Question

If $10,000 is invested at an annual rate of 11%, compounded quarterly, find the value of the investment after the given number of years.
a) 5 years b) 10 years c) 15 years

Answer 1

Answer:

Solution given;

principal [p]=$10000

rate[r]=11%

time[t]= 5 years

we have

compound amount quarterly =P(1+r/400)^4t

=$10000(1+11/400)^4×5=$17204.28431

Answer 2

Answer:

$a) A\simeq17204.28$

$b) A\simeq29598.74$

$c)A\simeq50922.51$

Step-by-step explanation:

The amount formula in compound interest is:

$A=P(1+\frac{r}{n} )^{nt}$

where:

P = principal amount

r = annual interest

n = number of compounding periods

t = number of years

We already know that:

P = $10000

$r = 11\% = \frac{11\%}{100\%}=0.11$

n = 4 (quarterly in a year)

a ) t = 5 years

$A=10000(1+\frac{0.11}{4} )^{(4)(5)}\\\\A=10000(1+\frac{0.11}{4} )^{20}\\\\A=17204.28431\\\\A\simeq17204.28$

b) t = 10 years

$A=10000(1+\frac{0.11}{4} )^{(4)(10)}\\\\A=10000(1+\frac{0.11}{4} )^{40}\\\\A=29598.73987\\\\A\simeq29598.74$

c) t = 15 years

$A=10000(1+\frac{0.11}{4} )^{(4)(15)}\\\\A=10000(1+\frac{0.11}{4} )^{60}\\\\A=50922.51361\\\\A\simeq50922.51$