Question

Brian invests $10,000 in an account earning 4% interest, compounded annually for 10 years. Five years after Brian's initial investment, Chris invests $10,000 in an account earning 7% interest, compounded annually for 5 years. Given that no additional deposits are made, compare the balances of the two accounts after the interest period ends for each account. (round to the nearest dollar) A) Chris has $766 more in his account than Brian. B) Brian has $766 more in his account than Chris. C) Chris has $1,593 more in his account than Brian. D) Brian has $1,593 more in his account than Chris.

Answer 1

B) Brian has $766 more in his account than Chris

Explanation:

Compound interest formula is
`A = P(1+r)^(2)`

P - principal amount

r - rate of interest

t - number of years

Brian invests $10,000 in an account earning 4% interest, compounded annually for 10 years

P = 10,000

r= 4% = 0.04

t = 10

Plug in all the factors

`Brian = 10000(1+0.04)^(10)= $14,802`

Chris invests $10,000 in an account earning 7% interest, compounded annually for 5 years.

P = 10,000

r= 7% = 0.07

t =5

Plug in all the factors

`Chris = 10000(1+.07)^(5 ) = $14,026`

$14,802 - 14,026 = $776

Answer 2

Brian has $776 more account in his account than Chris.

Explanation:

Compound interest Formula:

`A=P(1+r)^t`

`I`= A-P

A= Amount after t years

P= Initial amount

r= Rate of interest

t= Time in year

Given that,

Brian invests $10,000 in an account earning 4% interest, compounded annually for 10 years.

Here P = $10,000 , r= 4%=0.04, t=10 years

The amount in his account after 10 years is

`A=10000(1+0.04)^(10)`

=$14802.44

≈$14802

Five years after Brian's investment,Chris invests $10,000 in an account earning 7% interest, compounded annually for 5 years.

Here P = $10,000 , r= 7%=0.07, t=5 years

The amount in his account after 5 years is

`A=10000(1+0.07)^(5)`

=$14025.51

≈$14026

From the it is cleared that Brian has $(14802-14026)=$776 more account in his account than Chris.