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If w, x, y, and z are positive integers and w/x<y/z<1, what is the pro

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If w, x, y, and z are positive integers and w/x<y/z<1, what is the pro [#permalink] New post   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, \(\frac{z}{y}\), \(\frac{x+z}{w+y}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\), \(\frac{x}{w}\)
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Re: If w, x, y, and z are positive integers and w/x<y/z<1, what is the pro [#permalink] New post   14 Nov 2018, 19:02
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topper97 wrote:
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, \(\frac{z}{y}\), \(\frac{x+z}{w+y}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\), \(\frac{x}{w}\)



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
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Re: If w, x, y, and z are positive integers and w/x<y/z<1, what is the pro [#permalink] New post   15 Nov 2018, 02:15
topper97 wrote:
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, \(\frac{z}{y}\), \(\frac{x+z}{w+y}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\), \(\frac{x}{w}\)




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..
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Re: If w, x, y, and z are positive integers and w/x<y/z<1, what is the pro [#permalink] New post   01 Sep 2021, 23:18
Hi, chetan2u

I have a doubt here. I also considered the same fractions: 1/2 < 3/4 < 1. Everything else matched except (x+z) / (w+y). I thought since w/x = 1/2 and y/z = 3/4, therefore, (x+z) / (w+y) = (2+4)/ (1+3) = 6/4 = 3/2. Could you please let me know where I could have gone wrong here? Thank you.
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Re: If w, x, y, and z are positive integers and w/x<y/z<1, what is the pro [#permalink] New post   02 Sep 2021, 02:33
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sam12rawat wrote:
Hi, chetan2u

I have a doubt here. I also considered the same fractions: 1/2 < 3/4 < 1. Everything else matched except (x+z) / (w+y). I thought since w/x = 1/2 and y/z = 3/4, therefore, (x+z) / (w+y) = (2+4)/ (1+3) = 6/4 = 3/2. Could you please let me know where I could have gone wrong here? Thank you.



The fractions will not give you the exact value.
For example: \(\frac{w}{x}=\frac{1}{2}=\frac{1*a}{2*a}\) means w = a and x =2a, so you cannot say anything because of the variable’a’, and similarly for other fraction.

\(\frac{(x+z) }{ (w+y)} = \frac{(2a+4b)}{ (a+3b)}\)
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Re: If w, x, y, and z are positive integers and w/x<y/z<1, what is the pro [#permalink] New post   02 Sep 2021, 06:48
EncounterGMAT wrote:
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, \(\frac{z}{y}\), \(\frac{x+z}{w+y}\), \(\frac{xz}{wy}\), \(\frac{x^2}{w^2}\), \(\frac{x}{w}\)


let us assume some values that being x=60 , w=10 , z= 30 , y=15
A and E can be eleminated straight wawy from last eqn

NOW between C and D C triumphs with the last eqn
with the 3 rd eqn we can nail down B

Therefore IMO B
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Re: If w, x, y, and z are positive integers and w/x<y/z<1, what is the pro [#permalink] New post   10 Jun 2023, 00:44
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