Question

Two years ago, Ajay's age was seven times his son's age and five years later, he will be six times his son's present
age What is the sum of their present ages?​

Answer 1

Given:

Two years ago, Ajay's age was seven times his son's age and five years later, he will be six times his son's present age.

To find:

Sum of present ages.

Calculation:

Let Ajay's present age be x and his son's present age be y ;

Two years earlier:

• Ajay's age = x -2
• Son's age = y - 2

$\therefore \: (x - 2) = 7(y - 2)$

$= > \: x - 2 = 7y - 14$

$= > \: x - 7y + 12 = 0 \: \: \: \: .......(1)$

5 years later:

• Ajay's age = x + 5
• Son's age = y + 5

$\therefore \: (x + 5) = 6(y + 5)$

$= > \: x + 5= 6y + 30$

$= > \: x - 6y - 25 = 0 \: \: \: \: \: .......(2)$

Subtraction of eq.(1) from eq.(2)

$= > (x - 6y - 25) - (x - 7y + 12) = 0$

$= > y - 37 = 0$

$= > y = 37 \: years$

Putting value of y ;

$= > \: x - 6y - 25 = 0$

$= > \: x - 6(37) - 25 = 0$

$= > \: x - 222 - 25 = 0$

$= > \: x = 247 \: years$

So, sum = 37 + 247 = 284 years.

Answer 2

Let the present age of father be x years & son be y years.

$\sf \red{ According \: to \: the \: question}$

$x - 2 = 3(y - 2 {)}^{2} ..........1$

$\: \: \: \: \: \: \: \: \: \: \: x + 3 = 4(y + 3 {)}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:x + 3 = 4y + 12$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 4y + 9........2$

$\sf \red{Substituting \: 2 \: in \: 1 }$

$4y + 9 - 2 = 3( {y}^{2} + 4 - 4y)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: {3y}^{2} - 16y + 5 = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: {3y}^{2} - 15y - y + 5 = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: 3y(y - 5) - 1(y - 5) = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: (3y - 1)(y - 5) = 0$

$\red{ \therefore \: y \: \cancel{ = } \: \frac{1}{3} }$

$\red{ \therefore \: y = 5}$

$x + 3 = 4( + 3)$

$x = 4(5 + 3) - 3$

$=> 4 \times 8 - 3 = 32 - 3 = 29$

Age of Father = $\red{29\: years}$

Age of sun = $\red{5\: years}$