EXPLANATION:
Let the present age of Ashima be X years
Then the present age of Sunita = 2X years
Since Sunita is as twice as old as Ashima
If 6 years are subtracted from Ashima' s age then her age will be (X-6) years
If 4 years are added to Sunita's age then her age will be (2X+4) years
According to the given problem
If 6 years are subtracted fr Ashima' s age and 4 years are added to Sunita' s age then Sunita will be four times of Ashima.
→2X+4 = 4×(X-6)
→2X+4 = 4X-24
→4+24 = 4X-2X
→28 = 2X
→2X = 28
→X = 28/2
→X = 14 years
And 2X = 2×14 = 28 years
Ashima's present age = 14 years
Two years ago her age = 14-2 = 12 years
Sunita's present age = 28 years
Two years ago her age = 28-2 = 26 years
Answer: Two years ago the ages of Ashima and Sumita are 12 years and 26 years respectively.
Check: Two years ago the ages of Ashima and Sumita are 12 years and 26 years respectively.
Ashima' s present age = 12+2=14 years
Sunita's present age = 26+2 = 28 years
Sunita's present age = Twice of Ashima's present age
If 6 years subtracted from 14 years = 14-6 = 8 years
If 4 years are added to 28 = 28+4 = 32 years
→4×8 years
Sunita's age= 4 times the age of Ashima
Hence, verified.
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