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Question (Harder) for DS

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Question (Harder) for DS

by houstonrockets16 » Sun Jun 28, 2009 6:18 pm
Marta bought several pencils, if each pencil was either a 23 cent or 21 cent pencil, how many 23 cent pencils did she buy?

1) marta bought 6 pencils
2) the total value of the pencils she bought was 130 cents.
[spoiler]
The answer is B, statement 2 is sufficient. I do not see why. Please help; I am sure its because algerbraically there is only way these numbers add up but how can i do that for a different problem on the test!? THANKS![/spoiler]
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by ssmiles08 » Sun Jun 28, 2009 6:43 pm
houstonrockets16 wrote:Marta bought several pencils, if each pencil was either a 23 cent or 21 cent pencil, how many 23 cent pencils did she buy?

1) marta bought 6 pencils
2) the total value of the pencils she bought was 130 cents.
[spoiler]
The answer is B, statement 2 is sufficient. I do not see why. Please help; I am sure its because algerbraically there is only way these numbers add up but how can i do that for a different problem on the test!? THANKS![/spoiler]
Rats! I picked C the first time too...but looking at it more carefully you can see that only 2 numbers fit into the puzzle to make 130.

let x = number of 23 cent pencils
let y = number of 21 cent pencils

23x + 21y = 130

I looked at the units digit, to see what combinations would make a 0 at the end.

you know the total number of pencils have to equal 6 b/c the statements should never contradict one another.

so 3(2) + 1(4) = 0

23*2 + 21*4 = 130.

these are the only two numbers that fit. The number of 23 cent pencils would be 2.

Of course, my method was by trial and error, I would like to see other's opinions on this as well.
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by Ian Stewart » Sun Jun 28, 2009 7:26 pm
You might notice that if she buys only 5 pencils, the most she could spend is 5*23 = 115 cents, while if she buys 7 pencils, she must spend at least 7*21 = 147 cents. So if she spent more than 115 cents but less than 147, she must have bought 6 pencils. There could only be one combination of six 23 cent and 21 cent pencils that will add to 130 cents, since the more 23 cent pencils she buys, the more she'll spend - there's no need to actually find the number. So Statement 2 is sufficient alone.
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by shibal » Wed Jul 01, 2009 6:04 pm
i did some similar ds questions and is ALWAYS B....
maybe i'm judging, but i think they give you a x+y=number in the I stmt and a weird eq in the II stmt to rush you into C.....
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by tryin700 » Wed Jul 01, 2009 9:16 pm
I took a cue from another problem i looked earlier in PS section.

Statement 2 says --

23x+21y=130 (x is no of 23 cents pencil and y is no of 21 cents pencils)

multiples of 23: 23, 46, 69, 92 etc
multiples of 21: 21, 42, 63, 84, 105 etc

here 46+84=130

hence y=2 which gives us the answer, hence B
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by abcdefg » Sun Jul 19, 2009 8:12 pm
I understand the logic of this problem and know that B is the answer but statement 2 has 2 unknowns and 1 equation, isn't that insufficient to arrive at a unique solution? What am I missing here?
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by Ian Stewart » Mon Jul 20, 2009 4:06 am
abcdefg wrote:I understand the logic of this problem and know that B is the answer but statement 2 has 2 unknowns and 1 equation, isn't that insufficient to arrive at a unique solution? What am I missing here?
There is a *very* important restriction in this question: the unknowns must be positive integers. I've seen several recent GMAT questions in a format similar to the one above, and they seem designed to trap people who simply count unknowns and count equations, without understanding when that should be done. You might consider a simpler example: if x + y = 2, then of course we can't find x or y; x could be 1.5 and y could be 0.5, or x could be 100 and y could be -98. But if x and y need to be positive integers, there is only one solution: x = 1 and y = 1.

If you have one equation and two unknowns, but there are restrictions on your unknowns (they must be positive integers, for example), then there may be only one way to make the equation work - you have to check.
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by axat » Mon Jul 20, 2009 4:22 am
Wow! Thanks, Ian. This was an eye opener of an input. I have been following your explainations fairly regularly, and I am often stunned by your approaches to the questions. Your logic is always simple, and yet sublime.

Cheers.

Ian Stewart wrote:
abcdefg wrote:I understand the logic of this problem and know that B is the answer but statement 2 has 2 unknowns and 1 equation, isn't that insufficient to arrive at a unique solution? What am I missing here?
There is a *very* important restriction in this question: the unknowns must be positive integers. I've seen several recent GMAT questions in a format similar to the one above, and they seem designed to trap people who simply count unknowns and count equations, without understanding when that should be done. You might consider a simpler example: if x + y = 2, then of course we can't find x or y; x could be 1.5 and y could be 0.5, or x could be 100 and y could be -98. But if x and y need to be positive integers, there is only one solution: x = 1 and y = 1.

If you have one equation and two unknowns, but there are restrictions on your unknowns (they must be positive integers, for example), then there may be only one way to make the equation work - you have to check.
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by sayanpaul » Tue Apr 09, 2013 9:39 pm
Let the nos. of 23 cents pencils = A
let the nos. of 21 cents pencils = B

(1) From the first statement, she bought 6 pencils. From this we cannot determine the value of A.
Hence, insufficient.

(2) From the second statement, the total costs 130 cents.
Hence, 23A + 21B = 130.
Since A and B are real positive integers ( as they are the nos. of pencils ), let's determine the multiples of 23 and 21.
multiples of 23: 23,46,69,92,115 (not to exceed 130)
multiples of 21: 21,42,63,84,105,126
From these we can say that it is 46(23*2) and 84(21*4) which sums up to 130.
23(A)+21(B) = 130
or 23(2)+21(4) = 130
or A=2.
Hence statement (2) is sufficient.
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