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07 Dec 2017, 18:44
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Difficulty:

45% (medium)

Question Stats:

64% (00:58) correct 36% (01:06) wrong based on 36 sessions

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A shopkeeper purchased footballs at $10 per football and sold them at a price equal to the purchase price plus a markup. If the markup was 10% of the selling price, what was the shopkeeper's profit in dollars, to the nearest cent, on each football? A. 0.90 B. 1.00 C. 1.11 D. 2.00 E. 11.11 [Reveal] Spoiler: OA Manager Joined: 23 Oct 2017 Posts: 64 Re: A shopkeeper purchased footballs at$10 per football [#permalink]

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07 Dec 2017, 19:48
anhht13 wrote:
A shopkeeper purchased footballs at $10 per football and sold them at a price equal to the purchase price plus a markup. If the markup was 10% of the selling price, what was the shopkeeper's profit in dollars, to the nearest cent, on each football? A. 0.90 B. 1.00 C. 1.11 D. 2.00 E. 11.11 => Answer is C. Purchase price = 10; let Markup by x; SP = 10 + x; given: x = 10%(of SP); on solving x = 10/9 Profit = SP - Purchase Price = 10 + x- 10 = x =10/9 =1.11 A potential source of error could be when you take markup as 10 % of 10 SC Moderator Joined: 22 May 2016 Posts: 1566 A shopkeeper purchased footballs at$10 per football [#permalink]

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09 Dec 2017, 12:45
anhht13 wrote:
A shopkeeper purchased footballs at $10 per football and sold them at a price equal to the purchase price plus a markup. If the markup was 10% of the selling price, what was the shopkeeper's profit in dollars, to the nearest cent, on each football? A. 0.90 B. 1.00 C. 1.11 D. 2.00 E. 11.11 "Purchase price" and "selling price" could confuse. BOTH often mean what the last buyer has to pay. But here "purchase price" = "cost price (CP)" Rewrite: He paid$10 per football. He sold each football at a price equal to what he paid plus 10 percent of the selling price (SP). Selling price, SP, is on both sides of the equation, because markup = (.10)SP, see #3

1) Selling Price = Cost Price + markup

2) SP = CP + markup

3) SP = $10 + (.10)SP 4) (.90)SP =$10

5*) SP = $$(\frac{10}{.90}) \approx{11.11}$$

6) Profit per football:
(SP - CP) = ($11.11 -$10.00) = $1.11 Answer C *This arithmetic can also be SP = CP + markup SP = $$10 + \frac{1}{10}SP$$ $$\frac{9}{10}$$SP = $$10$$ SP$$=(\frac{10}{9})10$$ SP $$\approx{11.11}$$ _________________ At the still point, there the dance is. -- T.S. Eliot Formerly genxer123 A shopkeeper purchased footballs at$10 per football   [#permalink] 09 Dec 2017, 12:45
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