Question

The sum of the ages of a father and his 2 sons who are twins is 48. Ten years hence his age will be twice the sum of the son's ages. Find the present the age of one son.

Answer 1

Sum of Twins age = x

Father's age = 48 - x

after 10 years

Sum of Twin age = x + 20

Father's age = 48 - x + 10   or    2(x+20)

58 - x  = 2x + 40

3x = 18

x = 6

Present age of one son = 6/2 = 3years

Answer 2

Answer:

The present age of the one son is 3 years.

Step-by-step explanation:

We are given the sum of ages of a father and his 2 sons who are twins is 48.

Let the the age of each twin be x years. Then the sum of ages of two twins is $2x$.

Father's age is $48-2x$

After ten years father's age is twice the sum of the son's ages

Father's age after 10 years$= 48-2x+10$

Sum of ages of sons after ten years = $2x+10+10=2x+20$

Use the given condition that is father's age is twice the sum of ages of son's

$48-2x+10=2(2x+20)$

Perform subtraction on Left Hand Side

$58-2x=2(2x+20)$

Perform mulyiplication on RHS

$58-2x=4x+40$

$58-40=4x-(-2x)$

$18=4x+2x$

Perform addition on RHS

$18=6x$

Divide with 6 on both sides

$x=3$

Therefore, the present age of son is 3 years.