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Pencils and Markers

Expert replies
Source: — Data Sufficiency |

by Geva@EconomistGMAT » Tue Jan 25, 2011 1:49 am
Your overall goal in this question (as in most DS questions) is to show that the statements are insufficient - that they allow both a yes and no.
One insight that can make your life easier here is to translate the question: in the move from 5 and 3 to 4 and 4, we are actually swapping one notepad for one marker. So the question is asking "can you swap one for one and still stay under $10?"
This angle allows you to focus on the difference between the two prices, and compare that to the distance of the total from $10.

Stat. (1): less than $1 is a very broad definition. We can have the notepads at $0.5 each, so 5 notepads are $2.5, and have the markers cost $2.5, so that 3 markers cover the remaining $7.5 up to $10. In this case, swapping one notepad for one marker will definitely push you over $10, as the the procedure means adding a difference of 2.5-0.5 = $2 to your total, pushing the costs up to $12. So the answer in this scenario is no. But it is always no? Can you find a counter example where the answer is yes?

To find a counter example, two things need to happen:
a. The difference between notepad and marker needs to be made smaller.
b. the total needs to be under $10, so that the added difference due to swapping will not push you over $10.

Try notepad = $0.9 (so that 5 notepads are $4.5), and marker = $1.5, (so that 3 markers cost $4.5). The total in this case would be 4.5+4.5=$9, which would allow you to squeeze another $0.9 notepad in (violating the 5 and 3 demand), but that is easily fixed with a little tweak: slightly increasing notepads to $1.55 each brings the total up to $9.15.
In this case, swapping one notepad for one marker means adding a difference of 1.55-0.9 = $0.65 to the total, which, when added to the $9.15 total, will still be under $10. So the answer is "yes".

It takes a while to think through the reasoning and find these examples, but they satisfy both stat. (1) and (2), meaning that both statements, alone or combined, allow both a yes and a no answer. Thus, the answer is E.
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by ashforgmat » Tue Jan 25, 2011 3:14 am
THanks a lot Geva...
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by eyelikecheese » Tue Jan 25, 2011 11:14 am
The way I set the problem us is as follows:

N=notepad price
M=markers price
The problem states that 5N+3Y<=10

So its asking, is 4N+4Y<=10?

1.) N<1 INS
2.) N<1.1 INS

It offers no solution about M, and the 2 stems essentially state the same information. Is my logic correct?
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by Geva@EconomistGMAT » Tue Jan 25, 2011 11:32 pm
eyelikecheese wrote:The way I set the problem us is as follows:

N=notepad price
M=markers price
The problem states that 5N+3Y<=10

So its asking, is 4N+4Y<=10?

1.) N<1 INS
2.) N<1.1 INS

It offers no solution about M, and the 2 stems essentially state the same information. Is my logic correct?
If stat. (1) would've said N<0.5, would it still be insufficient? even though it says nothing about M?
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by fskilnik@GMATH » Wed Jan 26, 2011 8:54 am
Hi there!

I guess eyelikecheese´s approach is really nice, especially if considered graphically (please follow my arguments in the figure):

01. Both x- and y- axis are dotted because we are not allowed to accept x and y (dollars) negative nor zero ("no free lunch", right?)

02. The line segment in blue is related to the condition imposed in the question stem: 5x+3y <= 10 , in the sense that points (x,y) that respect that (and the property mentioned in 01) are contained in the triangle OBD, where O = (0,0) , B = (0, 10/3) and D = (2,0).

03. The line segment in red is important to understand that from the triangle OBD (item 02), only the part contained in quadrilateral OACD will answer "yes" to the problem posed (x+y <= 5/2), the part contained in triangle ABC will answer "no". (Points A and E are (0, 2.5) and (2.5, 0), of course.)

From the fact that we can choose positive xp (I mean "particular x") arbitrarily near the origin(*) (therefore less than 1 and also less than 10/11) and, with such xp, we are able to find a certain point I and point II (see figure), we are sure (without calculations) that the answer is "E".

Explicitly:

> Take xp = 1/100 and y = 1/100 (that is, xp = y = 1 cent) as point I
> Take xp = 1/100 and y = 3 (that is, xp = 1 cent and y = 3 dollars) as point II

[It is easy to find point C but it is needless...]

Regards,
Fabio.

(*) P.S.: the very-careful reader would say minimum xp should be 1/100 (that is, 1 cent) because we have money-value constraints... and he/she is right! :)

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