Question

Sunita is twice as old as Ashima. Four years from now, Sunita will be four times as old as
Ashima was six years ago. Find the sum of their present ages.

Answer 1

Answer:

Let Ashimas’s present age = A years

According to the first given condition i.e., Sunita is twice as old as Ashima.

Therefore, S = 2A …(i)

Given, if six years is subtracted from Ashima’s age and four years added to Sunita’s age, then Sunita will be four times that of Ashima’s age.

Therefore,

$\sf4×(A–6)=S+44×(A–6)=S+4$

$\sf⇒4A–24=S+4$

$\sf ⇒4A–S=28$

Now, putting value of S from equation (i) in equation (ii), we get

$\sf \: 4A –2A = 28$

$\sf \: ⇒2A=28$

Thus, Ashima’s present age is 14 years.

Putting value of A in equation (i), we have

$\sf \: S=2×14=28years$

Sunita’s present age is 28 years.

Ashima’s age 2 years ago = (14 – 2) years = 12 years

Sunita’s age 2 years ago = (28 – 2) years = 26 years

Sum of Ashima’s age and Sunita’s age two years ago = (12 + 26) years = 38 years

So, the correct answer is “38 years”.