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An engineer invested

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An engineer invested \$\\$10,\!000\$ in a six-month savings certificate that paid a simple annual interest rate of \$12\%\$. After six months, she invested the total value of her investment in another six-month certificate. After six more months, the investment was worth \$\\$11,\!130\$. What was the annual interest rate of the second certificate?

Sep 17, 2017
edited by michaelcai  Sep 17, 2017

#1
+1

This is the formula you would use to compare the 2 investments:

FV = PV x [1 + R]^N

FV =\$50,000 x [1 + 0.04]^2

FV =\$50,000 x      1.0816

FV =\$54,080 - Jose's investment.

FV =\$50,000 x [1 + 0.04/4]^(2*4)

FV =\$50,000 x       1.01^8

FV =\$50,000 x        1.0828567056280801

FV =\$54,142.84 - Patricia's investment.

\$54,142.84 - \$54,080 =~\$63 - extra that Patricia's investment earned.

Sep 17, 2017
#2
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What happened to your Jose's and Patricia's question that I just answered? Did you delete it and why?

Sep 17, 2017
#3
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I figured it out

Edit: 63 was correct

michaelcai  Sep 17, 2017
edited by michaelcai  Sep 17, 2017
#4
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\$10,000 x [12%] / 2 =\$10,000 x 0.06 =\$600 interest earned for the 1st 6 months.

\$10,000 + \$600 =\$10,600 principal + interest for the 1st 6 months.

\$11,130 / \$10,600 =1.05

[1.05 -1] x 100 =5% This is the effective annual interest on the 2nd certificate. Or you can:

1.05^1/2 =[1.0246950 - 1] x 100=2.4695% semi-annual copound rate. Or:

2.4695 x 2 =4.939% - This is called nominal annual rate compounded semi-annually, which is equivalent to 5% effective annual rate.

Sep 17, 2017
edited by Guest  Sep 17, 2017