Practical Business Math Procedures Chapter 7 Challenge Problems and Summary Practice Set
Challenge Problems
7-49
I. Given
a) Honda Insight Original Price $17,995.00
b) Honda Insight Dealer Price $16,495.00
c) Terms: 1/15 n/30
II. Unknown
a. How much is the rebate?
b. What percent is the rebate?
c. What is the amount of the discount if the dealer pays within 15 days?
d. What is the dealer’s final price?
e. What is the dealer’s total savings?
III. Equation
a. Rebate = Original Price – Dealer Price
b. % Rebate = (Rebate / Original Price) x 100%
IV. Solving and Solution
a. Rebate = Original Price – Dealer Price
= $17995.00 – $16495.00
= $1,500.00 rebate
b. % Rebate = (Rebate / Original Price) x 100%
= ($1500.00 / $174995.00) x 100%
= 8.34% rebate
c. 15day discount = Dealer Price x Discount
= $16495.00 x 1%
= $164.95 15day discount
d. Final Price = Dealer Price – Discount
= $16495.00 – [$16495.00 x (1%)]
= $16495.00 – $164.95
= $16330.05 final dealer price
e. Total Savings = Rebate + 15day discount
= ($1500.00 rebate) + ($164.95 15day discount)
= $1,664.95 total savings
7-50
V. Given
d) Invoice dated March 28 received March 30
e) 50 TV x $125.00
f) 10/4/2 chain discount
g) FOB shipping point = $70
h) 2/10 EOM (April)
i) 3 defective items (50 – 3 = 47)
j) April 8 $150.00 partial payment
k) Payment of full on May 6
VI. Unknown
a. Final Payment on May 6 w/o taxes
VII. Equation
a. (50 TV units less 3 defective) x $125.00 each = $5,875.00
b. Chain discount = 10/4/2
= (1.00 – 0.10) x (1.00 – 0.04) x (1.00 – 0.02)
= 0.90 x 0.96 x 0.98
= 0.847
= 1.00 - 0.847 = 0.153 ~ 15.33%
= $5875 x 15.33% discount
= $900.52 Chain discount
c. FOB Shipping Point = additional $70 to payment
d.
e. $150 partial payment to 2/10 EOM
= 100% - 2% = .98
$150 = $153.06 credited to amount
.98
VIII. Solving
Total Amount = $ 5875.00 Less: Chain Discount = $ 900.52 Add: FOB Shipping Point = $ 70.00 Less: Partial Payment Credit = $ 153.06
Payment Due on May
7-49
I. Given
a) Honda Insight Original Price $17,995.00
b) Honda Insight Dealer Price $16,495.00
c) Terms: 1/15 n/30
II. Unknown
a. How much is the rebate?
b. What percent is the rebate?
c. What is the amount of the discount if the dealer pays within 15 days?
d. What is the dealer’s final price?
e. What is the dealer’s total savings?
III. Equation
a. Rebate = Original Price – Dealer Price
b. % Rebate = (Rebate / Original Price) x 100%
IV. Solving and Solution
a. Rebate = Original Price – Dealer Price
= $17995.00 – $16495.00
= $1,500.00 rebate
b. % Rebate = (Rebate / Original Price) x 100%
= ($1500.00 / $174995.00) x 100%
= 8.34% rebate
c. 15day discount = Dealer Price x Discount
= $16495.00 x 1%
= $164.95 15day discount
d. Final Price = Dealer Price – Discount
= $16495.00 – [$16495.00 x (1%)]
= $16495.00 – $164.95
= $16330.05 final dealer price
e. Total Savings = Rebate + 15day discount
= ($1500.00 rebate) + ($164.95 15day discount)
= $1,664.95 total savings
7-50
V. Given
d) Invoice dated March 28 received March 30
e) 50 TV x $125.00
f) 10/4/2 chain discount
g) FOB shipping point = $70
h) 2/10 EOM (April)
i) 3 defective items (50 – 3 = 47)
j) April 8 $150.00 partial payment
k) Payment of full on May 6
VI. Unknown
a. Final Payment on May 6 w/o taxes
VII. Equation
a. (50 TV units less 3 defective) x $125.00 each = $5,875.00
b. Chain discount = 10/4/2
= (1.00 – 0.10) x (1.00 – 0.04) x (1.00 – 0.02)
= 0.90 x 0.96 x 0.98
= 0.847
= 1.00 - 0.847 = 0.153 ~ 15.33%
= $5875 x 15.33% discount
= $900.52 Chain discount
c. FOB Shipping Point = additional $70 to payment
d.
e. $150 partial payment to 2/10 EOM
= 100% - 2% = .98
$150 = $153.06 credited to amount
.98
VIII. Solving
Total Amount = $ 5875.00 Less: Chain Discount = $ 900.52 Add: FOB Shipping Point = $ 70.00 Less: Partial Payment Credit = $ 153.06
Payment Due on May