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Unit 3 Selected Review Problems Solutions

# Unit 3 Selected Review Problems Solutions - Unit 3 Selected...

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Unit 3 Selected Review Problems Solutions Section 5.2 Added question: 360(.7)(.875)(.95) = \$209.475 209.475-199 = 10.475÷209.475 = .0050005967 = 5.00% 4. Solution: # List price ( L ) Series of discounts rates Equivalent discount rate ( d ) Net price ( N ) Amount of discount ( L – N ) a. \$850.00 10%, 5% 14.50% \$726.75 \$123.25 b. \$697.82 5%, 4%, 3% 11.54% \$617.32 \$80.50 c. \$2022.71 %, 2%, 1% 8.32% \$1854.50 \$168.21 d. \$625.00 5%, 4%, d % 10.62% \$558.60 \$66.40 a. N = 850 (1 – 0.10)(1 – 0.05) = \$726.75 Amount of discount = L – N = 850 – 726.75 = \$123.25 d e = 1 – (1 – d 1 )(1 – d 2 ) = 1 – (1 – 0.10)(1 – 0.05) = 1 – (0.90)(0.95) = 0.145 = 14.50% b. Amount of discount = L – N = \$80.50 Therefore, L = 80.50 N . We also know, N = L (1 – d 1 ) (1 – d 2 ) (1 – d 3 ) = L (1 – 0.05) (1 – 0.04) (1 – 0.03) = L (0.88464) Substituting for ‘ N ’ in the previous equation, we get, L = 80.50 L (0.88464), L – L (0.88464) = 80.50 L (1 – 0. 88464) = 80.50 Solving for L , we get, L = 697.815534... = \$697.82 Now, amount of discount = 697.815534 ...– N = 80.50 N = \$617.315534… = \$617.32 d e = 1 – (1 – d 1 )(1 – d 2 ) = 1 – (1 – 0.05) (1 – 0.04) (1 – 0.03) = 1 – (0.95)(0.96)(0.97) = 0.11536 = 11.54 c. N = L (1 – d 1 ) (1 – d 2 ) (1 – d 3 ) = L (1 – 0.055)(1 – 0.02) (1 – 0.01) = L (0.916839) 1854.50 = L (0.916839) Solving for L , we get, L = \$2022.710639... = \$2022.71 Amount of discount = L – N = \$2022.710639... – 1854.50 = \$168.210639... = \$168.21 d e = 1 – (1 – d 1 )(1 – d 2 ) = 1 – (1 – 0.055) (1 – 0.02) (1 – 0.01) = 0.083161 = 8.32% d. Amount of discount = L – N = \$625 – 558.60 = \$66.40 N = L (1 – d 1 ) (1 – d 2 ) (1 – d 3 ) 558.60 = 625(1 – 0.05) (1 – 0.04) (1 – d %) 0.98 = 1 – d %

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d % = 0.02 d = 2% d e = (1 – d 1 ) (1 – d 2 ) (1 – d 3 ) = 1 – (1 – 0.05)(1 – 0.04)(1 – 0.02) = 0.10624 = 10.62% #10. Solution: L = N = L (1 – d ) = 408,000 (1 – 0.05) = \$387,600.00 Therefore, the list price and net price of the clothes were \$408,000.00 and \$387,600.00 respectively. 18. Solution: N = L (1 – d 1 ) (1 – d 2 ) (1 – d 3 ) N = 2000 (1 – 0.15) (1 – 0.10) (1 – 0.05) = \$1453.50 Amount of discount = L – N = 2000 – 1453.5 = \$546.50 Therefore, we would pay \$1453.50 for the leather jacket and the total amount of discount would be \$546.50. Section 5.3 4. Solution: 6% discount received on payment made on or before Mar 01 (1 st discount period) 3% discount received on payment made after Mar 01 but on or before Mar 16 (2 nd discount period) Net payment due by Mar 31 (credit period) (a) If paid on Feb 14, They would receive 6% discount on their invoice as it falls within the first 15 days of the invoice date. Therefore, they would have to pay: 260,800 (1 – 0.06) = \$245,152.00 (b) If paid on Feb 28, They would receive 6% discount on their invoice as it falls within the first 15 days of the invoice date. Therefore, they would have to pay: 260,800 (1 – 0.06) = \$245,152.00 (c) If paid on Mar 02, They would receive 3% discount on their invoice as it falls after 15 days and before 30 days of the invoice date. Therefore, they would have to pay: 260,800 (1 – 0.03) = \$252,976.00 (d) If paid on Mar 31, Net price has to be paid as it falls after the 30 days of the invoice date.
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Unit 3 Selected Review Problems Solutions - Unit 3 Selected...

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