1.

Sunita is twice as old as ashima.if six years is subtracted from ashima age and 4 years added to aunitas age then Sunita will be four time ashima age .How old were they two years ago​

Answer»

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


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