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Values of x and Y vary with Z

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Values of x and Y vary with Z

by ispolef1 » Mon Sep 28, 2009 7:10 am
The Values of x and y vary with the value of z so that each additive increase of 2 in the value of z corresponds to the value of x increasing by a factor of 2 and the value of y increasing by a factor of 3. if x and y are positive for each z>0, what is the value of x/(x+y) when z = 12?

1. When z = 6, x = 5y

2. When z = 0, x = y+1

oa is a
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by vaibhav.iit2002 » Tue Sep 29, 2009 9:55 am
as its a DS ques. we just need to know if a solution is possible.

1. lets ques. asks value of x/(x+y) for z=8.
when z is 6 lets y=a hence x=5a
goes from 6 to 8, x = 10a, y=3a
x/(x+y) = 10a/(13a) = 10/13

similarly we can have an answer for z=12

2. lets ques. asks value of x/(x+y) for z=2.
when z is 0 lets y=a hence x=a+1
goes from 0 to 2, x = 2(a+1), y=3a
x/(x+y) = (2a+2)/(2a+2+a) => ans. in terms of a

Therefore no solution

Hence A.
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by Nina1987 » Sat Dec 19, 2015 10:13 pm
I first rewrote x/(x+y) to (x+y)/x -> 1+y/x? -> y/x?

(1) When z=6, x=5y so at z=6 if y=p, x will be 5p
Since we know for every 2 in z, x grow by factor 2 and y by factor 3, at z=12, y= 3^3p and x=2^3*5*p. we know we can calculate y/x. SUFF

(2) z=0, x =y+1. Let y=a and so x=a+1. So at z=12, y=3^6*a and x=2^6(a+1)
Now y/x = (3^6*a) / (2^6(a+1)). No solution unless we know a. INSUF

Ans: A
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