Question

Radha is 6yrs younger than sudha. Product of their ages after 4 yrs will be same as 16 times the ages of sudha at that time. How old is Radha at present
a) 15 yrs
b) 10 yrs
c) 18 yrs
d) 16 yrs
e) 12 yrs ​

Answer 1

Answer :-

Radha's age = 12 years.

Option e.) is correct.

Solution :-

Let Radha's age be x and Sudha's age be y.

Case - I :

Condition 1 - Radha is 6 yrs younger than Sudha.

Equation :

=> x = y - 6      .... eq. i.)

Case - II :

Condition 2 - Product of their ages after 4 yrs will be same as 16 times the ages of Sudha at that time.

=> (x + 4) x (y + 4) = 16 (y + 4)

Substituting the value of x from eq. i.)  

$\tt{=> (y - 6 + 4) \times (y + 4) = 16 (y + 4)}\\ \\ => (y -2)\times (y + 4) = 16 (y + 4)\\ \\ => y^{2} + 4y - 2y - 8 = 16 (y + 4)\\ \\ => y^{2} + 2y - 8 = 16y + 64\\ \\ => y^{2} = 16y - 2y + 64 + 8\\ \\ => y^{2} = 14y + 72\\ \\ => y^{2} - 14y - 72 = 0$

Using quadratic formula to solve it further,

=> y² - 14y - 72 = 0

Here,

a = 1,

b = - 14,

c = -72

First finding discriminant,

$\tt{Discriminant = b^{2} -4ac}\\\\\tt{Discriminant = (-14)^{2} -4\times 1\times -72}\\\\\tt{Discriminant = 196 +288}\\\\\tt{Discriminant = 484}$

Now, finding the roots of y

$=>y=\dfrac{-b+\sqrt{Discriminant}}{2\times a}\:\:\:\:or\:\:\:\:=\dfrac{-b-\sqrt{Discriminant}}{2\times a}\\\\\\=>y=\dfrac{-(-14)+\sqrt{484}}{2\times 1}\:\:\:\:or\:\:\:\:=\dfrac{-(-14)-\sqrt{484}}{2\times 1}\\\\\\=>y=\dfrac{14+\sqrt{484}}{2\times 1}\:\:\:\:or\:\:\:\:=\dfrac{14-\sqrt{484}}{2\times 1}\\\\\\=>y=\dfrac{14+22}{2\times 1}\:\:\:\:or\:\:\:\:=\dfrac{14-22}{2\times 1}\\\\\\=>y=\dfrac{14+22}{2}\:\:\:\:or\:\:\:\:=\dfrac{14-22}{2}\\\\\\=>y=\dfrac{36}{2}\:\:\:\:or\:\:\:\:=\dfrac{-8}{2}$

$=>y=18\:\:\:\:or\:\:\:\:=-4$

As, age cannot be negative. Therefore, y = 18

Sudha's age = y = 18 years

Substituting the value of y in eq. i.)

=> x = y - 6

=> x = 18 - 6

=> x = 12

Radha's age = x = 12 years.