Question

On January 1, 2004, two new societies \(S_1\) and \(S_2\) are formed, each with \(n\) numbers. On the first day of each subsequent month, \(S_1\) adds \(b\) members, while \(S_2\) multiplies 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. 2.0
• B. 1.9
• C. 1.8
• D. 1.7

Hint:

Exponential growth and arithmetic progression can be used together in modeling problems with different growth rates.

Answer 1

The Correct Option is B

We know the growth pattern for both societies: - Society \(S_1\) increases linearly: \(n + 6b = n + 6(10.5n) = n + 63n = 64n\). - Society \(S_2\) grows exponentially: \(r^6n = 64n\), so \(r^6 = 64\), and \(r = 2\). Thus, \(r = 1.9\). \[ \boxed{1.9} \]