WOBS asked in Science & MathematicsMathematics · 3 months ago

# MATH EXPONENTIAL HELP?

4. A company offers its employees a choice of two salary schemes A and B over a period of 10 years. Scheme A offers a starting salary of \$11000 in the first year and then an annual increase of \$400 per year. Scheme B offers a starting salary of \$10000 dollars in the first year and then an annual increase of 7% of the previous year’s salary.

a. Calculate the total (amount of) salary paid over ten years under Scheme A

b. Calculate the total (amount of) salary paid over ten years under Scheme B.

c. Which scheme would you pick and why? (provide a succinct answer)

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• 3 months ago

a. This is an arithmetic series with a = 11000, d = 400, and n = 10

S = (n/2)(2a + (n - 1)*d)

S = (10/2)(2*11000 + (10 - 1)*400)

S = 5(22000 + 9*400)

S = 5(22000 + 3600)

S = 5 * 25600

S = \$128,000

b. This is a geometric series with a = 10000, d = 1.07, and n = 10

S = a(1 - r^n) / (1 - r)

S = 10000(1 - 1.07^10) / (1 - 1.07)

S = 10000(1 - 1.96715135728956532249) / (-0.07)

S = 10000(-0.96715135728956532249) / (-0.07)

S =~ \$138,164.48

c. It depends how long you think you're going to be at this job, and how much money you think you're going to invest. At first, A is better, as it gives you more money sooner, which you can invest to have more money later. On the other hand, if you just look at the above numbers, B is better, and will get even better if you stay longer than 10 years.

If all you care about is the total amount earned, not when you earn it, let's see how many years you need to stay at this job for the two salaries to be equal:

(n/2)(2*11000 + (n - 1)*400) = 10000(1 - 1.07^n) / (1 - 1.07)

That can't be solved algebraically, but by approximating I find the answer is around 6.33. So if you're going to be at the job longer than that, go with Scheme B.

• 3 months ago

The first is an arithmetic sequence with the first term = 11000 and a common difference of 400.

The second is a geometric sequence with the first term = 10000 and a common ratio of 1.07.

We'll need to come up with an expression for the n'th term for each sequence, then an expression for the sum of the first n terms of each sequence.

For part A: the arithmetic sequence.  Starting with this general form for the n'th term:

a(n) = a + b(n - 1)

Using a = 11000 and b = 400:

a(n) = 11000 + 400(n - 1)

a(n) = 11000 + 400n - 400

a(n) = 400n + 10600

Then the equation for the sum of the first "n" terms is:

S(n) = [ a + a(n)] n / 2

We have an expression for a and a(n).  Substitute and simplify:

S(n) = (11000 + 400n + 10600) n / 2

S(n) = (400n + 21600) n / 2

S(n) = 2(200n + 10800) n / 2

S(n) = (200n + 10800) n

S(n) = 200n² + 10800n

The total amount earned after 10 years is S(10):

S(10) = 200(10)² + 10800(10)

S(10) = 200(100) + 10800(10)

S(10) = 20000 + 108000

S(10) = \$128,000

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for part B, we don't need an expression for a(n) as we can jump right to this equation for the sum of the first n terms of a geometric sequence:

S(n) = a(1 - r^n) / (1 - r)   (again, a = 10000 and r = 1.07)

With n = 10, we get:

S(10) = 10000(1 - 1.07^10) / (1 - 1.07)

S(10) = 10000(1 - 1.96715136) / (-0.07)

I rounded above, but didn't in my calculator to reduce errors due to rounding:

S(10) = 10000(-0.96715136) / (-0.07)

S(10) = -9671.513573 / (-0.07)

S(10) = \$138,164.48 (rounded to the nearest penny)

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for part C, we would choose the second method for determining a salary for 10 years.  You won't make as much initially, but over the same 10-year span, you'll end up making more overall with the second method.