Simple Interest and Compound Interest
Simple Interest Formula
The Formula for simple interest helps you find the interest amount if the principal amount, rate of interest and time periods are given.
Simple interest formula is given as:
SI = (P × R ×T) / 100
Where SI = simple interest
P = principal
R = interest rate (in percentage)
T = time duration (in years)
In order to calculate the total amount, the following formula is used:
Amount (A) = Principal (P) + Interest (I)
Where,
Amount (A) is the total money paid back at the end of the time period for which it was borrowed.
Difference Between Simple Interest and Compound Interest:
There is another type of interest called compound interest. The major difference between simple and compound interest is that simple interest is based on the principal amount of a deposit or a loan whereas the compound interest is based on the principal amount and interest that accumulates in every period of time. Let’s see one simple example to understand the concept of simple interest.
Simple Interest Problems
Let us see some simple interest examples using simple interest formulas in maths.
Example 1:
Rishav takes a loan of Rs 10000 from a bank for a period of 1 year. The rate of interest is 10% per annum. Find the interest and the amount he has to the pay at the end of a year.
Solution:
Here, the loan sum = P = Rs 10000
Rate of interest per year = R = 10%
Time for which it is borrowed = T = 1 year
Thus, simple interest for a year, SI = (P × R ×T) / 100 = (1000× 100 ×1) / 100 = Rs 1000
Amount that Rishav has to pay to the bank at the end of the year = Principal + Interest = 10000 + 1000 = Rs 11,000
Example 2:
Namita borrowed Rs 50,000 for 3 years at the rate of 3.5% per annum. Find the interest accumulated at the end of 3 years.
Solution:
P = Rs 50,000
R = 3.5%
T = 3 years
SI = (P × R ×T) / 100 = (50,000× 3.5 ×3) / 100 = Rs 5250
Example 3:
Mohit pays Rs 9000 as an amount on the sum of Rs 7000 that he had borrowed for 2 years. Find the rate of interest.
Solution:
A = Rs 9000
P = Rs 7000
SI = A – P = 9000 – 7000 = Rs 2000
T = 2 years
R = ?
SI = (P × R ×T) / 100
R = (SI × 100) /(P× T)
R = (2000 × 100 /7000 × 2) =14.29 %
Thus, R = 14.29%
Q.1. The simple interest on a certain sum of money for 2(1/2) years at 12% per annum is Rs. 40 less than the simple interest on the same sum for 3(1/2) years at 10% per annum. Find the sum.
(A) Rs. 600
(B) Rs. 666
(C) Rs. 780
(D) Rs. 800
Solution
Then we can write: [{x*10*7}/{100*2}] – [{x*12*5}/{100*2}] = 40.
This can be written as: 7x/20 – 3x/10 = 4o. Therefore we have x = Rs. 800
Hence the sum is Rs. 800.
Q.2. A certain amount becomes Rs. 5760 in 2 years and Rs. 6912 in 3 years. What is the principal amount and the rate of interest?
(A) 5000
(B) 1152
(C) 5760
(D) 4000
Solution
Therefore, Rate of interest for 1 year = 100*1152/5760*1 = 20%
Let the principal be p.
Then, Principal = p[1+ 20/100]2 = 5760
Solving which gives Principal = Rs. 4000
Q.3. What is the rate of simple interest?
I. The total interest earned was Rs. 4000.
II. The sum was invested for 4 years.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
E. Both I and II are necessary to answer
Solution
Now, I gives, S.I. = Rs. 4000.
II gives, T = 4 years.
But, P is unknown. So, we cannot find R.
So, given data is insufficient to get R
Q.4. Find the compound interest on Rs. 3000 at 5% for 2 years, compounded annually.
(A) 307.5
(B) 3307.5
(C) 3000.5
(D) 3100
Solution
Therefore, CI = 3307.5 – 3000 = Rs. 307.5
Q.5. A man took a loan from a bank at the rate of 12 % p.a. simple interest. After three years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:
(A) Rs. 12000
(B) Rs. 11000
(C) Rs. 14000
(D) Rs. 15000
Solution
the principal = Rs. [{100×5400}/{12×3}]
= Rs. 15000
Q.6. The difference between SI and CI compounded annually on a certain sum of money for 2 years at 8% per annum is Rs. 12.80. Find the principal.
(A) 1000
(B) 2000
(C) 3000
(D) 4000
Solution
SI = x * 2 * 8 / 100 = 4x/25
CI = x[1+ 8/100]2 – x → 104x/625
Therefore, 104x/625 – 4x/25 = 12.80
Solving which gives x, Principal = Rs. 2000.
Q.7. What percentage of simple interest per annum did Anand pay to Deepak?
I. Anand borrowed Rs. 8000 from Deepak for four years.
II. Anand returned Rs. 8800 to Deepak at the end of two years and settled the loan.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
E. Both I and II are necessary to answer
Solution
I gives, P = Rs. 8000 and T = 4 years.
II gives, S.I. = Rs. (8800 – 8000) = Rs. 800.
⸫ R = [(100 x S.I.)/(P x T)]
= [(100 x 800)/(8000 x 4)]%
= 2 1/2 % p.a.
Thus, I and II both are needed to get the answer
Q.8. A certain principal amounts to Rs. 15000 in 2.5 years and to Rs. 16500 in 4 years at the same rate of interest. Find the rate of interest.
(A) 7%
(B) 6%
(C) 9%
(D) 8%
Solution
Simple interest for (4-2.5) years = 16500 – 15000
Therefore, SI for 1.5 years = Rs. 1500.
SI for 2.5 years = (1500/1.5) * 2.5 = 2500
Principal amount = 15000 – 2500 = Rs. 12500.
Rate of Interest = {(2500 * 100) / 12500} * 2.5 → R = 8%.
Q.9. A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?
A. Rs. 35
B. Rs. 245
C. Rs. 350
D. Cannot be determined
E. None of these
Solution
Since the principal is not given, so data is inadequate
Q.10. Find the compound interest on Rs. 10000 at 12% rate of interest for 1 year, compounded half-yearly.
(A) 12360
(B) 1236
(C) 11236
(D) 10025
Solution
Therefore, CI = 11236 – 10000 = Rs. 1236