Question

riaz was 4 times as old as his son 8 years ago.after 8 years, riaz will be twice as old as his son. what are their present ages.

Answer 1

Answer:

8 years before,

Riaz’s son → x years old

Riaz → 4x

Presently,

Riaz → 4x + 8

Riaz’s son → x + 8

8 years hence,

(As per the Question)

Riaz’s age = 2(His son’s age)

= 4x + 8 = 2(x + 8)

= 4x + 8 = 2x + 16

= 2x = 8

= x = 4

Therefore, their present ages are:

Riaz - 24 years

Riaz’s son - 12 years

Answer 2

HII DEAR MATE

BEFORE 8 YEARS-

SON'S AGE BE- X

SO, RIAZ AGE BE - 4X........(1)

AFTER 8 YEARS -

SON'S AGE - X+8

RIAZ AGE - 2(X+8)..............(2)

COMPARING THE EQUATIONS (1) & (2) -

4X+8 = 2(X+8)

4X +8 = 2X + 16

4X -2X = 16-8

2X. = 8

X. = 4

SO, NOW PRESENT AGE OF SON IS- X+8 i.e. 4+8= 12

AND AGE OF RIAZ IS - 2(X+8)

- 2(4+8)

- 2× 12

-24

HOPE THIS HELPS YOU