# Compound Interest

- The amount which is lent / deposited is called Principal
- The money that the principal generates is called Interest. This is the money generated as a result of borrowing/lending.
- Compound Interest is the interest calculated on the cumulative amount, rather than being calculated on the principal amount only.
- Amount, A = P [1 + (R / 100)]
^{n}, where P is the principal, R is the rate of interest per unit time period and n is the time period. - Compound Interest, CI = Amount – Principal
- If compounding period is not annual, rate of interest is divided in accordance with the compounding period. For example, if interest is compounded half yearly, then rate of interest would be R / 2, where ‘R’ is the annual rate of interest.
- If interest is compounded daily, rate of interest = R / 365 and A = P [ 1 + ( {R / 365} / 100 ) ]
^{T}, where ‘T’ is the time period. For example, if we have to calculate the interest for 1 year, then T = 365. For 2 years, T = 730. - If interest is compounded monthly, rate of interest = R / 12 and A = P [ 1 + ( {R / 12} / 100 ) ]
^{T}, where ‘T’ is the time period. For example, if we have to calculate the interest for 1 year, then T = 12. For 2 years, T = 24. - If interest is compounded half yearly, rate of interest = R / 2 and A = P [ 1 + ( {R / 2} / 100 ) ]
^{T}, where ‘T’ is the time period. For example, if we have to calculate the interest for 1 year, then T = 2. For 2 years, T = 4.

- If interest is compounded daily, rate of interest = R / 365 and A = P [ 1 + ( {R / 365} / 100 ) ]
- For finding the time period in which a sum of money will double itself at R % rate of compound interest compounded annually, we generally use either of the following two formulas :
- Time, T = 72 / R Years
- Time, T = 0.35 + (69 / R) Years

- When rate of interest is different for different years, say R1, R2, R3 and so on, the amount is calculated as A = P [1 + (R1 / 100)] [1 + (R2 / 100)] [1 + (R3 / 100)] …

### Sample Problems

**Question 1 : **Find the compound interest on Rs. 10,000 at 10% per annum for a time period of three and a half years.

**Solution : **Time period of 3 years and 6 months means for 3 years, the interest is compounded yearly and for the remaining 6 months, the interest is compounded compounded half yearly. This means that we have 3 cycles of interest compounded yearly and 1 cycle of interest compounded half yearly.

So, Amount = P [1 + (R / 100)]^{3} [1 + ( {R/2} / 100 )]

=> Amount = 10000 [1 + 0.1]^{3} [1 + 0.05]

=> Amount = 10000 (1.1)^{3} (1.05)

=> Amount = Rs. 13975.50

=> Compound Interest, CI = Amount – Principal = 13975.50 – 10000

Therefore, CI = Rs. 3975.50

**Question 2 : **If Rs. 5000 amounts to Rs. 5832 in two years compounded annually, find the rate of interest per annum.

**Solution : **Here, P = 5000, A = 5832, n = 2

A = P [1 + (R / 100)]^{n}

=> 5832 = 5000 [1 + (R / 100)]^{2}

=> [1 + (R / 100)]^{2} = 5832 / 5000

=> [1 + (R / 100)]^{2} = 11664 / 10000

=> [1 + (R / 100)] = 108 / 100

=> R / 100 = 8 / 100

=> R = 8 %

Thus, the required rate of interest per annum in 8 %

**Question 3 : **The difference between the SI and CI on a certain sum of money at 10 % rate of annual interest for 2 years is Rs. 549. Find the sum.

**Solution : **Let the sum be P.

R = 10 %

n = 2 years

SI = P x R x n / 100 = P x 10 x 2 / 100 = 0.20 P

CI = A – P = P [1 + (R / 100)]^{n} – P = 0.21 P

Now, it is given that CI – SI = 549

=> 0.21 P – 0.20 P = 549

=> 0.01 P = 549

=> P = 54900

Therefore, the required sum of money is Rs. 54,900

**Question 4 : **A sum of Rs. 1000 is to be divided among two brothers such that if the interest being compounded annually is 5 % per annum, then the money with the first brother after 4 years is equal to the money with the second brother after 6 years.

**Solution : **Let the first brother be given Rs. P

=> Money with second brother = Rs. 1000 – P

Now, according to the question,

P [1 + (5 / 100)]^{4} = (1000 – P) [1 + (5 / 100)]^{6}

=> P (1.05)^{4} = (1000 – P) (1.05)^{6}

=> 0.9070 P = 1000 – P

=> 1.9070 P = 1000

=> P = 524.38

Therefore, share of first brother = Rs. 524.38

Share of second brother = Rs. 475.62

**Question 5 : **A sum of money amounts to Rs. 669 after 3 years and to Rs. 1003.50 after 6 years on compound interest. Find the sum.

**Solution : **Let the sum of money be Rs. P

=> P [1 + (R/100)]^{3}= 669 and P [1 + (R/100)]^{6}= 1003.50

Dividing both equations, we get

[1 + (R/100)]^{3} = 1003.50 / 669 = 1.50

Now, we put this value in the equation P [1 + (R/100)]^{3}= 669

=> P x 1.50 = 669

=> P = 446

Thus, the required sum of money is Rs. 446

**Question 6 : **An investment doubles itself in 15 years if the interest is compounded annually. How many years will it take to become 8 times?

**Solution : **it is given that the investment doubles itself in 15 years.

Let the initial investment be Rs. P

=> At the end of 15 years, A = 2 P

Now, this 2 P will be invested.

=> Amount after 15 more years = 2 x 2 P = 4 P

Now, this 4 P will be invested.

=> Amount after 15 more years = 2 x 4 P = 8 P

Thus, the investment (P) will become 8 times (8 P) in 15 + 15 + 15 = 45 years

### Problems on Compound Interest | Set-2

This article has been contributed by **Nishant Arora**

Please write comments if you have any doubts related to the topic discussed above, or if you are facing difficulty in any question or if you would like to discuss a question other than those mentioned above.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

## Recommended Posts:

- Compound Interest | Set-2
- Program to find compound interest
- Program to find the rate percentage from compound interest of consecutive years
- Simple Interest | Set-2
- Simple Interest
- Program to find simple interest
- Times required by Simple interest for the Principal to become Y times itself
- Check if a Sequence is a concatenation of two permutations
- Check if original Array is retained after performing XOR with M exactly K times
- Check if an Array is a permutation of numbers from 1 to N
- Count of interesting primes upto N
- Check if an Octal number is Even or Odd
- Count number of binary strings without consecutive 1’s : Set 2
- Runge-Kutta 2nd order method to solve Differential equations