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A shopkeeper purchased footballs at $10 per football and sold them at

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A shopkeeper purchased footballs at $10 per football and sold them at  [#permalink]

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New post 06 Nov 2019, 06:33
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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

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

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New post 06 Nov 2019, 08:39
1
Assume SP of the football=x

\(10+\frac{x}{10}=x\)

x=\(\frac{100}{9}\)

Profit= \(\frac{10}{9}\)=1.11

Bunuel 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

Are You Up For the Challenge: 700 Level Questions
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Re: A shopkeeper purchased footballs at $10 per football and sold them at  [#permalink]

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New post 06 Nov 2019, 09:07
CP = 10$
SP = x
x= 10+10*x/100
x= 100/9
profit = 100/9-10 ; 10/9 ; 1.11
IMO C




Bunuel 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

Are You Up For the Challenge: 700 Level Questions
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Re: A shopkeeper purchased footballs at $10 per football and sold them at  [#permalink]

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New post 07 Nov 2019, 19:52
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Bunuel 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

Are You Up For the Challenge: 700 Level Questions


The markup is the difference between the selling price and the purchase price. If we let the selling price be p, we can create the equation:

p - 10 = 0.1p

0.9p = 10

p = 10/0.9 = 100/9 ≈ 11.11

Thus, the profit is approximately 11.11 - 10 = $1.11.

Answer: C
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Re: A shopkeeper purchased footballs at $10 per football and sold them at  [#permalink]

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New post 20 Nov 2019, 02:29
Bunuel 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

Are You Up For the Challenge: 700 Level Questions


assume markup = x.
SP = 10 + x.
now we are given,
x = \(\frac{1}{10}\)(10+x)
we get 9x = 10
and x = 10/9.
Hence, the markup becomes 1.11 which is the profit on each football.

OA - C
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Re: A shopkeeper purchased footballs at $10 per football and sold them at   [#permalink] 20 Nov 2019, 02:29
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