If \(w, x, y,\) and \(z\) are positive integers and \(\dfrac{w}{x}<\dfrac{y}{z}<1,\) what is the proper order, increasin

[Gmat_mission]

If \(w, x, y,\) and \(z\) are positive integers and \(\dfrac{w}{x}<\dfrac{y}{z}<1,\) what is the proper order, increasing from left to right, of the following quantities: \(\dfrac{x}{w}, \dfrac{z}{y}, \dfrac{x^2}{w^2}, \dfrac{xz}{wy}, \dfrac{x+z}{w+y}, 1?\)

(A) \(1, \dfrac{z}{y}, \dfrac{x}{w}, \dfrac{x+z}{w+y}, \dfrac{x^2}{w^2}, \dfrac{xz}{wy}\)

(B) \(1, \dfrac{z}{y}, \dfrac{x+z}{w+y}, \dfrac{x}{w}, \dfrac{xz}{wy}, \dfrac{x^2}{w^2}\)

(C) \(1, \dfrac{z}{y}, \dfrac{x}{w}, \dfrac{x+z}{w+y}, \dfrac{xz}{wy}, \dfrac{x^2}{w^2}\)

(D) \(1, \dfrac{z}{y}, \dfrac{x}{w}, \dfrac{xz}{wy}, \dfrac{x+z}{w+y}, \dfrac{x^2}{w^2}\)

(E) \(1, \dfrac{z}{y}, \dfrac{x+z}{w+y}, \dfrac{xz}{wy}, \dfrac{x^2}{w^2}, \dfrac{x}{w}\)

Answer: B

Source: Manhattan GMAT

[Scott@TargetTestPrep]

If \(w, x, y,\) and \(z\) are positive integers and \(\dfrac{w}{x}<\dfrac{y}{z}<1,\) what is the proper order, increasing from left to right, of the following quantities: \(\dfrac{x}{w}, \dfrac{z}{y}, \dfrac{x^2}{w^2}, \dfrac{xz}{wy}, \dfrac{x+z}{w+y}, 1?\)

(A) \(1, \dfrac{z}{y}, \dfrac{x}{w}, \dfrac{x+z}{w+y}, \dfrac{x^2}{w^2}, \dfrac{xz}{wy}\)

(B) \(1, \dfrac{z}{y}, \dfrac{x+z}{w+y}, \dfrac{x}{w}, \dfrac{xz}{wy}, \dfrac{x^2}{w^2}\)

(C) \(1, \dfrac{z}{y}, \dfrac{x}{w}, \dfrac{x+z}{w+y}, \dfrac{xz}{wy}, \dfrac{x^2}{w^2}\)

(D) \(1, \dfrac{z}{y}, \dfrac{x}{w}, \dfrac{xz}{wy}, \dfrac{x+z}{w+y}, \dfrac{x^2}{w^2}\)

(E) \(1, \dfrac{z}{y}, \dfrac{x+z}{w+y}, \dfrac{xz}{wy}, \dfrac{x^2}{w^2}, \dfrac{x}{w}\)

Answer: B

Solution:

Let x = 2, y = 3, z = 4 and w = 1. Notice that w/x = 1/2 and y/z = 3/4; so w/x < y/z < 1 is satisfied.

With these values for x, y, z and w:

x/w = 2/1 = 2

z/y = 4/3 ≈ 1.33

x^2/w^2 = 2^2/1^2 = 4

xz/wy = 8/3 ≈ 2.67

(x + z)/(w + y) = 6/4 = 3/2 = 1.5

We see that the correct ordering of the quantities is 1 < z/y < (x + z)/(w + y) < x/w < xz/wy < x^2/w^2.

Answer: B