Question

please help me!!!!!!!!!!!​

Answer 1

Step-by-step explanation:

Given :-

Sunita is as twice as old as Ashima. 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.

To find :-

How old were they two years ago?

Solution:-

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

Verified the given relations in the given problem