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 = XB'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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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.
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