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Seven years ago, Lisa was three times as old as Christine...

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by BTGmoderatorLU » Thu Jan 16, 2020 3:39 pm
Source: Veritas Prep

Seven years ago, Lisa was twice as old as Christine. In two years, Lisa will be twice as old as Christine. Lisa is how many years older than Christine?

A. 9
B. 14
C. 18
D. 21
E. 27

The OA is C
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Re: Seven years ago, Lisa was three times as old as Christine...

by Jay@ManhattanReview » Thu Jan 16, 2020 9:58 pm
The correct question is:

Seven years ago, Lisa was thrice as old as Christine. In two years, Lisa will be twice as old as Christine. Lisa is how many years older than Christine?

Say Lisa's age is A and Christine's age is C years.

Thus, L – 7 = 3(C – 7) => L = 3C – 14

Again, L + 2 = 2(C + 2) => L = 2C + 2

Thus, 3C – 14 = 2C + 2 => C = 16 years

=> L = 2C + 2 = 2*16 + 2 = 34 years

=> L – C = 34 – 16 = 18 years

The correct answer: C

Hope this helps!

-Jay
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Re: Seven years ago, Lisa was three times as old as Christine...

by Scott@TargetTestPrep » Sat Jan 18, 2020 1:01 pm
BTGmoderatorLU wrote: ↑
Thu Jan 16, 2020 3:39 pm
 Source: Veritas Prep

Seven years ago, Lisa was twice as old as Christine. In two years, Lisa will be twice as old as Christine. Lisa is how many years older than Christine?

A. 9
B. 14
C. 18
D. 21
E. 27

The OA is C
(Note: the posted question should read “thrice” instead of “twice.”)

We can let L = Lisa’s current age and C = Christine’s current age. We can create the equation:

L - 7 = 3(C - 7)

L - 7 = 3C - 21

L = 3C - 14

and

L + 2 = 2(C + 2)

L + 2 = 2C + 4

L = 2C + 2

Substituting, we have:

3C - 14 = 2C + 2

C = 16

So Lisa is 2(16) + 2 = 34 years old and she is 34 - 16 = 18 years older than Christine.

Answer: C

Scott Woodbury-Stewart
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