Compound Interest!

[RSK]

Hi guys,

Can someone help me understand how this problem can be solved quickly?

If $1000 was invested at an annual interest rate of 5.6% compunded annually, which of the following represents the amount the investment was worth after 3 years?

1000 (1.056)(3)
1000 (3+1.056)
1000 [1+3(0.056)]
1000 [1+(0.056)^3]
1000 (1.056)^3

Thanks!! =)

[LalaB]

it is a formula.u have to know it by heart.

principal amount*(1+r/n)^t
1000(1+0.056/1)^3
1000(1.056)^3

[shankar.ashwin]

Simple Interest = Pnr/100

Compound Interest = P[1+r/100]^n

where, P - principal
n - number of years and r - rate of interest.

given, P = 1000,n=3 and r = 5.6

Scan the answer choices, only E is of the form of the CI formula.

[ronnie1985]

Use the binomial expansion rule to make the approximation
(1+x)^n = 1 + nC1*x + nC2 x^2 + ...
Now for x <<1 x^2 and higher powers can be ignored
Here x = 0.056.

[ronnie1985]

But one of the answer choices is the same as the compound interest formula for amount which is option (E)

[RSK]

What is the correct formula? The formula posted by Shankar is different from yours and does not take into account whether the interest is compounded annually, biannually, etc.

it is a formula.u have to know it by heart.

principal amount*(1+r/n)^t
1000(1+0.056/1)^3
1000(1.056)^3

[him1985]

Hi RSK,

Correct formula is: Compound Interest = P[1+r/100]^n

where, P - principal
n - number of years and r - rate of interest.

given, P = 1000,n=3 and r = 5.6 . Correct answer is E.

[RSK]

A=P(1+r/q)^(nq)

The above is correct because it allows us to calculate the amount even if the interest is compounded twice or thrice a year, etc.

A= amount accumulated after n years including the interest.
r = interest rate as a decimal
n = no. of years
q = no. of times it is compounded