# Moderate Algebra Solved QuestionAptitude Discussion

 Q. On January 1, 2004 two new societies $s_1$ and $s_2$ are formed, each $n$ numbers. On the first day of each subsequent month, $s_1$ adds $b$ members while $s_2$ multiples its current numbers by a constant factor $r$. Both the societies have the same number of members on July 2, 2004. If $b = 10.5n$, what is the value of $r$?
 ✔ A. 2 ✖ B. 1.9 ✖ C. 1.8 ✖ D. 1.7

Solution:
Option(A) is correct

it is 6 months, from January 1, 2004 to July 2, 2004. So, increase will be 6 times.

No. of members $s_1$ will be in A.P.

On July 2nd , 2004,  $s_1$ will have $n+6b$ members

$=n+6×10.5n$

$=64n$

No. of members in $s_2$ will be in $G.P$.

On July 2nd, 2004 Number of members in  $s_2=nr^6$

$\Rightarrow 64n=nr^6$

$\Rightarrow r=2$

## (3) Comment(s)

Kartik
()

I think the solution is wrong as here we are taking increase in people 6 times. But it is given the no. of members increase subsequently so we must take the increase to be 5.

Pooja
()

why taken increase of member of 6 times..........

Mark
()

Since, from January 1, 2004 to July 2, 2004 it is 6 months. I guess solution mentions that.