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

9.4

Explanation:

The correct answer is actually 9.4. Delta Math corrected me afterwards.

Answer 2

10.1 years.

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+(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+(0.05)/(2))^(2t)`

`(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]`

`(\log(1.644))/(2\log(1.025))=t`

`t=10.066`

`t=10.1`

Therefore, the required time is 10.1 years.