Question

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.


WRONG ANSWERS WILL BE REPORTED...!!!

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

Answer 1

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=shreyash. ....

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Answer 2

Answer:

50 and 20

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