Knowee
Questions
Features
Study Tools

nd the CI on Rs.6550 at 4% pa for 2 years compounded half yearly(approx) ?Rs.537.1Rs.527.2Rs.530.7Rs.547.1

Question

nd the CI on Rs.6550 at 4% pa for 2 years compounded half yearly(approx) ?Rs.537.1Rs.527.2Rs.530.7Rs.547.1

🧐 Not the exact question you are looking for?Go ask a question

Solution

To find the compound interest (CI) on Rs.6550 at 4% per annum for 2 years compounded half yearly, we need to use the formula for compound interest:

CI = P(1 + r/n)^(nt) - P

where: P = principal amount (the initial amount of money) = Rs.6550 r = annual interest rate (in decimal) = 4/100 = 0.04 n = number of times that interest is compounded per year = 2 (since it's compounded half yearly) t = time the money is invested for in years = 2

First, convert the annual interest rate from a percentage to a decimal by dividing by 100:

r = 4/100 = 0.04

Next, substitute the values into the formula:

CI = 6550(1 + 0.04/2)^(22) - 6550 CI = 6550(1 + 0.02)^4 - 6550 CI = 6550(1.02)^4 - 6550 CI = 65501.08243216 - 6550 CI = 7087.83 - 6550 CI = Rs.537.83

So, the compound interest on Rs.6550 at 4% per annum for 2 years compounded half yearly is approximately Rs.537.83. The closest answer to this is Rs.537.1.

This problem has been solved

Similar Questions

Find the CI on Rs.6550 at 4% pa for 2 years compounded half yearly(approx) ?Rs.537.1Rs.527.2Rs.530.7Rs.547.1Clear ResponseSave & Next

The CI on Rs.9500 for 3 years at 5% for first year, 7% for second year, 10% for the third year will be?Rs.2560.48Rs.2340.67Rs.2240.57Rs.2241.67

At what rate of CI per annum will be a sum of Rs. 1000 becomes Rs. 1040.4 in 2 years?

If 40% increase in an amount in 4 years at SI. What will be the CI of Rs. 10000 after 2 years at the same rate?

In what time will Rs. 64,000 amount to Rs.68921 at 5% per annum interest being compounded half yearly?

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.