Question

Four years ago , Ali was thrice as old as his son. Eight years later , he will be twice as old as his son. Find their present ages.

Answer 1

Answer:

Present age of Ali is 40 years and

present age of his son is 16 years.

Step-by-step explanation:

Consider,

• Present age of Ali = x years
• Present age of his son = y years

4 years ago,

• Age of Ali = (x-4) years
• Age of his son = (y-4) years

According to the 1st condition :-

• Four years ago , Ali was thrice as old as his son.

$\to\sf{x-4=3(y-4)}$

$\to\sf{x-4=3y-12}$

$\to\sf{x=3y-12+4}$

$\to\sf{x=3y-8...............(i)}$

According to the 2nd condition :-

• Eight years later , he will be twice as old as his son

8 years later,

• Age of Ali = (x+8) years
• Age of his son = (y+8) years

$\to\sf{x+8=2(y+8)}$

$\to\sf{x+8=2y+16}$

$\to\sf{3y-8+8=2y+16\:[put\:x=3y-8]}$

$\to\sf{3y-2y=16}$

$\to\sf{y=16}$

Therefore,

★ Present age of his son = 16 years

★ Present age of Ali = 3×16-8 = 40 years.

Answer 2

Answer:

Let the present age of Nuri =x year

And present age of Sonu =y year

Five years ago

Age of Nuri =x–5years

Age of Sony =y–5years

Nuri was thrice as old as Sonu

X−5=3(y–5)

X–5=35–15

X–3y=−15+5

X–3y=−10 ………..(1)

Ten years later,

Age of Nuri =x+10

Age of Sonu =y+10

Nuri will be twice as old as Sonu.

X+10=2(y+10)

X+10=2y+20

X–2y=10 ………..(2)

X–3y=−10 ………..(1)

Subtracting equation (1) from equation (2) we get

Y=20

Plug this value in equation first we get

X−3∗20=−10

X=60–10

X=50

Hence age of Nuri =50 years and age of Sonu =20 years