Math, asked by Anonymous, 4 months ago

Q. Ten years hence, a man’s age will be twice the age of his son. Ten years ago, the man

was four times as old as his son. Find their present ages.

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Class 10th , LINEAR EQUATION, CBSE...​

Answers

Answered by Varun9097M
1

Answer:

Let the age of a man = x years And the age of his son = y years

Ten years hence,

Man's age =(x+10) years

Son's age =(y+10) years

According to the question, (x+10)=2(y+10)

⇒x+10=2y+20

⇒x=2y+20−10

⇒x=2y+10…(i)

Ten years ago,

Father's age =(x−10) years

Son's age =(y−10) years

According to the question, (x−10)=4(y−10)

⇒x−10=4y−40

⇒x=4y−30…(ii)

From Eq. (i) and (ii), we get 2y+10=4y−30

⇒2y−4y=−30−10

⇒−2y=−40

⇒y=20

On putting the value of y=20 in Eq. (i), we get x=2y+10

⇒x=2(20)+10

⇒x=50

Hence the age of the man is 50 yrs and the age of his son is 20 yrs.

Answered by sumanagarwala254
1

Answer:

Hence the age of the man is 50 yrs and the age of his son is 20 yrs.

Step-by-step explanation:

let the age of a man = x years And the age of his son = y years  

Ten years hence,

Man's age =(x+10) years

Son's age =(y+10) years

According to the question, (x+10)=2(y+10)

⇒x+10=2y+20

⇒x=2y+20−10

⇒x=2y+10…(i)

Ten years ago,  

Father's age =(x−10) years  

Son's age =(y−10) years  

According to the question, (x−10)=4(y−10)

⇒x−10=4y−40

⇒x=4y−30…(ii)

From Eq. (i) and (ii), we get 2y+10=4y−30

⇒2y−4y=−30−10

⇒−2y=−40

⇒y=20

On putting the value of y=20 in Eq. (i), we get x=2y+10

⇒x=2(20)+10

⇒x=50

Hence the age of the man is 50 yrs and the age of his son is 20 yrs.

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