Question

Nina made two investments: Investment \text{A}A has a value of \$50$50 at the end of the first year and increases by 8\%8% per year. Investment \text{B}B has a value of \$60$60 at the end of the first year and increases by \$3$3 per year. Nina checks the value of her investments once a year, at the end of the year. What is the first year in which Nina sees that investment \text{A}A's value exceeded investment text{B}B's value?

Answer 1

7

Explanation:

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Answer 2

year 7

Explanation:

If we assume that investment A earns interest compounded annually, its value can be modeled by the equation ...

A = 50·(1+0.08)^(t-1) . . . . . where t is the year number

The second investment earns $3 per year, so its value can be modeled by the equation ...

B = 60 + 3(t -1) . . . . . . . . . where t is the year number

We are interested in finding the minimum value of t such that ...

A > B

50·1.08^(t-1) > 60 +3(t-1)

This is a mix of exponential and polynomial terms for which no solution method is available using the tools of Algebra. A graphing calculator shows the solution to be ...

t > 6.552

The value at the end of year 1 is found for t=1, so the values of interest are seen after 6.55 years, in year 7.