Sunita is twice as old as Ashima. 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. Find the sum of their ages two years ago.

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40 years

42 years

36 years

38 years

answer is D.

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Detailed Solution

Concept- 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,

4 \times (A– 6) = S + 4

\Rightarrow 4 A– 24 = S + 4

\Rightarrow 4 A–S = 28 \dots (ii)

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

4 A – 2 A = 28

\Rightarrow 2 A = 28 \Rightarrow A = 14

Thus, Ashima’s present age is 14 years.
Putting value of A in equation (i), we have

S = 2 \times 14 = 28

years
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”.
Hence, the correct option is 4.

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