benejo wrote:

10 years ago, Jack was twice as old a Renee. Two years from now, Jack will be 1.5 time as old as Renee. What was the sum of the ages of Jack and Renee 5 years ago?

A. 28

B. 36

C. 46

D. 50

E. 56

Source: ExpersGlobal

Solve the first two equations. "Five years ago" can be handled at the end.

Ten years ago, Jack was twice as old as Renee:

J - 10 = 2(R -10)

J - 10 = 2R - 20

J = 2R - 10

Two years from now, Jack will be 1.5 times as old as Renee.

J + 2 =

\(\frac{3}{2}\)(R + 2)

J + 2 =

\(\frac{3}{2}\)R + 3

Substitute J = 2R - 10

(2R - 10) + 2 =

\(\frac{3}{2}\)R + 3

2R -

\(\frac{3}{2}\)R = 10 - 2 + 3

\(\frac{4}{2}\)R -

\(\frac{3}{2}\)R = 11

\(\frac{1}{2}\)R = 11

R = 22

J = 2R - 10

J = 34

R + J = (22 + 34) = 46

FIVE YEARS AGO is the same.

(Do the arithmetic. There is a reason it works. If you see it, great. If not, arithmetic is simple.)

R = (22-5) = 17

J = (34-5) = 29

R + J = 46

Answer C

**I left 1.5 = \(\frac{3}{2}\) until near the end. That fraction is not hard. If you wish, at second equation, multiply all by 2 to clear the fraction: J + 2 = \(\frac{3}{2}\)(R + 2)

2J + 4 = 3(R + 2) . . .finish
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