Algebra, Inequalities, Number Properties

[swerve]

If \(x\) and \(y\) are positive numbers such that \(x+y=1\), which of the following could be the value of \(100x+200y?\)

I. \(80\)
II. \(140\)
III. \(199\)

A. II only
B. III only
C. I and II
D. I and III
E. II and III

The OA is E

Source: Official Guide

[Scott@TargetTestPrep]

If \(x\) and \(y\) are positive numbers such that \(x+y=1\), which of the following could be the value of \(100x+200y?\)

I. \(80\)
II. \(140\)
III. \(199\)

A. II only
B. III only
C. I and II
D. I and III
E. II and III

The OA is E

Source: Official Guide

Solution:

We are given that x and y are positive numbers and that x + y = 1. We must determine possible values of 100x + 200y. An easy way to determine whether 80, 140, or 199 could be values of 100x + 200y is to use the given fact that x + y = 1 to determine the possible range of 100x + 200y. Since 200 is greater than 100, the high end of our range will be when y is the largest, and the low end of our range will be when x is the largest.

High end of range:

y = 1 and x = 0

100x + 200y = 100(0) + 200(1) = 200

Low end of range:

y = 0 and x = 1

100(1) + 200(0) = 100

Finally we must remember that x and y both must be positive, which means neither x nor y can be zero. They must each be a decimal between zero and one. Thus, the low end of the range cannot actually be 100, and the high end of the range cannot actually be 200. Therefore, we can create the following inequality:

100 < 100x + 200y < 200

Only 140 and 199 are greater than 100 and less than 200.

Answer: E