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Q.

Ten years hence, a man's age will be twice the age of his son. Ten years ago, the man was four times as old as his son. What are their present ages?


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a

The present age of the man is 53 years and the present age of his son is 20 years.

b

The present age of the man is 50 years and the present age of his son is 28 years.

c

The present age of the man is 50 years and the present age of his son is 20 years.

d

The present age of the man is 45 years and the present age of his son is 19 years. 

answer is C.

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Detailed Solution

Given that, 10 years hence, a man's age will be twice the age of his son.
Ten years ago, the man was four times as old as his son.
Let the man be referred to as A and the boy be referred to as B.
Let the age of A be x years and the age of B be y years.
Five years ago:
A's age = x5  yrs B's age = y5  yrs   According to the given information, five years ago, A was thrice as old as B.
x5 = 3 y5 x5 = 3y15 x3y =15+5 x3y =10 1    Ten years later,
A's age =  x + 10 B's age =  y + 10   According to the given information, ten years later, A shall be twice as old as B.
x+10=2(y+10) x+10=2y+20 x2y=2010 x2y=10......(2)   From equation 1,
x=3y10  …(3)
Substituting the value of x=3y10   in the second equation and solving it, we get, x - 2y=10 3y10  - 2y=10 y=10+10 y=20  
Putting the value of y = 20 in equation 3 to find the value of x, we get,
x=3y10 x=3 20 10 x = 60 10 x=50   Hence, the present age of A is 50 years and the present age of B is 20 years.
Thus, option 3 is correct.
 
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