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

Vincen wrote: ↑

Tue Jan 05, 2021 3:13 am

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

Answer: E

Source: Official Guide

Given: x + y = 1
Given: x and y are positive numbers. So, if x + y = 1, then x and y are each less than 1

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

Since y is a POSITIVE number and since y < 1, we know that: 0 < 100y < 100
So, 100 < 100 + 100y < 200
In other words 100 + 100y (aka 100x + 200y) can have any value between 100 and 200

Answer: E

NOTE: If anyone needs more convincing, consider these two cases:
case a: If x = 0.6 and y = 0.4, then 100x + 200y = 60 + 80 = 140 (value II)
case b: If x = 0.01 and y = 0.99, then 100x + 200y = 1 + 198 = 199 (value III)