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A shopkeeper purchased footballs at $10 per football

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Intern
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S
Joined: 12 Oct 2017
Posts: 39

Kudos [?]: 2 [0], given: 15

CAT Tests
A shopkeeper purchased footballs at $10 per football [#permalink]

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New post 07 Dec 2017, 17:44
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Difficulty:

  45% (medium)

Question Stats:

62% (01:10) correct 38% (01:12) wrong based on 29 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

Kudos [?]: 2 [0], given: 15

Manager
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Joined: 23 Oct 2017
Posts: 63

Kudos [?]: 2 [0], given: 4

Re: A shopkeeper purchased footballs at $10 per football [#permalink]

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New post 07 Dec 2017, 18:48
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

Kudos [?]: 2 [0], given: 4

VP
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Joined: 22 May 2016
Posts: 1245

Kudos [?]: 457 [0], given: 679

Premium Member CAT Tests
A shopkeeper purchased footballs at $10 per football [#permalink]

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New post 09 Dec 2017, 11:45
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

Kudos [?]: 457 [0], given: 679

A shopkeeper purchased footballs at $10 per football   [#permalink] 09 Dec 2017, 11:45
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