Question: Stacie is choosing between two jobs. Working for company A, Stacie will make $45,000 the first year…

Stacie is choosing between two jobs. Working for company A, Stacie will make $45,000 the first year and will receive an additional $1,500 per year for each year worked. Working for company B, Stacie will make $42,000 the first year, and she will receive an additional $1,800 per year for each year worked. The expressions representing each company’s total pay over 30 years are shown.

Total Pay over 30 Years at Two Companies

How much more will company B pay over 30 years?

• $40,500
• $49,200
• $49,500
• $12,000

Solution

Stacie is considering two job options. The problem provides formulas for calculating total pay over 30 years with Companies A and B. We need to determine how much more Company B will pay compared to Company A over 30 years. Company A: \[ \sum_{n=1}^{30} [45,000 + 1,500 \times (n - 1)] \] This is an arithmetic series where: First term \(a = 45,000\) and common difference \(d = 1,500\). Total pay for Company A over 30 years is \[ S_n = \frac{n}{2} (2a + (n - 1) \times d) \] Substituting the values: \[ S_{30} = \frac{30}{2} [2 \times 45,000 + (30 - 1) \times 1,500] \] \[ S_{30} = 15 [90,000 + 29 \times 1,500] \] \[ S_{30} = 15 [90,000 + 43,500] \] \[ S_{30} = 15 \times 133,500 \] \[ S_{30} = 2,002,500 \] Company B: \[ \sum_{n=1}^{30} [42,000 + 1,800 \times (n - 1)] \] This is another arithmetic series where: First term \(a = 42,000\) and common difference \(d = 1,800\). Total pay for Company B over 30 years is \[ S_n = \frac{n}{2} (2a + (n - 1) \times d) \] Substituting the values: \[ S_{30} = \frac{30}{2} [2 \times 42,000 + (30 - 1) \times 1,800] \] \[ S_{30} = 15 [84,000 + 29 \times 1,800] \] \[ S_{30} = 15 [84,000 + 52,200] \] \[ S_{30} = 15 \times 136,200 \] \[ S_{30} = 2,043,000 \] The difference in total pay over 30 years: \[ 2,043,000 - 2,002,500 = 40,500 \] Thus, Company B will pay $40,500 more than Company A over 30 years.