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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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Marta bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, how many 23-cent pencils did Marta buy?

1). Marta bought a total of 6 pencils.
2). The total value of the pencils Marta bought was 130 cents.

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Re: Math-DS: So subtle [#permalink]

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New post 02 May 2008, 11:54
lexis wrote:
Marta bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, how many 23-cent pencils did Marta buy?

1). Marta bought a total of 6 pencils.
2). The total value of the pencils Marta bought was 130 cents.


seems like C..

x+y=6

23x+21y=130

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New post 02 May 2008, 13:50
lexis wrote:
Marta bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, how many 23-cent pencils did Marta buy?

1). Marta bought a total of 6 pencils.
2). The total value of the pencils Marta bought was 130 cents.


I would say C.

There are no other possible combinations of 21 cents (#4) and 23 cents(#2).

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Re: Math-DS: So subtle [#permalink]

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New post 02 May 2008, 18:20
lexis wrote:
Marta bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, how many 23-cent pencils did Marta buy?

1). Marta bought a total of 6 pencils.
2). The total value of the pencils Marta bought was 130 cents.


B

This is C-trap

C-trap question will lure you to choose C, but actually B is correct.
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Re: Math-DS: So subtle [#permalink]

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New post 03 May 2008, 11:59
I tried getting the answer employing only statement B, but I can't seem to get it!
How do you get around it?

Is the OA B or C?

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Re: Math-DS: So subtle [#permalink]

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New post 03 May 2008, 12:16
Well I suppose if you were to find all values that would fit the equation:
23x+21y=130

If you subtract 130-23 =107, check if this is divisible by 21, it is NOT DIVISIBLE
Next try 130-2(23) = 130-46 =84, check if divisible by 21, it is since 21*4 = 84

Therefore we can conclude that x=2 and y=4 satisfy the above equation and B is sufficient

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New post 04 May 2008, 00:34
2) 23x+21y=130
x,y are both positive integers. y is equal to or less than 6 (~130/21). If you try all possible values of y, you will find merely y=4, x=2 as the correct answer.
So we do not need x+y as 1 shows to find the final answer.

Statement 2 alone is sufficient. OA is B.

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Re: Math-DS: So subtle   [#permalink] 04 May 2008, 00:34
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