# Find the number of children of the parents from the following?

## Presently the sum of the ages of parents is six times the sum of the ages of children. If two years ago sum of the ages of parents was ten times the sum of the ages of children and six years hence the sum of the ages of parents would be three times the ages of children, what is the number of children?

Oct 29, 2017

3 children

#### Explanation:

Let a, b and x be the sum of the ages of man and his his wife, sum of the ages of their children and number of their children respectively:
$a = 6 b$ =>eq-1
$a - 4 = 10 \left(b - 2 x\right)$ =>eq-2
$a + 12 = 3 \left(b + 6 x\right)$ =>eq-3
substitute 6b for a in eq-2 and eq-3:
$6 b - 4 = 10 b - 20 x$
$6 b + 12 = 3 b + 18 x$
simplify the above equations:
$4 b - 20 x = - 4$
$3 b - 18 x = - 12$
or:
$b - 5 x = - 1$
$b - 6 x = - 4$ => multiply by -1:

$b - 5 x = - 1$
$- b + 6 x = 4$ => add the two equations b's cancel solve for x:
$x = 3$

Oct 29, 2017

The couple have $3$ children.

#### Explanation:

Let the number of children be $n$ and sum of the ages of children at present be $c$ and

the sum of the ages of parents at present be $p$

Hence $p = 6 c$ ................(A)

two years ago the sum of the ages of children would have been $c - 2 n$ and the sum of the ages of parents would have been $p - 4$ and as latter is $10$ times the former, we have

$p - 4 = 10 \left(c - 2 n\right)$ and putting value of $p$ from (A), we have

$6 c - 4 = 10 c - 20 n$ or $20 n - 4 c = 4$ ................(1)

Similarly six years hence the sum of the ages of children would be $c + 6 n$ and the sum of the ages of parents would be $p + 12$ and as latter is $3$ times the former, we have

$p + 12 = 3 \left(c + 6 n\right)$ and putting value of $p$ from (A), we have

$6 c + 12 = 3 c + 18 n$ or $18 n - 3 c = 12$ ................(2)

Multiplying (1) by $3$ and subtracting the product from $4$ times (2), we get

$72 n - 12 c - \left(60 n - 12 c\right) = 12 \times 4 - 4 \times 3$

or $12 n = 36$

or $n = 3$

Hence, the couple have $3$ children.