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# Marta bought several pencils. If each pencil was either a

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Marta bought several pencils. If each pencil was either a [#permalink]

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16 Nov 2009, 21:52
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Marta bought several pencils. If each pencil was either a \$0.23 pencil or a \$0.21 pencil, how many \$0.23 pencils did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of the pencils Marta bought was \$1.30

OA will be posted in 24 hours

Kudos [?]: 116 [0], given: 2

Manager
Joined: 13 Aug 2009
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Kudos [?]: 125 [0], given: 16

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16 Nov 2009, 23:34
let x = number of \$0.23 pencils
let y = number of \$0.21 pencils

Statement 1 gives us:
x+y=6
one equation, one unknown... INSUFFICIENT

Statement 2 gives us:
.23x+.21y=1.30
one equation, one unknown... INSUFFICIENT

Statements 1 and 2:
two equations, two unknowns... SUFFICIENT

Kudos [?]: 125 [0], given: 16

Manager
Joined: 11 Sep 2009
Posts: 129

Kudos [?]: 420 [0], given: 6

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17 Nov 2009, 02:31

Let's represent the number of \$0.21 pencils bought as: m
Let's represent the number of \$0.23 pencils bought as: n

Statement 1: Marta bought a total of 6 pencils.

The combination of 6 pencils could be anything. Clearly, insufficient.

Statement 2: The total value of the pencils Marta bought was \$1.30.

This can be solved intuitively, but I'll show the process that I go through mentally:

0.21m + 0.23n = 1.30
0.21(m+n) + 0.02n = 1.30

n = [1.30 - 0.21(m+n)] / 0.02

Clearly, n must be less than m+n, but also greater than or equal to 0. Also, (m+n) must be an even number, otherwise n is not an integer.

(m+n) = 2, n = 44 (NOT POSSIBLE)
(m+n) = 4, n = 23 (NOT POSSIBLE)
(m+n) = 6, n = 2 (OKAY)
(m+n) = 8 n = -19 (NOT POSSIBLE)

Therefore, sufficient.

Kudos [?]: 420 [0], given: 6

GMAT Tutor
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Kudos [?]: 1952 [0], given: 6

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17 Nov 2009, 08:46
chicagocubsrule wrote:
Marta bought several pencils. If each pencil was either a \$0.23 pencil or a \$0.21 pencil, how many \$0.23 pencils did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of the pencils Marta bought was \$1.30

OA will be posted in 24 hours

You can also look at this as follows: if she bought 7 pencils, she would spend at least 7*0.21 = \$1.49. If she bought 5 pencils, she would spend at most 5*0.23 = \$1.15. So if she spent \$1.30, as we learn from Statement 2 alone, she must have bought exactly 6 pencils. So Statement 2 is sufficient alone (once we know that she bought 6 pencils, there can only be one combination of \$0.23 and \$0.21 pencils that give a total cost of \$1.30, since the more \$0.23 pencils she buys, the greater the cost will be).
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Kudos [?]: 1952 [0], given: 6

Manager
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17 Nov 2009, 09:07
Quote:
You can also look at this as follows: if she bought 7 pencils, she would spend at least 7*0.21 = \$1.49. If she bought 5 pencils, she would spend at most 5*0.23 = \$1.15. So if she spent \$1.30, as we learn from Statement 2 alone, she must have bought exactly 6 pencils. So Statement 2 is sufficient alone (once we know that she bought 6 pencils, there can only be one combination of \$0.23 and \$0.21 pencils that give a total cost of \$1.30, since the more \$0.23 pencils she buys, the greater the cost will be).

Good approach.
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Kudos [?]: 1620 [0], given: 18

Re: Pencils   [#permalink] 17 Nov 2009, 09:07
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