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

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19 Feb 2014, 22:36

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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]

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19 Feb 2014, 22:59

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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]

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25 May 2015, 04:57

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]

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25 May 2015, 20:25

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. _________________

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

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02 Aug 2015, 06:55

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VeritasPrepKarishma your solution is excellence. I am wondering, how do we know that one form of fraction (you mentioned it as case) is easier to manipulate than another? I tried to solve this problem using the original form of fraction but I ended up nowhere. Do you have the solution using original form of the fraction?

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

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02 Aug 2015, 23:30

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evdo wrote:

VeritasPrepKarishma your solution is excellence. I am wondering, how do we know that one form of fraction (you mentioned it as case) is easier to manipulate than another? I tried to solve this problem using the original form of fraction but I ended up nowhere. Do you have the solution using original form of the fraction?

When you need to separate the variables, multiple terms in the numerator are easy to handle since you can split them:

(x+y)/ x = x/x + y/x = 1 + y/x

But if they are in the denominator, you cannot separate them.

You can start with the equations as they are but you will eventually cross multiply to simplify the denominator. If you do different things with the equations, you will get different but equivalent expressions. So a case can be made to do the question by number plugging instead. _________________

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