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

Question

Seven years ago, Lisa was three times 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

GMATinsight · Answer

Lisa current age = L
Christine current age = C

Seven years ago lisa = L-7
L-7 = 3(C-7)
L = 3C -14

Two years hence
L+2 = 2 (C+2)
L = 2C + 2

Now we know

3C -14 = 2C + 2
i.e. C = 16
i.e. L = 34

Differnece = 34-16 = 18

Answer: Option C

sashiim20 · Answer

Let present ages of Lisa's be L and Christine's be C respectively.
Seven years ago; 
L - 7 = 3(C-7) = 3C - 21
3C - L = 21 - 7
3C - L = 14 ----------- (i)

In Two years from now; 
L + 2 = 2(C + 2) = 2C + 4
2C - L = 2 - 4
2C - L = -2 ------------ (ii)

Subtracting (i) from (ii), we get;
C = 14 + 2 = 16

Substituting value of C in (ii), we get;
2(16) - L = -2
L = 32 + 2 = 34

Therefore difference of age between Lisa and Christine is = 34 - 16 =18. Answer C...

tlaw · Answer

We are trying to solve for L-C.

We know that L-7 = 3(C-7), and L+2 = 2(C+2).
Solving for these two equations, we get C=16 and L=34.

L-C is therefore 18 ( option C ).

ashudhall · Answer

1)X-7 =3(y-7)
X=3y-14
2) X +2=2(y+2)
X=2y+2
3y-14=2y+2
Y=16
X=34
Hence difference =18
C

Sent from my ONEPLUS A3003 using  GMAT Club Forum mobile app

ScottTargetTestPrep · Answer

We can let Lisa’s current age = L and Christine’s current age = C.
 
Seven years ago, Lisa was (L - 7) years old and Christine was (C - 7) years old, and Lisa was three times as old as Christine:
 
L - 7 = 3(C - 7)
 
L - 7 = 3C - 21
 
L = 3C - 14
 
In two years, Lisa will be (L + 2) years old and Christine will be (C + 2) years old, and Lisa will be twice as old as Christine:
 
L + 2 = 2(C + 2)
 
L + 2 = 2C + 4
 
L = 2C + 2
 
Equating our two equations, we have:
 
3C - 14 = 2C + 2
 
C = 16
 
Since C = 16, L = 2(16) + 2 = 34.

Thus, Lisa is 34 - 16 = 18 years older than Christine.
 
Answer: C

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