Question

Sunita is as twice as old as ashima. If six years is subtracted from ashima age and 4 year is added to sunita age, then sunita will be four times ashima age. How old were they two years ago

Answer 1

Answer:

Given:

• Sunita is as twice as old as ashima

• six years is subtracted from ashima age and 4 year is added to sunita age, then sunita will be four times ashima age

To find:

• The ages of the both sunitha and ashima.

Solving Question:

 Let 'x' be sunitha's age and 'y' be that of ashima's age .

Therefore,

Sunita is as twice as old as ashima

⇒ x = 2y [from given] .....equ(1)

six years is subtracted from ashima age

⇒ (y-6)

and 4 year is added to sunita age

⇒ (x+4)

then sunita will be four times ashima age

⇒ (x+4)= 4(y-6)

or, x+4= 4y - 24

or, x - 4y = -28 .......equ(2)

Solution:

Take equ(1) and (2).

x = 2y

x - 4y = -28

substitute values of equ(1) in equ(2)

⇒ 2y - 4y = -28

or, -2y = -28

or, y = 14

substitute it in equ(1)

x = 2y

⇒ x = 2*14

or, x = 28

∴ The present ages of sunitha is 28 years and that of ashima is 14 years.

Answer 2

Let the age if Sunita be 2x

∴ Ashima = x

∴  2x+4=4(x-6)

∴ 2x+4=4x-24

∴ 24+4=4x-2x

∴28=2x

∴x=$\frac{28}{2}$

∴x=14

∴ Ashima=14 years and Sunita= 28 years

∴ Ashima was 12 years old 2 years ago and Sunita was 24 years old 2 years ago.

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