Question

Ishita is twice as old as Rhea. If six years is
subtracted from Prea's age and four years added
to Ishita's age, then Ishita will be four times
thea's age. How old were they two
years ago?​

Answer 1

EXPLANATION.

Let the age of Rhea be = x years.

Let the age of ishita be = y years.

=> y = 2x ........(1)

if 6 is subtracted from rhea age and

4 is added to ishita age.the ishita

will be 4 times rhea age.

=> Let the present age of rhea = ( x - 6 )

=> Let the present age of ishita = ( y + 4 )

=> 4 ( x - 6 ) = ( y + 4 ) .......(2)

put the value of equation (1) in (2) we get,

=> 4 ( x - 6 ) = ( 2x + 4 )

=> 4x - 24 = 2x + 4

=> 2x = 28

=> x = 14

Therefore,

2 years ago,

Age of rhea = x = 14 years.

=> 14 - 2 = 12 years.

Age of ishita = y = 2x = 2 X 14 = 28

=> 28 - 2 = 26 years.

Age of rhea = 12 years.

Age of ishita = 26 years.

Answer 2

Step-by-step explanation:

Let the age of Rhea be x years.

Therefore, age of Ishita will be 2x years.

It is given that, If six years is subtracted from Reha's age and four years added to Ishita's age :]

• Age of Rhea = x - 6

• Age of Ishita = 2x + 4

According to Question now,

➳ 4(x - 6) = 2x + 4

➳ 4x - 24 = 2x + 4

➳ 4x - 2x = 4 + 24

➳ 2x = 28

➳ x = 28/2

➳ x = 14 years

Therefore,

• Present age of Rhea = x = 14 years

• Present age of Ishita = 2x = 2(14) = 28 years

So,

• Age of Rhea 2 years ago = x - 2 = 14 - 2 = 12 years.

• Age of Ishita 2 years ago = 2x - 2 = 28 - 2 = 26 years