The difference between the compound interest and simple interest on a certain sum of money in 3 years at 5% p.a. is Rs. 6,100. What is the initial amount?
Rs. 8,00,000
Calculating the Initial Amount from CI and SI Difference
This problem involves finding the initial principal amount given the difference between the compound interest (CI) and simple interest (SI) earned over a specific period at a given interest rate.
Let's first understand the concepts:
- Simple Interest (SI): Calculated only on the principal amount for the entire duration. The interest remains constant each year.
- Compound Interest (CI): Calculated on the principal amount plus the accumulated interest from the previous periods. The interest grows each year.
The difference between CI and SI arises because compound interest includes interest on interest, while simple interest does not.
For a principal amount \(P\), interest rate \(R\) per annum, and time period \(n\) years, the formulas are:
- Simple Interest \(SI = \frac{P \times R \times n}{100}\)
- Compound Interest \(CI = P \left(1 + \frac{R}{100}\right)^n - P\)
The difference between CI and SI is \(CI - SI\).
For a period of 3 years, the difference \(CI - SI\) can be calculated using the formula:
\[ CI - SI = P \left[ \left(1 + \frac{R}{100}\right)^3 - 1 - \frac{3R}{100} \right] \]
This formula simplifies to:
\[ CI - SI = P \left[ 3\left(\frac{R}{100}\right)^2 + \left(\frac{R}{100}\right)^3 \right] \]
We are given:
- Difference (CI - SI) = Rs. 6,100
- Interest Rate (R) = 5% p.a.
- Time period (n) = 3 years
We need to find the Principal amount \(P\).
Using the formula for the difference for 3 years:
\[ 6100 = P \left[ 3\left(\frac{5}{100}\right)^2 + \left(\frac{5}{100}\right)^3 \right] \]
\[ 6100 = P \left[ 3\left(\frac{1}{20}\right)^2 + \left(\frac{1}{20}\right)^3 \right] \]
\[ 6100 = P \left[ 3\left(\frac{1}{400}\right) + \left(\frac{1}{8000}\right) \right] \]
\[ 6100 = P \left[ \frac{3}{400} + \frac{1}{8000} \right] \]
To add the fractions, find a common denominator, which is 8000:
\[ 6100 = P \left[ \frac{3 \times 20}{400 \times 20} + \frac{1}{8000} \right] \]
\[ 6100 = P \left[ \frac{60}{8000} + \frac{1}{8000} \right] \]
\[ 6100 = P \left[ \frac{60 + 1}{8000} \right] \]
\[ 6100 = P \left[ \frac{61}{8000} \right] \]
Now, solve for \(P\):
\[ P = \frac{6100 \times 8000}{61} \]
\[ P = 100 \times 8000 \]
\[ P = 800000 \]
Therefore, the initial amount (principal) is Rs. 8,00,000.
This matches one of the given options.
Important Questions from Simple and Compound Both
X took a loan of Rs.1400 with simple interest for as many years as the rate of interest. If he paid Rs.1210 as interest at the end of that period, what was the rate of interest?
The S.I. on a certain sum of money for 4 years at 4 percent per annum exceeds the C.I. on the same sum for 3 years at 5 percent per annum by Rs. 57. Find the approximate sum.
Compound interest on a certain sum of money for 2 years at a rate of 'r' per cent per annum (compounding annually) is Rs. 8385. Simple interest on the same sum at the same rate for 2 years is Rs 7800. What is the value of r?
Find the difference between compound interest and simple interest on a sum of Rs. 5000 at 10% per annum for 3 years.
A sum of money becomes two times of itself in 8 years at simple interest, and it becomes four times of itself in 2 years at compound interest, when interest is compounded annually. Find the ratio of the rate of simple interest to the rate of compound interest offered per year.
