Carcass wrote:
If x is the number on the number line between 5 and 15 that is twice as far from 5 as from 15, then x is
A. \(5 \frac{2}{3}\)
B. \(10\)
C. \(11 \frac{2}{3}\)
D. \(12 \frac{1}{2}\)
E. \(13 \frac{1}{3}\)
Since x is BETWEEN 5 and 15, we know that:
15 - x = the distance from x to 15
x - 5 = the distance from x to 5
GIVEN: x is twice as far from 5 as x is from 15In other words, the distance from x to 5 is TWICE the distance from x to 15.
So, we can write:
x - 5 = 2(
15 - x)
Expand: x - 5 = 30 - 2x
Rearrange values to get: 3x = 35
Solve: x = 35/3 = 11 2/3
Answer: C
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep
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