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14 Nov 2018, 09:19
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If w, x, y, and z are positive integers and \(\frac{w}{x}\)<\(\frac{y}{z}\)<1, what is the proper order, increasing from left to right, of the following quantities: \(\frac{x}{w}\), \(\frac{z}{y}\), \(\frac{x^2}{w^2}\), \(\frac{xz}{wy}\), \(\frac{x+z}{w+y}\), 1?
(A) 1, \(\frac{z}{y}\), \(\frac{x}{w}\), \(\frac{x+z}{w+y}\), \(\frac{x^2}{w^2}\), \(\frac{xz}{wy}\)
(B) 1, \(\frac{z}{y}\), \(\frac{x+z}{w+y}\), \(\frac{x}{w}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\)
(C) 1, \(\frac{z}{y}\), \(\frac{x}{w}\), \(\frac{x+z}{w+y}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\)
(D) 1, \(\frac{z}{y}\), \(\frac{x}{w}\), \(\frac{xz}{wy}\), \(\frac{x+z}{w+y}\), \(\frac{x^2}{w^2}\)
(E) 1, \(\frac{z}{y}\), \(\frac{x+z}{w+y}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\), \(\frac{x}{w}\)
(A) 1, \(\frac{z}{y}\), \(\frac{x}{w}\), \(\frac{x+z}{w+y}\), \(\frac{x^2}{w^2}\), \(\frac{xz}{wy}\)
(B) 1, \(\frac{z}{y}\), \(\frac{x+z}{w+y}\), \(\frac{x}{w}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\)
(C) 1, \(\frac{z}{y}\), \(\frac{x}{w}\), \(\frac{x+z}{w+y}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\)
(D) 1, \(\frac{z}{y}\), \(\frac{x}{w}\), \(\frac{xz}{wy}\), \(\frac{x+z}{w+y}\), \(\frac{x^2}{w^2}\)
(E) 1, \(\frac{z}{y}\), \(\frac{x+z}{w+y}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\), \(\frac{x}{w}\)
14 Nov 2018, 19:02
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topper97
Take them as w/x<y/z<1. =>1/2<3/4<1..
So
\(\frac{x}{w}=\frac{2}{1}\),
\(\frac{z}{y}=\frac{4}{3}\),
\(\frac{x^2}{w^2}=\frac{4}{1}\),
\(\frac{xz}{wy}=8/3\),
\(\frac{x+z}{w+y}=\frac{6}{3}\),
1<4/3<(2/1 & 6/3)<8/3<4
So 1<z/y<(X/w or (X+z)/(w+y))<xz/wy<x^2/w^2
So Ans is either B or C..
Let us check between X/w and (X+z)/(w+y) by different set of numbers
Let w/x =3/4 and y/z=5/6
So x/z=4/3 and (X+z)/(w+y)=(4+6)/(3+5)=10/8=5/4
4/3>5/4 so X/w > (X+z)/(w+y)
Thus answer is B
Archit3110
Major Poster
15 Nov 2018, 02:15
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topper97
Let w=1 ,x=2, y=3 and z=4 , z>y>x>w
Upon using the relations given,
we would end up finding that relation in B is correct..
