Question

$A$ is $25$ years older than $B$. In $15$ years, $A$ will be twice of $B$. Find the present ages of $A$ and $B$.

• A. Present age of $A$ is $40$ years and
  Present age of $B$ is $15$ years
• B. Present age of $A$ is $37$ years and
  Present age of B is $12$ years
• C. Present age of $A$ is $35$ years and
  Present age of $B$ is $10$ years
• D. Present age of $A$ is $45$ years and
  Present age of $B$ is $20$ years

Answer 1

Answer: (C)

Present age of $A$ is $35$ years and

Present age of $B$ is $10$ years

Let the present age of $A$ be $x$ and the present age of $B$ be$y$.

It is given that $A$ is $25$ years older than $B$ that is:

$x=y+25$

$x-y=\mathrm{25..........}\left(1\right)$

Also, In $15$ years, $A$ will be twice of $B$ that is:

$x+15=2(y+15)$

$x+15=2y+30$

$x-2y=30-15$

$x-2y=\mathrm{15..........}\left(2\right)$

Now subtract equation 1 from equation 2 to eliminate $x$, because the coefficients of $x$ are same. So, we get

$(x-x)+(-2y+y)=15-25$

i.e. $-y=-10$

i.e. $y=10$

Substituting this value of $y$ in (1), we get

$x-10=25$

i.e. $x=25+10=35$

Hence, the present age of $A$ is $35$ years and the present age of $B$ is$10$ years.