Question

Sunita is twice as old as ashima if 6 years is subtracted from ashima age and 4 years is added to Sunita age then sunita will be 4times ashima age. how old were the two years ago

Answer 1

Here is your solution

Let ,

Ashimas age be x

Sunita,s age = 2x

If 6 years is subtracted from ashima age and 4 years added to sunita age .

then sunita will be 4 times of ashima,s age.

A/q

=>  4 (x - 6) = (2x + 4) 

=>  4x - 24 = 2x + 4

=>  4x - 2x = 4 + 24 

=> 2x = 28

=> x = 28/2 

=> x = 14

Hence

Present Ashima,s age = x = 14 years 

present Sunita,s age = 2x = 2 × 14 = 28 years

2 years ago age

=>Ashima,s age = x = 14 -2=12 years 

=>Sunita,s age = 2x = 2 × 14 = 28-2 =26 years

Hope it helps you

Answer 2

let x be the age of Ashima.

so, age of Sunita = 2x

and, according to the question,

if 6 years are subtracted from Ashima ageand 4 years is added to Sunita age :-

ashima's age = x - 6

sunita's age = 2x + 4

then , another condition applies here such that :-

4 ( x - 6 ) = 2x + 4

=> 4x - 24 = 2x + 4

take the terms which can be simplified to one side such that :

=> 4x - 2x = 4 + 24

=> 2x = 28

=> x = $\frac{28}{2}$

=> x = 14

so, present age of ashima = x = 14 yrs .

and, present age of Sunita = 2x = 2 × 14 = 28 yrs.

thus,

age of ashima 2 yrs ago = 14 - 2 = 12 yrs.

and, age of Sunita 2 yrs ago = 28 - 2 = 26 yrs.