Calvin deposits $400 in a savings account that accrues 5% interest compounded monthly. after c years, calvin has $658.80. makayla deposits $300 in a different savings account that accrues 6% interest compounded quarterly. after m years, makayla has $613.04. what is the approximate difference in the number of years that calvin and makayla have their money invested?

The approximate difference in the number of years that Calvin and Makayla have their money invested is 2 years.

Given information:

Calvin deposits $400 in a savings account that accrues 5% interest compounded monthly.

After c years, Calvin has $658.80.

Makayla deposits $300 in a different savings account that accrues 6% interest compounded quarterly. After m years, Makayla has $613.04.

Using the formula of compound interest, the value of c in years can be calculated as,

So, Calvin has invested $400 for about 10 years.

Similarly, the value of m in years can be calculated as.

So, the approximate difference in the number of years that Calvin and Makayla have their money invested will be 12-10=2 years.

Therefore, the approximate difference in the number of years that Calvin and Makayla have their money invested is 2 years.

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You will need a special formula to compute this.
Years = log (total/principal) / [n * log * (1 + rate / n)]

Part A) Calvin $400 5% monthly 658.80 Time = ?
Monthly compounding "n" = 12
Years = log(658.80/400) / [12 * log(1+ (.05/n))
Years = log ( 1.647 ) / (12 * log ( 1.0041666667 )
Years = 0.21669359917 / 12 * 0.0018058008777
Years = 0.21669359917 / 0.0216696105
Years = 9.999884362


Part B) Makayla 300 6% quarterly  613.04Time=?
Quarterly compounding
n = 4
Years = log (total/principal) / [n * log * (1 + rate / n)]
Years = log (613.04/300) / [4 * log (1 + .06/4)]
Years = log ( 2.0434666667 ) / 4 * log (1.015)
Years = 0.31036755784 / 4 * 0.0064660422492
Years = 0.31036755784 / 0.025864169
Years = 11.9999044949

So, the difference is roughly 3 years.

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