ronit sharma
ronit sharma
8 years ago

Three times the age of A is 11 more than four times the age of B.If five years ago, A was twice as old as B, then what is the sum of their present ages? Assume all ages are in years.

A. 28

B. 34

C. 39

D. 48

Answer: Option B

Solution(By Examveda Team)

Let A's present age = X
B's present age = Y
Now,
3X = 4Y + 11
3X - 4Y = 11 -------------------- (1)
5 years ago,
(X - 5) = 2(Y-5)
X - 5 = 2Y - 10
X - 2Y = -5 ------------------------------ (2)
Solving equation (1) and (2)
3X - 4Y - 3X + 6Y = 11+15
2Y = 26
Y = 13.
3X = 11 + 4*13
3X = 63
X = 21.
A's age = 21 years.
B's age = 13 years.
Sum of the age = 21 + 13 = 34 years.

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Comments ( 2 )

  1. Kumar Chandan
    Kumar Chandan :
    8 years ago

    3X - 4Y = 11 -------------------- (1)
    X - 2Y = -5 ------------------------------ (2)
    Solving equation (1) and (2)
    3X - 4Y - 3X + 6Y = 11+15
    2Y = 26
    Y = 13.
    3X = 11 + 4*13
    3X = 63
    X = 21.
    Sum of the age = 21 + 13 = 34 years.

  2. Ronit Sharma
    Ronit Sharma :
    8 years ago

    reply soon

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