Math, asked by raghunathji99, 9 months ago

1. Hari is twice as old as his son. Twenty
years ago, he was four times as old as his
son. Find their present age.​

Answers

Answered by pandaXop
43

Hari = 60 Years

Son = 30 Years

Given:

  • Hari is twice as old as his son.
  • Twenty years ago he was four times as old his son.

To Find:

  • What are their present ages ?

Solution: Let the present age of son be x years. Therefore,

➼ Hari's age = 2 times of x

➼ Hari's age = 2x years

[ Now, 20 years ago their ages were ]

  • Son's age = (x – 20) years
  • Hari's age = (2x – 20) years

A/q

  • Twenty years ago hari was four times as old his son.

\implies{\rm } (2x 20) = 4(x 20)

\implies{\rm } 2x 20 = 4x 80

\implies{\rm } 20 + 80 = 4x 2x

\implies{\rm } 60 = 2x

\implies{\rm } 60/2 = x

\implies{\rm } 30 = x

So, present ages of

➙ Son is x = 30 Years

➙ Hari is 2x = 2(30) = 60 Years

Answered by BrainlyNisha001
16

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Given :-

♦Hari is twice as old as his son

♦Twenty

years ago, he was four times as old as his

son.

To find :-

Present age of Father and son.

Solution :-

♦Let the present age of Hari be x yrs

♦And present age of son be y yrs

ATQ,

x = 2y --------------------(i)

20 yrs ago,

Hari's age = (x-20) yrs

Son's age = (y-20) yrs

=> (x-20) = 4(y-20)

=> x-20 = 4y-80

Putting the value of x

2y-20 = 4y-80

=> 2y-4y = -80+20

=> -2y = -60

=> 2y = 60

=> y = 30

x = 2y

= 2×30

= 60

Hence ,

=> Hari's age = 60

Son's age = 30

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