Question

five years ago a man was old as his son, five years hence the father will be three times old as his son. Find their present ages.




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Answer 1

Answer:

Let Five years ago the age of son be x years and age of father be 7x years

Present age of son =x+5

Present age of father =7x+5

5 years later their age will (x+10) and (7x+10)

∴7x+10=3(x+10)

7x−3x=20

4x=20

x=5

So, the present age of son=x+5=5+5=10years

and the present age of father=7x+5=35+5=40years

Answer 2

➲CORRECT QUESTION :

• Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son . Find their present ages.

☯GIVEN :

• Five years ago, the man was seven times as old as his son.

• Five years hence, the father will be three times as old as his son.

☯TO FIND :

• The present age of Father.

• The present age of his son.

➲SOLUTION :

✞ Let the present age of father and his son be a years and b years.

According to the First Condition :

• Five years ago , the man was seven times as old as his son.

Five years ago,

• Age of father = (a-5) years

• Age of his son = (b-5) years

$\implies{\sf{a-5=7(b-5)}}$

$\implies{\sf{a-5=7b-35}}$

$\implies{\sf{a=7b-35+5}}$

$\implies\bf{a=7b-30....(i)}$

According to the Second Condition :

• 5 years hence, the father will be 3 times as old as his son .

Five years hence ,

• Age of father = (a+5) years

• Age of son = (b+5) years

$\implies\sf{a+5=3(b+5)}$

$\implies\sf{a+5=3b+15}$

• Substituting the value of eq(1) :

$\implies\sf{7b-30+5=3b+15}$

$\implies\sf{7b-25=3b+15}$

$\implies\sf{7b-3b=15+25}$

$\implies\sf{4b=40}$

$\implies\sf{b=\dfrac{40}{4}}$

$\implies \underline{\boxed{\bf{\red{b=10}}}}$

• Substituting the value of b in eq(1) :

$\implies\sf{a=7b-30}$

$\implies\sf{a=7\times10-30}$

$\implies\sf{a=70-30}$

$\implies\underline{\boxed{\bf{\red{a=40}}}}$

$\huge{ \pink{ \therefore}}$ Present age of father = 40 years

$\huge{ \red{ \therefore}}$Present age of his son = 10 years