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# Practice Problems

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### Practice Problems

1. 1. PRACTICE PROBLEMS Demand
2. 2. <ul><li>Problem: 1 </li></ul><ul><li>Construct a demand schedule for a product X for alternative prices Re.1, 2, 3, 4,and 5 given its demand function : Dx = 90 – 2Px. Draw the demand curve for the same? What is its mathematical attribute? </li></ul>
3. 3. <ul><li>Solution:1 </li></ul><ul><li>P = 1 … Dx = 90 – 2 Px = 90 – 2 x 1 = 90 – 2 = 88 </li></ul><ul><li>P = 2 ….. Dx = 90 – 2 x 2 = 90-4 = 86 </li></ul><ul><li>P = 3 ….. Dx = 90 – 2 x 3 = 90-6 = 84 </li></ul><ul><li>P = 4 … Dx = 90 – 8 = 82 </li></ul><ul><li>P = 5 … Dx = 90 – 10 = 80 </li></ul><ul><li>Mathematical attribute = Dx = f (Px), ∆ Dx < 0 </li></ul><ul><li> ∆ Px </li></ul>
4. 4. <ul><li>Problem: 2 </li></ul><ul><li>The demand curve for a commodity X is represented by Qx = 160000 – 1000Px. Construct the demand schedule assuming initial price to be Rs.100 and consequent increase by Rs.10 upto Rs.150. Plot the demand curve. </li></ul>
5. 5. <ul><li>Solution:2 </li></ul><ul><li>P = 100 Dx = 160000 – 1000 Px = 160000 - 1000 x 100 </li></ul><ul><li>= 160000 – 100000 = 60000 </li></ul><ul><li>P = 110 Dx = 160000 – 1000 x 110 = 160000 – 110000 = 50000 </li></ul><ul><li>P = 120 Dx = 40000 </li></ul><ul><li>P = 130 Dx = 30000 </li></ul><ul><li>P = 140 Dx = 20000 </li></ul><ul><li>P = 150 Dx = 10000 </li></ul>
6. 6. <ul><li>Problem – 3 </li></ul><ul><li>Suppose the demand function for Komal butter in a town is estimated to be Qd = 600 – 5P when Qd is the quantity demanded of butter (in ‘000 kgs per week) and P stands for price, </li></ul><ul><li>a) estimate at what price demand would be zero. </li></ul><ul><li>b) Draw a demand curve at alternative prices Rs.25, 35, 50, 60 and 80. </li></ul><ul><li>c) What is the statistical characteristics of this demand curve. – Downward sloping showing demand steadily falls as price increases. </li></ul>
7. 7. <ul><li>Solution: </li></ul><ul><li>a) estimate at what price demand would be zero. </li></ul><ul><li>Qd = 600 – 5P </li></ul><ul><li>Put Qd = 0, then </li></ul><ul><li>0 = 600 – 5P, </li></ul><ul><li>5P = 600 </li></ul><ul><li>P = 600/5 = 120. </li></ul><ul><li>When demand is zero, price is Rs.120. </li></ul>
8. 8. <ul><li>b) Draw a demand curve at alternative prices Rs.25, 35, 50, 60 and 80. </li></ul><ul><li>P = 25, Qd = 600 – 5P = 600 – 5 x 25 = 600 – 125 = 475 </li></ul><ul><li>P = 35, Qd = 600 - 5P = 600 – 5 x 35 = 600 – 175 = 425 </li></ul><ul><li>P = 50 Qd = 600 – 5 x 50 = 600 – 250 = 350 </li></ul><ul><li>P = 60 Qd = 600 – 5 x 60 = 600 – 300 = 300 </li></ul><ul><li>P = 80 Qd = 600 - 5 x 80 = 600 – 400 = 200 </li></ul>Price Quantity 25 35 50 60 80 200 300 350 425 475
9. 9. <ul><li>c) What is the statistical characteristics of this demand curve. – Downward sloping showing demand steadily falls as price increases. </li></ul><ul><li>Downward sloping showing demand steadily falls as price increases. </li></ul>
10. 10. <ul><li>Problem : 4 </li></ul><ul><li>Central Plaza conducted a study of the demand for men’s ties. It found that the average monthly demand (D) in terms of price (P) is given by the equation Dx = 800 – 5P. </li></ul><ul><li>(a)How many ties per day can its store expect to sell at a price of Rs.100 per tie? </li></ul><ul><li>(b) If the store wants to sell 500 ties per month, what price it should charge? </li></ul>
11. 11. <ul><li>Solution: </li></ul><ul><li>(a)How many ties per day can its store expect to sell at a price of Rs.100 per tie? </li></ul><ul><li>Demand Equation for ties is Dx = 800 – 5P. </li></ul><ul><li>If P = 100, Dx = 800 – 5 x 100 = 300 </li></ul><ul><li>Demand is 300 ties per month, or 10 per day. </li></ul>
12. 12. <ul><li>Solution: </li></ul><ul><li>(b) If the store wants to sell 500 ties per month, what price it should charge? </li></ul><ul><li>Demand Equation is DX = 800 - 5Px </li></ul><ul><li>If Dx = 500, </li></ul><ul><li>500 = 800 – 5P, </li></ul><ul><li>5P = 800 – 500 = 300, </li></ul><ul><li>P = 300/5 = 60. </li></ul><ul><li>To sell 500 ties, the price will be Rs.60. </li></ul>
13. 13. <ul><li>Problem – 5: </li></ul><ul><li>The demand function for beer in a city is given as </li></ul><ul><li>Qd = 400 – 4P, </li></ul><ul><li>Where Qd = quantity demanded of beer (in ‘000 bottles per week), P = price of beer per bottle. </li></ul><ul><li>a) Construct a demand curve assuming price Rs.10, 12, 15, 20, 25 per bottle. </li></ul><ul><li>b) At what price would the demand be zero. </li></ul><ul><li>c) If the producer wants to sell 380,000 bottles per week, what price should he charge? </li></ul>
14. 14. <ul><li>Demand Schedule: </li></ul><ul><li>P = 10 : Qd = 400 – 4 x 10 = 360 </li></ul><ul><li>P = 12 : Qd = 400 – 4 x 12 = 352 </li></ul><ul><li>P = 15 : Qd = 400 – 4 x 15 = 340 </li></ul><ul><li>P = 20 : Qd = 400 – 4 x 20 = 320 </li></ul><ul><li>P = 25 : Qd = 400 – 4 x 25 = 300 </li></ul><ul><li>This data has been plotted on a graph paper </li></ul><ul><li>as under: </li></ul>
15. 15. <ul><li>P </li></ul><ul><li>25 - </li></ul><ul><li>20 - </li></ul><ul><li>15 – </li></ul><ul><li>10 – </li></ul><ul><li>0 </li></ul><ul><li> 300 310 320 330 340 350 360 Qd </li></ul>
16. 16. b) At what price would the demand be zero. <ul><li>In the equation : Qd = 400 – 4P, </li></ul><ul><li>let us put Qd = 0 </li></ul><ul><li>400 – 4P = 0 , 4P = 400, P = 100. </li></ul><ul><li>That is to say, at the price of Rs.100 per bottle, the demand for beer will be zero. </li></ul>
17. 17. <ul><li>c) If the producer wants to sell 380,000 bottles per week, what price should he charge? </li></ul><ul><li>380,000 bottles is 380 (‘000 omitted) </li></ul><ul><li>In the given equation Qd = 400 – 4P, </li></ul><ul><li>Put Qd = 380, </li></ul><ul><li>380 = 400 – 4P, </li></ul><ul><li>4P = 400 – 380 </li></ul><ul><li>4P = 20, </li></ul><ul><li>P = 5 </li></ul><ul><li>The producer should fix the price at Rs.5 per bottle, in order to sell 380,000 bottles per week. </li></ul>
18. 18. PRACTICE PROBLEMS Method of Least Squares
19. 19. <ul><li>Problem: </li></ul><ul><li>The annual sales of a company are as follows: </li></ul><ul><li>Year 1988 1989 1990 1991 1992 </li></ul><ul><li>Sales 45 56 78 46 75 </li></ul><ul><li>By the method of least squares, find the trend values for each of the five years. Also estimate the annual sales of 1993. </li></ul>
20. 20. <ul><li>Solution: </li></ul> y = n.a. + b  x ……………….  1) Substituting the values  xy = a  x + b  x2 ……………. (2) we get,  xy  x 2  x  y n = 5 70 375 25 5 75 1992 65 184 16 4 46 1991 60 234 9 3 78 1990 55 112 4 2 56 1989 50 45 1 1 45 1988 (6) (5) (4) (3) (2) (1) Trend value of sales(‘000) Xy X 2 X Sales (Rs.’000) Year
21. 21. <ul><li> y = n.a. + b  x ……………….  1) </li></ul><ul><li> xy = a  x + b  x 2 ……………. (2) </li></ul><ul><li>Substituting the values of  x,  x 2 ,  xy ,  y , n , we get, </li></ul><ul><li>= 5a + 15b …………………..(3) </li></ul><ul><li>= 15a + 55b …………………..(4) </li></ul><ul><li>Multiplying (3) x 3 we get </li></ul><ul><li>= 15a + 45b </li></ul><ul><li>= 15a + 55b </li></ul><ul><li>Therefore, 10b = 50, b = 5 </li></ul><ul><li>Substituting the value of b in equation (3) above, </li></ul><ul><li>= 5a + 15 x 5 Y = a + bx = 45 + 5x </li></ul><ul><li>5a + 75 = 300 Y1988 = 45 + 5(1) = 50 </li></ul><ul><li>5a = 300 – 75 Y1989 = 45 + 5(2) = 55 </li></ul><ul><li>5a = 225 Y1990 = 45 + 5(3) = 60 </li></ul><ul><li>a = 45 Y1991 = 45 + 5(4) = 65 </li></ul><ul><li>Y1992 = 45 + 5(5) = 70 </li></ul><ul><li>Estimating the annual sales of Y1993 = 45 + 5(6) = 75. </li></ul>