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

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

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30 May 2005, 17:23
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266. 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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30 May 2005, 18:06
WinWinMBA wrote:
266. 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.

Going with B.

A. Insufficient
B.

130 = 23x + 21y

Only one solution is there for this equation ie (2,4)
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ash
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30 May 2005, 19:19
B too

y=23x+21y

23-cent pencil : 0,23,46,69,92,115
21-cent pencil : 0,21,42,63,84,105,126

if you want to obtain 130, only one solution is possible : 46+84 so 2*23-cent pencil + 4*21-cent pencil
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30 May 2005, 19:39
good problem

When I was doing OG I remember I screwed it up even that I solved it. I got the number of pencils from 2) (total was 6 as in 1) ) but somehow I instinctively connected it to 1) because it said 6 pencils and mistakingly answered C.
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31 May 2005, 06:58
This should have been an easy question, but I think most people would choose C because:

1. x + x = 6
2. 21x + 23y = 130

Two variable, two questions.

But how do you know, just by looking at equation 2 that it's solvable even though there are 2 variable in the same equation ? Do you have to keep plug in numbers ? There are many "legitimate" DS questions that have C as the answer for similar problems. So how do you tell them apart ?
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31 May 2005, 09:09
WinWinMBA wrote:
This should have been an easy question, but I think most people would choose C because:

1. x + x = 6
2. 21x + 23y = 130

Two variable, two questions.

But how do you know, just by looking at equation 2 that it's solvable even though there are 2 variable in the same equation ? Do you have to keep plug in numbers ? There are many "legitimate" DS questions that have C as the answer for similar problems. So how do you tell them apart ?

Only way is - be aware that this is a COMMON Gmat trap. So, double check the stmt to see if there is a unique solution set first. Plug in values if necessary, stop as soon as it becomes obvious there is more than one solution set. Now see if the info in the other stmt will help to establish a unique solution.
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31 May 2005, 10:00
ssap wrote:
WinWinMBA wrote:
This should have been an easy question, but I think most people would choose C because:

1. x + x = 6
2. 21x + 23y = 130

Two variable, two questions.

But how do you know, just by looking at equation 2 that it's solvable even though there are 2 variable in the same equation ? Do you have to keep plug in numbers ? There are many "legitimate" DS questions that have C as the answer for similar problems. So how do you tell them apart ?

Only way is - be aware that this is a COMMON Gmat trap. So, double check the stmt to see if there is a unique solution set first. Plug in values if necessary, stop as soon as it becomes obvious there is more than one solution set. Now see if the info in the other stmt will help to establish a unique solution.

I too vote for B.

For a change I didn't use formulae, though I'd use "street logic".

This is how I solved it.
Martha had to spend at least 130 cents , so she would have to spend at least 126 cents buying only 21 cent pencils. Now that she has 4 cents more she might exchange 2 21-cent pencils with 2 23-cent pencils along with the 4 cents she has. so she would end up with 2 23 cent pencils.

HMTG.
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31 May 2005, 13:55
WinWinMBA wrote:
This should have been an easy question, but I think most people would choose C because:

1. x + x = 6
2. 21x + 23y = 130

Two variable, two questions.

But how do you know, just by looking at equation 2 that it's solvable even though there are 2 variable in the same equation ? Do you have to keep plug in numbers ? There are many "legitimate" DS questions that have C as the answer for similar problems. So how do you tell them apart ?

One reason why I plugged in numbers for equation 2 was because 21 and 23 are big numbers and finding a unique soln if it exists is not so bad/painful.

Suppose EQ 2 was
2x+3y = 130

Then I wouldnt try this method

The other point is you learn from your mistakes. Now that you know this is a common trap, you'll be wary and will think harder before choosing C.
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ash
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I'm crossing the bridge.........

Re: Tricky !   [#permalink] 31 May 2005, 13:55
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