Question

Two years ago, ajay was thrice as old as his daughter and six years later , he will be four years older than twice her age. How old are they now?

Answer 1

Answer:

x=38,y=14.

Step-by-step explanation:

let Ajay's age=x

His daughter's age=y

Given equations are...

=}x-2=3(y-2)

=}x-2=3y-6

=}x=3y-6+2

=}x=3y-4

=}x-3y=-4. ----(1)

=}x+6=2(y+6)+4

=}x+6=2y+12+4

=}x+6=2y+16

=}x=2y+16-6

=}x=2y+10

=}x-2y=10 ----(2)

Subtracting equation (2) from (1)

x-3y=-4

- x-2y=10

----------------

-y =-14

=}y=14

putting this value in equation (2)

=}x-2y=10

=}x-2*14=10

=}x-28=10

=}x=10+28

=}x=38

therefore,x=38,y=14.

Answer 2

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Let the age of Raju be ‘ x ’ and the age of daughter = ‘ y ’ both in year

Then 2 years back their ages are x - 2 and y - 2

∴ By the given condition

️ ➭ x - 2 = 3(y - 2)

️ ➭ x - 3y = -4 ………(i)

️ Again six years later their age will be x + 6 and y + 6 respectively.

∴ By the given condition

️ ➭ x + 6 = 2(y + 6) + 4

➭ x - 2y = 10 ………(ii)

Subtraction from (ii) from (i) , we get

➭ -y = -14

➭ y = 14

Put y = 14 in equation (i) , we get

️➭ x - 3 × 14 = -4

️➭ x = 38

∴ x = 38 ; y = 14

So, Raju's present age = 38 years

Raju's daughter age = 14 years

Hence Verified