Question

The sum of the ages of a man and his wife is $6$ times the sum of the ages of their children. Two years ago the sum of their ages was $10$ times the sum of the ages of their children at that time. After six years the sum of their ages will be $3$ times the sum of the ages of their children. How many children do they have?

• A. $2$ children
• B. $3$ children
• C. $6$ children
• D. $7$ children

Answer 1

The correct option is B $3$ children

Let the couple have $x$ children.
Let the age of the man be M years and the age of wife of W years and the total age of the children be C years.
By the first condition, $M+W=6C$ ...(1)
Two years ago, man's age $=(M-2)$.
wife's age $=(W-2)$
and total age of the children, $=(C-2x)$ as there are $x$ children,
$(M-2)+(W-2)=10(C-2x)$
$M+N-4=10C-20x$ ...(2)
From, equation $1and2$,
$6C-10C+20x=4$
$-4C+20x=4$
$=>-C+5x=1$ ...(3)
After $6$ years, man's age $=(M+6)$
wife's age $=(W+6)$
and total age of children, $=(C+6x)$ as there as $x$ children
Given, $(M+6)+(W+6)=3(C+6x)$
$=>M+W+12=3C+18x$ ...(4)
From equation 1 and 4,
$6C+12=3C+18x$
$=>C-6x=-4$ ...(5)
Solving equation $3$ and $5$, we get
$x=3$
Hence, the man and wife have $3$ children.