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10 years ago, Jack was twice as old a Renee. Two years from now, Jack

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10 years ago, Jack was twice as old a Renee. Two years from now, Jack  [#permalink]

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New post 07 Apr 2018, 02:01
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Difficulty:

  45% (medium)

Question Stats:

69% (02:44) correct 31% (02:39) wrong based on 25 sessions

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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
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Re: 10 years ago, Jack was twice as old a Renee. Two years from now, Jack  [#permalink]

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New post 07 Apr 2018, 02:49
j-10=2(r-10)
j+2= 3/2(r+2)
j+r-10 is answer i.e. 46
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10 years ago, Jack was twice as old a Renee. Two years from now, Jack  [#permalink]

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New post 07 Apr 2018, 06:20
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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10 years ago, Jack was twice as old a Renee. Two years from now, Jack &nbs [#permalink] 07 Apr 2018, 06:20
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