision most? a. what if initial license fee was $5,000 instead of $10,000? b. what if more than 5 companies entered the market? c. what if a new format "captured the market" in 4 years?
- 60. Analysis & Probability/Algebra: R&D Program Manager Hewlett-Packard Company Job Description: Develop full business models for development programs and work with all functional areas (R&D, marketing, finance, manufacturing, etc.) to bring new products to market. Solution: (See problem for details.) 1. During which year will the original development earn money, disregarding the $1,000,000 R&D investment? Use a chart to organize income vs expenses: Before Year 1 Year 1 Year 2 Year 3 Year 4 Expenses $-1,000,000 $ -200,000 $ -50,000 $ -50,000 $ -50,000 License Fees $20,000 (2·$10,000) $30,000 (3·$10,000) - 0 - - 0 - Income Per Tape Fees $100,000 (2·10·5,000) $200,000 (2·10·10,000) $150,000 (3·10·5,000) $500,000 (5·10·10,000) $500,000 (5·10·10,000) Yearly Income $120,000 $380,000 $500,000 $500,000 Yearly Gain/Loss $-80,000 $330,000 $450,000 $450,000 Answer: During Year 2, the company begins to earn more than it spends. Yearly Income - Yearly Expense = Yearly Gain or Loss (Earnings) ($380,000 - $50,000 = $250,000) 2. After "market is mature" no new companies buying a license, and all companies are at final volume. [ (# of license companies) x (tape cartridges per year) x ($ fee/cartridge) ] - expenses by original company [(5) x ($10,000) x ($10)] - $50,000 = $450,000 3. Was this a good investment? Yes. By the end of Year 4, the original company has regained its investment plus $150,000 and adds another $450,000 per year. 4. Effects of "variation" - sensitivity analysis. a. License only $5,000 → only drops by $25,000 / minor change b. More than 5 license companies → adds $100,000/year at "maturity" / major change c. New format "captures market" in 4 years → might make investment a bad decision / difficult change
- 61. Numbers and Operations: IT Software Specialist Hewlett Packard Job Description: Provide operating system support and consulting to our worldwide customers. Problem: Our data center has a total of 65 computers. Each quarter we need to report to our customers the percentage of uptime for the data center. If we are reporting for the months August, September, and October and there are 49 computers that are up 100% of the time, what is the percentage of computers available 100%. If the total downtime of all 65 computers totals 2,435 minutes, what is the average downtime per computer per month? What is the total uptime percentage?
- 62. Numbers and Operations: IT Software Specialist Hewlett Packard Job Description: Provide operating system support and consulting to our worldwide customers. Problem: Our data center has a total of 65 computers. Each quarter we need to report to our customers the percentage of uptime for the data center. If we are reporting for the months August, September, and October and there are 49 computers that are up 100% of the time, what is the percentage of computers available 100%. If the total downtime of all 65 computers totals 2,435 minutes, what is the average downtime per computer per month? What is the total uptime percentage? Solution: (49 ÷ 65) = 0.7538 x 100 = 75.38% computers available 100% (2435 ÷ 65) ÷ 3 = 12.49 min. average downtime/computers per month (60 min. x 24 hrs.) = total minutes in a day = 1440 minutes (31 days + 30 days + 31 days) = total days in 3 months = 92 days 1440 x 92 x 65 = 8,611,200 total possible computing minutes 8,611,200 - 2,435 = 8,608,765 total actual computing minutes (8,608,765 ÷8,611,200) = .9997 x 100 = 99.97% uptime
- 63. Measurements/Geometry: Handyman – Concrete Home Improvement & Repairs Job Description: Manage individuals who market, sell, and support products that are currently on market (the problem is submitted as a homeowner). Problem: A man needs to make a cement driveway from his garage to the street—a distance of 135 feet. The driveway needs to be 4 inches thick. Cement costs $55/cubic yard with a 15% discount for cash. How many cubic yards of cement will he need for a driveway with a width of 15 feet? How much will he pay if he uses cash?
- 64. Measurements/Geometry: Handyman – Concrete Home Improvement & Repairs Job Description: Manage individuals who market, sell, and support products that are currently on market (the problem is submitted as a homeowner). Problem: A man needs to make a cement driveway from his garage to the street—a distance of 135 feet. The driveway needs to be 4 inches thick. Cement costs $55/cubic yard with a 15% discount for cash. How many cubic yards of cement will he need for a driveway with a width of 15 feet? How much will he pay if he uses cash? Solution: 1 cubic yard = 27 cubic feet 4 inches = 4/12 foot Volume of driveway: 135 ft. x 15 ft. x 4/12 ft. = 675 cu. ft. 675 cu. ft. ÷ 27 cu. ft.= 25 cu. yd. 25 cu.yds. x $55/cu. yd. = $1,375 If bill is paid in cash, take 15% of $1,375. Deduct $206.25 from the cost and customer.
- 65. Measurements/Geometry: Handyman – Remodeling Home Improvement & Repair Job Description: Provide cost estimates to customers for home projects. Problem: The cost of carpet is $13.00/square yard and the cost to install the carpet is $5.00/square yard. The cost of hardwood flooring is $7.00/square foot and the installation cost is $2.00/square foot. Jon's family room measures 18 feet x 24 feet. 1) What is the cost for carpet, including installation? 2) What is the cost for hardwood flooring, including installation? 3) What is the total cost for hardwood, including installation, around a perimeter four feet wide; and carpet, including installation, in the remaining middle area of the family room floor?
- 66. Measurements/Geometry: Handyman – Remodeling Home Improvement & Repair Job Description: Provide cost estimates to customers for home projects. See problem page for details. Solution: 1. What is the cost for carpet, including installation? carpet = $13/sq. yd. Installation = $5/sq .yd. hardwood = $7/sq. ft. installation = $2/sq. ft. 24 ft. x 18 ft. = 432 sq. ft of carpet needed $13 (carpet) + $5 (installation) = $18/sq .yd. installed carpet 432 sq. ft. / 9 sq. ft. = 48 sq x $18/sq .yd. = $864 for carpet 2. What is the cost of hardwood flooring including installation? 432 sq. ft. x $9/sq. ft. = $3,888 for hardwood floor 3. What is the total cost for the hardwood and carpet as described. (hardwood with carpet) carpet = $13/sq .yd. installation = $5/sq .yd. hardwood = $7/sq. ft. installation = $2/sq. ft. (24 ft. x 18 ft.) - (16 ft. x 10 ft.) = 272 sq. ft. of hardwood needed $7 (hardwood) + $2 (installation) = $9/sq .yd. installed hardwood 272 sq. ft. x $9/sq. ft. = $2,448 for hardwood (16 ft. x 10 ft.) / 9 sq. ft. = 17.77 sq .yd. of carpet needed 17.77 sq .yd. x $18/sq .yd. = $320 for carpet $2,448 + $320 = $2,768 total for hardwood and carpet
- 67. Geometry: Job Estimator Idaho Barns Job Description: Sell and install pre-fabricated barns and buildings. Price entire projects including labor and materials. Problem: A family wants to buy a barn to keep their animals in. The barn is 36 feet wide and 48 feet long. The foundation requirements call for concrete pillars placed every 12 feet through the structure. The pillars are to be 2.5 feet deep and 12 inches in diameter. How much concrete will be used for this project? The final amount needs to be calculated in cubic yards because that is how concrete is sold.
- 68. Geometry: Job Estimator Idaho Barns Job Description: Sell and install pre-fabricated barns and buildings. Price entire projects including labor and materials. Problem: A family wants to buy a barn to keep their animals in. The barn is 36 feet wide and 48 feet long. The foundation requirements call for concrete pillars placed every 12 feet through the structure. The pillars are to be 2.5 feet deep and 12 inches in diameter. How much concrete will be used for this project? The final amount needs to be calculated in cubic yards because that is how concrete is sold. Solution: 20 pillars are required for the building. Calculate how much concrete is needed for each pillar and multiply by 20. Circle area = πr2 = 3.14 x 6 x 6 = 113.04 square inches Cylinder volume = (πr2 ) x ht 2.5 feet = 30 inches (113.04 sq in) x (30 in) = 3391.2 cubic inches Convert to cubic feet 3391.2 cubic in ÷1,728 in3 /ft3 = 1.9625 cubic ft x 20 = 39.25 cubic feet 27 cubic feet = 1 cubic yard Convert to cubic yards 39.25 ft3 ÷ 27ft3 /yd3 = 1.45 cubic yards of concrete needed
- 69. Geometry: Farm Owner / Manager Idaho Farm Owner Job Description:Operate farms, ranches, greenhouses, nurseries, timber tracts, or other agricultural production establishments which produce crops, horticultural specialties, livestock, poultry, finfish, shellfish, or animal specialties. May plant, cultivate, harvest, perform post-harvest activities, and market crops and livestock; may hire, train, and supervise farm workers or supervise a farm labor contractor; may prepare cost, production, and other records. May maintain and operate machinery and perform physical work. Problem: Betsy is tethered to the barn at one corner by a 100 ft rope. A fence keeps her out of the garden. Find, to the nearest square foot, the area in which Betsy can graze. Area of triangle = base x height / 2 Area of circle = πr2 Area of sector = Area of circle x sector angle 360°
- 70. Geometry: Farm Owner / Manager Idaho Farm Owner Job Description: See problem page for details See problem for details. Solution: Break the diagram into 4 different areas. Begin with exact values and calculate the approximation at the end. 1. Area of Section I is contained in a right isosceles triangle. Area of Section I: (25 ft x 25 ft) ÷ 2 = 312.50 ft2 2. Area of Section II is contained in a sector of a circle with a sector angle of 45°, a radius of 100 ft - diagonal of Section I isosceles triangle: 100 ft – ft = 100 ft - ft = 100 ft - 35.3553 ft = 64.6447 ft Area of Section II: π(64.6447 ft)2 ÷360°/45° = 13128.5 ft2 ÷8 = 1,641.1 ft2 3. Area of section III is contained in a sector of a circle with a sector angle of 225° and a radius of 100 ft. Area of Section III: π(100 ft)2 ÷ 360°/225° = 19,635 ft2 4. Area of Section IV is contained in a sector of a circle with a sector angle of 90° and a radius of 100 ft - 80 ft = 20 ft. Area of Section IV: π(20 ft)2 ÷ 360° / 90° = 1,256.64 ft2 ÷ 4 = 314.16 ft2 5. Area in which Betsy can graze = sum of areas of Sections I - IV (312.50 + 1,641.1 + 19,635 + 314.16) ft2 = 21,903 ft2
- 71. Numbers and Operations: Bank Teller Idaho Independent Bank Job Description: Help customers cash checks, make deposits, balance different accounts and ledgers. Problem: A buyer wants to buy a used car at a cost of $10,000.00 financed for 5 years or 60 months. The buyer would like to know the savings on the finance charge if he puts 10% down vs. 20% down against the loan at the time it is taken out. Assumptions: The 60 monthly payments are: a. with 0% down: $207.55 b. with 10% down: $186.80 c. with 20% down: $166.04
- 72. Numbers and Operations: Bank Teller Idaho Independent Bank Job Description: Help customers cash checks, make deposits, balance different accounts and ledgers. Problem: A buyer wants to buy a used car at a cost of $10,000.00 financed for 5 years or 60 months. The buyer would like to know the savings on the finance charge if he puts 10% down vs. 20% down against the loan at the time it is taken out. Assumptions: The 60 monthly payments are: a. with 0% down: $207.55 b. with 10% down: $186.80 c. with 20% down: $166.04 Solution: Cost of Car % Down Original Loan Amount Payment x Months = Total Total Loan Amount Finance Charge $10,000 0 % $10,000.00 $207.55 x 60 = $12,453.00 $12,453.00 $2,453.00 $10,000 10% $9,000.00 $186.00 x 60 = $11,208.00 $11,208.00 $2,208.00 $10,000 20% $8,000.00 $166.04 x 60 = $ 9,962.40 $9,962.40 $1,962.40 1. By putting 10% down at the time the loan is taken out, the buyer saves: $2,453.00 - $2,208.00 = $245.00 2. By putting 20% down at the time the loan is taken out, the buyer saves: $2,453.00 - $1,962.40 = $490.60
- 73. Measurements: Physician Idaho Physical Medicine & Rehabilitation Problem: You are the physician caring for an 8 year old boy in the hospital. He weighs 62 pounds. You need to calculate a flow rate for his IV fluids. You use the following rule of thumb to determine the appropriate flow rate: 4 cc per hour per Kg for the first 10 Kg of bodyweight. (1 Kg = 2.2 pounds) Plus 2 cc per hour per Kg for the next 10 Kg of bodyweight Plus 1 cc per hour per Kg for bodyweight over 20 Kg. How fast should this boy's IV run in cc per hour?
- 74. Measurements: Physician Idaho Physical Medicine & Rehabilitation Problem: You are the physician caring for an 8 year old boy in the hospital. He weighs 62 pounds. You need to calculate a flow rate for his IV fluids. You use the following rule of thumb to determine the appropriate flow rate: 4 cc per hour per Kg for the first 10 Kg of bodyweight. (1 Kg = 2.2 pounds) Plus 2 cc per hour per Kg for the next 10 Kg of bodyweight Plus 1 cc per hour per Kg for bodyweight over 20 Kg. How fast should this boy's IV run in cc per hour? Solution: Convert 62 pounds to Kg. 62 lbs. ÷ 2.2 lbs./Kg = 28 Kg First 10 Kg: @ 4cc/hKg = 40 cc/h Second 10 Kg: @ 2cc/hKg = 20 cc/h Final 8 Kg: @ 1cc/hKg = 8 cc/h Answer: 68 cc/h
- 75. Numbers and Operations: Family Physician Idaho Physicians Associates Job Description: Educate patients on wellness issues, diagnose and treat patients as problems occur. Problem: A child has a middle ear infection requiring an antibiotic that is dosed at 60 mg per kg a day and divided into 3 doses per day. The medication is supplied as a suspension, 250 mg / 5 ml. 1. How many teaspoons is each dose for a child weighing 48 pounds? 2. How many ounces of medicine should the pharmacist dispense for an 8 day course of treatment? Facts needed: 2.2 lbs = 1 kg 1 oz = 6 teaspoons 1 teaspoon = 5 ml
- 76. Numbers and Operations: Family Physician Idaho Physicians Associates Job Description: Educate patients on wellness issues, diagnose and treat patients as problems occur. Problem: A child has a middle ear infection requiring an antibiotic that is dosed at 60 mg per kg a day and divided into 3 doses per day. The medication is supplied as a suspension, 250 mg / 5 ml. 1. How many teaspoons is each dose for a child weighing 48 pounds? 2. How many ounces of medicine should the pharmacist dispense for an 8 day course of treatment? Facts needed: 2.2 lbs = 1 kg 1 oz = 6 teaspoons 1 teaspoon = 5 ml Solution: 1. How many teaspoons per dose? 48 lb ÷ 2.2 lbs/kg = 21.8181 kg x 60 mg/kg = 1309.09 mg per day 1309.09 mg ÷ 3 doses = 436.36 mg per dose 250 mg of medication in 5 ml / 1 tsp = 5 ml = 250 mg of medication 436.36 mg ÷ 250 mg/tsp = 1.7454545 = 1.75 teaspoons per dose 2. How many ounces of medicine for 8 days of treatment? 3 doses/day x 8 days x 1.75 tsp/dose = 42 tsp of medicine 42 tsp ÷ 6 tsp/ounce = 7 ounces of medicine
- 77. Algebra/Geometry: Utility Facility Representative Idaho Power Job Description: Design and layout overhead and underground power lines. Problem: Determine the downward force (V) exerted on the pole by a down guy holding 3 conductors and determine the tension on the guy wire. V (Vertical Force) = (#of conductors) x T Tan ß G (Guy Wire Tension) = (# of conductors) x T Sin ß T = Tension of each conductor = 4,073 lbs H = Height of attachment = 35 ft L = Length of anchor lead = 21 ft Because this is a static system, the sum of the horizontal forces = 0 and the sum of the vertical forces = 0.
- 78. Algebra/Geometry: Utility Facility Representative Idaho Power Job Description: Design and layout overhead and underground power lines. See problem for details. Solution: Step one: Determine β: tan β = opposite = 21’ = 0.6 adjacent 35’ β = Tan-1 (0.6) = 31° Step two: Determine the guy wire tension G We can get G by summing the horizontal forces. Σ F (horizontal direction) = 0 because the system is static. -G sin β + 3 (4073 lbs) = 0 G = 3 x 4073 lbs = 12219 sin β sin 31° G = 23,750 lbs Step three: Determine the vertical force V: We can get the compression force on the pole by summing the forces in the vertical direction. Σ F (vertical direction) = 0 because the system is static. -G cos β + V = 0 V= G cos β = 23,750 lbs * cos 31° = 20,358 lbs
- 79. Geometry: GIS Analyst Idaho State Tax Commission Job Description: Perform geographical analysis on land values, land types, and determine fair market values over different geographical areas. Problem: Jim, an appraiser, needs to find the value of a lot in a subdivision. He knows that land prices in this particular area are selling at $6.25 per square foot. Jim has the following lot to appraise, but he must first find the square footage to determine the appraised value of the lot.
- 80. Geometry: GIS Analyst Idaho State Tax Commission Job Description: Perform geographical analysis on land values, land types, and determine fair market values over different geographical areas. Problem: Jim, an appraiser, needs to find the value of a lot in a subdivision. He knows that land prices in this particular area are selling at $6.25 per square foot. Jim has the following lot to appraise, but he must first find the square footage to determine the appraised value of the lot. Solution: Area of circle = πr2 Area of ½ circle = (πr2 )/2 = (π(25)2 )/2 = 981.75 sq. ft. Area of square = length x width = 75 ft. x 50 ft. = 3,750 sq. ft. Square + half circle = 3,750 + 981.75 = 4,731.75 sq. ft. $6.25 x 4,731.75 = $29,573.44
- 81. Numbers and Operations: Tax Analyst Idaho State Tax Commission Job Description: Analyzes and audits tax returns prepared by State of Idaho residents. Problem: A person has $20,000 that is taxable at a tax rate of 7%. She files the return 35 days after the due date. The interest rate for late payment is 9% a year, assessed on a daily basis. The late payment penalty is 5% per month for each month or partial month after the due date, to a maximum of 25%, or $10.00 whichever is greater. 1. What is the total tax? 2. What is the penalty? 3. What is the interest due? 4. What is the total due?
- 82. Numbers and Operations: Tax Analyst Idaho State Tax Commission Job Description: Analyzes and audits tax returns prepared by State of Idaho residents. Problem: A person has $20,000 that is taxable at a tax rate of 7%. She files the return 35 days after the due date. The interest rate for late payment is 9% a year, assessed on a daily basis. The late payment penalty is 5% per month for each month or partial month after the due date, to a maximum of 25%, or $10.00 whichever is greater. 1. What is the total tax? 2. What is the penalty? 3. What is the interest due? 4. What is the total due? Solution: 1. The total tax due is found by multiplying the taxable amount ($20,000) times the tax rate (7%). $20,000 x .07 = $1,400 tax 2. The total penalty due is calculated by multiplying the taxable amount ($20,000) times the tax rate (7%) times the interest rate (9%) times the number of days late (35) divided by the number of days in the year (365). ($20,000 x .07 x .09 x 35 days) ÷ 365 days/yr = $12.08 penalty 3. The total interest due is calculated by multiplying the taxable amount ($20,000) times the tax rate (7%) times the months late (2) times the payment penalty (5%). $20,000 x .07 x 2 months x .05 = $140 interest 4. Therefore, the total tax, penalty and interest due is calculated by adding these three amounts together. $1,400.00 + $12.08 + $140.00 = $1,552.08 total This type of calculation would be done by anyone filing tax returns and by Tax Commission personnel in positions ranging from clerical staff to auditors or compliance officers. The salary range of those positions ranges from $14,000 to $35,000.
- 83. Algebra: Financial Analyst Interstate Food Processing Corporation Job Description: Provide analyses as required to ensure the organization is meeting its financial objectives. Functions include preparation of budgets, analysis of profitability by customer and product, and the estimation of new products costs and prices. Problem: Businesses generally incur two types of expenses: fixed and variable. Fixed expenses are those that do not change with the amount of business. For example, in our business we have certain costs such as rent, electricity, and some employee wages (customer service representatives and billing clerks) that do not change whether we make one sale or one hundred sales. Variable costs, however, increase directly with the amount of sales. In our business, these costs would include items such as raw products, packaging materials (bags and boxes), and the wages of employees that produce the items. Each time we introduce a new product, we estimate how many cases per month we must sell in order to pay for the fixed costs of the operation and begin to make a profit. This estimate is called a Break-Even Analysis and is the point where total sales revenue equals total (fixed and variable) costs. The difference between the Selling Price per case and the Variable Costs per case is called the Contribution Margin. Determine the number of cases that must be sold to reach the break even point. The following information is available to solve the problem: Selling price per case (P) $8.95 Variable cost per case (V) $6.45 Fixed Costs per month (F) $24,000.00 Quantity of cases (Q) ? Break even Quantity (QBE) QBE = F / (P-V)
- 84. Algebra: Financial Analyst Interstate Food Processing Corporation Job Description: Provide analyses as required to ensure the organization is meeting its financial objectives. Functions include preparation of budgets, analysis of profitability by customer and product, and the estimation of new products costs and prices. See problem for details. Solution: The breakeven point is reached when costs equals profits. F + V • Q = P • Q $24,000.00 + 6.45 • Q = 8.95 • Q QBE = F / (P-V) QBE = $24,000.00 / ($8.95 - $6.45) = 9,600 cases
- 85. Measurements: Property Manager (1) J. D. Property Job Description: Owns, operates and manages rental properties. Oversees or does all maintenance and/or repairs on houses and property. Problem: You need to pour a new sidewalk that is 4' wide, 16' long, and 4" thick. Concrete costs $70.00/cubic yard. 1) How many cubic yards of concrete are required? 2) How much will the concrete cost?
- 86. Measurements: Property Manager (1) J. D. Property Job Description: Owns, operates and manages rental properties. Oversees or does all maintenance and/or repairs on houses and property. Problem: You need to pour a new sidewalk that is 4' wide, 16' long, and 4" thick. Concrete costs $70.00/cubic yard. 1) How many cubic yards of concrete are required? 2) How much will the concrete cost? Solution: 1 yard = 3 feet 1 cubic yard = 3 ft x 3 ft x 3ft 1 cubic yard = 27 cubic feet (ft³) 4 inch = 1/3 foot 4 ft x 16 ft x 1/3 ft = 21.3 ft³ 1) 21.3 ft³ x 1yd³ ÷ 27 ft³ = 0.79 cubic yards needed 2) .79 cubic yards x $70.00 / cubic yard = $55.30
- 87. Measurements: Property Manager (2) J. D. Property Job Description: Owns, operates and manages rental properties. Oversees or does all maintenance and/or repairs on houses and property. Problem: You need to build a wall using 2x4s covered with 4' x 8' sheet rock. The wall is 20 feet long and 8 feet high. 1) How many 2x4s are required if the 2x4s are on 16" centers, spaced 16” apart? 2) How many sheets of sheet rock will be needed?
- 88. Measurements: Property Manager (2) J. D. Property Job Description: Owns, operates and manages rental properties. Oversees or does all maintenance and/or repairs on houses and property. Problem: You need to build a wall using 2x4s covered with 4' x 8' sheet rock. The wall is 20 feet long and 8 feet high. 1) How many 2x4s are required if the 2x4s are on 16" centers, spaced 16” apart? 2) How many sheets of sheet rock will be needed? Solution: 1) 20 feet x 12 inches/foot = 240 inches 240 inches ÷ 16 inch centers = 15 spaces + 1 = 16 (8-foot 2x4s) 2) 20 feet ÷ 4-foot sheets = 5 sheets of sheet rock
- 89. Measurements: Property Manager (3) J. D. Property Job Description: Owns, operates, and manages rental properties. Oversees or does all maintenance and/or repairs on houses and property. Problem: You need to re-roof a house with composite shingles. Each side of the roof is 20 feet by 40 feet. How many packages of shingles are required and how much will it cost? Each package of shingles covers 12 square feet (ft²) and costs $7 plus another $12 to have it put on the roof (for labor and other materials such as nails).
- 90. Measurements: Property Manager (3) J. D. Property Job Description: Owns, operates, and manages rental properties. Oversees or does all maintenance and/or repairs on houses and property. Problem: You need to re-roof a house with composite shingles. Each side of the roof is 20 feet by 40 feet. How many packages of shingles are required and how much will it cost? Each package of shingles covers 12 square feet (ft²) and costs $7 plus another $12 to have it put on the roof (for labor and other materials such as nails). Solution: 1. 20 ft x 40 ft = 800 ft² x 2 = 1600 ft² roof 1600 ft² ÷ 12 ft² coverage/pkg = 133.3 or 134 packages of shingles 2. 134 packages x $7 per package = $938 for shingles 134 packages x $12 = $1608 for labor/materials $938 + $1608 = $2546 roof installed
- 91. Measurements/Geometry: Architect James Gipson Associates Problem: #1 John and Joan are planning a new home. They want as much window area as possible. The local energy code permits a maximum window area of 17% of the house floor area. The windows John and Joan will use are each 3 ft. x 5 ft. and the floor area of the house is 1,720 square feet. How many windows can they put into their new house? #2 Karen is figuring how many bundles of shingles to order for the roof on a garage. The garage is 24 ft. x 24 ft., with a single roof ridge down the middle, plus a 2 ft. overhang all around. Three bundles of shingles are required for each 100 sq. ft. of roof. How many bundles of shingles should Karen order?
- 92. Measurements/Geometry: Architect James Gipson Associates Problem: #1 John and Joan are planning a new home. They want as much window area as possible. The local energy code permits a maximum window area of 17% of the house floor area. The windows John and Joan will use are each 3 ft. x 5 ft. and the floor area of the house is 1,720 square feet. How many windows can they put into their new house? #2 Karen is figuring how many bundles of shingles to order for the roof on a garage. The garage is 24 ft. x 24 ft., with a single roof ridge down the middle, plus a 2 ft. overhang all around. Three bundles of shingles are required for each 100 sq. ft. of roof. How many bundles of shingles should Karen order? Solution: #1 17% of 1,720 sq. ft. = 292.4 sq. ft. Each window is 3' x 5' or 15 sq. ft. 292.4 ÷ 15 = 19.49 Therefore, John and Joan can have at most 19 windows. #2 Consider the end view of the roof as two right triangles, each with a base of 12' + 2' = 14' and a height of 7'. a² + b² = c² 7² + 14² = c² = 245 sq. ft. c= 15.65 ft. Area of half the roof = 15.65' x (24'+ 4') = 438.2 sq. ft. Area of the roof = 438.2 x 2 = 876.4 sq. ft. The number of shingle bundles required = area of roof ÷ 33 1/3 876.4 ÷ 33 1/3 = 26.29 Therefore, Karen should order 27 bundles.
- 93. Geometry: Excavator John Sawyer Excavation Job Description: Digging foundations for businesses and houses Problem: The following problem is used to make sure houses and buildings are "square" (sides form right angles). If the length of one side of your house is 44 feet and the adjacent side is 68 feet, show that all corners form 90-degree angles. The diagonals of a rectangle have equal measure. Show that the diagonals of the following figures are equivalent.
- 94. Geometry: Excavator John Sawyer Excavation Job Description: Digging foundations for businesses and houses Problem: The following problem is used to make sure houses and buildings are "square" (sides form right angles). If the length of one side of your house is 44 feet and the adjacent side is 68 feet, show that all corners form 90-degree angles. The diagonals of a rectangle have equal measure. Show that the diagonals of the following figures are equivalent. Solution: Use the Pythagorean Theorem. 44² + 68² = x² 6560 = x 80.99 = x 81 ft. = x If this number is equal to the length of your diagonals when measured, you will have 90-degree angles.
- 95. Algebra: Finance J.R. Simplot Company Job Description: Indebtedness compliance reporting, prospective lender correspondence, asset leasing, and various other financial analysis projects. Problem: Financial Indebtedness Covenant (this is an example problem. Amounts are not actual, as the company does not release financial information). A Company needs cash to expand its current production facilities and has decided to borrow from the bank. As one of the conditions of borrowing from the bank, the Company agrees that its leverage ratio (total borrowings to total shareholders' equity) will not exceed 1.25 to 1.0. The Company has no current borrowings. Shareholders' equity is composed of shareholders' contributions of cash and accumulated company earnings. 1. If current shareholders' equity is 300, what is the maximum amount the Company can currently borrow? 2. If the Company wishes to reserve 50% of its maximum available borrowing for future needs, how much will it borrow currently? 3. If the Company needs to borrow 425 for the expansion and earns 10 per month, how many months must it accumulate earnings before qualifying for the loan of 425? (Remember current shareholders' equity).
- 96. Algebra: Finance J.R. Simplot Company Job Description: Indebtedness compliance reporting, prospective lender correspondence, asset leasing, and various other financial analysis projects. Problem: Financial Indebtedness Covenant (this is an example problem. Amounts are not actual, as the company does not release financial information). A Company needs cash to expand its current production facilities and has decided to borrow from the bank. As one of the conditions of borrowing from the bank, the Company agrees that its leverage ratio (total borrowings to total shareholders' equity) will not exceed 1.25 to 1.0. The Company has no current borrowings. Shareholders' equity is composed of shareholders' contributions of cash and accumulated company earnings. 1. If current shareholders' equity is 300, what is the maximum amount the Company can currently borrow? 2. If the Company wishes to reserve 50% of its maximum available borrowing for future needs, how much will it borrow currently? 3. If the Company needs to borrow 425 for the expansion and earns 10 per month, how many months must it accumulate earnings before qualifying for the loan of 425? (Remember current shareholders' equity). Solution: 1. Maximum borrowing = 1.25 x Shareholders' equity (Financial Covenant) 375 = 1.25 x 300 2. Current borrowing = 50% x Maximum borrowing 187.5 = 50% x 375 3. Borrowing needs = 1.25 x [Current Shareholders' Equity + (Monthly Earnings x Number Months)] 425 = 1.25 x [300 + (10 x number of months)] 425 = 375 + 12.5 x (number of months) (425-375) ÷ 12.5 = number of months 4 = number of months
- 97. Numbers and Operations: Financial Services Vice President Key Bank Job Description: Providing commercial loans, leases, derivative products and cash management. Problem: Determine the gain on the investment. Stock is purchased @ $21 1 ⁄4 per share Stock is sold @ $34 1 ⁄8 per share Shares owned: 1,280
- 98. Numbers and Operations: Financial Services Vice President Key Bank Job Description: Providing commercial loans, leases, derivative products and cash management. Problem: Determine the gain on the investment. Stock is purchased @ $21 1 ⁄4 per share Stock is sold @ $34 1 ⁄8 per share Shares owned: 1,280 Solution: 34 1 ⁄8 - 21 1 ⁄4 = 34.125 - 21.25 = 12.875 (the difference between selling price and buying price/share) $12.875 x 1280 shares of stock = $16,480 gain
- 99. Geometry/Numbers & Operations/Measurements: Glass Installer Klassic Glass & More Job Description: Cut and install glass - both auto and residential. Problem: You need to cut a mirror, but all of the measurements are in inches instead of feet. Find the length of each side in feet to cut a piece of glass and find the total square feet in the mirror.
- 100. Geometry/Numbers & Operations/Measurements: Glass Installer Klassic Glass & More Job Description: Cut and install glass - both auto and residential. Problem: You need to cut a mirror, but all of the measurements are in inches instead of feet. Find the length of each side in feet to cut a piece of glass and find the total square feet in the mirror. Solution: 48 inches ÷ 12 in/ft = 4 feet (width) 84 inches ÷ 12 in/ft = 7 feet (length) 4 ft x 7 ft = 28 square feet in mirror
- 101. Numbers and Operations: Director of International Finance Lamb Weston Job Description: Directs accounting functions in Holland, Turkey and India for the sale, manufacturing, and distribution of Frozen French Fries. Problem: Company A is a manufacturer and seller of French Fries and owns 40% of a similar company in Germany (Company G). Company G sells French Fries throughout Europe. Company G sells 30,000,000 pounds of French Fries per year in Belgium alone. The customers in Belgium pay Company G in Belgian Franks. Company G keeps their financial records in German Marks; however, Company G reports profits to Company A in U.S. Dollars. How much money will Company A make in one year from sales of Company G's fries to Belgium, given the financial information below? Currency Conversion Table 1 U.S. Dollar = 1.50 German Marks 1 German Mark = 20 Belgian Franks 1 U.S. Dollar = 30 Belgian Franks 1. Average sales price per pound in Belgium: 10 Belgian Franks/pound. 2. Average cost to produce-per-pound in Germany: 0.40 German Marks (cents)/pound.
- 102. Numbers and Operations: Director of International Finance Lamb Weston Job Description: Directs accounting functions in Holland, Turkey and India for the sale, manufacturing, and distribution of Frozen French Fries. See problem page for details. Solution: Revenue: 30,000,000 pounds x 10 Belgian Franks = 300,000,000 Convert to dollars: 300,000,000 ÷ 30 Franks/dollar = $10,000,000 Cost: 30,000,000 pounds x 0.40 German Marks = 12,000,000 Convert to dollars: 12,000,000 ÷ 1.50 Marks/dollar = $8,000,000 Revenue - Cost = Profit $10,000,000 - $8,000,000 = $2,000,000 (Company G) 40% of profit belongs to Company A. Therefore, $2,000,000 x 0.40 = $800,000 is the profit of Company A from sales in Belgium.
- 103. Geometry: Contractor Lamm Construction Job Description: Residential construction Problem: A builder needs to place a house on a lot that has no parallel sides. How would he place the house on the lot and make sure the house is square on the lot?
- 104. Geometry: Contractor Lamm Construction Job Description: Residential construction Problem: A builder needs to place a house on a lot that has no parallel sides. How would he place the house on the lot and make sure the house is square on the lot? Solution: Find the starting point marked (D). Measure the width of the house to find the distance (DE). Measure the length to find distance (EF). If DF² = (FE)² + (DE)² , then the house is square on the lot as proved by the Pythagorean theorem.
- 105. Algebra/Geometry: Manufacturer Lawn Mower Factory Job description: Manufacture lawn mower parts & lawn mowers Problem: A worker in a lawnmower factory is asked to make the part shown below for a new lawnmower design. To program a machine to cut the part out, the worker must find the slope and length of edge D.
- 106. Algebra/Geometry: Manufacturer Lawn Mower Factory Job description: Manufacture lawn mower parts & lawn mowers Problem: A worker in a lawnmower factory is asked to make the part shown below for a new lawnmower design. To program a machine to cut the part out, the worker must find the slope and length of edge D. SOLUTION:
- 107. Numbers and Operations: Property Development Manager L.B. Industries, Inc. Job Description: Locate potential sites, negotiate purchases, coordinate analysis of attorneys and engineers, coordinate pre-purchase due diligence, determine lot configuration and pricing, oversee development of sites, and negotiate sales. Problem: The Eagle Land Company has the opportunity to purchase 76.80 acres of land in Boise, near the Cole/I-84 Interchange, for $20,000 an acre. The property is zoned for light industrial and retail use. Developed property of this sort is currently selling for $3.50 a square foot. As configured, it appears that 415,000 square feet of the land will need to be dedicated as non-sellable roadways, sidewalks and median strips. Engineering fees and improvements (including sewer lines, telephone, electrical power, curbs and gutters) will cost the company an additional $1,250,000.00. To receive an adequate profit after salaries, interest charges, realtors' fees and other fixed costs, the Eagle Land Company must receive a mark up of at least 150% above the dollar amount of its purchase and development costs (250% of the total cost of the development.) Commercial lots are generally sold on a square footage basis. 1. What is the minimum price Eagle Land Company must receive, on a square footage basis, for the land in this subdivision? (Hint: An acre contains 43,560 square feet.) 2. Does it make economic sense for Eagle Land Company to purchase this property? Support your conclusion.
- 108. Numbers and Operations: Property Development Manager L.B. Industries, Inc. Job Description: Locate potential sites, negotiate purchases, coordinate analysis of attorneys and engineers, coordinate pre-purchase due diligence, determine lot configuration and pricing, oversee development of sites, and negotiate sales. See problem page for details. Solution: 1. What is the minimum price Eagle Land Company must receive, on a square footage basis, for the land in this subdivision? Cost of land:76.80 acres x $20,000/acre = $1,536,000 Total cost to develop the subdivision equals cost of land plus cost of engineering and improvements: $1,536,000 + $1,250,000 = $2,786,000 The total minimum price equals total cost to develop the subdivision times 250%: $2,786,000 x 2.50 = $6,965,000 Total square footage in the development equals total acres times square footage per acre: 76.80 acres x 43,560 sq ft/acre = 3,345,408 square feet Total sellable square footage equals total square footage in the development less square footage to be dedicated as roadways, sidewalks and median strips: 3,345,408 sq ft - 415,000 sq ft = 2,930,408 sq ft The minimum sales price per square foot equals the total minimum sales price for the subdivision divided by the sellable square footage in the development: $6,965,000.00 ÷ 2,930,408 sq ft = $2.3768021 or $2.38/sq ft 2. Does it make economic sense for Eagle Land Company to purchase this property? Yes, this land can be sold for $3.50 per square foot and $2.38 per square foot will insure an adequate profit. If the land is sold for $3.50 per square foot, the Eagle Land Company would receive an additional profit of $1.12 per square foot above the amount it deems to be an adequate profit. At 2,930,408 square feet of sellable land in the entire development, the additional $1.12 per square foot would result in an extra $3,282,056.90 in profit on property that cost $2,786,000 to develop.
- 109. Measurements/Geometry: Homemaker – Seamstress Lori Purcell Job Description: All around handywoman: painter, carpenter, seamstress, cook, and yard and home maintenance. Problem: A homemaker wants to save money by making bedroom curtains rather than buying them. Using the following information, determine how much fabric she will need and how much the fabric will cost. Window dimensions: 6' wide, 4' long. The finished width of the curtains will be 3 times the window width. The curtains will have two gathered sections with a 3" hem at the bottom and a 1" hem on each side. The uncut fabric is 54" wide and costs $15.99 per yard. Fabric is sold by the yard with partials in 1/2, 1/4, or 1/3 yard. You will need to add 1/2 inch for seam allowances on sides, top and bottom. A seam is where the fabric is sewn together; the allowance is the material that is folded back on either side of the seam. Add 6" for curtain-rod pocket at the top.
- 110. Measurements/Geometry: Homemaker – Seamstress Lori Purcell Job Description: All around handywoman: painter, carpenter, seamstress, cook, and yard and home maintenance. See problem for details. Solution: 1. Find the dimensions of the window in inches. Window width: 6 x 12 = 72" Window length: 4 x 12 = 48" 2. Curtain width: the curtains will be 3 times the width of the window. 72" wide x 3 = 216" total width of the curtains 216" ÷ 2 sections = 108" width of each section 3. Add the side hems and seam allowances to the width of one section. 108" + 2" (2 side hems) + 1" (2 seam allowances) = 111" 111" represents the cut fabric width 4. Because the fabric width is 54" long, 3 fabric widths will be required for one section (111 inches). This will add another seam or 2 additional seam allowances. The cut fabric width will now be 111" + 1" = 112" 5. To make the curtains, 5 fabric widths will be required. (2 widths plus a few inches for each section.) Rather than buying 3 widths for each curtain, buy 2 for each curtain and share the extra width. 6. Curtain length = window length + hem + seam allowance + rod pocket 48" + 3" + 1" + 6" = 58" (fabric length per curtain) 7. For the curtains, 5 lengths of fabric will be needed. Each length will be 58" long. 58" x 5 = 290" ÷ 36"/yd = 8 yards + 2 inches She must purchase 8.25 yards, since .25 is the smallest division 8. Cost of fabric: 8.25 yards x $15.99 (cost per yard) = $131.92
- 111. Numbers and Operations: Salesperson – Shoe Department Macy's Job Description: Helps the customer purchase a pair of shoes. Problem: The wage rate for a salesperson at Macy's is $8.60 per hour for a 40 hour week and 1.5 times the base pay ("time and a half") for overtime. If a salesperson works 48 hours in one week, how much is the gross salary she or he earns?
- 112. Numbers and Operations: Salesperson – Shoe Department Macy's Job Description: Helps the customer purchase a pair of shoes. Problem: The wage rate for a salesperson at Macy's is $8.60 per hour for a 40 hour week and 1.5 times the base pay ("time and a half") for overtime. If a salesperson works 48 hours in one week, how much is the gross salary she or he earns? Solution: 1. Find the salary for the first 40 hours. $8.60 x 40 = $344.00 2. Find the salary for the 8 hours overtime. 1.5 x $8.60 = $12.90/hour x 8 hours = $103.20 3. Add the two amounts together. $344.00 + $103.20 = $447.20
- 113. Algebra: Medical Technologist Mercy Medical Center Job Description: Performs testing on blood and body fluids to aid in diagnosis and treatment of disease. Problem: Creatinine is a waste product of muscle metabolism normally found at relatively constant levels in the blood. By measuring the plasma and urine levels of creatinine, in a specific period of time, the creatinine clearance or glomerular filtration rate can be calculated. The creatinine clearance is used to determine the extent of nephron damage in known cases of renal disease, to monitor effectiveness of treatment and to determine the feasibility of administering medications which can build up to toxic levels if the glomerular filtration rate is markedly reduced. Using a urine creatinine of 96 mg/dL (U), plasma creatinine of 1.3 mg/dL (P), and urine volume of 125 mL from a 2 hour specimen (V), calculate the glomerular filtration rate (creatinine clearance [C].) The standard formula used to calculate the milliliters of plasma creatinine cleared per minutes [C] is: C = UV/P. Where V = urine volume in milliliters per minute. U = urine creatinine concentration in milligrams per deciliter. P = plasma creatinine concentration in milligrams per deciliter. Hint: Convert dL to mL
- 114. Algebra: Medical Technologist Mercy Medical Center Job Description: Performs testing on blood and body fluids to aid in diagnosis and treatment of disease. See problem for details. Solution: Calculate urine volume in ml/min. V = 125mL/120 min. = 1.04 mL/min Convert U and P from dL to mL: U=(96mg/dL )(0.1dL/mL) = 9.6mg/mL P=(1.3mg/dL)(0.1dL/mL) = 0.13mg/mL C = [9.6 mg/mL (U) / 0.13 mg/mL (P)] x 1.04 mL/min. (V) C = 76.9 mL/min Teacher notes: * normal ranges for creatinine clearance: men - 107 to 139 mL/min and women - 87 to 107 mL/min
- 115. Numbers and Operations: Elementary Teacher Meridian School District Job Description: Give a solid educational foundation and instill a love of learning, teach the basics such as math, science, reading, writing, language, art and history plus personal safety, nutrition, drug awareness and morals. Problem: In a five day work week, a typical elementary class can be taught in .5-hour blocks. If the school day is 6.5 hours long, and each .5-hour requires a written lesson plan except for the six .5-hour blocks that are taught by specialists such as music [2,] PE [2,] library and computer. Deduct 45 minutes for lunch each day and .5 hour a day for recess. Deduct 3 minutes a day for flag salute, roll call, etc. How many lesson plans is a teacher writing each week?
- 116. Numbers and Operations: Elementary Teacher Meridian School District Job Description: Give a solid educational foundation and instill a love of learning, teach the basics such as math, science, reading, writing, language, art and history plus personal safety, nutrition, drug awareness and morals. Problem: In a five day work week, a typical elementary class can be taught in .5-hour blocks. If the school day is 6.5 hours long, and each .5-hour requires a written lesson plan except for the six .5-hour blocks that are taught by specialists such as music [2,] PE [2,] library and computer. Deduct 45 minutes for lunch each day and .5 hour a day for recess. Deduct 3 minutes a day for flag salute, roll call, etc. How many lesson plans is a teacher writing each week? Solution: 6.5 hours/school day - .5 hr/recesses = 6 hours a day 6.0 hours - .75 hr/lunch = 5.25 hours a day 5.25÷.5 = 10 half-hour blocks/day (plus .25 hour or 15 minutes) 10 half-hour blocks x 5 days = 50 half-hour blocks/week 50 blocks – 6 half-hour special classes = 44 half-hour blocks 44 blocks or lesson plans needed each week with 15 minutes left for the flag salute, roll call, etc
- 117. Algebra: Accountant Micron Technology, Inc. Job Description: Responsible for the development, implementation and maintenance of the company's benefits programs including time off, sick pay, unemployment, retirement plans, disability, and group and life insurance. Problem: The following financial data was published in the Company's annual report. (Amounts in millions, except for per share and employee data) Fiscal year 2000 1999 1998 Net sales $ 7,336.3 $ 3,764.0 $ 3,025.3 Cost of goods sold 3,957.1 2,950.4 2,744.9 Gross Profit $ 3,379.2 $ 813.6 $ 280.4 Operating income (loss) 2,293.3 (47.0) (516.4) Net income (loss) 1,504.2 (68.9) (247.1) Earnings (loss) per share: $2.73 $(0.13) $(0.57) Cash and liquid investments $ 2,466.4 $ 1,613.5 $ 649.6 Property, plant and equipment, net 4,257.6 3,799.6 3,035.3 Total assets 9,631.5 6,965.2 4,703.5 1. If you were creating this spread sheet, what would be the formula for calculating Gross Profit? 2. Knowing the Net Income and Earnings per Share, how would you calculate the number of “outstanding shares” (number of shares of stock available for trading). 3. Create a formula for calculating the miscellaneous assets not itemized on the spread sheet. 4. Create a formula for calculating the annual percentage increase or decrease in the company's Net Sales. 5. Create a formula for calculating annual increase or decrease of the Net Income as a percentage of Sales
- 118. Algebra: Accountant Micron Technology, Inc. Job Description: Responsible for the development, implementation and maintenance of the company's benefits programs including time off, sick pay, unemployment, retirement plans, disability, and group and life insurance. See problem for details. Solution: 1. Gross Profit = Net Sales - Cost of Goods 2. Outstanding Shares = net income ÷ earnings per share 3. Miscellaneous Assets = total assets - investments + property 4. Difference in Sales = sales2nd yr - sales1st yr Percentage increase (decrease) = difference ÷ sales1st yr 5. Percentage = net income ÷ sales Difference = percentage2nd yr - percentage1st yr Increase (decrease) = difference ÷ percentage1st yr
- 119. Numbers and Operations/Algebra: Benefits Manager Micron Technology, Inc. Job Description: Responsible for the development, implementation and maintenance of the company's benefits programs including time off, sick pay, unemployment, retirement plans, disability, and group and life insurance. Problem: An employee's 401(k) contributions plus his employer's matching contributions and Return On Equity (ROE) contributions must be less than or equal to 25% his gross pay (minus insurance premiums and 401 (k) contributions.) Employees whose percentage is greater than 25% are entitled to a refund. If my salary and contributions match those listed below, what is my actual percentage? What should my 401(k) refund be to decrease the percentage to 25%? IRS formula: K + M + R = .25 (G - K - P) K = $3,000 401(k) contributions M = $1,500 Employer's matching contributions R = $400 ROE contributions G = $20,000 Gross Pay P = $800 Pre-tax Insurance Premiums
- 120. Numbers and Operations/Algebra: Benefits Manager Micron Technology, Inc. Job Description: Responsible for the development, implementation and maintenance of the company's benefits programs including time off, sick pay, unemployment, retirement plans, disability, and group and life insurance. Problem: An employee's 401(k) contributions plus his employer's matching contributions and Return On Equity (ROE) contributions must be less than or equal to 25% his gross pay (minus insurance premiums and 401 (k) contributions.) Employees whose percentage is greater than 25% are entitled to a refund. If my salary and contributions match those listed below, what is my actual percentage? What should my 401(k) refund be to decrease the percentage to 25%? IRS formula: K + M + R = .25 (G - K - P) K = $3,000 401(k) contributions M = $1,500 Employer's matching contributions R = $400 ROE contributions G = $20,000 Gross Pay P = $800 Pre-tax Insurance Premiums Solution: Actual Percentage = (3000 + 1500 + 400) / (20000 - 3000 - 800) = 4900/16200 = .302 (30.2% actual percentage) Allowable Contributions = (x + 1500 + 400) / (20000 - x - 800) = .25 x = $2,320 Refund = Actual Contributions - Allowable Contributions Refund = K - x 3000 - 2320 = $680 Refund
- 121. Algebra: Chemist Micron Technology, Inc. Job Description: Analyzes chemicals, tools, and wafers for the fabrication process. Analyzes waste products resulting from the memory chip fabrication process. Problem: The government requires that companies analyze and report the amount of ethyl lactate present in waste sent to a waste disposal company. The ethyl lactate sample area is 6,821,193 counts. An ethyl lactate standard has a "concentration" of 10.16 wt% and a peak area of 10,617,862 counts. What is the concentration (amount) of ethyl lactate in a solvent sample from gas chromatography data?
- 122. Algebra: Chemist Micron Technology, Inc. Job Description: Analyzes chemicals, tools, and wafers for the fabrication process. Analyzes waste products resulting from the memory chip fabrication process. Problem: The government requires that companies analyze and report the amount of ethyl lactate present in waste sent to a waste disposal company. The ethyl lactate sample area is 6,821,193 counts. An ethyl lactate standard has a "concentration" of 10.16 wt% and a peak area of 10,617,862 counts. What is the concentration (amount) of ethyl lactate in a solvent sample from gas chromatography data? Solution: Since the relationship is linear, use a ratio: Concentrate of standard (X1) is to counts of standard (c1) as concentrate of sample (X2) is to counts of sample (c2) Standard Sample X1 = X2 c1 c2 10.16% = X2 wt% 10,617,862 6,821,193 X2 wt% = 10.16% * 6,821,193 = 10.16% * 0.64243 = .0653 = 6.53% 10,617,862 Concentration of sample = 6.53 wt %
- 123. Algebra: Semiconductor Engineer Micron Technology, Inc Job Description: Evaluate production lots suspected of quality or reliability issues and ensure lots are held until all concerns have been addressed. Identify the root cause of failure and respond with appropriate corrective actions. Problem: Micron's manufacturing areas function 24 hours per day, 7 days a week. Nitride deposition is the capacity bottleneck in the fab. The fab has 3 nitride furnaces and runs the following nitride recipes with total run time per number of wafers run as indicated below: Recipe # Wafers Run Total Time per Run X 100 3.7 hrs Y 50 2.3 hrs Z 100 4.8 hrs The fab runs the following part types with potential revenue and profit as indicated below: Part Type Nitride Deps Potential Revenue per Wafer Potential Profit per Wafer Required Ships per Week 3 X, Y $500 $100 1000 5 Z $300 $75 1500 7 X, Y, Z $525 $100 500 11 X, Z $475 $125 3000 How many wafer starts of each part type (including the required ships per week) will maximize revenue? How many wafer starts of each part type (including the required ships per week) will maximize profit?
- 124. Algebra: Semiconductor Engineer Micron Technology, Inc Solution: (See problem for details) Calculate time per wafer per recipe: tr = time per wafer (each recipe) hr = hours per run (each recipe) Wr = number of wafers run (each recipe) tr = hr⁄Wr tX = hX ⁄ WX = 3.7⁄100 =.037 hrs per wafer (recipe X) tY = hY ⁄ WY = 2.3⁄50 = .046 hrs per wafer (recipe Y) tZ = hZ⁄ WZ = 4.8⁄100 = .048 hrs per wafer (recipe Z) Calculate the time needed per required ship for each part type: Tp = time per wafer (each part type) tp = time per recipe (each part type) Sp = ships required per week (each part type) Hp = hours per week per required ship (each part type) Hp = (Σtr) (Sp) H3 = (.037 + .046) = .083 hrs x 1000 = 83.0 hrs per week H5 = (.048) = .048 hrs x 1500 = 72.0 hrs per week H7 = (.037 + .046 + .048) = .131 hrs x 500 = 65.5 hrs per week H11 = (.037 + .048) = .085 hrs x 3000 = 255.0 hrs per week Total Required Hours 475.5 hrs per week Calculate number of hours left to maximize profit/revenue X= (hrs/day)(days/wk)(furnaces) - required hours X = 24 x 7 x 3 = 504 - 475.5 = 28.5 extra hours per week Calculate potential revenue and profit per hour for each part type: rp = revenue per wafer (each part type) Rp = revenue per hour (each part type) pp = profit per wafer (each part type) Pp = profit per hour (each part type) Rp = rp ⁄hp Pp = pp⁄hp R3 = $500⁄.083 hrs = $6,024 revenue per hour P3 = $100⁄.083 hrs = $1,205 profit per hour R5 = $300⁄.048 hrs = $6,250 revenue per hour P5 = $75⁄.048 hrs = $1,563 profit per hour R7 = $525⁄.131 hrs = $4,008 revenue per hour P7 = $100⁄.131 hrs = $ 763 profit per hour R11 = $475⁄.085 hrs = $5,588 revenue per hour P11 = $125⁄.085 hrs = $1,471 profit per hour Running part type 5 will maximize revenue and profits from extra capacity. Calculate the number of extra part type 5 wafers that can be run in the extra hours: n = extra hrs⁄hrs/wafer n = 28.5 hrs⁄.048 hrs/wafer = 593 additional wafers of part type 5 Number of wafer starts of each part type (including required ships) to maximize revenue and profit: Part Type Wafer Starts 3 1,000 5 1500 + 593 = 2,093 7 500 11 3,000
- 125. Numbers and Operations: Graphic Designer Micron Technology, Inc. Job Description: Designs and produces various artwork and layouts for publications, presentations, advertising, and Web sites. Problem: When a graphic designer formats a document, he needs to make the text as easy to read as possible. The length of a line of text in a paragraph affects the readability. If the line is too long, the reader's eye has trouble tracking back to the beginning of the next line. If the line is too short, the eye shifts too often, and many words have to be hyphenated-also interfering with readability. For optimum readability, typesetters agree that the length of a line of text in a document should be no longer (in picas) than 2.5 times the point size of the typeface--nor shorter (in picas) than 1 times the point size--unless the line spacing (leading) is greater than 120% of the point size, which is the default leading of many desktop publishing programs. 12 points = 1 pica 6 picas = 1 inch What is the widest and the narrowest column (or line length) in inches the writer should use for optimum readability with 9-point type on 10.5 point leading?
- 126. Numbers and Operations: Graphic Designer Micron Technology, Inc. Job Description: Designs and produces various artwork and layouts for publications, presentations, advertising, and Web sites. Problem: When a graphic designer formats a document, he needs to make the text as easy to read as possible. The length of a line of text in a paragraph affects the readability. If the line is too long, the reader's eye has trouble tracking back to the beginning of the next line. If the line is too short, the eye shifts too often, and many words have to be hyphenated-also interfering with readability. For optimum readability, typesetters agree that the length of a line of text in a document should be no longer (in picas) than 2.5 times the point size of the typeface--nor shorter (in picas) than 1 times the point size--unless the line spacing (leading) is greater than 120% of the point size, which is the default leading of many desktop publishing programs. 12 points = 1 pica 6 picas = 1 inch What is the widest and the narrowest column (or line length) in inches the writer should use for optimum readability with 9-point type on 10.5 point leading? Solution: 9 point text x 2.5 = 22.5 picas 22.5 picas ÷ 6 picas/inch = 3.75 inches, maximum length 9 picas ÷ 6 picas/inch= 1.5 inches, minimum length 120% leading of 9 point type = 9 x 1.2 = 10.8 points. Note: Increasing the leading to 12 or 13 points would allow the designer to use wider columns.
- 127. Algebra: Manufacturing Engineer Micron Technology, Inc Job Description: Develop wafer level test strategies and test programs. Provide failure analysis reporting. Monitor device yields, failure rates, and repair rates. Interact with various engineering and product groups to optimize device yields and minimize costs. Problem: Following the fabrication process of memory chips, wafers are tested for functionality. Wafers are tested with voltage that stresses performance causing those devices that do not meet specifications to fail. "Yield" refers to the number or percentage of acceptable units produced on each wafer compared to the maximum possible. Yield vs. Parameter – Use graph paper to create an x-y graph of yields vs. electrical parameters measured in volts. Parameter Yield 5v 50% 10v 60% 15v 70% 1. How does the parameter relate to the yield? (Positive vs. negative correlation) 2. Where do we want the parameter to be for the highest yield? 3. What’s the equation of the line? Y = mx+b 4. We did more tests, raising the voltage for the parameter. Plot both the old and new data. Now, where does the graph say to set the volts? Old Data New Data Parameter Yield Parameter Yield 5v 50% 20v 80% 10v 60% 30v 30% 15v 70% 40v 20%
- 128. Algebra: Manufacturing Engineer Micron Technology, Inc Job Description: Develop wafer level test strategies and test programs. Provide failure analysis reporting. Monitor device yields, failure rates, and repair rates. Interact with various engineering and product groups to optimize device yields and minimize costs. Solution: Use graph paper to create an x-y graph of yields vs. electrical parameters measured in volts. Parameter Yield 5v 50% 10v 60% 15v 70% 1. How does the parameter relate to the yield? (Positive vs. negative correlation) Positive correlation - As voltage increases, the yield percentage increases 2. Where do we want the parameter to be for the highest yield? 15v would be the parameter that produces the highest yield. 3. What is the equation of the line? Y = mx + b 50 = mx 5+b 60 = mx 10+b m=2, b=40 Y=2x+40 4. We did more tests, raising the voltage for the parameter. Plot both the old and new data. Now, where does the graph say to set the volts? Based on the new data, set the volts at 20 Old Data New Data Parameter Yield Parameter Yield 5v 50% 20v 80% 10v 60% 30v 30% 15v 70% 40v 20%
- 129. Numbers and Operations: Manufacturing Manager Micron Technology, Inc. Job Description: Manage an engineering and manufacturing group that designs and manufactures PC motherboards. Problem: An FCT (Functional Circuit Tester) takes 6 minutes to test a board. In a month, 30,000 boards need to be tested. How many testers are needed if the testing personnel work 20 hours a day and 22 days per month.
- 130. Numbers and Operations: Manufacturing Manager Micron Technology, Inc. Job Description: Manage an engineering and manufacturing group that designs and manufactures PC motherboards. Problem: An FCT (Functional Circuit Tester) takes 6 minutes to test a board. In a month, 30,000 boards need to be tested. How many testers are needed if the testing personnel work 20 hours a day and 22 days per month. Solution: 22 days/month x 20 hrs/day = 440 hours/month 60 min/hr ÷6 min/board = 10 boards/hr by each tester 10 boards/hr x 440 hrs/month = 4,400 boards/month by each tester 30,000 boards ÷4,400 boards/tester = 6.8 testers or 7 testers needed
- 131. Numbers and Operations: Manufacturing Technician Micron Technology, Inc. Problem: If the total number of "die" (memory chips before assembly and packaging) on an 8-inch wafer is 714, and 370 of those die fail, what is the "yield" (the percent of good die) on the wafer? If the yield on another 8-inch wafer is 67%, how many die have failed?
- 132. Numbers and Operations: Manufacturing Technician Micron Technology, Inc. Problem: If the total number of "die" (memory chips before assembly and packaging) on an 8-inch wafer is 714, and 370 of those die fail, what is the "yield" (the percent of good die) on the wafer? If the yield on another 8-inch wafer is 67%, how many die have failed? Solution: 714 total - 370 failed = yield = good die ÷ total die = 344 ÷ 714 = .4817 or 48.2% yield on 1st wafer 714 total x 67% yield = 478 good die 714 total - 478 good die = 236 failed die/second wafer
- 133. Numbers and Operations: Payroll Clerk Micron Technology, Inc. Job Description: Performs standard accounting clerical duties in the areas of accounts payable, accounts receivable, cost accounting, payroll or credit and collections. Problem: If my biweekly gross is $900, and I get a $5,000 per year raise, how much will my biweekly take-home paycheck increase? Taxes, social security, insurance and retirement contributions take approximately $27% of my gross; but with my raise, the deductions will increase to 32% of my gross. What is the difference between my old hourly wage and my new hourly wage?
- 134. Numbers and Operations: Payroll Clerk Micron Technology, Inc. Job Description: Performs standard accounting clerical duties in the areas of accounts payable, accounts receivable, cost accounting, payroll or credit and collections. Problem: If my biweekly gross is $900, and I get a $5,000 per year raise, how much will my biweekly take-home paycheck increase? Taxes, social security, insurance and retirement contributions take approximately $27% of my gross; but with my raise, the deductions will increase to 32% of my gross. What is the difference between my old hourly wage and my new hourly wage? Solution: 1. $900 x 26 paychecks = $23,400 (annual salary w/o overtime) $23,400 + $5,000 = $28,400 (annual salary with raise) $28,400 ÷ 26 paychecks = $1,092.30 (biweekly gross with raise) 32% of $1092.30 = $1092.30 x .32 = $349.54 $1092.30 - $349.54 = $742.76 (biweekly take home with raise) 27% of $900 = $900 x .27 = $243 $900 - $243 = $657 (biweekly take home before raise) $742.76 - $657.00 = $85.76 (increase in biweekly take home) 2. $900÷80 hours biweekly = $11.25 per hour (before raise) $1,092.30÷80 hours biweekly = $13.65 per hour (after raise) $13.65 - $11.25 = $2.40 per hour raise (difference)
- 135. Geometry: Plumber (1) Micron Technology, Inc. Job Description: Performs a wide variety of skilled plumbing duties in the construction, maintenance, repair and alteration of facilities. Problem: A plumber needs to run three 2-inch lines around the mechanical room. He must offset the pipe around the air handlers in the corners. The outside line is to be 8 inches from the wall and 5 inches from the corner. The spreads between the lines are to be 9 inches and the angle is 45°. Pieces A and B are 3.75 inches and 7.5 inches respectively. What should be the length of pieces C, D, and E?
- 136. Geometry: Plumber (1) Micron Technology, Inc. Job Description: Performs a wide variety of skilled plumbing duties in the construction, maintenance, repair and alteration of facilities. Problem: A plumber needs to run three 2-inch lines around the mechanical room. He must offset the pipe around the air handlers in the corners. The outside line is to be 8 inches from the wall and 5 inches from the corner. The spreads between the lines are to be 9 inches and the angle is 45°. Pieces A and B are 3.75 inches and 7.5 inches respectively. What should be the length of pieces C, D, and E? Solution: Use Pythagorean Theory: a² + b² = c² Special case for 45° right triangles: Since a=b, equation becomes 2a² = c² Length C: side a1 = b1 = 12 + 2 = 14" 2a² = c² 2 x 14² = C² 2 x 196 = C² 392 = C² = 19.8" = Length C Length D: side a2 = 9 + 2 + 12 + 2 - 3.75 (piece A) = 21.25"" 2a² = D² 2 x 21.25² = D² 2 x 451.5 = D² 903 = D² = 30.05" = Length D Length E: side a3 = 9 + 2 + 9 + 2 + 12 + 2 - 7.5 (piece B) = 28.5" 2a² = E² 2 x 28.5² = E² 2 x 812.25 = E² 1624.5 = E² = 40.31" = Side E
- 137. Geometry: Plumber (2) Micron Technology, Inc. Job Description: Performs a wide variety of skilled plumbing duties in the construction, maintenance, repair and alteration of facilities. Problem: The pipe fitters have to be able to calculate the length of pipe needed, including the lengths for corners and bends. What is the length of pipe for the expansion bend shown in the illustration which has a radius (R) of 24 inches, two 10 inch tangents (T), and distance in degrees (D) as shown. Total L = LA + LB + LC + 2LT
- 138. Geometry: Plumber (2) Micron Technology, Inc. Job Description: Performs a wide variety of skilled plumbing duties in the construction, maintenance, repair and alteration of facilities. Problem: The pipe fitters have to be able to calculate the length of pipe needed, including the lengths for corners and bends. What is the length of pipe for the expansion bend shown in the illustration which has a radius (R) of 24 inches, two 10 inch tangents (T), and distance in degrees (D) as shown. Total L = LA + LB + LC + 2LT Solution: Length of the bend: L = 2 π R · D ÷ 360 (R=radius, D=distance in degrees) LA = 2 π 24 · 135 ÷ 360 = 56.538" LB = 2 π 24 · 270 ÷ 360 = 113.076" LC = 2 π 24 · 135 ÷ 360 = 56.538" 56.538 (A) + 113.076 (B) + 56.538 (C) = 226.152 inches Length of the pipe: L + 2T = 226.152 + 20 = 246.152 inches
- 139. Algebra/Geometry: Process Engineer – Cost Micron Technology, Inc Job Description: Responsible for all aspects of product development including design verification and circuit debug, device characterization, text methodology, yield optimization, and cost reduction. Problem: Memory chips are fabricated on silicon wafers. Efforts are made to "shrink" the chip size so that more chips can be made per wafer. Original chip size: x = .5 inch / y = 1 inches - 100 possible chips per wafer The process engineers tell us that we can shrink the chip by 20%... NEW chip size: x = .4 inch / y = .8 inch - 157 possible chips per wafer Wafer diameter = 8 inches Area of a circle = πr2 Assuming that each processed wafer costs $500 ... 1. What is the cost per chip to process the original chip? 2. What is the cost per chip to process the new chip? 3. How much cheaper is it to produce the new chip? 4. If we sell 1,000,000 of these chips every month, how much more money will we make each month with the new chip? 5. The cost to develop the shrink of this chip was $7,000,000. How long will it take to pay for the shrink? Assuming the company has a 10% profit sharing program (this means that every quarter--three-month period--10% of the profits are divided among the employees equally) and employs 5,000 people... 6. How much would your quarterly profit-sharing check increase with this shrink? 7. How much would you have to shrink the chip to be able to buy a $225 digital camera with your profit-sharing check?
- 140. Algebra/Geometry: Process Engineer – Cost Micron Technology, Inc Assuming that each processed wafer costs $500 ... 1. What is the cost to process the original chip? Cost of processed wafer ÷ # of chips $500 ÷ 100 chips = $5 per chip 2. What is the cost to process the new chip? Cost of processed wafer ÷ # of chips $500 ÷ 157 chips = $3.18 per chip 3. How much cheaper is it to produce the new chip? $5 - $3.18 = $1.82 4. If we sell 1,000,000 of these chips every month, how much more money will we make each month? Savings per chip x # of chips sold = additional income / month $1.82 x 1,000,000 chips = $1,820,000 / month 5. The cost to develop the shrink of this chip was $7,000,000. How long will it take to pay for the shrink? (Assume 30 days per month) Cost of development ÷ additional income / month $7,000,000 ÷ $1,820,000 = 3.84 months 3 months + (30 days x .84) = 3 months 25 days 6. (See problem for details.) How much would your quarterly profit-sharing check increase with this shrink? increased income/month x 3 months x 10% = additional profits to share $1,820,000 x 3 x .10 = $546,000 profits to share ÷ #employee = $ for each employee $546,000 ÷ 5000 employees = $109.20 increase for each employee 7. If our chips are selling steadily at $5.00 and our customers are purchasing 1 million parts per month, how much would the company have to shrink the chip to be able to buy a $225 digital camera with your profit-sharing check? Work the problem in reverse: $ needed x #employee = amount needed to split $225 x 5000 = $1,125,000 $1,125,000 x 10 = $11,250,000 total profits needed (share 10% of profits) $11,250,000 ÷3 months = $3,750,000 per month $3,750,000 ÷ 1,000,000 chips = $3.75 profit per chip needed $5 (original cost) - $3.75 = $1.25 cost per chip needed cost per wafer ÷ cost per chip = #chips $500 (cost per wafer) ÷ $1.25 = 400 chips needed per wafer 50.24 (area of wafer) ÷ 400 chips = .125 sq inches (area needed) needed area ÷ original area = % percent of original area .125÷ 5 = .25 or 25% of original area (% of dimensions) 2 = (% of area) √ .25 = .5 or 50% of original dimensions or 50% reduction CHECK: .5 (original width) x 50% = .25 inches 1 (original length) x 50% = .5 inches new chip area = .25 x .5 = .125 square inches Solution
- 141. Algebra/Geometry: Process Engineer – Design Micron Technology, Inc Job Description: Responsible for all aspects of product development including design verification and circuit debug, device characterization, text methodology, yield optimization, and cost reduction. Problem: In the semiconductor industry, shrink refers to the reduction in die (or chip) size. Reducing the size of the die can reduce the cost of manufacturing per die, since you get more die on each wafer. Computers and other devices that require memory are getting smaller and smaller, yet they need more and more memory. Therefore, the memory chips need to be smaller. The smaller the die, the smaller or shorter the circuitry. This means the electrical charges can travel faster from one point to the next. Therefore, shrinking the die can also increase its speed. Wafer diameter = 8 inches Area of a circle = πr2 (pie x radius2 ) Original chip size: x = .5 inch y = 1 inch 1. What is the area of the wafer? 2. What is the area of the chip? 3. Find the number of possible chips per wafer? The process engineers tell us that we can shrink the chip by 20%... 4. What are the new x and y dimensions of the chip? 5. What is the new area of the chip? 6. The x and y dimensions are reduced by a factor of 0.8. By what percentage did the area decrease? 7. Find the number of possible chips per wafer with the new chip size? ORIGINAL chip size: ______ good chips per wafer. New chip size: ______ good chips per wafer.
- 142. Algebra/Geometry: Process Engineer – Design Micron Technology, Inc Job Description: Responsible for all aspects of product development including design verification and circuit debug, device characterization, text methodology, yield optimization, and cost reduction. See problem for details. Solution: Original chip size: x = .5 inch y = 1 inch A 1 = area of Wafer, A2 = area of Chip 1. What is the area of the wafer? area = πr2 = 3.14 x 42 inches = 3.14 x 16 = 50.24 sq inches 2. What is the area of the original chip? area = x * y = .5 inch x 1 inch = .5 square inch 3. Find the number of possible chips per wafer? A 1 ÷ A2 = # of chips 50.24 ÷ .5 = 100.48 or 100 possible chips The process engineers tell us that we can shrink the dimensions of the chip by 20%... 4. What are the new x and y dimensions of the chip? 100%- 20% = 80% = .5 x 80% = .4 inches 1inch x 80% = .8 inches 5. What is the new area of the chip? area = length * width = .4 inch x .8 inch = .32 square inches 6. The new x and y dimensions are 80% of the original. By what percentage did the area decrease? (% of length) x (% of width) = (% of area) = 80% x 80% = 64% of original or % of original = new area = .32 square inches = 64% original area .5 square inches 100% - 64% = 36% decrease 7. Find the number of possible chips per wafer with the new chip size? A 1 ÷ A2 = # of chips, 50.24 ÷ .32 = 157 possible chips ORIGINAL chip size: _100__ good chips per wafer. NEW chip size: _157__ good chips per wafer.
- 143. Measurements: Production Area Supervisor Micron Technology, Inc. Job Description: Supervise, organize and monitor production in the work area while maintaining quality and efficiency. Motivate and evaluate personnel and maintain accurate documentation. Facilitate problem solving and quality improvement activities. Problem: Our company tests computer chips. Each chip must go through both a HOT test and a COLD test. Your goal for the 30-day month is 25,070,216 chips tested through the HOT and COLD tests. It takes 472 seconds to test 32 chips through the HOT step and 208 seconds to test 32 chips through the COLD step. Expect to have to retest 5% due to failure to complete test. The plant is open 24 hours a day, 7 days a week. How many HOT and COLD testers do you need to move an even amount each day?
- 144. Measurements: Production Area Supervisor Micron Technology, Inc. Job Description: Supervise, organize and monitor production in the work area while maintaining quality and efficiency. Motivate and evaluate personnel and maintain accurate documentation. Facilitate problem solving and quality improvement activities. Problem: Our company tests computer chips. Each chip must go through both a HOT test and a COLD test. Your goal for the 30-day month is 25,070,216 chips tested through the HOT and COLD tests. It takes 472 seconds to test 32 chips through the HOT step and 208 seconds to test 32 chips through the COLD step. Expect to have to retest 5% due to failure to complete test. The plant is open 24 hours a day, 7 days a week. How many HOT and COLD testers do you need to move an even amount each day? Solution: 25,070,216 chips + 5% failures = 26,323,726 chips (originally tested) 26,323,726 chips ÷ 30 days = 877,458 chips per day 60 seconds/minute x 60 minutes/hour = 3600 seconds/hr HOT TEST: 3600 seconds ÷ 472-second test = 7.62 x 32 chips per tester = 244 chips per hour 244 chips/hour x 24 hrs/day = 5,856 chips per day per tester 877,458 chips/day ÷ 5,856 chips per tester = 150 HOT testers per day COLD TEST: 3600 seconds ÷ 208-second test = 17.3 x 32 chips per tester = 554 chips per hour 554 chips/hour x 24 hrs/day = 13,296 chips per day per tester 877,458 chips/day ÷ 13,296 chips per tester = 66 COLD testers per day
- 145. Numbers and Operations: Purchasing Agent Micron Technology, Inc. Job Description: Order and maintain sufficient parts to complete customer orders. Assemble and test print heads. Responsible for obtaining materials, components, equipment, and services and evaluate vendor reliability. Problem: "Bare" wafers (wafers waiting to be processed) are currently being stored in two warehouses off site. The cost per month in Warehouse A is $.45 per square foot. The wafers are taking up a 1,200 sq/ft area. Warehouse B as 2,000 sq/ft of wafers, and they cost $.43 per square foot to store each month. What is the total annual cost for storing wafers in the two warehouses? How much could be saved by storing all the wafers in Warehouse B?
- 146. Numbers and Operations: Purchasing Agent Micron Technology, Inc. Job Description: Order and maintain sufficient parts to complete customer orders. Assemble and test print heads. Responsible for obtaining materials, components, equipment, and services and evaluate vendor reliability. Problem: "Bare" wafers (wafers waiting to be processed) are currently being stored in two warehouses off site. The cost per month in Warehouse A is $.45 per square foot. The wafers are taking up a 1,200 sq/ft area. Warehouse B as 2,000 sq/ft of wafers, and they cost $.43 per square foot to store each month. What is the total annual cost for storing wafers in the two warehouses? How much could be saved by storing all the wafers in Warehouse B? Solution: $.45 x 1200 sq ft = $540 per month in Warehouse A $.43 x 2000 sq ft = $860 per month in Warehouse B $540 + $860 = $1,400 x 12 months = $16,800 annually (Warehouses A & B) 1200 + 2000 = 3200 sq ft total storage area 3200 sq ft x .43 = $1376 x 12 months = $16,512 annually (Warehouse B) $16,800 - $16,512 = $288 savings if all stored in Warehouse B
- 147. Analysis & Probability/Algebra: Quality Control (1) Micron Technology, Inc Job Description: Develop wafer level test strategies and test programs. Provide failure analysis reporting. Monitor device yields, failure rates, and repair rates. Interact with various engineering and product groups to optimize device yields and minimize costs. Problem: To create a method control chart for quality control, the following numbers were collected while monitoring a fabrication process. What is the mean of the following set of numbers? What is the standard deviation (STD)? 3.05 3.02 3.03 2.97 2.98 3.10 2.94 3.06 STD = – Where Xi is given numbers M1 is mean n is number of values
- 148. Analysis & Probability/Algebra: Quality Control (1) Micron Technology, Inc Job Description: Develop wafer level test strategies and test programs. Provide failure analysis reporting. Monitor device yields, failure rates, and repair rates. Interact with various engineering and product groups to optimize device yields and minimize costs. See problem for details. Solution: Mean (M1) = sum (Σ) of readings ÷ # of readings M1 = Σ ÷ n = (3.05 + 3.02 + 3.03 + 2.97 + 2.98 + 3.10 + 2.94 + 3.06) ÷ 8 = 24.15÷8 = 3.02 Standard Deviation is a statistical measure of the range of variance or deviation from the average. It describes uniformity: the smaller the number, the more uniform the readings; the larger the number, the greater the deviation. Deviation = difference between the reading (X) and the average of the readings (M1) Dev = X - M1 Standard Deviation (STD) = square root of the mean(M2) of the squares of the difference or deviation of each reading with the mean (M1) of the readings (Xi) STD = – Reading Xi Deviation Xi - M1 Squared Deviation (Xi - M1) 2 3.05 3.05 - 3.02 = .03 .0009 3.02 3.02 - 3.02 = 0 0 3.03 3.03 - 3.02 = .01 .0001 2.97 2.97- 3.02 = -.05 .0025 2.98 2.98 - 3.02 = -.04 .0016 3.10 3.10 - 3.02 = .08 .0064 2.94 2.94 - 3.02 = -.08 .0064 3.06 3.06 - 3.02 = .04 .0016 M2 = (.0009 + 0 + .0001 + .0025 +.0016 + .0064 + .0064 + .0016) / n-1= .0195 / 7 = .0028 STD = √M2 = .053 This means that on average, the readings varied only .053 (higher or lower) than the average of the readings.
- 149. Analysis & Probability/Algebra: Quality Control (2) Micron Technology, Inc Job Description: Develop wafer level test strategies and test programs. Provide failure analysis reporting. Monitor device yields, failure rates, and repair rates. Interact with various engineering and product groups to optimize device yields and minimize costs. Problem: The CTE (Center to Edge) range of the readings taken of the center of a wafer and various points on the edge of a wafer after Chemical Mechanical Planarization (CMP) must be calculated and compared with a critical value to determine if the process is accurate. The critical value is 1000. Determine whether Process A and Process B are accurate (absolute value is less than 1000). X1 X2 X3 X4 X5 Process A 22000 23500 21000 24000 21500 Process B 17000 19000 20000 21000 19700
- 150. Analysis & Probability/Algebra: Quality Control (2) Micron Technology, Inc Job Description: Develop wafer level test strategies and test programs. Provide failure analysis reporting. Monitor device yields, failure rates, and repair rates. Interact with various engineering and product groups to optimize device yields and minimize costs. Problem: The CTE (Center to Edge) range of the readings taken of the center of a wafer and various points on the edge of a wafer after Chemical Mechanical Planarization (CMP) must be calculated and compared with a critical value to determine if the process is accurate. The critical value is 1000. Determine whether Process A and Process B are accurate (absolute value is less than 1000). X1 X2 X3 X4 X5 Process A 22000 23500 21000 24000 21500 Process B 17000 19000 20000 21000 19700 Solution: The CTE range is calculated by finding the difference between the average or mean of the readings taken on the edge of the wafer with the reading taken in the center. Process B is accurate because it has a CTE range less than the critical value of 1000. X1 X2 X3 X4 X5 Process A 22000 23500 21000 24000 21500 Process B 17000 19000 20000 21000 19700
- 151. Algebra: Tax Specialist Micron Technology, Inc Job Description: Prepare state tax returns. Gather information to prepare and calculate apportionment percentages. Research state tax laws. Problem: Micron does business in both Idaho and California. When it is time to file tax returns, California requires a company to apportion its income to California. To apportion the income, the company needs to determine its apportionment percentage. The apportionment percentage is made up of three factors: sales, property, and payroll. In calculating the apportionment percentage, the sales factor is doubled. Each factor is calculated by dividing the activity within California by the activity everywhere. Once the factors are calculated (remembering to double the sales factor), the factors are added together and divided by 3 (the number of factors). What is a company's taxable income in California if the taxable income of the company is $10 million. Taxable income is an apportionment percentage of the total income. California Everywhere Sales 40 100 Property 2 10 Payroll 1 5
- 152. Algebra: Tax Specialist Micron Technology, Inc Job Description: Prepare state tax returns. Gather information to prepare and calculate apportionment percentages. Research state tax laws. See problem for details. Solution: A = Apportionment Percentage A = 2 x sales/sales everywhere + property/property everywhere + payroll/payroll everywhere Apportionment Percentage Calculations: Sales: 40/100 x 2 = .8 Property: 2/10 =.2 Payroll: 1/5 = .2 .8 + .2 + .2 = 1.2 / 3 = .4 apportionment percentage Taxable Income: $10 million (taxable income) x .4 (apportionment %) = $4 million (CA taxable income)
- 153. Algebra/Geometry: Technical Writer Micron Technology, Inc Job Description: Responsible for writing content, editing, and formatting the corporate newsletters as well as working with various departments and outside agency to produce the Annual Report. Problem: The latest issue of the company newsletter needs to include an architect’s rendering of the new facility. The drawing (which will be scanned by the printer because it is too large for the in-house scanners) is 31/2 feet wide by 16 inches high. In preparing the newsletter for the printer, however, the writer must create a placeholder in the newsletter for the drawing. The writer wants the graphic to lay across 2 of the 3 columns of text. The newsletter is a standard 81/2 by 11 inches format, with 1/2-inch margins all around and 1/4-inch “gutters” between columns. What should the measurements be for the placeholder? What is the percent of reduction?
- 154. Algebra/Geometry: Technical Writer Micron Technology, Inc Job Description: Responsible for writing content, editing, and formatting the corporate newsletters as well as working with various departments and outside agency to produce the Annual Report. Problem: The latest issue of the company newsletter needs to include an architect’s rendering of the new facility. The drawing (which will be scanned by the printer because it is too large for the in-house scanners) is 31/2 feet wide by 16 inches high. In preparing the newsletter for the printer, however, the writer must create a placeholder in the newsletter for the drawing. The writer wants the graphic to lay across 2 of the 3 columns of text. The newsletter is a standard 81/2 by 11 inches format, with 1/2-inch margins all around and 1/4-inch “gutters” between columns. What should the measurements be for the placeholder? What is the percent of reduction? Solution: Paper width - margins - gutters ÷3 columns = Column Width 8 1/2 - (2 x 1/2) - (2 x 1/4) = 7 ÷3 = 2 1/3 inches wide 2 columns + gutter = Image Width (2 x 2 1/3) + 1/4 = 4 2/3 + 1/4 = 4 11/12 or 4.916 inches wide Reduced size ÷ Original Size = Percent Reduction 4.916 inches ÷ (3.5 ft x 12 in) = 11.7% reduction Original Height x Percent Reduction = Image Height 16 inches x 11.7% = 1.87 inches High
- 155. Numbers & Operations/Analysis & Probability: Test Engineer Micron Technology, Inc. Job Description: Develop wafer level test strategies and test programs. Provide failure analysis reporting. Monitor device yields, failure rates, and repair rates. Interact with various engineering and product groups to optimize device yields and minimize costs. Problem: Following the fabrication process of memory chips, wafers are tested for functionality. "Yield" refers to the number or percentage of acceptable units produced on each wafer compared to the maximum possible. Backend yields examples: Test1 => 95% Test2 => 90% Test3 => 98% Test4 => 99% In Out In Out In Out In Out 1000 950 950 855 855 838 838 830 1500 1750 2150 500 1. Given the number of parts sent into the first test and the yield of this test, calculate the number of parts out of test one. (Round to the nearest whole part.) 2. Calculate the number of parts out after the 2nd, 3rd and last tests. 3. What is the % yield of all 4 tests together? 4. How many parts do I need to send into the 1st test to get 1000 after the last test? 5. Calculate the yield for each test. Compare the yields of process A with those of process B. Which is better? Which test saw the biggest difference? Test1 Test2 Test3 Test4 Process In Out Yield In Out Yield In Out Yield In Out Yield A 500 415 415 403 403 395 395 392 B 450 385 385 312 312 309 309 306
- 156. Numbers & Operations/Analysis & Probability: Test Engineer Micron Technology, Inc. Job Description: Develop wafer level test strategies and test programs. Provide failure analysis reporting. Monitor device yields, failure rates, and repair rates. Interact with various engineering and product groups to optimize device yields and minimize costs. See problem for details. Solution: Backend yields examples: 1. Given the number of parts sent into the first test and the yield of this test, calculate the number of parts out of test one. (Round to the nearest whole part.) Test1 => 95% In Out 1000 • .95 950 1500 • .95 1425 1750 • .95 1663 2150 • .95 2043 500 • .95 475 2. Calculate the number of parts out after the 2nd , 3rd and last tests. (Answers in the above solution.) 3. What is the % yield of all 4 tests together? .95 x .90 x .98 x .99 = .8295 = 82.95% 4. How many parts do I need to send into the 1st test to get 1000 after the last test? 1000 parts 82.95% = 1206 parts 5. Calculate the yield for each test. Out ÷ In = %Yield Test1 Test2 Test3 Test4 Process In Out Yield In Out Yield In Out Yield In Out Yield A 500 415 83% 415 403 97% 403 395 98% 395 392 99% B 450 385 86% 385 312 81% 312 309 99% 309 306 99% Compare the yields of process A with those of process B. Which is better? Process A: Process A: = Out (Test4) ÷ In (Test1) = 392 ÷ 500 = 78.4% Process B: = Out (Test4) ÷ In (Test1) = 306 ÷ 450 = 68% Which test saw the biggest difference? Test 2: from 97% to 81% Test1 => 95% Test2 => 90% Test3 => 98% Test4 => 99% In Out In Out In Out In Out 1000 950 -> 950 • .9 855 -> 855 • .98 838 -> 838 • .99 830 1500 1425 -> 1425 • .9 1283 -> 1283 • .98 1227-> 1257 • .99 1244 1750 1663 -> 1663 • .9 1497 -> 1497 • .98 1467-> 1467 • .99 1452 2150 2043 -> 2043 • .9 1839 -> 1839 • .98 1802-> 1802 • .99 1784 500 475 -> 475 • .9 428 -> 428 • .98 419-> 419 • .99 415
- 157. Measurements/Geometry: Administrative Assistant & Sales Midgley-Huber, Inc. Job Description: Count all equipment we sell from commercial building plans. Figure the cost of supplying the equipment, profit waived, add freight expense and quote to contractors. Problem: #1: The following louver order needs to have a Kynar finish put on at the factory. The added price is $600 minimum or $7.50 a square foot. 17 each - 16" x 16" 2 each - 42" x 18" 3 each - 30" x 24" What cost will be added to the order for the Kynar finish? #2: To pick the right size of exhaust fan for a room you must first decide how many air changes per hour you want. One air change is equal to the volume of air in the room. Exhaust fans are rated by how many cubic feet per minute (CFM) they move. How many CFMs of air must an exhaust fan move if we want 5 air changes per hour in a room measuring 24' x 24' with a 12' high ceiling?
- 158. Measurements/Geometry: Administrative Assistant & Sales Midgley-Huber, Inc. Job Description: Count all equipment we sell from commercial building plans. Figure the cost of supplying the equipment, profit waived, add freight expense and quote to contractors. Problem: #1: The following louver order needs to have a Kynar finish put on at the factory. The added price is $600 minimum or $7.50 a square foot. 17 each - 16" x 16" 2 each - 42" x 18" 3 each - 30" x 24" What cost will be added to the order for the Kynar finish? #2: To pick the right size of exhaust fan for a room you must first decide how many air changes per hour you want. One air change is equal to the volume of air in the room. Exhaust fans are rated by how many cubic feet per minute (CFM) they move. How many CFMs of air must an exhaust fan move if we want 5 air changes per hour in a room measuring 24' x 24' with a 12' high ceiling? Solution: #1: 16" x 16" = 256 sq.in. x 17 pieces = 4,352 sq. in. 42" x 18" = 756 sq. in. x 2 pieces = 1,512 sq. in. 30" x 24" = 720 sq. in. x 3 pieces = 2,160 sq. in. Total: = 8,024 sq. in. 12" x 12" = 144 in. or 1 sq. ft. 8024 sq. in. ÷ 144 in. = 55.72 sq. ft. 55.72 sq. ft. x $7.50 = $417.90 Since this amount is under the $600 minimum, $600 will be added to the cost. #2: Volume of room = 24' x 24' x 12' = 6912 ft³ Volume of air moved in 5 air changes per hour = 6912 ft³ x 5/hr = 34,560 ft³/hr. Fan rating in CFM = 34,560 ft³/hr x (1 hr ÷ 60 min.) = 576 CFM
- 159. Measurements: Construction Supervisor MONROC Job Description: Responsible for coordinating subcontractors' work and arrival of materials. Insures work is done correctly, safely, and on time. Problem: How many cubic yards of concrete will it take to make a bridge girder, called a voided flat slab? The girder is 48 inches wide, 26 inches tall, and 54 feet long. There are four 16 inch diameter voids within the slab – each is 24 feet long. On each side of the girder is a 1" by 6" keyway.
- 160. Measurements: Construction Supervisor MONROC Job Description: Responsible for coordinating subcontractors' work and arrival of materials. Insures work is done correctly, safely, and on time. Problem: How many cubic yards of concrete will it take to make a bridge girder, called a voided flat slab? The girder is 48 inches wide, 26 inches tall, and 54 feet long. There are four 16 inch diameter voids within the slab – each is 24 feet long. On each side of the girder is a 1" by 6" keyway. Solution: 1. Figure the total possible volume of the voided flat slab. VOL = W x H x L 48" width x 26" height x (54' x 12") length = 808,704 in³ 808,704 in³ ÷ (12" x 12" x 12") = 468 ft³ (volume) 2. Calculate the 1" x 6" keyways on each side of the girder. Girder length in inches = 54’ x 12”/ft = 648” 1" x 6" x 648" = 3,888 in³ ÷ 12" x 12" x 12" = 2.25 ft³ 2.25 ft³ x 2 keyways = 4.5 ft³ for both keyways 3. Calculate the 16" diameter voids. Volume of cylinder = πr² x length r = 16 ÷ 2 = 8 Length of cylinder in inches = 24’ x 12”/ft = 288” 3.14 x (8" x 8”) x 288" = 57,876.48 in³ ÷ (12in/ft) ³ = 33.499 ft³ 33.49 ft³ x 4 voids = 133.97 ft³ 4. Subtract the volume of the keys and voids from the total volume of concrete. 468 ft³ - (4.5 + 133.97) ft³ = 329.53 ft³ ÷ (3ft/yd)³ = 12.21 yd³ This is a common geometry problem for engineers and construction supervisors. It requires math and geometry as well as analytical thought.
- 161. Numbers and Operations: Area Sales Manager Nagel Beverage Company Job Description: Manage sales people in the field and evaluate and train sales staff. Problem: Calculate employee contributions. The company will match retirement contributions by 2% if the employee will contribute 1% of their wages. If an employee's wage for the year is $30,000, how much must the employee contribute to achieve maximum company matching? How much money will the company match?
- 162. Numbers and Operations: Area Sales Manager Nagel Beverage Company Job Description: Manage sales people in the field and evaluate and train sales staff. Problem: Calculate employee contributions. The company will match retirement contributions by 2% if the employee will contribute 1% of their wages. If an employee's wage for the year is $30,000, how much must the employee contribute to achieve maximum company matching? How much money will the company match? Solution: $30,000 x .01 = $300 employee contribution $30,000 x .02 = $600 company matching Total contribution: $300 + $600 = $900
- 163. Numbers and Operations: Licensed Practical Nurse NCS Health Care Job Description: Provides medical care and assists physicians in administering medication and treatments. Problem: A doctor has ordered an injection of Demerol to be given to her patient. She ordered 35 mg be given. How much fluid (ml) would be injected to give the 35 mg ordered? The syringe of Demerol is 50 mg for 2 ml.
- 164. Numbers and Operations: Licensed Practical Nurse NCS Health Care Job Description: Provides medical care and assists physicians in administering medication and treatments. Problem: A doctor has ordered an injection of Demerol to be given to her patient. She ordered 35 mg be given. How much fluid (ml) would be injected to give the 35 mg ordered? The syringe of Demerol is 50 mg for 2 ml. Solution: 35 mg required ÷ 50 mg/2ml = .70 (70% of syringe containing 2 ml) .70 x 2 ml = 1.40 ml fluid to be injected, leaving .6 ml
- 165. Numbers and Operations: Certified Public Accountant O'Neil & Associates CHTD. Job Description: Review business transactions, prepare financial statements, business, tax and financial planning and reporting. Problem: I want to pay a $1,000 bonus to an employee. The IRS requires me to withhold 20% for income taxes and 7.5% for Social Security and Medicare taxes. 1. What should the amount of the bonus be to be certain the employee receives $1,000? 2. As the employer, I have to match the 7.5% Social Security/Medicare tax and send that to the IRS as well. How much money will I have to send to the IRS?
- 166. Numbers and Operations: Certified Public Accountant O'Neil & Associates CHTD. Job Description: Review business transactions, prepare financial statements, business, tax and financial planning and reporting. Problem: I want to pay a $1,000 bonus to an employee. The IRS requires me to withhold 20% for income taxes and 7.5% for Social Security and Medicare taxes. 1. What should the amount of the bonus be to be certain the employee receives $1,000? 2. As the employer, I have to match the 7.5% Social Security/Medicare tax and send that to the IRS as well. How much money will I have to send to the IRS? Solution: 1. b = the bonus amount 20% + 7.5% = 27.5% or .275 b - .275 b = $1,000 .725 b = $1,000 b = $1000 ÷ .725 b = $1,379.31 (bonus amount) 2. bonus - net bonus = taxes withheld $1,379.31 bonus - $1,000 net bonus = $379.31 taxes withheld $1,379.31 x 7.5% = $103.45 Social Security/Medicare tax $379.31 withheld + $103.45 employer match = $478.76 sent to IRS
- 167. Measurements/Geometry: Carpenter (1) Pacific Star Cabinetry Job Description: Carpenter Problem: Determine the lumber cost of a 12' x 16' redwood deck for a hot tub. Solid lumber is sold by the board footage. A board foot is defined as a piece of lumber 1 inch thick by 12 inches wide by 12 inches long, or 144 cubic inches. The 2" x 6" premium-grade redwood used for decking is $1.35 per board foot and is cut 16' long to lay across treated framing. The seven 6" x 6" treated framing are $1.25 per board foot and are cut 13 feet long. How far apart should the framing be?
- 168. Measurements/Geometry: Carpenter (1) Pacific Star Cabinetry Job Description: Carpenter See problem for details. Solution: Use the following formula to determine the amount of board feet in the deck: T x W x L" ÷ 144" = board feet T = thickness in inches W = width in inches L = length in inches Decking redwood = 2" thick x 6" wide (12' x 12") width of deck ÷ 6" width of board = 24 boards (2" x 6" x (16' x 12")) length of deck ÷ 144" = 2,304" ÷ 144" = 16 board feet each 16 x 24 boards = 384 board feet x $1.35 = $518.40 for decking Framing = 6" x 6" thick x 13' 6" x 6" x (13' x 12") ÷ 144" = 5,616 in ÷ 144" = 39 board feet each 39 x 7 boards = 273 board feet x $1.25 = $341.25 for treated posts 518.40 + 341.25 = $859.65 total cost of lumber Calculate length of space between posts: Number of spaces x (width of board +width of space) + extra width of board 6(6" + s) + 6" = 36” + 6s + 6” = 192” 6s + 42” = 192” 6s = 192” – 42” = 150” Length of space = 150”÷ 6 = 25" between boards
- 169. Measurements/Geometry: Carpenter (2) Dan Roman Construction Job Description: Builds residential homes. Problem: A customer would like a bonus room to be added to an existing home. The new room is to be 26' x 24' with an 8' ceiling and a 2' roof overhang. The ridge of the roof is to be centered over the 24 foot wall and 5 feet above the top of the wall of the bonus room. Assuming the builder uses standard 4' x 8' plywood sheets, determine the following: 1. How many plywood sheets will be needed to cover the walls of the bonus room (not accounting for doors or windows). 2. How many plywood sheets will be necessary to cover the roof over the bonus room.
- 170. Measurements/Geometry: Carpenter (2) Dan Roman Construction Job Description: Builds residential homes. Problem: A customer would like a bonus room to be added to an existing home. The new room is to be 26' x 24' with an 8' ceiling and a 2' roof overhang. The ridge of the roof is to be centered over the 24 foot wall and 5 feet above the top of the wall of the bonus room. Assuming the builder uses standard 4' x 8' plywood sheets, determine the following: 1. How many plywood sheets will be needed to cover the walls of the bonus room (not accounting for doors or windows)? 2. How many plywood sheets will be necessary to cover the roof over the bonus room? Solution: 1. To find how many sheets will be necessary for the 4 walls, divide the area of the walls by the area of a plywood sheet (not allowing for doors or windows). [(2 walls) (26' x 8') + (2 walls) (24' x 8')] ÷ 4' x 8' plywood sheet [(2 x 208 sq ft) + (2 x 192 sq ft) ] ÷ 32 sq ft = 25 sheets of plywood for the walls. 2. To find how many sheets will be necessary for the roof, divide the total area of the roof (2 equal sides) by the area of a plywood sheet. The ridge of the roof is 26 feet and the overhang is 2 feet. The height of the roof is 5 feet. Find the area of each side of the roof by calculating the length from the ridge to the edge (a2 + b2 = c2 ), adding the overhang, and multiplying the total length by the width (ridge). 52 + 122 = c2 25 + 144 = c2 = 169 c = 13 ft = length ridge to edge Roof area: (2' + 13') x 26' = 390 sq ft x 2 (both sides) = 780 sq ft Now divide by the plywood sheet dimension: 780 sq ft÷4'x8' sheets = 24.375 or 25 sheets of plywood for the roof
- 171. Measurements: Painter Painting & Repairs Done Right Problem: If paint covers 300 square feet per gallon, how many gallons will it take to paint a home with a flat roof measuring 63 feet long, 47 feet wide, and 10 feet high? If the paint is $17 a gallon, what will the paint cost be? (Assume no door and window areas)
- 172. Measurements: Painter Painting & Repairs Done Right Problem: If paint covers 300 square feet per gallon, how many gallons will it take to paint a home with a flat roof measuring 63 feet long, 47 feet wide, and 10 feet high? If the paint is $17 a gallon, what will the paint cost be? (Assume no door and window areas) Solution: Gallons needed = Area of walls ÷ ft² covered by one gallon = 2(L x H) + 2(W x H) 2(63 x 10) + 2(47 x 10) --------------------------- = ------------------------------ = 7.3 gallons of paint 300 300 8 gallons x $17 = $136 (must buy nearest full gallon)
- 173. Measurements/Geometry: Electrical Contractor Power Plus, Inc. Job Description: Run electrical jobs, including putting up lights and signal poles. Install lights in buildings. Problem: An electrician has to pour a concrete signal base 4' in diameter, 14' deep with two 6" conduits coming up from the bottom and centered in the base. How much concrete does he need to order? Concrete is ordered by the cubic yard. Solid lumber is sold by the board footage. A board foot is defined as a piece of lumber 1 inch thick by 12 inches wide by 12 inches long, or 144 cubic inches.
- 174. Measurements/Geometry: Electrical Contractor Power Plus, Inc. Job Description: Run electrical jobs, including putting up lights and signal poles. Install lights in buildings. Problem: An electrician has to pour a concrete signal base 4' in diameter, 14' deep with two 6" conduits coming up from the bottom and centered in the base. How much concrete does he need to order? Concrete is ordered by the cubic yard. Solid lumber is sold by the board footage. A board foot is defined as a piece of lumber 1 inch thick by 12 inches wide by 12 inches long, or 144 cubic inches. Solution: Subtract the area in cubic feet of the two conduits from the area in cubic feet of the base and translate to cubic yards. diameter of base = 4 ft diameter of each conduit = .5 ft Formula for concrete needed = (area of base ft³) - 2 (area of conduit ft³) yd³ (π r² · h) - 2 (π r² · h) = (3.14 x 2² x 14) - 2 (3.14 x 0.25² x 14) yd³ 3 ft x 3 ft x 3 ft = 175.84 ft³ - 5.495 ft³ 27 ft³ = 6.3 cubic yards
- 175. Numbers and Operations: Screen Printer Pullover Prints Job Description: Customize designs for corporations, business and individuals on textiles. Problem: A dryer cures ink on shirts. The length of the dryer includes 5 feet going in, 5 feet coming out, and the heat chamber of 10 feet. The dryer is 5 feet wide. A shirt load equals 3 shirts across. The average space occupied by shirts is 18" x 20" with about 2 inches spacing between. A length of material can travel from one end to the other of the dryer in about 1 minute, 30 seconds. How many shirts would be on a full belt load? How many shirts can be cured every hour?
- 176. Numbers and Operations: Screen Printer Pullover Prints Job Description: Customize designs for corporations, business and individuals on textiles. Problem: A dryer cures ink on shirts. The length of the dryer includes 5 feet going in, 5 feet coming out, and the heat chamber of 10 feet. The dryer is 5 feet wide. A shirt load equals 3 shirts across. The average space occupied by shirts is 18" x 20" with about 2 inches spacing between. A length of material can travel from one end to the other of the dryer in about 1 minute, 30 seconds. How many shirts would be on a full belt load? How many shirts can be cured every hour? Solution: 5 ft + 10 ft + 5 ft = 20 ft x 12 in/ft = 240 inches (length of dryer) 240 in ÷ (20 + 2) = 10 shirts x 3 shirts across = 30 shirts per load 30 shirts can be completed every 1.5 minutes. 60 minutes÷1.5 min per load x 30 shirts/load = 1,200 shirts/hour
- 177. Numbers and Operations: Realtor R.T. NAHAS Co. Problem: A property owner sells 11.3 acres of a 16.4 acre parcel of land to a retail store. The date of closing of the sales is August 7, 1998. The seller owes taxes on the entire parcel (16.4 acres) from January 1, 1998 to August 7, 1998 plus taxes on 5.1 acres from August 8, 1998 to December 31, 1998. The buyer owes taxes from August 8, 1998 to December 31, 1998 on 11.3 acres. How much of the tax bill is owed by the buyer and how much by the seller of the property? Assumptions: August 7 is the 219th day of the year. Property taxes on the entire parcel are $16,290 per year.
- 178. Numbers and Operations: Realtor R.T. NAHAS Co. Problem: A property owner sells 11.3 acres of a 16.4 acre parcel of land to a retail store. The date of closing of the sales is August 7, 1998. The seller owes taxes on the entire parcel (16.4 acres) from January 1, 1998 to August 7, 1998 plus taxes on 5.1 acres from August 8, 1998 to December 31, 1998. The buyer owes taxes from August 8, 1998 to December 31, 1998 on 11.3 acres. How much of the tax bill is owed by the buyer and how much by the seller of the property? Assumptions: August 7 is the 219th day of the year. Property taxes on the entire parcel are $16,290 per year. Solution: 1. Seller's taxes = taxes owed through August plus the portion of the tax bill due on the remaining acres 219 days / 365 days = .60 = 60% of the year to August 7th $16,290 x 60% year = $9,774 taxes owed through August 7th 16.4 acres - 11.3 acres = 5.1 acres 5.1 acres / 16.4 acres = 31.1% of the property retained Taxes owed on the remaining 5.1 acres from August 8 through December 31 (40% of the year) $16,290 x 40% year x 31.1% property = $2,026.46 $9,774 + 2,026.46 = $11,800.46 TOTAL taxes owed by the seller 2. Buyer's taxes = the difference between the seller's taxes and the total tax bill. $16,290 - $11,800.46 = $4,489.54 taxes on 11.3 acres after sale
- 179. Numbers and Operations: Dental Assistant Reed K Jarvis, DDS PA Job Description: Assists dentist and orders supplies. Chair-side duties allowed by Idaho code include alginate impressions and coronal polish. Problem: A dental assistant needs to order a year's supply of cavity-detecting x-ray film. The practice consists of 800 adults and 700 children who are supposed to come in every 6 months to have checkups. 90% of the patients follow this procedure. Adults need 4 films for Bitewing x-rays. Children need 2 films for Bitewing x-rays. Kodak distributes the film in packages of 100 units. How many packages should the dental assistant order?
- 180. Numbers and Operations: Dental Assistant Reed K Jarvis, DDS PA Job Description: Assists dentist and orders supplies. Chair-side duties allowed by Idaho code include alginate impressions and coronal polish. Problem: A dental assistant needs to order a year's supply of cavity-detecting x-ray film. The practice consists of 800 adults and 700 children who are supposed to come in every 6 months to have checkups. 90% of the patients follow this procedure. Adults need 4 films for Bitewing x-rays. Children need 2 films for Bitewing x-rays. Kodak distributes the film in packages of 100 units. How many packages should the dental assistant order? Solution: 800 adults x 2 visits/year x 90% show-up rate = adult visits 1,600 visits x .90 = 1,440 adult visits 1,440 adult visits x 4 films = 5,760 films for adults 700 children x 2 visits/year x 90% show-up rate = child visits 1,400 visits x .90 = 1,260 child visits 1,260 child visits x 2 films = 2,520 films for children 5,760 + 2,520 = 8,280 films needed 8,280 films ÷100 (films per package) = 82.8 = 83 packages
- 181. Numbers and Operations: Lead Estimator RRR Construction Company Job Description: Management and technical consulting. Problem: An important customer has asked you to bid on a construction project. You estimate labor costs at $100,000; materials and supplies at $50,000; permits and related costs at $10,000; overhead at $20,000; and taxes at $30,000. You know from experience that you should project a contingency of 15% for cost overruns. How much should you bid to complete the project and make a 20% profit? What percent profit would you make if no cost overruns are experienced?
- 182. Numbers and Operations: Lead Estimator RRR Construction Company Job Description: Management and technical consulting. Solution: (See problem page for details.) 1. Your total estimated project costs are $210,000. Personnel $100,000 Materials and supplies $50,000 Permitting & related costs $10,000 Overhead $20,000 Taxes $30,000 TOTAL $210,000 Your total estimated project costs with contingency for cost overruns are $241,500. Total estimated project costs $210,000 x 15 % Contingency = $31,500 Total additional contingency amount TOTAL w/ contingency (cost + contingency) = $241,500 3. Your total bid to the customer should be $289,800. Total estimated project costs with contingency $241,500 x 20% Profit = $48,300 Total additional profit Total bid (costs + contingency + profit) = $289,800 4. Your percent profit if no cost overruns are experienced would be 27%. Total bid with profit $289,800 - $210,000 Total cost with no overruns = $79,800 Total profit with no overruns profit÷ total bid = % profit $79,800÷ $289,800 = .27536 = ~27.5% profit Logic Check: $289,800 (total bid) - 27.563% (profit w/o overruns) = $210,000.63 (cost w/o overruns)
- 183. Analysis & Probability/Numbers & Operations: Salesman Sawtooth Lumber Sales Job Description: Arranges for the transfer (sale) of manufactured lumber from the manufacturer(sawmill) to a user for a fee/commission. Problem: A lumberyard in Salt Lake City wants to buy a truckload of lumber to replenish their inventory. They need two different sizes of dimensional lumber in various lengths. The lumber is shipped from Coeur d'Alene, Idaho. Your problem is to give a quote to the lumberyard on the cost of the shipment. Lumber is sold by a standard unit of measure called the board foot. A board foot is described as a piece of lumber 12 inches wide by 1 inch thick by 1 foot long. Various sizes of lumber are converted to this common unit of measure by use of a factor. The factor is found by using the formula: thickness x width / 12 = the board footage in a 1-foot piece. Therefore, a 2 x 4 x 1-foot piece would equal .6667 board foot (2 x 4 / 12.) For 2 x 10, the factor would be 1.6667. (2 x 10 / 12) We work in quantities of 1,000 board feet. The mill is asking $380.00 per thousand board feet for the 2 x 4 boards and $440.000 per thousand for the 2 x 10 boards. This cost does not include the cost of transportation or the broker’s profit. The broker's profit is determined at 4% of the delivered cost. A standard semi can haul 48,000 lbs. of lumber. The lumber ordered weighs 2,000 lbs per thousand board feet. Thus, the customer is in effect asking for approximately 24,000 board feet. The customer wants the order divided 50% of 2 x 10 boards and 50% of 2 x 4 boards, or approximately 12,000 board feet of each size. The truck needs $1.25 per loaded mile to haul the load. It is 650 miles from Coeur d'Alene, Idaho to Salt Lake City. The customer wants the following lengths: 2 x 4s 200/8' 200/10' 200/12' 400/14' 400/16' 2 x 10s 80/8' 80/10' 160/14' 160/16' 1. What is the total number of board feet being delivered? 2. What will the total cost be to haul the wood? 3. What is the total cost for the wood? 4. What will the brokers fee be? 5. What price should the broker quote that will also include her profit?
- 184. Analysis & Probability/Numbers & Operations: Salesman Sawtooth Lumber Sales Job Description: Arranges for the transfer (sale) of manufactured lumber from the manufacturer(sawmill) to a user for a fee/commission. Solution: (See problem page for details.) 1. Shipment composition: Knowing how many lengths the customer wants, it is necessary to do a running total of the lineal footage, then multiply the total by the factor for the board size to get the number of board feet. 2 x 4s: (200 x 8) + (200 x 10) + (200 x 12) + (400 x 14) + (400 x 16) = 18,000 18,000 feet x .6667 (factor for 2x4s) ≈ 12,000 board feet 2 x 10s: (80 x 8) + (80 x 10) + (160 x 14) + (160 x 16) = 6240 6,240 feet x 1.6667 (factor for 2x10s) = 10,500 board feet Total board feet = 12,000 + 10,500 = 22,500 From this, it is determined that what the customer wants will fit on one truck. 2. Figuring the freight: The truck needs $1.25 per loaded mile to haul the load. It is 650 miles. 650 miles x $1.25/loaded mile = $812.50 $812.50 / 22.5 (thousands of board feet) = $36.11 or ~ $36.00 per thousand board feet. 3. Cost of the wood: The price is determined by the cost of the material at the source plus the freight plus the broker's profit. 2X4: 12,000/1,000 = 12 (Board feet in thousands) 12 X $380.00 = $4560.00 2X10: 10500/1,000= 10.5 (Board feet in thousands) 10.5 X $440.00 = $4620.00 Total Cost of Wood = $4,560 + 4,620 = $9180.00 4. Broker’s Fee: The broker’s fee is 4% of the cost of the wood with freight charges added. (812.50 + 9180.00) X .04 = (9992.50)(.04)= $399.70 5. Quoting a price: The price is determined by the cost of the material at the source plus the freight plus the broker's profit. 812.50 + 9180.00 + 399.70 = $10,392.70 Verbal quotes are given over the phone 40 to 50 times a day. Once the price is agreed upon verbally, the deal is considered made. The ability to estimate accurately is very important. Quote too low and you lose money; too high and you're not competitive. The size of the broker profit leaves little room for error.
- 185. Numbers and Operations: Mobile Veterinarian for Cats and Dogs Mobile Small Animal Clinic Job Description: Provide medical/surgical care for dogs and cats. Problem: A cat has been brought into the clinic to be spayed. Ketamine anesthetic is needed to anesthetize the cat during the surgery. How large should the dosage be in order to safely anesthetize the cat? The cat weighs 10 pounds. The usual dosage given is 10 mg/pound.
- 186. Numbers and Operations: Mobile Veterinarian for Cats and Dogs Mobile Small Animal Clinic Job Description: Provide medical/surgical care for dogs and cats. Problem: A cat has been brought into the clinic to be spayed. Ketamine anesthetic is needed to anesthetize the cat during the surgery. How large should the dosage be in order to safely anesthetize the cat? The cat weighs 10 pounds. The usual dosage given is 10 mg/pound. Solution: 10 lb cat x 10 mg/lb = 100 mg Ketamine.
- 187. Numbers and Operations: Government Agency Manager Social Security Administration Job Description: Administers a Federal social insurance program which pays retirement, survivors, disability and health insurance. Problem: A man or woman retiring in 1999 at age 65 would be eligible for a full retirement benefit. This benefit would be computed based on their average wages over a period of 35 years. The full retirement benefit is reduced by 5/9 of 1% for each month under the age of 65 that a person elects to retire and receive a reduced benefit. If a person who would receive a full benefit of $1,000 per month at age 65 chooses to retire at age 62, how much would that person receive each month?
- 188. Numbers and Operations: Government Agency Manager Social Security Administration Job Description: Administers a Federal social insurance program which pays retirement, survivors, disability and health insurance. Problem: A man or woman retiring in 1999 at age 65 would be eligible for a full retirement benefit. This benefit would be computed based on their average wages over a period of 35 years. The full retirement benefit is reduced by 5/9 of 1% for each month under the age of 65 that a person elects to retire and receive a reduced benefit. If a person who would receive a full benefit of $1,000 per month at age 65 chooses to retire at age 62, how much would that person receive each month? Solution: 65 years - 62 years = 3 years x 12 = 36 months earlier retirement 36 months x 5/9 x 1% = 19.99 or 20% reduction 20% reduction = $1,000 x 0.2 = $200 $1,000 full benefit - $200 reduction = $800.00 reduced benefit
- 189. Numbers and Operations: Operating Room Nurse St. Alphonsus Hospital Problem: The average person has 2,000 cc of blood in their normal circulation. If they lose more than 20% of their volume they need a transfusion. During surgery, blood is lost on sponges. Dry sponges weigh 75 gm. Each cc of blood weighs 1 gm. After a surgery, if there are 10 sponges weighing 1,250 gm, how much blood did the patient lose? Would the patient need a transfusion?
- 190. Numbers and Operations: Operating Room Nurse St. Alphonsus Hospital Problem: The average person has 2,000 cc of blood in their normal circulation. If they lose more than 20% of their volume they need a transfusion. During surgery, blood is lost on sponges. Dry sponges weigh 75 gm. Each cc of blood weighs 1 gm. After a surgery, if there are 10 sponges weighing 1,250 gm, how much blood did the patient lose? Would the patient need a transfusion? Solution: 2000 cc x 20% = 400 cc (blood loss that requires transfusion) 10 sponges x 75 gm = 750 gm (weight of 10 dry sponges) 1250 gm (sponges after surgery) - 750 gm (dry sponge) = 500 gm of blood in sponges 500 gm = 500 cc which is greater than 400 cc (20% of volume), so YES, a transfusion is needed.
- 191. Measurements: Registered Nurse St. Alphonsus Hospital Job Description: Recovery Room nurse monitors and cares for post-operative patients. Dispenses medication, IV fluids, oxygen, etc. Problem: A 220 pound male patient needs an intravenous infusion of dopamine. Dosage range of 2 - 20 mg/kg/minute is titrated. If you begin at a rate of 5 mg/kg/minute with a concentration of 3200mg/CC (mL), what is the rate of infusion at cc/hour for this patient? 1 kg = 2.2 lbs
- 192. Measurements: Registered Nurse St. Alphonsus Hospital Job Description: Recovery Room nurse monitors and cares for post-operative patients. Dispenses medication, IV fluids, oxygen, etc. Problem: A 220 pound male patient needs an intravenous infusion of dopamine. Dosage range of 2 - 20 mg/kg/minute is titrated. If you begin at a rate of 5 mg/kg/minute with a concentration of 3200mg/CC (mL), what is the rate of infusion at cc/hour for this patient? 1 kg = 2.2 lbs Solution: How much does this patient weight in kg? 220 lbs ÷ 2.2 lbs = x ÷ 1 2.2 x = 220 x = 100 kg How much does this patient get in an hour? 5 mg x 100 kg x 60 minutes = 30,000 mg / hour Concentration is 3,200 mg / cc 3200 mg ÷ 30,000 mg = 1 cc ÷ x 3200 x = 30,000 x = 30,000 / 3200 x = 9.375 or ~9 cc/hr
- 193. Numbers and Operations: Telemarketing Manager Telemarketing Job Description: Manage a group of phone telemarketers and help them reach the office averages. Interview, hire and train new employees. Problem: During the year 1998, 109 telemarketers spent 8,943 hours talking to prospective customers. There were 180,317 presentations and 34,751 sales. 1. What was the average number of hours each telemarketer talked to customers? 2. What was the average number of presentations per hour? Sales? 3. What was the conversion percentage of sales to presentations?
- 194. Numbers and Operations: Telemarketing Manager Telemarketing Job Description: Manage a group of phone telemarketers and help them reach the office averages. Interview, hire and train new employees. Problem: During the year 1998, 109 telemarketers spent 8,943 hours talking to prospective customers. There were 180,317 presentations and 34,751 sales. 1. What was the average number of hours each telemarketer talked to customers? 2. What was the average number of presentations per hour? Sales? 3. What was the conversion percentage of sales to presentations? Solution: 1. 8,943 hours ÷109 telemarketers = 82.04 average hours for each telemarketer 2. 180,317 presentations ÷8,943 hours = 20.16 presentations per hour 34,751 sales÷8,943 hours = 3.89 sales per hour 3. 34,751 sales÷180,317 presentations = .1927 or 19.27% conversion of sales to presentations
- 195. Numbers and Operations: Railroad Conductor Union Pacific Railroad Job Description: Moving freight trains from Nampa, Idaho, to LaGrande, Oregon Problem: Today's freight train is mixed freight for transport from Chicago to Portland. I need to determine the speed I will be allowed to travel. To do this, I need to determine tons per operative brake (TOBs). If under 80 TOBs, I can go 70 mph. If between 80 - 100 TOBs, I can travel at 65 mph. If over 100 TOBs, —60 mph. Train data/information: The train is 6110 feet long consisting of 45 cars loaded with freight and 18 empty cars. The total weight is 5813 tons. Each car has 1 operative brake. Total operative brakes: 63 What speed can I travel?
- 196. Numbers and Operations: Railroad Conductor Union Pacific Railroad Job Description: Moving freight trains from Nampa, Idaho, to LaGrande, Oregon Problem: Today's freight train is mixed freight for transport from Chicago to Portland. I need to determine the speed I will be allowed to travel. To do this, I need to determine tons per operative brake (TOBs). If under 80 TOBs, I can go 70 mph. If between 80 - 100 TOBs, I can travel at 65 mph. If over 100 TOBs, —60 mph. Train data/information: The train is 6110 feet long consisting of 45 cars loaded with freight and 18 empty cars. The total weight is 5813 tons. Each car has 1 operative brake. Total operative brakes: 63 What speed can I travel? Solution: TOB = x x = 5813 ÷ 63 If x < 80, speed is 70 miles per hour If x > 80 but < 100, speed is 65 miles per hour If x > 100 then speed is 60 miles per hour. On today's train, x is 92.27. Thus, I can travel 65 miles per hour.
- 197. Analysis & Probability/Geometry: Geologist U.S. Department of the Interior Job Description: Administer the minerals (precious metals) on public lands. Problem: A company has a contract to remove 15,000 cubic yards (truck volume) of pit run (unprocessed) volcanic cinders from Horse Butte Cinder Pit. The cinders are being placed on a new road in a nearby subdivision. The contractor's truck drivers are required to place a signed and dated load ticket in a ticket box located at the pit entrance each time a load of cinders is hauled out of the pit. You, the inspector, have measured his bottom-dumping trucks to determine the volume they hold. They all have the following shape and measurements: When you once visited the pit, you noticed that trucks heaped to overflowing sailed right by the ticket box without leaving tickets on several occasions. (Your presence was not observed). You mention this to the contractor, but still observe some truckers not stopping. After the contractor has finished hauling cinders, you count all load tickets. There are 479. You are concerned and decide to make measurements of the road in the subdivision to determine the volume of cinders hauled. Cinders were placed on 6.6 miles of road. These are your measurements taken at several places along the road: Distance (mi) Width (ft) Thickness (in) 1.0 24 8 2.0 25 10 3.0 23 9 4.0 24 6 5.0 23 9 6.0 25 7 The cinders on the road were partially compacted by the haul trucks and traffic from the subdivision. Was the contractor in violation of the permit (figure 10% compacted)? If so, how much?
- 198. Analysis & Probability/Geometry: Geologist U.S. Department of the Interior Job Description: Administer the minerals (precious metals) on public lands. See problem for details. Solution: Calculations for truck volume problem 24 ft. x 6 ft. x 4 ft = 576 cu. ft. [ (18 ft. x 3 ft. + 24 ft. x 6 ft) ÷ 2 ] (2) = 198 cu. ft. Total truck volume = 576 cu. ft. + 198 cu. ft. = 774 cu. ft. 27 cu. ft. in 1 cu. yd. Convert to cu. yd. = 774 cu. ft. ÷ 27 cu. ft./yd. = 28.67 cu. yd. 479 loads x 28.67 cu. yd. = 13,731.3 cu. yd. Heaped Truck: 774 cu. ft. + 198 cu. ft. = 972 cu. ft. Convert to cu. yd. = 972 cu. ft. ÷ 27 cu. ft./yd = 36 cu. yd. Distance = 6.6 miles; convert to feet = 5280ft/mi x 6.6mi = 34848 ft Average width = 24 ft. Average thickness = 8.2 inches; convert to feet 8.2 in. ÷ 12 in/ft. = 0.683 ft Total hauled = (total road length) x (road width) x (road thickness) = (34848 ft)(24 ft)(0.683 ft) = 571,507 cu. ft. = 21,166 cu. yd. Material compacted about 10%: 21,166 ÷ .9 = 23,518 cu. yd. The contract volume is 15,000 cu.yds. Therefore, the contractor is in violation (+8,518 cu.yds) = 23,518 cu. yd. – 15,000 cu. yd. *This problem demonstrates a case of having more information given than is actually needed.
- 199. Numbers and Operations: Mortgage Loan Officer Washington Mutual Bank Problem: A house costs $120,000. The buyer finances (borrows) $100,000. His payment of principal and interest is $699.22 a month. He decides to pay $100 extra per month. By paying extra, his loan is paid off in 245 months (or 20+ years) versus 360 months (or 30 years). How much did our borrower save over the life of the loan by paying the extra $100 each month?
- 200. Numbers and Operations: Mortgage Loan Officer Washington Mutual Bank Problem: A house costs $120,000. The buyer finances (borrows) $100,000. His payment of principal and interest is $699.22 a month. He decides to pay $100 extra per month. By paying extra, his loan is paid off in 245 months (or 20+ years) versus 360 months (or 30 years). How much did our borrower save over the life of the loan by paying the extra $100 each month? Solution: $699.22 x 360 months = $251,719.20 $799.22 x 245 months = $195,808.90 $251,719.20 - $195,808.90 = $55,910.30 saved
- 201. Algebra: Equipment Engineer Zilog, Inc. Job Description: Repair and maintenance of robotics in the photolithography (optics) department. Problem: In the last ten years, we have seen microchips become larger to accommodate a wide range of electronic functions. At the same time the physical geometries of the chips had to be reduced to improve the electrical characteristics and reduce the overall size. These advances in microchip technology occurred because of the capabilities of the photolithographic equipment. To increase the area of exposure, lenses were made with larger numerical apertures. As the apertures increase, the depth of focus decreases. Depth of focus is a distance measured from the optimum focus plane in which the quality of the printed image does not change. Ten years ago, lenses had a depth of focus of 10 microns and today the normal depth of focus is 1 to 2 microns. As we expose two images on the wafers, it is extremely important that the entire wafer be as close to the optimum focal plane as possible. If the wafer is tilted, a portion of the image may be outside the depth of focus. To insure this does not happen, we run a test called uneven focus. . To calculate the offsets necessary to correct for x and y tilt, substitute the best focus values in the following equations. Compare results to the specification of <25ppm (parts per million). Graphic representation of uneven focus along the y axis
- 202. Algebra: Equipment Engineer Zilog, Inc. Job Description: Repair and maintenance of robotics in the photolithography (optics) department. See problem for details. Solution: The following are the calculations required to determine the tilt correction: This value exceeds the 25 ppm specification. Therefore, a software offset must be made.
- 203. Algebra: Process Engineer Zilog, Inc. Job Description: Maintains, improves, and develops new processes for the microscopic printing process of computer chips. Must carry out experiments for testing possible improvements and use statistics, charts, graphs, and text to report findings. Problem: In the computer industry, the photo process involves printing the small features on the computer chips. One basic unit of length is the micron (µ). 1µ = 1/1,000,000 meter. Printing the small features is difficult to do and difficult to measure. How well a computer chip performs depends largely on the size of the features; measuring them accurately is very important. A Scanning Electronic Microscope (SEM) is used and magnifies the image around 50,000 times. The SEM may measure 5% high or 5% low. How can the SEM measurements be calibrated? These small measurements are called Critical Dimensions (CDs). A 'Line' is a printed feature of material. A 'Space' is where material was removed in between Lines. The 'Pitch' is the sum of a Line and Space. The pitch is always constant whether the Line is bigger or smaller than normal. From design drawings the actual Pitch (AP) is 1.5µ. The SEM CD measured Pitch (MP) is 1.45µ. The SEM CD measured Line (ML) is 0.85µ. What is the actual Line (AL) width?
- 204. Algebra: Process Engineer Zilog, Inc. Job Description: Maintains, improves, and develops new processes for the microscopic printing process of computer chips. Must carry out experiments for testing possible improvements and use statistics, charts, graphs, and text to report findings. Problem: In the computer industry, the photo process involves printing the small features on the computer chips. One basic unit of length is the micron (µ). 1µ = 1/1,000,000 meter. Printing the small features is difficult to do and difficult to measure. How well a computer chip performs depends largely on the size of the features; measuring them accurately is very important. A Scanning Electronic Microscope (SEM) is used and magnifies the image around 50,000 times. The SEM may measure 5% high or 5% low. How can the SEM measurements be calibrated? These small measurements are called Critical Dimensions (CDs). A 'Line' is a printed feature of material. A 'Space' is where material was removed in between Lines. The 'Pitch' is the sum of a Line and Space. The pitch is always constant whether the Line is bigger or smaller than normal. From design drawings the actual Pitch (AP) is 1.5µ. The SEM CD measured Pitch (MP) is 1.45µ. The SEM CD measured Line (ML) is 0.85µ. What is the actual Line (AL) width? Solution: Proportional fractions The ratio of Actual Pitch (AP) / Measured Pitch = Actual Line (AL) / Measured Line (ML) 1.5µ / 1.45 = AL / 0.85 AL = (0.85µ) x (1.5÷ 1.45) = 0.88µ

Be the first to comment