Question

A business owner opens one store in town A. The equation p(x) = 10,000(1.075) represents the anticipated profit after t years. The business owner opens a store in town B six months later and predicts the profit from that store to increase at the same rate. Assume that the initial profit from the store in town B is the same as the initial profit from the store in town A. At any time after both stores have opened, how does the profit from the store in town B compare with the profit from the store in town A?

Answer 1

The answer to your question is 96%

I just took the test.

Answer 2

The rigth equation to anticipate the profit after t years is p(t) = 10,000 (1.075)^t

So, given that both store A and store B follow the same equations but t is different for them, you can right:

Store A: pA (t)  10,000 (1.075)^t

Store B: pB(t'): 10,000 (1.075)^t'

=> pA(t) / pB(t') = 1.075^t / 1.075^t'

=> pA(t) / pB(t') = 1.075 ^ (t - t')

And t - t' = 0.5 years

=> pA(t) / pB(t') = 1.075 ^ (0.5) = 1.0368

or pB(t') / pA(t) = 1.075^(-0.5) = 0.964

=> pB(t') ≈ 0.96 * pA(t)

Which means that the profit of the store B is about 96% the profit of store A at any time after both stores have opened.