M is what percent of X?

[Mark]

If m > 0, y > 0, and x is m percent of 2y, then, in terms of y, m is what percent of x?
a) y/200
b) 2y
c) 50y
d) 50/y
e) 5000/y

I may be going crazy, but I don't think the answer to this question is correct in all cases. The requirement is met if x = 5, m = 25, and y = 10. Therefore, m is 500 percent of x. In terms of y, this could be 50y (50 x 10 = 500) OR 5000/y (5000/10 = 500).

The answer could be C OR E.

It doesn't seem like this is a fair or accurate question. Please let me know if I am incorrect.

[Saurav]

If m > 0, y > 0, and x is m percent of 2y, then, in terms of y, m is what percent of x?

from the statement

X= (m/100) * 2y

we need to find m / X * 100 (m is what percent of x)

interchange x and 2y on either side, we get

1 / 2y = (m/100) * (1 / X)

multiply both sides by 100 * 100

(100 * 100) / 2y = (100 * 100) * (m/100) * (1 / X)

5000 / 2y = 100 * m / X (= m percent of X)

answer is E.

[MSR]

Saurav,
Thank you for responding. I completely understand the algebraic solution. My question addresses the fact that if x=5, m=25, and y=10, the answer is 5000/y OR 50y. Will the GMAT ever put in a situation where 2 answers work? If so, what should I do? Test other #s?
Thanks.

[Guest]

Consider X=90, m=45, Y=100, hence m is 50 percent of X (=90). In terms of Y, this is 5000 / Y. Only one solution. So it would be better to take another case.

[Saurav]

I dont think GMAC would ever put two possible solutions to a problem solving question. It will go against their instuctions. They have enough to make us work harder through their DS questions.

[RonPurewal]

Saurav,
Thank you for responding. I completely understand the algebraic solution. My question addresses the fact that if x=5, m=25, and y=10, the answer is 5000/y OR 50y. Will the GMAT ever put in a situation where 2 answers work? If so, what should I do? Test other #s?
Thanks.

yes, you should test other numbers.

remember, these are your numbers. if you happen to encounter a situation in which 2 or more answer choices give the "correct" number, then that's pure coincidence.
don't forget that the problem is stated in terms of variables. this means, among other things, that the correct answer must work for ALL choices of values for those variables.

you don't have to go back to square one, though: if you've eliminated an answer, then it stays out of contention. you only need to test whichever answer choices worked the first time around.

[RonPurewal]

I dont think GMAC would ever put two possible solutions to a problem solving question. It will go against their instuctions. They have enough to make us work harder through their DS questions.

no, they won't.
and we don't, either.

but that's not what happened in this problem. what happened here is that the poster plugged in his/her own numbers, and those numbers happened, coincidentally enough, to produce the desired number from 2 answer choices.
this does not indicate a fault in the problem; it's pure chance, an artifact of the numbers chosen as plug-ins.
and it can DEFINITELY happen on official problems, too. proof: get out your purple og quant supplement, open it up to problem solving question #111, try plugging in k = 1, and watch what happens. (don't post details of that problem here)

[saliqk]

Question:

If m > 0, y > 0, and x is m percent of 2y, then, in terms of y, m is what percent of x?

Guys,

For some reason I am not seeing a connection from one part of the problem to the next. I am getting confused right from the start and would appreciate your input.

I am trying to create a formula based upon the words listed in the question, and then trying to solve the problem.

x is m percent of 2y: x = (m/100)*(2y)
in terms of y, m is what percent of x: m=(z/100)*(x)
(Is the above correct at all?)

[sambhavverman]

Answer is E ...
x/2y*100=m

hence m/x*100= 100*100/2y ( by simple manipulation)

= 5000/y...

[saliqk]

sambhavverman:

I am not seeing it. I am sure this problem isn't overly difficult, but I'm have a hard time seeing a connection.

Could someone post a step by step to this problem because I am not sure where I am messing up.

You mentioned m=(x/2y)*100. Not sure how you go there.
x is m percent of 2y: x=(m/100)(2y) is this correct?
then the are asking for: in terms of y, m is what percent of x: m=(z/100)x. is this part correct?

I think my issue is understanding the actual problem.

[sambhavverman]

Answer is E ...
x/2y*100=m

hence m/x*100= 100*100/2y ( by simple manipulation)

= 5000/y...

ok sidiq..
lets put values and justify the answer itself...

x=10 and y= 50

hence m= 10

m is what % of x
i.e.
10 is what % of 10

= (20/10)*100= 100
implies m is equal to (100% of) x....

if u put values directly in 5000/y u will get m=100 again....

i hope that will help//

[tim]

Question:

If m > 0, y > 0, and x is m percent of 2y, then, in terms of y, m is what percent of x?

x is m percent of 2y: x = (m/100)*(2y)
in terms of y, m is what percent of x: m=(z/100)*(x)
(Is the above correct at all?)

This is good. Now you're trying to find z in terms of y.

z = 100m/x from your last equation
since z is what we're looking for, now we need to get a 100m/x from the first equation
x = m/100 * 2y
m = 100x/2y
100m = 10000x/2y
100m/x = 10000/2y = 5000/y

[lancequas]

probably a stupid question but why do we multiply both sides by 100 before manipulating?

[tim]

Lance,
Multiplying by 100 IS manipulating, so no one is multiplying by 100 BEFORE manipulating. But the bottom line is you can multiply by 100 anytime you feel like it as long as you do it to both sides of the equation. You’ll notice that I multiplied both sides by 100 twice in my solution. Most frequently this will be done to clear out fractions or to get you what you need. Remember in this problem I needed to get 100m/x, so at some point I HAD to multiply by 100..