Problem Solving

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!! =)

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

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.

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.

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

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.

LalaB wrote: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

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.

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