1.Gold Company invested $1,491,420.00 for 11 years in bonds. What was the effective rate of interest earned on the investment if the bonds matured to $4,019,013.00?
2.If $73,376.39 was the interest earned on lending $101,000 for 7 years, what was the quarterly compounding nominal interest rate charged on the loan.
+23254 The compound interest formula: A = P(1 + r/n)^(n·t)
A = final amount P = beginning amount r = rate (as a decimal)
n = number of times compounded per year t = number of years
1) 4 019 013 = 1 491 420(1 + r/1)^(1·11) ---> 4 019 013 = 1 491 420(1 + r)^(11)
---> (divide by 1 491 420) ---> 2.694 756(1 + r)^11
---> (take the 11th root) ---> 1.0943 = 1 + r ---> r = 0.094304365
2) If $73,376.39 was the interest earned on lending $101,000, this means that the final amount is $73,376.39 + $101,000 = $174,376.39:
174,376.39 = 101 000(1 + r/4)^(4·7) ---> 174,376.39 = 101 000(1 + r/4)^28
---> (divide by 101 000) ---> 1.726498911 = (1 + r/4)^28
---> (take the 28th root) ---> 1.019694849 = 1 + r/4
---> 0.019694849 = r/4 ---> r = 0.0787793946
3) a) Interest of $227,708.21 on a loan of $59,000 loan gives a final amount of $286,708.21.
b) 286 708.21 = 59 000(1 + 0.074/2)^(2·n)
---> 4.859461186 = (1.037)^(2n)
---> (find the log of both sides) ---> log( 4.859461186 ) = log[ 1.037^(2n) ]
---> (exponents come out as multipliers) ---> log( 4.859461186 ) = 2n · log[ 1.037 ]
---> (divide by 2· log[ 1.037 ] ) ---> 21.7567247 years
+23254 The compound interest formula: A = P(1 + r/n)^(n·t)
A = final amount P = beginning amount r = rate (as a decimal)
n = number of times compounded per year t = number of years
1) 4 019 013 = 1 491 420(1 + r/1)^(1·11) ---> 4 019 013 = 1 491 420(1 + r)^(11)
---> (divide by 1 491 420) ---> 2.694 756(1 + r)^11
---> (take the 11th root) ---> 1.0943 = 1 + r ---> r = 0.094304365
2) If $73,376.39 was the interest earned on lending $101,000, this means that the final amount is $73,376.39 + $101,000 = $174,376.39:
174,376.39 = 101 000(1 + r/4)^(4·7) ---> 174,376.39 = 101 000(1 + r/4)^28
---> (divide by 101 000) ---> 1.726498911 = (1 + r/4)^28
---> (take the 28th root) ---> 1.019694849 = 1 + r/4
---> 0.019694849 = r/4 ---> r = 0.0787793946
3) a) Interest of $227,708.21 on a loan of $59,000 loan gives a final amount of $286,708.21.
b) 286 708.21 = 59 000(1 + 0.074/2)^(2·n)
---> 4.859461186 = (1.037)^(2n)
---> (find the log of both sides) ---> log( 4.859461186 ) = log[ 1.037^(2n) ]
---> (exponents come out as multipliers) ---> log( 4.859461186 ) = 2n · log[ 1.037 ]
---> (divide by 2· log[ 1.037 ] ) ---> 21.7567247 years
