Question

The sum of the ages of two friends A and B 18 years ago was half of the sum of their ages today.
Presently. A is twice as old as B. What is the
present age of A (in years)?​

Answer 1

GIVEN :-

• At present , A is twice old as B .

• 18 years ago , sum of ages of two friends A and B was half the sum of

       their ages today .

TO FIND :-

• Present age of A .

SOLUTION :-

    Let ,

     Present age of A = x

     Present age of B = y

   18 years before ,

     Age of A = x - 18

     Age of B = y - 18

  According to the first condition ,

   $\longrightarrow\sf x = 2y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .................(1)$

  According to the second condition ,

   $\longrightarrow \sf (x-18)+(y-18) = \dfrac{x+y}{2} \\\\\longrightarrow x+y-36 = \dfrac{x+y}{2} \\\\\longrightarrow 2(x+y-36) = x+y \\\\\longrightarrow 2x+2y-72 = x+y \\\\\longrightarrow 2x-x+2y-y = 72 \\\\\longrightarrow x+y = 72 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .............(2)$

  Substitute x = 2y in equation (2) ,

   $\longrightarrow \sf 2y+y = 72 \\\\\longrightarrow 3y = 72 \\\\\longrightarrow y = 24$

  Present age of A = 2y

                             = 2 × 24

                            = 48 years