The ages of two children are 11 and 8 years. in how many years' time will th e product of their age be 208?

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted on

A more "low-level" solution for anyone who hasn't reached algebra yet:


Their ages differ by three years, and of course they always will, at any future time. So we look for two numbers that multiply to 208 and differ by three. The factors of 208 are 1, 2, 4, 8, 13, 16, 26, 52, 104, 208.

13 and 16 differ by three and multiply to 208, so those must be their ages at the desired future time. That's five years from now.

Trial and error also works well. Try 12 times 9, then 13 times 10, and so on. It doesn't take long to get 16 times 13.

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

The ages of two children are 11 years and 8 years.

Let after x years' time the product of their ages will be 208 years.

Hence, `(11+x)(8+x)=208`

`rArr x^2+19x+88=208`

`rArr x^2+19x-120=0`

`rArr x^2+24x-5x-120=0`

`rArr x(x+24)-5(x+24)=0`

`rArr (x+24)(x-5)=0`

Time cannot be negative, so we will discard `x=-24` .

Hence, `x=5` .

Therefore, after 5 years' time the product of their ages will be 208 years.

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