Question

a person invest in 9000 in a bank. the bank pays 5% interest compounded semi annually. to the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 14800 dollars

Answer 1

Answer:

10.1 years.

Step-by-step explanation:

It is given that,

Principal = 9000

Rate of interest = 5%

No. of times interest compounded = 2 times in an year

Amount after certain time = 14800

The formula for amount:

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

where, P is principal, r is rate of interest, n is no. of times interest compounded in an year and t is time in years.

Substitute the given values in the above formula.

$14800=9000(1+\frac{0.05}{2})^{2t}$

$\frac{14800}{9000}=(1+0.025)^{2t}$

$1.644=(1.025)^{2t}$

Taking log both sides.

$\log(1.644)=\log(1.025)^{2t}$

$\log(1.644)=2t\log(1.025)$    $[\because \log a^b=b\log a]$

$\frac{\log(1.644)}{2\log(1.025)}=t$

$t=10.066$

$t=10.1$

Therefore, the required time is 10.1 years.