### Compound Interest | Formula | Problem

Here we are learning some essential tricks required to solve the compound interest problems in competitive exams.

### What is compound interest ?

Interest for the first period, add it to the total, and then calculate the interest for the next period, and so on. The process is repeated until the amount for the last period has been found. The difference between the original principal and the final amount is called Compound Interest (CI). In another way, it can be described as an interest in interest. It makes a deposit grow faster as compared to simple interest (SI). The interest at which compound interest accumulates depends on the frequency of compounding: more the number of compounding periods, the greater the compound interest.

Amount = P * [1 + r100]t

Where,
P is the principal (the initial amount you borrow or deposit)
r is the annual rate of interest
t is the number of years the amount is deposited or borrowed for.
A is the amount of money accumulated after n years, including interest.

Case 1: when interest is compound annually:

Amount = P * [1 + r100]t

Case 1: When interest is compounded half-yearly:

Amount = P * [1 + r/2100]2t = P * [1 + r200]2t

Case3: When interest is compounded quarterly:

Amount = P * [1 + r/4100]4t = P * [1 + r400]4t

Case4: When rate of interest is r1%,r2% and r3% for 1st year, and 2nd year and 3rd year respectively,
Then, Amount = P[1 + r1100] * P[1 + r2100] * P[1 + r3100]

The above-mentioned formula is not new for you. We think that all of you know their uses. When dealing with the above formula, some mathematical calculations become lengthy and take more time. To simplify the calculations and save the valuable time we are giving some extra information. Study the following sections carefully and apply them during your calculations.

The problem is generally asked for up to a period of 3 years and the rates interest is 10%,5%, and 4%.

Amount = P * [1 + r100]t

We have the basic formula:
Amount = P * [1 + r100]t

If the principal is Rs. 1, the amount for first, second and third years will be…..
(1+(r/1000)),(1+(r/100))2 and(1+(r/100))3 respectively.

And, if the rate of interest is 10%,5% and 4% these values will be
(11/10) ,(11/10)2,(11/10)3
(21/20), (21/20)2 ,(21/20)3
(26/25), (26/25)2, (26/25)3

(1). Rs. 7500 is borrowed at CI at the rate of 4% per annum. What will be the amount to be paid after 2 years?
Solution: As the rate of interest is 4% per annum and the time is 2 years,
r = 4% per annum, t = 2 years, P = Rs. 7500
Then, by the use of formula
Amount = P * [1 + r100]t
= 7500 * [1+4100]2
= 7500 * 676625 = 8112
So, after 2 years. Rs. 7500 will produce Rs. 8112.
Another useful method of calculating CI.
1). For 2 years
We know that compound interest and simple interest remain the same for the first year. In the second year, they differ because we also count simple interest on the year’s simple interest to get the compound interest. So, mathematically if our rate of interest is r%, then for 2 years

Simple interest (SI) = 2*r =2r% of capital. (with the help of basic formula of
SI = PRT/100, where P= capital, R=r, T= 2)
And Compound interest (CI) = (2r+(r2/100)%of capital
{with the help of CI formula (1+(r/100))2
here we apply the [(a + b)2 = a2 +b2 + 2ab]
then we get
(2r+(r2 /100))% of capital…( of capital means that principal ,principal amount will be given in the question)}

Therefore, the difference is r2/100% of capital.
Now take the above solved example. We can calculate the SI mentally. That is
2 x 4 = 8% of 7500 =Rs. 600
Difference is (16/100 x 100) x 7500 =Rs. 12
So, CI = Rs 612 required value = Rs(7500 + 612) = Rs 8112

2). For 3 years: If the rate of interest is r% for 3 years

Total SI = 3*r = 3r% of capital
CI = (3r+(3r 2 /100)+(r 3 /100 2 ))% of capital.
Explanation
{with the help of CI formula (1+(r/100))3 ,
here we apply the [(a + b)3 = a3 + b3 + 3ab(a + b)] then we get ((3r2 /100)+(r3 /100 2 ))% of capital……( of capital
means that principal , principal amount will be given in the question)}
Therefore, difference = ((3r2 /100)+(r3 /1002 ))% of the capital.
Now see the following example:
Find the compound interest on Rs. 8000 at 10% per annum for 3 years?
Solution: In this question we use the difference of SI and CI for 3 years =
((3r2 /100)+(r3 /1002 ))% of the capital.
Hence: SI = 30% of 8000 = Rs. 2400
Difference = [ {(3*102 )/100} + { 103 /1002 }]% of 8000
= (3 + 0.1)% of 8000
= 3.1% of 8000 = Rs. 248
Therefore , CI = Rs.(2400 + 248) = Rs. 2648.

(2). A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is?
Solution: use the formula of CI
Amount = P * [1 + r100]t
Amount = Rs.[1600*(1+52×100)2 + 1600*(1+52×100) ]
= Rs.[1600 x 4140 x 4140 + 1600 x 4140]
= Rs.[1600 x 4140(1600 x 4140 + 1) ]
= Rs.[(1600x41x81)/(40×40)]
= Rs. 3321
C.I. = Rs. (3321 – 3200) = Rs. 121

(3). James invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at a compound interest rate 5 percent per annum. How much amount will Albert get on the maturity of the fixed deposit?
Solution: use the formula of CI
Amount = P x [1 + r100]t
Amount = 8000*[1 + 5100]2
= 8000 x 2120 x 2120
Amount = Rs. 8820

(4). Find the CI, if Rs 1000 was invested for 1.5 years at 20% p.a. compounded half-yearly?
Solution:
Here P = 1000 Rs. R = 20%, n = 1.5 year = 3 half-yearly
This said that the interest is compounded half-yearly.
So, the rate of interest will be halved and time will be doubled.
Amount = P x [1 + r/2100]2t (from above case 1)
CI = Amount – Principal
CI = P x [1 + r/2100]t – P
CI = 1000 x [1 + 20/2100]3 – 1000
CI = 1000 x [1 + 10100]3 – 1000
= 1000 x 3 – 1000
= 1000 x 110100 x 110100 x 110100 – 1000

Solving this then, we get
CI = Rs. 331

(5). Akshay invested a sum of money at CI. It amounted to Rs 2420 in 2 years and Rs 2662 in 3 years. Find the rate of percent per annum?
Solution:
Important Formula:
To find the difference between SI and CI for 2 years
we use the formula Difference = P[R/100]2
Last year interest = 2662 – 2420 = Rs 242
Therefore,
Rate as R = 100 x S.IP × T
Rate% = (242 x 100)/(2420 x 1)
R% = 10%

(6). A sum of money is put on CI for 2 years at 20%. It would fetch Rs 482 more if the interest is payable half-yearly than if it were payable yearly. Find the sum?
Solution:
t = 2 year, r =20%
Let the Principal P = Rs 100
Amount = P x [1 + r100]t
If compounded annually,
A = 100 [1 + 20100]2
If compounded half yearly,

Amount = P x [1 + r/2100]2t (from above case 1)
Amount = 100 x [1 + 20/2100]2×2
Amount = 100 x [1 + 10100]4
= 100 x 4
= 100 x 110100 x 110100 x 110100 x 110100
= 146.41
Difference,146.41 – 144 = 2.41
If difference is 2.41
Then,
Principal = Rs 100
If difference is 482
Then,
Principal = 100/2.41 × 482
P = Rs 20000.

(7). The CI on a sum of Rs 625 in 2 years is Rs 51. Find the rate of interest?
Solution:
Here P = Rs 625, t = 2 years, CI = Rs 51
We know that
A = P + CI
A = 625 + 51 = 676
Now use the formula:

Amount = P x [1 + r100]t
676 = 625 x [1 + r100]2
676625 = [1 + r100]2
We can see that 676 is the square of 26 and 625 is the square of 25
Therefore,
676625 = [1 + r100]2
(676625 )2 = [1 + r100]2
676625 = [1 + r100]
676625 – 1 = r100
R = 4%

(9). Sumit invests Rs. 20,000 in two banks. He invested half of the sum in a bank that pays compound interest and half in a bank that pays simple interest. The interest rate in both the banks is 10% per annum. How much interest will he make at the end of 2 years?
Solution:
The principal in the bank with simple interest is Rs. 20,0002 = Rs. 10,000
Simple Interest = P x R x T100
Interest collect =10,000 x 2 x 10100 = Rs. 2,000
The principal in the bank with compound interest is Rs.20,0002 = Rs. 10,000
Interest collect = 10,000∗1.12 – 10,000 = Rs. 2,100
Total interest collect = 2,000 + 2,100 = Rs. 4,100

(10). Vijay lends a sum of Rs. 1000 to Sanjay at a rate of 10% per annum for a period of two years.
(i): Compound Interest with compensation in two equal annual installments.
(ii): Simple Interest with compensation after two years In which of the two situations is the total amount repaid more and by how much?
Solution:
(i): Let the installment amount = x
Amount after 1 year = 1000 + 1000×10% = 1100
Amount for second year after compensation installment = 1100 – x Amount to be paid after 2 years = (1100 – x) + (1100-x)*10% = 1.1(1100-x)
This should be equal to x.
Hence,
We have 1.1(1100-x) = x => x = 576.20
Hence,
Total amount paid = 576.20 x 2 = 1152.4

(ii): By formula
Simple Interest = P x R x T100
Total interest =1000 x 10 x 2100 = Rs. 200
Hence,
A = P + I
Total amount compensation = 1000 + 200 = 1200
Hence,
The amount paid in situation 2 is more than the amount paid in situation 1 by around Rs. 48

I hope, this article will help you a lot to understand the Compound Interest | Formula | Problems. If you still have any doubts and problems with any topic of mathematics you can ask your problem in the Ask Question section. You will get a reply shortly.

Simple Interest