Math, asked by ayushihazarika5, 7 months ago

The present ages of the father and the sons are in the ratio 7:3 .Tens years ago,the father's age was four times of his son. Find their present ages.




plz help me.​

Answers

Answered by arpitafelix6040
1

Answer:

The Father's age is 42 years and the Son's age is 18years

Step-by-step explanation:

Since the present ages are in ratio 7:3

Therefore,

take father's age as 7x

take Sons age as 3x

∴(x-10)=4(3x-10) [10 years ago]

= 7x-10=12x-40

= 7x-12x= -40+10

= -5x= -30

= x=6

∴ Fathers age = 7*6 = 42 years

Sons age = 3*6 = 18 years

VERIFICATION:

42/18=7/3

⇒Hence Proved

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