A probability problem by Sanskar Tare

Let the present number of members of Club G be $n$ and the sum of their ages be $S$ . Then the present average age is $\dfrac{S}{n}$ . Therefore, we have:

$\begin{cases} \dfrac{S-5\times 9}{n-5} = \dfrac{S}{n} + 1 &...(1) \\ \dfrac{S + 5\times 17}{n + 5} = \dfrac{S}{n} + 1 &...(2) \end{cases}$

$\begin{aligned} \Rightarrow \dfrac{S-45}{n-5} & = \dfrac{S + 85}{n + 5} \\ (S-45)(n+5) & = (S+85)(n-5) \\ S(n+5-n+5) & = 45(n+5) + 85(n-5) \\ 10S & = 130n - 200 \\ S & = 13n - 20 \end{aligned}$

From $(1):$

$\begin{aligned} \dfrac{13n-20-45}{n-5} & = \dfrac{13n-20}{n} + 1 \\ (13n-65)n & = (14n-20)(n-5) \\ 13n^2 - 65n & = 14n^2 - 90n + 100 \\ n^2 -25n +100 & = 0 \\ (n-20)(n-5) & = 0 \end{aligned}$

Since $n > 5$ , $n = \boxed{20}$

Let the present no. of members be n Let the average age be y So,the present sum of ages becomes ny When 5 members,each 9 years old leave, the no. of members become n-5 and sum of ages becomes ny-45

giving -> ny-45= (y+1)(n-5) So, ny-45=ny+n-5y-5 So, 5y-n= 40 ............. (1)

When 5 members each 17 years old join the group,number of members becomes n+5 and sum of ages becomes ny+85

giving -> ny +85 =(n+5)(y+1) so, ny + 85 = ny +5y+n So, 5y +n= 80 ................(2)

ADDING EQUATION(1) AND (2)WE GET 10y=120 GIVING y =12 AND n=20 HENCE ANSWER IS 20