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Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
19 Feb 2014, 21:36

2

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MrWallSt wrote:

If x / (x+y) = n, and x / (x-y) = m, then x/y=? (|x| does not equal |y|, xy does not equal 0)

A. 3mn/2 B. (3m)/(2n) C. (n * (m+2))/2 D. 2nm / (m-n) E. (n^2 - m^2) / nm

You can do it either algebraically or by assuming numbers:

Algebra: Note that we are happier with (x+y)/x rather than x/(x+y) since in the former case we can make manipulations easily. So let's take the inverse of both n and m

Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
19 Feb 2014, 21:59

@Karishma thanks for the post. Your blogs are extremely helpful btw. I am not sure if you ever followed up about your CFA, but I hope that went well for you. _________________

Any and all kudos are greatly appreciated. Thank you.

with algebric solution, i am getting many answers in terms of m & n. so the problem needed to be solved by picking numbers. when we pick x = 3 & y = 1. I am getting B & D both as answer and but with x = 2 & y= 3. I am getting D as an answer. shouldnt it be unique for all numbers, since there is no restriction (like x, y = consequetive or x>y or viceversa) on picking numbers.

with algebric solution, i am getting many answers in terms of m & n. so the problem needed to be solved by picking numbers. when we pick x = 3 & y = 1. I am getting B & D both as answer and but with x = 2 & y= 3. I am getting D as an answer. shouldnt it be unique for all numbers, since there is no restriction (like x, y = consequetive or x>y or viceversa) on picking numbers.

When you take one of the numbers as 1 or 0, or when you take numbers to be equal, you could get multiple options satisfying your conditions.

For example, if n = 1 3n/(m+n) will give the same result as 3/(m+n).

Sometimes, you will get the same answer from multiple options if the options are written intelligently. So you could need to try out 2 or even 3 sets of values. _________________

Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
25 May 2015, 03:57

VeritasPrepKarishma wrote:

When you take one of the numbers as 1 or 0, or when you take numbers to be equal, you could get multiple options satisfying your conditions.

For example, if n = 1 3n/(m+n) will give the same result as 3/(m+n).

Sometimes, you will get the same answer from multiple options if the options are written intelligently. So you could need to try out 2 or even 3 sets of values.

well the question states xy != 0, so x or y cant be 0. also |x| != |y| so they cant be equal either.

I was hoping for a more bullet proof solution for this question. as 2 sets will take more time. we will get answer but at a huge expense of time.

Re: If x/(x+y) = n, and x/(x-y) = m, then x/y = ? (|x| does not [#permalink]
25 May 2015, 19:25

Expert's post

Alaukik wrote:

VeritasPrepKarishma wrote:

When you take one of the numbers as 1 or 0, or when you take numbers to be equal, you could get multiple options satisfying your conditions.

For example, if n = 1 3n/(m+n) will give the same result as 3/(m+n).

Sometimes, you will get the same answer from multiple options if the options are written intelligently. So you could need to try out 2 or even 3 sets of values.

well the question states xy != 0, so x or y cant be 0. also |x| != |y| so they cant be equal either.

I was hoping for a more bullet proof solution for this question. as 2 sets will take more time. we will get answer but at a huge expense of time.

Note that it is a generic comment about number plugging - not just specific to this question.

Also, I have given the algebra solution above - as "bulletproof" as you can get. _________________

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