Math, asked by jannatvirk, 12 days ago

five years ago Mathur age was twice his daughters age. if the sum of their present ages is 100 years find there ages is 5 year ago. ​

Answers

Answered by nagalakshmi369
1

answer :

let,

present age of Mathur = 'x' years

present age of Mathur's daughter = 'y' years

five years ago,

Mathur's age = (x-5) years

his daughter's age = (y-5) years

Mathur age was twice his daughter's age

Mathur age = 2×his daughter's age

(x-5) = 2×(y-5)

x-5 = 2y-10

x = 2y-10+5

x= 2y-5 ---------- ( 1 )

given that,

sum of their present ages = 100 years

x+y = 100

substitute equation 1 in the above equation

2y-5+y = 100

3y-5 = 100

3y = 100+5

y = 105/3

y = 35 ----------- ( 2 )

the present age of mathur's daughter = y

= 35 years

the present age of Mathur = x

= 2y-5 ( from eq ( 1 ) )

= 2(35)-5

= 70-5

= 65

the present age of Mathur = x

= 65 years

5 years ago,

Mathur age = (x-5) years

= 65-5

= 60 years

his daughter's age = (y-5) years

= 35-5

= 30 years

.

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