Math, asked by Sarojmimrot123, 1 year ago

2 years ago a father was 5 times as old as his son 2 years later his age will be 8 more than 3 times the age of the Son find the present age of father and son

Answers

Answered by Anonymous
7
Hi there !!
Here's your answer

Let the son's age 2 years ago be x years
Father's age 2 years ago = 5x years

Present ages of :
Son = x + 2 years
Father = 5x + 2 years

Ages of father and son 2 years later:
Son = x + 2 + x = x + 4 years
Father = 5x + 2 + 2 = 5x + 4 years

Given,
2 years later fathers age will be 8 more than 3 times the age of the Son

So,
the following balanced equation will be formed

3(x + 4) + 8 = (5x + 4)

3x + 12 + 8 = 5x + 4

3x + 20 = 5x + 4

20 - 4 = 5x - 3x

16 = 2x

x =  \frac{16}{2}

x = 8


So,
Present age of son = x + 2 = 8 + 2 = 10 years
Present age of father = 5x + 2 = 5(8)+2 = 42 years

Thus,
the present ages of father and son are 42 years and 10 years respectively


________________________________


Hope it helps !!

Anonymous: :-)
IshanS: :thumbs_up:
Anonymous: :-)
Answered by IshanS
3
Hi there!

Let the age of father is x years and that of son is y years.

ATQ,

x - 2 = 5(y-2)

⇒x - 5y = -10+2

⇒ x - 5y =  -8

⇒ x = 5y-8

x+2 = 3(y+2)+8

⇒ x - 3y = 6 + 8 - 2

⇒ 5y - 8 - 3y = 12

⇒ 2y = 12 + 8

⇒ y = 20/2

⇒ y= 10

So,
x = 5y - 8 = 50 - 8 = 42

Therefore,

The age of father is = x = 42 yrs.
The age of son is = y =10 yrs. 

[ Thank you! for asking the question. ]
Hope it helps

Anonymous: Awesome!✋
Sarojmimrot123: Yes it is
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