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
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.
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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