At her current job, Mary gets a 1.5% raise twice per year.

swerve wrote:At her current job, Mary gets a 1.5% raise twice per year. Which of the following choices represents Mary’s current income y years after starting the job at a starting salary of s?
$$A.\ s\left(1.5\right)^{2y}$$
$$B.\ s\left(0.015\right)^{2y}$$
$$C.\ s\left(1.015\right)^{2y}$$
$$D.\ s\left(1.5\right)\frac{y}{2}$$
$$E.\ s\left(1.015\right)\frac{y}{2}$$
The OA is C

Source: Manhattan Prep

One approach here is to apply the COMPOUND INTEREST formula.
However, if you didn't see that the question is analogous to a COMPOUND INTEREST question, we can also solve the question by looking for a pattern.

Let's try that:

Time elapsed (in years) | salary
0 | s
0.5 | (1.015)(s)
1 | (1.015)(1.015)(s)= (1.015)²(s) =
1.5 | (1.015)(1.015)(1.015)(s) = (1.015)³(s) =
2 | (1.015)�(s)
2.5 | (1.015)�(s)
3 | (1.015)�(s)
3.5 | (1.015)�(s)
4 | (1.015)�(s)
.
.
.
See the pattern???
.
.
.
y | (1.015)^2y(s)

Answer: C

Cheers,
Brent

swerve wrote:At her current job, Mary gets a 1.5% raise twice per year. Which of the following choices represents Mary’s current income y years after starting the job at a starting salary of s?
$$A.\ s\left(1.5\right)^{2y}$$
$$B.\ s\left(0.015\right)^{2y}$$
$$C.\ s\left(1.015\right)^{2y}$$
$$D.\ s\left(1.5\right)\frac{y}{2}$$
$$E.\ s\left(1.015\right)\frac{y}{2}$$
The OA is C

Source: Manhattan Prep

This problem tests the same concept as a compound interest problem. Recall that A = P(1 + r)^t for compound interest. Now we replace P with s, r with 1.5%, or 0.015, and t with 2y and obtain:

A = s(1 + 0.015)^(2y) = s(1.015)^(2y)

Alternate Solution:

At every raise, Alice’s salary gets multiplied by 1.015 corresponding to a raise of 1.5%. Thus, in one year, her salary becomes s(1.015)^2 and in y years, her salary becomes s((1.015)^2)^y = s(1.015)^(2y).

Answer: C