Question

Two years ago, father was three times as old as his son and two years hence twice
his age will be equal to five times that of his son. Find their present ages.​ Answer with explanation.

Answer 1

Solutions :-

Let the father's age be x
And the son's age be y

Given : Two years ago, father was three times as old as his son.

x - 2 = 3(y - 2)
=> x = 3y - 6 + 2
=> x = 3y - 4 ________(i)

Given : Two years hence twice
his age will be equal to five times that of his son.

2(x + 2) = 5(y + 2)
=> 2x + 4 = 5y + 10
=> 2x - 5y = 10 - 4
=> 2x - 5y = 6 ________(ii)


Putting the value of x in the equation (ii) we get,

2(3y - 4) - 5y = 6
=> 6y - 8 - 5y = 6
=> y = 6 + 8 = 14


Putting the value of y in the equation (i) we get,

x = 3(14) - 4
=> x = 42 - 4 = 38


Hence,
The father's age = 38 years
And the son's age = 14 years

Answer 2

Answer: present age of father = 38 y &

present age of his son = 14 y

Step-by-step explanation:-

Let the present age of father = X

Let the present age of son = Y

2 years ago, age of father = (X-2)

2 years ago,age of son = (Y-2)

After 2 year, age of father = X+2

after 2 year, age of son = Y+2

Acc. to ques. 1:-

X-2 = 3(Y -2)

X - 2 = 3Y -6

X - 3Y = 2-6

X - 3Y = -4 ___eq.(1)

Acc. to q. 2:-

2(X+2) = 5(Y +2)

2X + 4 = 5Y + 10

2X - 5Y = -4 + 10

2X -5Y = 6____eq.(2)

On multiplying equation first by 2 and using elimination method, we get:-

2X - 6Y = -8

2X - 5Y = 6

(- ) (+) (-)

___________

0 - Y = -14

Y = 14

put the value of Y in equation first:-

X -3×14 = -4

X -42 = -4

X = 42-4 = 38

So, present age of father = 38 years

present age of his son = 14 years.