Question

Deepa's mother is 4 years more than 3 times as old as deepa is now six years from now she will be 8 years more than twice as old as deepa will be then. How old is each of them now?​

Answer 1

Let,

• Present age of Deepa = x

• Present age of Mother = 3x+4

• Age of Deepa after 6 years = x+6

• Age of Mother after 6 years = 2(x+6)+8

Gives,

$\sf{\implies (3x+4)+6=2(x+6)+8}$

$\sf{\implies 3x+4+6=2x+12+8}$

$\sf{\implies 3x+10=2x+20}$

$\sf{\implies 3x-2x=20-10}$

$\sf{\implies x=10}$

$\sf{\implies Age \ of \ Deepa=10}$

$\boxed{\sf{\therefore Age \ of \ Deepa=10}}$

$\sf{\implies Age \ of \ Mother=3x+4}$

$\sf{\implies Age \ of \ Mother=(3\times10)+4}$

$\sf{\implies Age \ of \ Mother=30+4}$

$\sf{\implies Age \ of \ Mother=34}$

$\boxed{\sf{\therefore Age \ of \ Mother=34}}$

Answer 2

Step-by-step explanation:

Let deepa s age = x years [Present Age]

Deepas mother age = (4+3x) years [Present Age]

After 6 years ,,

deepas age = (x+6) years

deepas mother age = 2(x+6)+8 years

Therefore , (4+3x)+6 = 2(x+6)+8

● 4+3x+6 = 2x+12+8

● 3x + 10 = 2x+20

● x = 10

Now, age of deepa = 10 years

deepas mother age = 4+ 3x

= 4 + 3 (10)

= 4 + 30

= 34 years