SSC CGL 4 June 2019 Shift-2 Question 11

Question 11

10 years ago. a father's age was $$3\frac{1}{2}$$ times that of his son, and 10 years from now, the fathers age will be $$2\frac{1}{4}$$ times that of the son. What will be the sum of the ages of the father and the son at present?


Let the present age of the son is x years and father's age is y years.

10 Year before, the age of the son was =x-10 and age of father =y-10

As per the condition given in the question,


$$\Rightarrow 2y-20=7x-70$$

$$\Rightarrow 7x-2y=50------(i)$$

Now, The age of son after 10 year, will be $$x+10$$ and father's age $$=y+10$$

as per the given condition in the question,

$$\Rightarrow y+10=\dfrac{9(x+10)}{4}$$

$$\Rightarrow 4y+40=9x+90$$

$$\Rightarrow 9x-4y=-50 ------(ii)$$

From the equation (i) and (ii),

$$x=30years$$ and $$y=80years$$

Hence the sum of the age=$$80+30=110$$years

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