Question

two years ago Salim 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

Let Salim's present age be x years.
And his daughter's present age be y years.
Two years ago-
Salim's age = (x-2) years
His daughter's age = (y-2) years
ATP- x-2 = 3(y-2)
x -2 = 3y-6
x -3y = -4 .......1
Six years later -
Salim's age =(x+6) His daughter's age =(y+6) years
ATP- x+6 = 2(y+6) +4
x+6 = 2y+12+4
x+6 = 2y+16
x-2y = 16-6
x-2y = 10 .......2. Subtract equation 2 from equation 1
x-3y = -4
x-2y = 10
- + -
_________
-y = -16
y = 16 ......3
Present age of Salim's daughter is 16 years.
Put value from equation 3 in equation 1.
x-3(16) = -4
x-48 = -4
x = -4+48
x = 44
Salim's present age is 44 years.
HOPE IT WILL HELP YOU!!!

Answer 2

Answer:

14 years, 38 years

Step-by-step explanation:

Let the age of daughter two years be x years.

Two years ago:

Age of Salim = 3x

6 years later:

Age of daughter = x + 2 + 6 = x + 8

Age of Salim = 3x + 2 + 6 = 3x + 8.

Given that 6 years later he will be 4 years older than twice her age.

⇒ 3x + 8 = 2(x + 8) + 4

⇒ 3x + 8 = 2x + 16 + 4

⇒ 3x + 8 = 2x + 20

⇒ x = 12

Therefore:

Present age of daughter = x + 2 = 14

Present age of Salim = 3x + 2 = 38

Hope it helps!