(Q) The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than three times the age of his son. Find their present ages.

Ans: Let’s consider the present age of Father is X and present age of son is Y

Step 1: by given 1st condition, age of a man is twice the square of the age of his son

∴ X = 2 (Y)²

∴ X = 2 Y² ………. (i)

Step 2: by given 2nd condition, Eight years hence, the age of the man will be 4 years more than three times the age of his son

∴ X + 8 = 3 (Y + 8) + 4

∴ X + 8 = 3 Y + 24 + 4

∴ X = 3 Y + 20 ………(ii)

Step 3: By comparing both the equations (i) and (ii), we get:

2 Y² = 3 Y + 20

∴ 2 Y² – 3 Y – 20 = 0

∴ 2 Y² – 8 Y + 5 Y – 20 = 0

∴ 2 Y (Y – 4) + 5 (Y – 4) = 0

∴ (Y – 4) (2 Y + 5) = 0

∴ Y = 4 and Y = –

Here, we reject Y = – , as age can not be a negative; and accept Y = 4

∴ Y = 4 years

Step 4: By substituting the value of Y in Equation (ii), we get:

X = 3 Y + 20

∴ X = 3 (4) + 20 = 12 + 20 = 32

∴ X = 12 + 20 = 32 years

Therefore, the present age of Father is 32 years and present age of son is 4 years.

Check: by 1st condition, 32 = 2 x (4) . It satisfies the 1st condition.

and (32 + 8) = 4 + 3 (4 + 8). This condition also satisfies the 2nd condition.

Since it meets both of the given conditions, hence our answer is correct. 

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