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Patrick purchased 80 pencils and sold them at a loss

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Patrick purchased 80 pencils and sold them at a loss  [#permalink]

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New post 22 Jun 2014, 11:14
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E

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  45% (medium)

Question Stats:

61% (01:34) correct 39% (01:48) wrong based on 175 sessions

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Patrick purchased 80 pencils and sold them at a loss equal to the selling price of 20 pencils. The cost of 80 pencils is how many times the selling price of 80 pencils?

(A) 0.75
(B) 0.8
(C) 1
(D) 1.2
(E) 1.25

Source: Math bible Nova
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Posts: 47951
Re: Patrick purchased 80 pencils and sold them at a loss  [#permalink]

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New post 22 Jun 2014, 11:38
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1
Game wrote:
Patrick purchased 80 pencils and sold them at a loss equal to the selling price of 20 pencils. The cost of 80 pencils is how many times the selling price of 80 pencils?

(A) 0.75
(B) 0.8
(C) 1
(D) 1.2
(E) 1.25

Source: Math bible Nova


Say the cost price of 80 pencils was $80 ($1 per pencil) and the selling price of 1 pencil was p.

Selling at a loss: 80 - 80p = 20p --> p = 4/5.

(cost price)/(selling price) = 1/(4/5) = 5/4 = 1.25.

Answer: E.
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Re: Patrick purchased 80 pencils and sold them at a loss  [#permalink]

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New post 23 Jun 2014, 00:10
Let cost price of 80 pencils = 80c

Let selling price of 80 pencils = 80s ............ (1)

Pencils sold at a loss equal to the selling price of 20 pencils

80c - 80s = 20s

80c = 100s

\(s = \frac{4c}{5}\)

Selling price of 80 pencils \(= 80 * \frac{4c}{5} = 64c\) ........... From (1)

Require to find

Cost of 80 pencils = How many times Selling Price of 80 pencils

80c = 64c x ???

Answer \(= \frac{80}{64} = \frac{5}{4} = 1.25\)

Answer = E
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Patrick purchased 80 pencils and sold them at a loss  [#permalink]

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New post 23 May 2016, 12:13
1
Game wrote:
Patrick purchased 80 pencils and sold them at a loss equal to the selling price of 20 pencils.
The cost of 80 pencils is how many times the selling price of 80 pencils?

(A) 0.75
(B) 0.8
(C) 1
(D) 1.2
(E) 1.25


\(selling price = p\)
\(cost = c\)

since we have incurred a loss equal to the cost of 20 pencils

\(80p-80c=20c\)
\(80p=100c\)
and given that the problem is asking us how many times the cost is equal to the selling price

we can calculate p:

\(p = \frac{100}{80} c\) or -> \(p = \frac{5}{4} c\) -> \(p = 1,25 c\)

or Answer (E)
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Re: Patrick purchased 80 pencils and sold them at a loss  [#permalink]

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New post 09 Oct 2017, 17:24
Game wrote:
Patrick purchased 80 pencils and sold them at a loss equal to the selling price of 20 pencils. The cost of 80 pencils is how many times the selling price of 80 pencils?

(A) 0.75
(B) 0.8
(C) 1
(D) 1.2
(E) 1.25


We can let c = the cost per pencil and p = the selling price per pencil; thus:

80p - 80c = -20p

100p = 80c

The problem is asking for the ratio of 80c to 80p. Since 80c = 100p, 80c/80p = 100p/80p = 10/8 = 5/4 = 1.25.

Answer: E
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Re: Patrick purchased 80 pencils and sold them at a loss  [#permalink]

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New post 17 Jul 2018, 03:54
Game wrote:
Patrick purchased 80 pencils and sold them at a loss equal to the selling price of 20 pencils. The cost of 80 pencils is how many times the selling price of 80 pencils?

(A) 0.75
(B) 0.8
(C) 1
(D) 1.2
(E) 1.25

Source: Math bible Nova


Since the two blue values above are equal, the given question stem is the same as the following:
The cost of EACH PENCIL is how many times the selling price of EACH PENCIL?

Let the selling price of each pencil = $1.
Thus, the selling price of 80 pencils = $80.

Since Patrick's loss is equal to the selling price of 20 pencils, we get:
Loss = 20*1 = 20.

Since the 80 pencils are sold for $80 at a loss of $20, Patrick's cost for the 80 pencils must be $100.
Since 80 pencils have a total cost of $100, the cost per pencil \(= \frac{100}{80} = \frac{5}{4} = 1.25\).

Resulting ratio:
\(\frac{cost-per-pencil}{selling-price-per-pencil} = \frac{1.25}{1} = 1.25\).


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Re: Patrick purchased 80 pencils and sold them at a loss &nbs [#permalink] 17 Jul 2018, 03:54
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