Question

Account A and Account B each start out with $600. If Account A earns $50 each year and Account B earns 6% if its value each year, after how many years will Account B have more money?

Answer 1

To find out when Account B will surpass Account A in value, we compare the growth of Account A via simple interest ($600 + $50/year) with the growth of Account B using compound interest (600(1 + 0.06)^n). We seek the smallest integer n where Account B's value exceeds Account A's, which can be found by solving the inequality (1.06)^n > 1 + 5/6n.

To determine after how many years Account B will have more money, we can compare the growth formulas for each account. Account A uses simple interest, where the value grows by a fixed amount each year. Account B uses compound interest, where the value grows by a percentage each year.

• Account A's value after n years: A = 600 + 50n
• Account B's value after n years: B = 600(1 + 0.06)^n

We need to find the smallest integer n such that B > A:

1. Set up the inequality: 600(1 + 0.06)^n > 600 + 50n
2. Rewrite and simplify the inequality: (1 + 0.06)^n > 1 + ⅖n
3. Solve the inequality using trial and error, a graphing calculator, or algebraic methods to find the smallest n that satisfies it.

For example, if we try n = 10, Account A would have A = 600 + 50(10) = $1,100. Account B would have B = 600(1.06)^10 ≈ $1,070.86. Since B < A at 10 years, we try incrementing n until B > A.

At some point, due to the effect of compound interest, Account B will exceed the amount in Account A. The exact year needs to be calculated as per the steps outlined above.