Ch.5_Solution_Manual_Ed.1_v5_

Ch.5_Solution_Manual_Ed.1_v5_ - Exercises 5.2 Calculate the...

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Exercises 5.2 Calculate the missing values for questions 1 through 4: Exercise 5.2, Solution 1 : List price ( L ) Single discount rate ( d ) Amount of discount ( d × L ) Net price ( N ) a $1875 20% $375 $1500 b $230 12% $27.60 $202.40 c $800 54% $432 $368 d $500 2% $10 $490 a. Amount of discount = d L L = = $1875.00 N L dL 1875 375 $1500.00 b. L = N Amount of discount = 202.40 27.60 = $230.00 d L = 27.60 d = = 0.12 = 12.00% discount c. Amount of discount = L – N = 800 – 368 = 432 Amount of discount = d L d = = 0.54 = 54.00% d. N = L – Amount of discount = 500 – 10 = 490 Amount of discount = d L d = = 0.02 = 2.00% Exercise 5.2, Solution 3: List price ( L ) Series of discounts rates Equivalent discount rate ( d e ) Net price ( N ) Amount of discount ( L N ) a. $540.00 20%, 10% 28.00% $388.80 $151.20 b . $1970.19 22%, 12.5%, 10% 38.58% $1210.19 $760.00 c. $2768.09 , 3%, 1% 9.25% $2512.00 $256.09 d . 1200.00 10%, 6%, d % 17.94% $984.74 $215.26 a. N = 540 (1 – 0.20)(1 – 0.10) = $388.80 Amount of discount = L – N = 540 – 388.80 = $151.20 d e = 1 – (1 – d 1 )(1 – d 2 ) = 1 – (1 – 0.20)(1 – 0.10) = 1 – (0.80)(0.90) = 0.28 = 28.00 b. Amount of discount = L – N = $760 Therefore, L = 760 N. We also know, N = L (1 – d 1 ) (1 – d 2 ) (1 – d 3 ) = L (1 – 0.22) (1 – 0.125) (1 – 0.10) = L (0.61425) Substituting for ‘ N ’ in the previous equation, we get,
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L = 760 L (0.61425), L – L (0.61425) = 760 L (1 – 0.61425) = 760 Solving for L , we get, L = 1970.187946. .. = $1970.19 Now, amount of discount = 1970.187946. .. – N = 760 N = $1210.187946… = $1210.19 d e = 1 – (1 – d 1 )(1 – d 2 ) = 1 – (1 – 0.22)(1 – 0.125) (1 – 0.10) = 1 – (0.78)(0.875)(0.90) = 0.38575 = 38.58% c. N = L (1 – d 1 ) (1 – d 2 ) (1 – d 3 ) = L (1 – 0.055) (1 – 0.03) (1 – 0.01) = L (0.9074835) 2512 = L (0.9074835) Solving for L , we get, L = $2768.094406 = $2768.09 Amount of discount = L – N = 2768.094406 – 2512 = $ 256.094406 = $256.09 d e = 1 – (1 – d 1 )(1 – d 2 ) = 1 – (1 – 0.055) (1 – 0.03) (1 – 0.01) = 1 – (0.945)(0.97)(0.99) = 0.0925165 = 9.25% d. Amount of discount = L – N = 1200 – 984.74 = $215.26 N = L (1 – d 1 ) (1 – d 2 ) (1 – d 3 ) 984.74 = 1200(1 – 0.10) (1 – 0.06) (1 – d ) 0.969996. .. = 1 – d % d %= 0.030003. .. d = 3% Equivalent discount rate d e = 1 – (1 – d 1 )(1 – d 2 )(1 – d 3 ) = 1 – (1 – 0.1)(1 – 0.06)(1 – 0.03) = 1 – (0.9)(0.94)(0.96) = 0.17938 Therefore, d e = 17.94% Exercise 5.2, Solution 5: Amount of discount = d L = 0.15 110 = $16.50 Net price = L – Amount of discount = 110 – 16.50 = $93.50 Therefore, the amount of trade discount is $16.50 and the net price the retailer would pay for each pair of shoe is $93.50. Exercise 5.2, Solution 7:
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This note was uploaded on 04/04/2012 for the course MATH 1052 taught by Professor Kit during the Winter '12 term at Fanshawe.

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Ch.5_Solution_Manual_Ed.1_v5_ - Exercises 5.2 Calculate the...

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