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Martha bought several pencils. If each pencil was either a

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

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New post 20 Dec 2005, 08:01
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Martha bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, ho wmany 23-cent pencils did Marta buy?

(1) Martha bought a total of 6 pencils.

(2) The total value of the pencils Martha bouht was 130 cents.

Please explain.
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New post 20 Dec 2005, 08:08
i gives you

x+ y = 6 linear equation 2 variables - insuff

ii - 21x+ 23y = 130 linear equation 2 variables insuff


C
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New post 20 Dec 2005, 08:23
Thats what I thought oo, OA is B though :shock:
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New post 20 Dec 2005, 08:36
After the advice that OA is B it becomes clear, but in an actual test I would have made it wrong.

The only way to combine x pencils, which cost 23, and y pencils, which cost 21, with the result of 130, is x=2 and y=4
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New post 20 Dec 2005, 08:49
Good question.

How do we check for all possibilities in 2 mins?
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New post 20 Dec 2005, 08:53
Would be nice to know.

However I think the main problem is to know that such a solution exists.
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New post 20 Dec 2005, 08:56
If it looks too easy to be C, then the odds are the answer is either A, B, or D. "BAD" strategy by Veritas.

Statement one alone does not give us a weighted average, ratio, or total---> insuff.

Statement two alone gives a total, which with a little manipulation (total is very small) we can determine that .21*4+.23*2=1.30---> Suff.

(B)
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New post 20 Dec 2005, 12:01
The stem states Martha spend a total amount 21x+23y, where x and y are the number of pencils of each kind. Find x.

(1) gives an equation of x+y=6. 2 variables, only one equation. Insufficient.

(2) 21x+23y=130. Same problem as in st. (1)

Combining (1) and (2) 2 variables-2 equations suffice to find x.

C stands
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New post 20 Dec 2005, 12:06
Yup, trick question. Next time, I'll keep this trick in mind. I'm about to go straight to the sticky about Veritas strategy.
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Re: Martha and pencils [#permalink]

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New post 21 Dec 2005, 02:13
sandalphon wrote:
Martha bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, ho wmany 23-cent pencils did Marta buy?

(1) Martha bought a total of 6 pencils.

(2) The total value of the pencils Martha bouht was 130 cents.

Please explain.


I chose B. 23x+21y=130. Was sure I am going to try x to be max of 5. So with little working for x from 1 to 5, found that x to be 2 and y to be 4. Hence B
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New post 21 Dec 2005, 03:13
sandalphon wrote:
Martha bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, ho wmany 23-cent pencils did Marta buy?

(1) Martha bought a total of 6 pencils.

(2) The total value of the pencils Martha bouht was 130 cents.

Please explain.

I think it is C
Here is my reasoning
1)alone insuff we can't figure out anything but the quant of pencils!
2)same here
1&2 together
Let x be 23 cents pencils and y be 21 cents pencils so X+Y=6
From 2 statement 23X+21Y=130
Hence we have
X+Y=6
23X+21Y=130
X=6-Y substitute 23*(6-Y)+21Y=130--->138-23Y+21Y=130--->2Y=8--->Y=4,X=2
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New post 21 Dec 2005, 03:18
Oops !!Nice trap here))
The only good thing that i am not alone in this trap)) :wink:
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New post 21 Dec 2005, 06:28
let 23c pencil be x, 21c pencil be y

(1) x+y=6 --> not useful. Insufficient.

(2) 23x + 21y = 130 --> only one combination can lead to this: x = 2, y = 4

Ans B

It's pretty easy to manipulate. You can see that x cannot be 5-10 as that will result in y being fractional. You can then quickly work out x for values of 4, then 3, then 2... each time you do that, subtract it off from 130 and see if it's divisible by21. Sounds like a lot of work, but it's very easy and fast math really... :wink:
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New post 26 Feb 2011, 00:12
sandalphon wrote:
Martha bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, ho wmany 23-cent pencils did Marta buy?

(1) Martha bought a total of 6 pencils.

(2) The total value of the pencils Martha bouht was 130 cents.

Please explain.



By no means would I have picked B on the real test day. Its got no straight forward step-by-step approach :-(
In my mind it was a clear C. But boy was I wrong! I think the BAD strategy is going to be on my check list as well.
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Re: Martha and pencils [#permalink]

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New post 26 Feb 2011, 02:47
This question is discussed here: equations-100204.html

For more on the questions with Diophantine equation (equations whose solutions must be integers only) check:
car-dealer-data-sufficiency-105682.html
gmat-prep2-92785.html
help-me-to-solve-it-101966.html
data-sufficiency-word-problem-106212.html
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Re: Martha and pencils   [#permalink] 26 Feb 2011, 02:47
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