Unit 3 Selected Review Problems Solutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 Net price Amount of discount 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 %
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/04/2012 for the course MATH 1052 taught by Professor Kit during the Winter '12 term at Fanshawe.

Page1 / 11

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online