## Basic College Mathematics (10th Edition)

Published by Pearson

# Chapter 6 - Percent - 6.8 Compound Interest - 6.8 Exercises - Page 456: 43

(a) $\$254.48$(b) More than doubled (times 5). (c) See example. #### Work Step by Step (a) To find out the interest earned with compounded interest, we look at the table on page 453 in the textbook. We look up the column "6 years" and the row "6% interest" in the table. This gives us the value "1.4185". We multiply the principal ($\$4350$) by this number: $4350*1.4185=\$6170.48$And subtract the original principal to find the interest:$6170.48-4350=\$1820.48$ Next, we figure out the interest earned with simple interest: $I=Prt$ $I=4350*0.06*6=\$1566$We find the difference between the two interest amounts:$1820.48-1566=\$254.48$ (b) If the time is doubled from 6 to 12 years, the simple interest will also double because the interest scales proportionally with time (I=Prt). Thus the simple interest will be: $1566*2=\$3132$To find the compound interest, we look up "12 years" and "6% interest" in the table. We get "2.0122":$4350*2.0122-4350=4403.07$The new difference is:$4403.07-3132=\$1271.07$ The ratio of the new difference to the old one is: $1271.07/254.48=4.99\approx 5$ So it more than doubled. (c) Another example: Bob invests $\$1000$at 5% interest for 5 years. Compare the difference between simple and compounded interest for this time period. Then compare difference to interest over 10 years. Simple interest after 5 years:$I=PrtI=1000*0.05*5I=\$250$ For 10 years: $I=250*2=\$500$Compound interest for 5 years (use table):$1000*1.2763-1000=\$276.3$ For 10 years: $1000*1.6289-1000=\$628.9$Difference in interest for 5 year period:$276.3-250=26.3$Difference in interest for 10 year period:$628.9-500=128.5$Ratio of differences:$128.5/26.3=4.88\approx 5\$ This is a similar result to what we obtained in (b).

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