Question

two years ago, a man's age was three times the square of
his sons age in three years time his age will be four times his son's age. Find their present age​

Answer 1

Answer:

Let the man's present age be x

And the son's present age be y

Two years ago :

Man's age = x -2

son's age = y- 2

A/Q

x - 2 = 3 ( y-2 )^

x - 2= 3 ( y^ - 2.y.2 + 2^)

x - 2= 3( y^ - 4y + 4)

x -2 = 3y^ -12y + 12

x - 3y^ = -12y+ 2+12

x - 3y^= -12y+14......(i)

Three years time:

Man's age= x+ 3

Son's age= y+ 3

A/Q

x+3=4(y+3)

x+3=4y+12

x= 4y+12-3

x=4y+9......(ii)

Putting the value of (ii) in (i), we get,

(4y+9)-3y^=-12y+14

8y-3y^=14-9

8y-3y^=5

-3y^+8y-5=0

3y^-8y+5=0

3y^-(3+5)y+5=0

3y^-3y-5y+5=0

3y(y-1)-5(y-1) = 0

(3y-5) (y-1) =0

Therefore, either:

3y-5=0

y=5/3

Or,

Y-1=0

y =1

Putting the value of y in (ii), we get,

x =4y+9

x =4×1+9

× = 14.