- 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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Program to find compound interest
- Compound Interest | Set-2
- Program to find the rate percentage from compound interest of consecutive years
- Simple Interest
- Program to find simple interest
- Simple Interest | Set-2
- Times required by Simple interest for the Principal to become Y times itself
- Sum of products of all possible Subarrays
- Rearrange array to make Bitwise XOR of similar indexed elements of two arrays is same
- Longest subsequence having maximum GCD between any pair of distinct elements
- Maximum MEX from all subarrays of length K
- Count ways to arrange N distinct objects if all clockwise arrangements are considered the same
- Count of subarrays having sum equal to its length | Set 2
- Kth number exceeding N whose sum of its digits divisible by M