Carol is three times Alice’s age but only twice as old as Betty. Alice is twelve years younger than Carol. How old is Be

[BTGModeratorVI]

Carol is three times Alice’s age but only twice as old as Betty. Alice is twelve years younger than Carol. How old is Betty?

A. 6
B. 9
C. 12
D. 18
E. 24

Answer: B
Source: Veritas Prep

[Brent@GMATPrepNow]

Carol is three times Alice’s age but only twice as old as Betty. Alice is twelve years younger than Carol. How old is Betty?

A. 6
B. 9
C. 12
D. 18
E. 24

Answer: B
Source: Veritas Prep

Let B = Betty's age

Carol is twice as old as Betty.
So, 2B = Carol's age

Carol is three times Alice’s age
In other words, Alice's age is 1/3 of Carol's age
So, 2B/3 = Alice's age

Alice is twelve years younger than Carol
We can write: 2B/3 = 2B - 12
Multiply both sides by 3 to get: 2B = 6B - 36
Solve: B = 9

Answer: B

Cheers,
Brent

[swerve]

Carol is three times Alice’s age but only twice as old as Betty. Alice is twelve years younger than Carol. How old is Betty?

A. 6
B. 9
C. 12
D. 18
E. 24

Answer: B
Source: Veritas Prep

Let age of Alice \(= x\), so

\begin{array}{|c|c|c|}
\hline
\text{Carol} & \text{Alice} & \text{Betty} \\ \hline
3x & x & \dfrac{3x}{2} \\ \hline
\end{array}

Given that \(3x - 12 = x \Longrightarrow x = 6\)

Age of Betty \(=\dfrac{3\cdot 6}{2} = 9\)

[deloitte247]

Let Carol's age =c
Let Alice's age = a
Let Betty's age = b


c = 3a ....... eqn 1
c = 2b ........eqn 2
a = c - 12 .........eqn 3


From eqn 2; b = c/2
Where c = 3a and a = c - 12
c = 3 (c - 12)
c = 3c - 36
-3c + c = -36
-2c/-2 = -36/-2
c = 18

Betty (b) = c/2 where c = 18
b = 18/2 = 9
b = 9

Answer = B