Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

No Downloads

Total views

2,916

On SlideShare

0

From Embeds

0

Number of Embeds

458

Shares

0

Downloads

48

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Algebra 2 WARM-UP 1 WORD PROBLEM WARM-UP 1How long will it take 100 storks to catch 100 frogs,when five storks need five minutes to catch five frogs?Answer: 5 minutes.
- 2. Algebra 2 WARM-UP 2 WORD PROBLEM WARM-UP 1Logan, Vikrant and Emily differ greatly in height.Vikrant is 14” taller than Emily. The difference betweenVikrant and Logan is two inches less than between Loganand Emily. Vikrant at 6’-6” is the tallest of the three.How tall are Logan and Emily?Answer: If Vikrant is 6’-6”, then Emily must be 5’-4”,and Logan must be 6’-0” (8” greater than Emily and 6”less than Vikrant).
- 3. Algebra 2 WARM-UP 3 WORD PROBLEM WARM-UP 1Find the numbers that will replace letters a and b sothat the five-digit number will be divisible by 36: 19a 9b (Note: There are two possible solutions)Answer: 19692 and 19296. To be divisible by 36, anumber must be divisible by 9 and 4. To be divisible by9, the sum of the digits must be divisible by 9. The lasttwo digits must be divisible by 4. Therefore, b can beeither 2 or 6.
- 4. Algebra 2 WARM-UP 4 WORD PROBLEM WARM-UP 1On the way home from school, Tom found out that hegot only half the allowance that Mark got. Suzi is threeyears older and receives three times what Tom gets.Together, the three receive $144. How much is eachstudent getting?Answer: Divide the total by 6: 144/6 =24. ThereforeTom gets $24; Mark gets $48; and Suzi receives $72.
- 5. Algebra 2 WARM-UP 5 WORD PROBLEM WARM-UP 1Students in class with less than 30 students finishedtheir algebra test. 1/3 of the class received a “B”, ¼received a “B-”, and 1/6 received a “C”. 1/8 of the classfailed. How many students received an “A”.Answer: There were 3 “A’s”. Look for a commondenominator – the only one smaller than 30 is 24. Whenyou add up the known fractions, you have 21/24.
- 6. Algebra 2 WARM-UP 6 WORD PROBLEM WARM-UP 1On a road 75 miles long, two trucks approach eachother. Truck A is traveling at 55 mph while Truck B istraveling at 80 mph. What is the distance between thetwo trucks one minute before they collide?Answer: 2.25 miles. The trucks are approaching eachother at a speed of 135 mph (55 + 80). 135/60=2.25
- 7. Algebra 2 WARM-UP 7WORD PROBLEM WARM-UP 1Ten years more than three times Charlie’s age is twoyears less than five times his age. How old is he?Answer: 6 years.
- 8. Algebra 2 WARM-UP 8 WORD PROBLEM WARM-UP 1The average age of the three Wilson children is 7 years. If the two younger children are 4 years old and 7 years old, how many years old is the oldest child?Answer: 10 years.
- 9. Algebra 2 WARM-UP 9WORD PROBLEM WARM-UP 1A box of 100 personalized pencils costs $30. How many dollars does it cost to buy 2500 pencils?Answer: $750.
- 10. Algebra 2 WARM-UP 10 WORD PROBLEM WARM-UP 1Jeff has an equal number of nickels, dimes and quarters worth a total of $1.20. Anne has one more of each type of coin than Jeff has. How many coins does Anne have? If x y = 6 and x + y = 12, what is the value of y?Answer: 12 coins.
- 11. Algebra 2 WARM-UP 11 WORD PROBLEM WARM-UP 1Alex has fifteen nickels and dimes. He has seven morenickels than dimes. How many of each coin does he have?Answer: 11 nickels and 4 dimes. If x y = 6 and x + y = 12, what is the value of y? n d 15 n d 7 (d 7) d 15 2d 7 15 d 4
- 12. Algebra 2 WARM-UP 12 WORD PROBLEM WARM-UP 1Joel has two fewer quarters than dimes and a total offourteen dimes and quarters. How many of each coindoes he have?Answer: 8 dimes and 6 quarters. q d 2 If x y = 6 and x + y = 12, what is the value of y? q d 14 (d 2) d 14 2d 2 14 d 8
- 13. Algebra 2 WARM-UP 13 WORD PROBLEM WARM-UP 1Ten years more than three times Charlie’s age is two years less than five times his age. How old is Charlie?Answer: 6 years old. 3C 10 5C 2 If x y = 6 and x + y = 12, what is the value of y?
- 14. Algebra 2 WARM-UP 14 WORD PROBLEM WARM-UP 1When Alice is three times as old as she was five years ago, she will be twice her present age. How old is she?Answer: 15 years old. 3(A 5) 2A If x y = 6 and x + y = 12, what is the value of y?
- 15. Algebra 2 WARM-UP 15 WORD PROBLEM WARM-UP 1The sum of Gary’s and Vivian’s ages is twenty-threeyears. Gary is seven years older than Vivian. How old iseach person?Answer: Vivian – 8 years; Gary - 15 years. G V 23 G V 7 V (V 7) 23 V 2 7 23
- 16. Algebra 2 WARM-UP 16 WORD PROBLEM WARM-UP 1Brad is five years younger than Louise. The sum of theirages is thirty-one years. How old is each person?Answer: Brad – 13 years; Louise – 18 years. B L 5 B L 31 (L 5) L 31 2L 5 31
- 17. Algebra 2 WARM-UP 17 WORD PROBLEM WARM-UP 1The sum of the ages of Juan and Herman is twenty-fouryears. Juan is twice as old as Herman. How old is each?Answer: Juan – 16 years; Herman – 8 years. J H 24 J 2H (2H ) H 24 3H 24
- 18. Algebra 2 WARM-UP 18 WORD PROBLEM WARM-UP 1If Edith were five years older, she would be twiceFred’s age. If she were three years younger, she wouldbe exactly his age. How old is each one?Answer: Edith – 11 years; Fred – 8 years. E 5 2F E 3 F E 5 2(E 3) E 5 2E 6
- 19. Algebra 2 WARM-UP 19 WORD PROBLEM WARM-UP 1When Leonard is five years older than double hispresent age, he will be three times as old as he was ayear ago. How old is he?Answer: Leonard – 8 years. 2L 5 3(L 1) 2L 5 3L 3 8L
- 20. Algebra 2 WARM-UP 20 WORD PROBLEM WARM-UP 1If Karen were two years older than she is, she would betwice as old as Larry, who is eight years younger thanshe. How old is each?Answer: Karen – 18 years; Larry – 10 years. 2L 5 3(L 1) 2L 5 3L 3
- 21. Algebra 2 WARM-UP 21 WORD PROBLEM WARM-UP 1Yolanda has a total of thirty-seven nickels and dimes.The dimes come to 40¢ more than the nickels. How manyof each coin does she have?Answer: 22 nickels; 15 dimes. n d 37 10d 5n 40 n n 4 37 2
- 22. Algebra 2 WARM-UP 22 WORD PROBLEM WARM-UP 1Sue has a total of forty nickels and dimes. She has twomore dimes than nickels. If she had eleven more coins,she would have 90¢ more. How many nickels and dimesdoes she have?Answer: 19 nickels; 21 dimes. n d 40 d n 2 n (n 2) 40
- 23. Algebra 2 WARM-UP 23 WORD PROBLEM WARM-UP 1Lisa has a total of fifty-four nickels and dimes. If shehad three more nickels, the value of the coins would be$4. How many of each does she have?Answer: 31 nickels; 23 dimes. n d 54 5(n 3) 10d 400 or 5n 10d 400 15
- 24. Algebra 2 WARM-UP 24 WORD PROBLEM WARM-UP 1Amy has two more nickels than dimes and five moredimes than quarters. Her nickels, dimes, and quarterstotal $3.25. How many of each kind does she have?Answer: 13 nickels; 11 dimes; 6 quarters. n d 2 d q 5 5n 10d 25q 325
- 25. Algebra 2 WARM-UP 25 WORD PROBLEM WARM-UP 1Luke has three times as many nickels as dimes and fivetimes as many pennies as nickels. He has $2.80. Howmany of each coin does he have?Answer: 105 pennies; 21 nickels; 7 dimes. n 3d p 5n p 5n 10d 280
- 26. Algebra 2 WARM-UP 26 WORD PROBLEM WARM-UP 1If Eustace had twice as many nickels and half as manyquarters, he would have 60¢ less. Suppose he now hassixteen nickels and quarters. How many of each kinddoes he have?Answer: 105 pennies; 21 nickels; 7 dimes. n 3d p 5n p 5n 10d 280
- 27. Algebra 2 WARM-UP 27 WORD PROBLEM WARM-UP 1A rectangle whose perimeter is fifty feet is five feetlonger than it is wide. What are its dimensions? What isits area?Answer: w = 10 ft; l = 15 ft; A = 150 square feet P 2w 2l P 2w 2(w 5) P 4w 10
- 28. Algebra 2 WARM-UP 28 WORD PROBLEM WARM-UP 1You are given the formula A = bc.Rewrite the given equation to show the effect of eachstatement. If b is increased by 6 and c is… a. decreased by 2, then A increases by 15. b. increased by 2, then A doubles.Answer: a. A 15 (b 6)(c 2) b. 2A (b 6)(c 2)
- 29. Algebra 2 WARM-UP 29 WORD PROBLEM WARM-UP 1You are given the formula A = bc.What is the effect on A if… a. b is doubled and c is unchanged? b. b is doubled and c is halved? c. b is tripled and c is doubled?Answer: a. A is doubled b. A is unchanged c. A is six times as much
- 30. Algebra 2 WARM-UP 30 WORD PROBLEM WARM-UP 1You are given the formula for the area of a rectangle,A = lw, where l and w are in feet. Rewrite the givenequation to show the effect of each statement. a. If the length increases by 5 feet and the width is unchanged, then the area increases by 40 square feet. b. The width is two-thirds of the length.Answer: a. A 40 (l 5)w 2 2 b. A l 3
- 31. Algebra 2 WARM-UP 31 WORD PROBLEM WARM-UP 175% of the length of a rectangle and 20% of its widthare eliminated. How does the area of the resultingrectangle compare with the area of the originalrectangle?Answer: New area is 20% of original area A lw (.25l )(.80w ) .2(lw ) .20A
- 32. Algebra 2 WARM-UP 32 WORD PROBLEM WARM-UP 1The width of a rectangle is 40 cm less than itsperimeter. The rectangle’s area is 102 sq. cm. What arethe rectangle’s dimensions?Answer: 6 cm by 17 cm P 2(l w ) w P 40 102 lw
- 33. Algebra 2 WARM-UP 33 WORD PROBLEM WARM-UP 1A rectangle is three centimeters longer than it is wide.If its length were to be decreased by two centimeters,its area would decrease by thirty square centimeters.What is its area?Answer: 270 square centimeters l w 3 A lw A 30 (l 2)w
- 34. Algebra 2 WARM-UP 34 WORD PROBLEM WARM-UP 1Porter drove for 3 hours at 40 mph and for 2 hours at50 mph. What was her average speed during that time?Answer: 44 mph D rt sumof distances average speed= sumof times D 40(3); D 50(2) 1 2
- 35. Algebra 2 WARM-UP 35 WORD PROBLEM WARM-UP 1A car traveled from A to B at 50 mph, from B to C at 60mph, and returned (C to B to A) at 80 mph. What wasthe average speed on the round trip if the distancefrom A to B is 100 miles and from B to C is 120 miles?Answer: 65 5 mph 27 sumofdistances a average speed= sumof times D D 2(D D ) a 1 ; a 2 1 2 t t 1 2 t t t 1 2 3
- 36. Algebra 2 WARM-UP 36 WORD PROBLEM WARM-UP 1It took 3 hours and 40 minutes for a car traveling at 60mph to go from A to B.a) How long will the return trip take if the car travels at 80 mph?b) What must the car’s average speed be from B to A if the return trip is to be made in 2-1/2 hours?Answer: 2.75 hours D = 60 x 3-2/3 = 220 miles 220/80 = 2.75 hours
- 37. Algebra 2 WARM-UP 37 WORD PROBLEM WARM-UP 1A road runs parallel to a railroad track. A car travelingan average speed of 50 mph starts out on the road atnoon. One hour later, a train traveling an average speedof 90 mph in the same direction as the car passes thespot where the car started. If the car and the traincontinue to travel along parallel paths, at what time willthe train overtake the car?Answer: 2:15 pm. D = 50t = 90(t-1)
- 38. Algebra 2 WARM-UP 38 WORD PROBLEM WARM-UP 1A car traveling parallel to a railroad track at an averagespeed of 55 mph starts out on the road at noon. A traintraveling at an average speed of 95 mph in the samedirection also starts at noon. They both arrive at thesame spot at 2:15 pm. How far ahead of the train wasthe car when they both began?90 miles. Dt= 95(9/4); Dc=55(9/4)Difference = Dt- Dc
- 39. Algebra 2 WARM-UP 39 WORD PROBLEM WARM-UP 1Two planes fly at the same speed in still air. They leavethe airport at the same time and fly in the same aircurrent but in opposite directions. The plane going withthe air current is 1,470 miles from the airport 3 hoursafter takeoff. The plane flying against the air current is2,050 miles from the airport 5 hours after takeoff.What is the speed of the air current?40 mph. 1470 = (r + c)(3); 2050 = (r – c)(5)
- 40. Algebra 2 WARM-UP 40 WORD PROBLEM WARM-UP 1Two canoeists paddle the same rate in still water. Onecanoeist paddled upstream for 1-1/2 hours and was 18miles from the starting point. The other canoeistpaddled downstream for 2 hours and was 36 miles fromthe starting point. At what speed do the canoeistspaddle in still water?15 mph. 18 = (r – c)(1.5); 36 = (r + c)(2)
- 41. Algebra 2 WARM-UP 41 WORD PROBLEM WARM-UP 1It took Dana 6 minutes to circle a quarter-mile trackthree times. What was her average speed?7½ mph. 3(1/4) = r(6/60)
- 42. Algebra 2 WARM-UP 42 WORD PROBLEM WARM-UP 1A driver averaged a speed of 20 mph more for a tripfrom A to B than on the return trip. The return triptook one-and-a-half times as long. What was theaverage speed from a) A to B b) B to Aa) 60 mph; b) 40 mph; D = rt = (r – 20)(3/2t)
- 43. Algebra 2 WARM-UP 43 WORD PROBLEM WARM-UP 1A runner averaged 8 kph during a race. If she hadaveraged 1 kph more, she would have finished in 20minutes less. How long did it take her to finish therace?3 hours; D = 8t = (8 + 1)(t – 20/60)
- 44. Algebra 2 WARM-UP 44 WORD PROBLEM WARM-UP 1A driver drove at 80 kph for 20 minutes of a 1 hour trip.His average speed for the whole trip was 75 kph. Whatwas his average speed for the other 40 minutes of thetrip?72-1/2 kph; D = 75(1) = 80(20/60) + r(40/60)
- 45. Algebra 2 WARM-UP 45 WORD PROBLEM WARM-UP 1Nikita has already driven 1 mile at 30 mph. How fastmust she drive the second mile so that the averagespeed for her trip is 60mph?Answer: She cannot drive fast enough. She has already used up all of her time.
- 46. Algebra 2 WARM-UP 46 WORD PROBLEM WARM-UP 1Assume that all masons work at the same rate of speed.If it takes eight masons (all working at the same time)fifteen days to do a job, how long will it take for the jobto be done by ten masons?8 masons x 15 days = 120 mason-days. Therefore 10masons will take 12 days.
- 47. Algebra 2 WARM-UP 47 WORD PROBLEM WARM-UP 1Suppose the amount of water that can flow through twopipes is directly proportional to the squares of theirradii. Pipe A has a radius of 3 inches and water flowsthrough it at 150 gallons per second. At what rate willwater flow through Pipe B which has a radius of 4inches?R=kr 2. Therefore k 32 , so k 150 50 . 9 3
- 48. Algebra 2 WARM-UP 48 WORD PROBLEM WARM-UP 1Harold and Jem together can do a job in six days.Harold can do the job working alone in eight days. Howlong does it take Jem to do the job working alone?24 Days.1 1 1 ; H 8H J 6
- 49. Algebra 2 WARM-UP 49 WORD PROBLEM WARM-UP 1It takes four minutes to fill a bathtub if the water isfull open and the drain is closed. It takes six minutes toempty the tub if the drain is open and the water isturned off. How long will it take to fill the tub if thewater is fully turned on and the drain is open? 1 1 112 minutes 4 6 m
- 50. Algebra 2 WARM-UP 50 WORD PROBLEM WARM-UP 1Two bricklayers working together can do a job in 8 days.One of the bricklayers takes 12 days to do the jobalone. How long does it take the other bricklayer to dothe job?Answer: 24 days. Look at how much is accomplished per day. Together they complete 1/8 of the job in one day. One bricklayer would complete 1/12 of the job in one day. Therefore… 1/8 – 1/12 = 1/24
- 51. Algebra 2 WARM-UP 51 WORD PROBLEM WARM-UP 1A 15,000 gallon water tank can be filled in 20 minuteswith two intake pipes, one of which allows a 40% greaterflow than the other. At what rate does the water flowthrough each of the two pipes?Answer A = 312.5 gpm & B = 437.5 gpm pipe A + pipe B = 750 gallon/minute (gpm) Since pipe B = 1.40 A, we can say… 1.00 A + 1.40 A = 750 gpm A = 312.5 gpm & B = 437.5 gpm
- 52. Algebra 2 WARM-UP 52 WORD PROBLEM WARM-UP 1Jeff takes 40% longer than Ken to do a job. Jeff andKen working together can do the job in thirty-fivehours. How long does it take each of them working aloneto do the job?J 1.40K 1 1 1 J K 35

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment