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A dealer buys a boat at auction and pays 10 % below
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30 Jul 2017, 08:17
30 Jul 2017, 08:17
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A dealer buys a boat at auction and pays 10 % below list price. He then sells the boat at a profit of 30% of his cost of the boat. At the same selling price, what would the dealer's percent profit over his cost have been if he had bought the boat at list price?
Round your answer to the nearest whole percentage point before entering:
enter your value
Show: ::
17%
Re: A dealer buys a boat at auction and pays 10 % below
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25 Sep 2017, 06:04
25 Sep 2017, 06:04
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Let's use some numbers. Fix the original price of the boat at 100. Then, the dealer bought it at (1-0.10)*100=90 with the discount of 10%. Then, the profit is 30% of 90, i.e. 0.3*90=27. Thus, the selling price must be 90+27=117, given the formula profit = sell price - cost. Now, keeping the selling price at 117, if the dealer would have paid the boat fully, the cost would have been 100 and the profits 117-100=17. Then, the profits are 17/100*100% of the cost, i.e. 17%.
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Re: A dealer buys a boat at auction and pays 10 % below
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30 Mar 2021, 22:15
30 Mar 2021, 22:15
Carcass wrote:
A dealer buys a boat at auction and pays 10 % below list price. He then sells the boat at a profit of 30% of his cost of the boat. At the same selling price, what would the dealer's percent profit over his cost have been if he had bought the boat at list price?
Round your answer to the nearest whole percentage point before entering:
enter your value
Show: ::
17%
Let the List price be \(L\)
Buying Price \(= 0.9L\)
Selling Price \(= 1.3(0.9L) = 1.17L\)
Hence, the profit was \(17\)%
