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Que: A merchant bought 2,400 fans for $30 each. He sold 60 percent of the fans for $40 each and the rest for $35......

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by Max@Math Revolution » Sat Jan 09, 2021 9:18 pm

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B

C

D

E

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Que: A merchant bought 2,400 fans for $30 each. He sold 60 percent of the fans for $40 each and the rest for $35 each. What was the merchant’s average profit per fan?

(A) $6
(B) $8
(C) $9
(D) $10
(E) $12
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Re: Que: A merchant bought 2,400 fans for $30 each. He sold 60 percent of the fans for $40 each and the rest for $35....

by Max@Math Revolution » Mon Jan 11, 2021 9:16 pm
Solution: Cost price of 2,400 fans = $ (30 * 2400) = $72,000

Number of fans sold at $40 each = \(\frac{60}{100}\cdot2,400\ =\ 1,440\)

Thus, selling price of these 1,440 fans = $(1,440 * 40) = $57,600

Number of fans sold at $ 35 each = 2400 – 1440 = 960

Thus, selling of these 960 fans = $(960 * 35) = $33,600

Total selling price of 2,400 fans = $57,600 + $33, 600 = $91,200

Total profit: Selling price – Cost price = $91,200 - $72,000 = $19,200

Profit per fan: \(\frac{19,200}{2,400}\) = 8

Therefore, B is the correct answer.

Answer B
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Re: Que: A merchant bought 2,400 fans for $30 each. He sold 60 percent of the fans for $40 each and the rest for $35....

by Scott@TargetTestPrep » Tue Jan 19, 2021 6:21 am
Max@Math Revolution wrote: ↑
Sat Jan 09, 2021 9:18 pm
 Que: A merchant bought 2,400 fans for $30 each. He sold 60 percent of the fans for $40 each and the rest for $35 each. What was the merchant’s average profit per fan?

(A) $6
(B) $8
(C) $9
(D) $10
(E) $12
Solution:

We can use 10 fans as the number of fans bought and sold, instead of 2,400 fans, the actual number of fans bought and sold.

Therefore, the cost is 30 x 10 = $300 and the revenue is 40 x 6 + 35 x 4 = 240 + 140 = $380. We see that the total profit is 380 - 300 = $80 and the average profit per fan is 80/10 = $8.

Answer: B

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