Question

age of a father is three times the sum of Ages of his two sons five years hence age of father will be two times the sum of Ages of his two sons find present age of father​

Answer 1

Answer:

Step-by-step explanation:

Given :-

Age of a father is three times the sum of Ages of his two sons.

Five years hence age of father will be two times the sum of Ages of his two sons.

To Find :-

Present age of father.

Solution :-

Let the age of two sons be x and y.

And the age of father be 3(x + y).

5 years hence,

Age of 2 sons = x + y + 10

Age of father = 2(x + y + 10) = 3(x + y) + 5

According to the Question,

⇒ 2(x + y + 10) = 3(x + y) + 5

⇒ 2x + 2y + 20 = 3x + 3y + 5

⇒ 2x - 3x + 2y - 3y = - 20 + 5

⇒ - (x + y) = - 15

⇒ x + y = 15

Father's present age = 3(x + y) = 3(15) = 45 years.

Hence, the present age of father​ is 45 years.

Answer 2

Given :

• Age of a father is three times the sum of ages of his two sons.
• 5 years hence age of father will be two times the sum of ages of his two sons.

To find :

• Present age of father.

Solution :

Consider,

• Age of father = x years.
• Age of 1st son = y years.
• Age of 2nd son = z years.

According to the 1st condition :-

• Age of a father is three times the sum of ages of his two sons.

$\implies\sf{x=3(y+z)}$

$\implies\sf{x=3y+3z..............eq[1]}$

According to the 2nd condition :-

• 5 years hence age of father will be two times the sum of ages of his two sons.

After 5 years,

• Age of father = (x+5) years
• Age of 1st son = (y+5) years
• Age of 2nd son = (z+5) years

$\implies\sf{x+5=2[(y+5)+(z+5)]}$

$\implies\sf{x+5=2(y+z+10)}$

$\implies\sf{3y+3z+5=2y+2z+20\:[put\:x=3y+3z\: from\:eq(1)]}$

$\implies\sf{y+z=15}$

Now put y+z = 20 in eq[1].

$\implies\sf{x=3(y+z)}$

$\implies\sf{x=3\times\:15}$

$\implies\sf{x=45}$

Therefore, the present age of father is 45 years.