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

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

by M7MBA » Wed May 20, 2020 6:51 am

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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

[spoiler]OA=E[/spoiler]

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

by Scott@TargetTestPrep » Sun May 31, 2020 10:15 am
M7MBA wrote: ↑
Wed May 20, 2020 6:51 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

[spoiler]OA=E[/spoiler]

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



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

by Brent@GMATPrepNow » Mon Jun 01, 2020 8:29 am
M7MBA wrote: ↑
Wed May 20, 2020 6:51 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

[spoiler]OA=E[/spoiler]

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)
Brent Hanneson - Creator of GMATPrepNow.com
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