Video on Compound Interest - shortcuts, tips and tricks
Compound Interest
Points to Remember
1) Compound Interest: It is the interest added to the principal of a loan or deposit, so that the added interest also earns interest. This process is called compounding and can also be referred as “interest on interest”.
The interest rate and frequency of compounding which may be yearly, half yearly, quarterly, monthly, daily, etc. must be specified to accurately define the final amount to be paid with interest.
2) Effective annual rate: A sum of Rs. 100 is borrowed at R % per annum. If compound interest is reckoned quarterly or half yearly, then compound interest rate for 1 year is called as effective annual rate.
3) Principal: It is the money borrowed or lent out for a certain period of time.
4) Interest: Rate of money paid regularly for using money on lent.
Concept of compound interest
A person borrows Rs. 2000 at 10 % compound interest. Find the total amount paid by him after 3 years.
1) This concept can be easily understood with the diagram shown below. In this case, what happens is, when 10 % interest is applied on the principal, the amount for the first year increases to Rs. 2200.
2) As shown in the figure, finally after 3 years, the total amount to be paid is Rs. 2662.
3) The amount at the end of a 1st year is the principal for the second year and the amount at the end of the 2nd year is the principal for the 3rd year.
4) But in the case of simple interest, the amount after 3 years would be Rs. 2600. [Simple interest = (PRN/100)]
Important Formulae:
Principal = P, Rate = R % per annum, Time = n years
1. Amount = P | 1 + | R | n | - - - - - - - - - [Interest compounded annually] | ||
100 |
1. Amount = P | 1 + | (R/2) | 2n | - - - - - - - - - [Interest compounded Half-yearly] | ||
100 |
1. Amount = P | 1 + | (R/4) | 4n | - - - - - - - - - [Interest compounded quarterly] | ||
100 |
Quick Tips and Tricks
1) Numerical on population:
a) If population of a city is P1 and it increases by R % annually, then population after n years is given by:
P2 = P1 | 1 + | R | n | ||
100 |
b) If population of a city is P1 and it decreases by R % annually, then the population after n years is given by:
P2 = P1 | 1 – | R | n | ||
100 |
2) If interest is compounded annually and time is in fraction say 2 | 3 | years |
5 |
Amount = P | 1 + | (3/5)R | n | - - - - - - - - - [Interest compounded annually] | ||
100 |
3) If rates are different for different years (R1 % for 1st year, R2 % for 2nd year, R3 % for 3rd year---- Rn for n years)
Amount for n years = P | 1 + | R1 | 1 + | R1 | 1 + | R1 | ||||||
100 | 100 | 100 |
4) Present worth of sum of Rs. X due n years, hence R % p.a. compound interest is given by:
5) If difference between compound interest and simple interest is given for:
a) Two years
C.I. – S.I. = P | R | 2 | ||
100 |
b) Three years
C.I. – S.I. = P | R2 | × | (300 + R) | ||||
1002 | 100 |
6) Rule of 72: This rule is used to find:
a) Time required for an amount to double itself, at a given rate of interest.
Time = | 72 |
Rate of interest |
b) Rate at which an amount doubles itself in given time.
Rate of interest = | 72 |
Time |
Note: The rule of 72 is derived as follows:
We know the general formula to calculate total amount.
Future value = Present value | 1 + | R | n | ||
100 |
But when we have to calculate time required for an amount to double itself, at a given rate of interest, the above expression can be written as follows:
Future value (Final amount) = 2 Present value
Future value | = 2 |
Present value |
2 = | 1 + | R | n | ||
100 |
Using rule of logarithm, we get
Time = | 72 |
Rate of interest |
Types of numerical in this chapter:
In this chapter there are basically 4 types of numerical. Understanding and studying the concepts clearly will help in solving numerical on this chapter.
Type 1: Numerical based on population
Q 1. The population of a city increases 5 % annually but decreases by ¼ % due to emigration. Find the net increase in percent in 3 years.
A. 8.63 %
b. 11.89 %
c. 13.25 %
d. 14.93
%
View solution
Correct Option: (d)
Hint: If population of a city is P1 and it increases by R % annually, then population after n years is given by:
P2 = P1 | 1 + | R | n | ||
100 |
Solution:
We are given that, the population of a city increases 5 % annually but decreases by ¼ % due to emigration.
Assume original population of the city = 100
1) Increase in population = 5 %
2) Decrease in population due to emigration = ¼ %
Hence, net annual increase = 5 % - ¼ % = 19/4 %
Therefore,
Population in 3 years = P2 = P1 | 1 + | R | n | ||
100 |
= 100 | 1 + | 19 | 3 | ||
4 × 100 |
= 100 × | 419 | 3 | ||
400 |
=114.93
Population after 3 years will be 114.93 and at present it is 100. Therefore,
Increase in population = 114.93 – 100 = 14.93 %
Q 2. The value of a sewing machine depreciates at the rate of 10 % after every year. If at the end of 3 years, its value is Rs. 8748, then find its purchase price.
a. 8000
b. 10000
c. 12000
d. 15000
View solution
Correct Option: (c)
Hint: If cost of a machine is P1 and it decreases by R % annually, then the purchase price after n years is given by:
P2 = P1 | 1 – | R | n | ||
100 |
We are given that the value of a sewing machine depreciates at the rate of 10 % after every year. After 3 years, its value is Rs. 8748.
8748 = P1 | 1 – | 10 | 3 | ||
100 |
P1 =Rs.12000
Purchase price of the sewing machine = Rs. 12000
Type 2: Numerical to find compound interest (yearly, quarterly, and half-yearly)
Q 3. Find the compound interest on Rs. 5000 for 9 months at 6% per annum, if the interest is reckoned quarterly.
a. Rs. 218.98
b. Rs. 228.39
c. Rs. 250.69
d. Rs. 356.50
View solution
Amount = P | 1 + | (R/4) | 4n | - - - - - - - - - [Interest compounded quarterly] | ||
100 |
We are given:
Principal = Rs. 5000, Time = 9 months = 3 quarters, Rate = 6 % per annum
Substituting the given values, we get
Amount = P | 1 + | (6/4) | 3 | ||
100 |
Amount=Rs.5228.39
Therefore,
Compound interest = 5228.39 – 5000 = Rs. 228.39
Q 4. Find the compound interest on Rs. 20,000 in 2 years at 4 % per annum, the interest being compounded half-yearly.
a. Rs. 1648.64
b. Rs. 1596.32
c. Rs. 14826.56
d. Rs. 11563.99
View solution
Correct Option: (a)
Interest is compounded half-yearly, therefore,
Amount = P | 1 + | (R/2) | 2n | - - - - - - - - - [Interest compounded Half-yearly] | ||
100 |
Given: Principal = Rs. 20000, Rate = 2 % per half-year, Time = 2 years = 4 half- years
Amount = 20000 | 1 + | 2 | 4 | ||
100 |
Amount=Rs.21648.64
Compound Interest = Total amount – Principal
= 21648.64 – 20000
= Rs. 1648.64
Q 5. Find compound interest on Rs. 8500 at 4 % per annum for 2 years, compounded annually.
a. Rs. 752.6
b. Rs. 693.6
c. Rs. 553.6
d. Rs. 593.6
View solution
Correct Option: (b)
If interest is compounded annually, use the hint shown below:
Amount = P | 1 + | R | n | - - - - - - - - - [Interest compounded annually] | ||
100 |
We are given:
Principal = Rs. 8500, Rate = 4 % per annum, Time = 2 years
Amount = 8500 | 1 + | 4 | 2 | ||
100 |
= Rs. 9193.6
Compound Interest = Total amount – Principal
= 9193.6 – 8500
= 693.6
Compound Interest = Rs. 693.6
Type 3: Numerical to find rate of interest/years
Q 6. A amount of Rs. 500 amounts to Rs. 583.20 in two years if compounded annually. Find the rate of interest per annum.
a. 5.6 %
b. 6 %
c. 8 %
d. 9.2 %
View solution
Q 7. A certain sum amounts to Rs. 7000 in 2 years and to Rs. 8000 in 3 years. Find the sum.
a. Rs. 6959.37
b. Rs. 6459.37
c. Rs. 5359.37
d. Rs. 5759.37
View solution
Correct Option: (c)
We are given,
1) Amount = Rs. 7000, Time = 2 years
2) Amount = Rs. 8000, Time = 3 years
To calculate sum, we must first calculate rate of interest.
Simple interest = PRN / 100
S.I. on Rs. 7000 for 1 year = Rs. (8000 – 7000) = Rs. 1000
Let sum be Rs. X
7000 = P | 1 + | 100 | 2 | ||
7 × 100 |
P = 5359.37
Q 8. A sum of money doubles itself at compound interest in 10 years. In how many years will it be eight times?
a. 30 years
b. 28 years
c. 25 years
d. 22.5 years
View solution
Type 4: Numerical to find sum/rate of interest when difference between C.I. and S.I. is specified
Q 9. The difference between C.I. and S.I. on a certain sum at 10 % per annum for 2 years is Rs. 530. Find the sum.
a. 53000
b. 57500
c. 69800
d. 28090
View solution
Correct Option: (a)
Let the sum be Rs. P
Compound interest = Amount – Sum
C.I. = P | 1 + | 10 | 2 | – P | ||
100 |
Simple Interest = | P × R × N | = | P × 10 × 2 | = | P |
100 | 100 | 5 |
We are given that, the difference between C.I. and S.I. on a certain sum at 10 % per annum for 2 years is Rs. 530. Therefore,
C.I.– S.I.=530
P=Rs.53000
Hence, the required sum = Rs. 53,000
Q 10. The difference between C.I. and S.I. accrued on an amount of Rs. 20,000 in 2 years was Rs. 392. Find the rate of interest per annum.
a. 11.5 %
b. 13 %
c. 14 %
d. 12 %
View solution
Q 11. The compound interest on a certain sum at 50/3 % for 3 years is Rs. 127. Find simple interest on same sum for same period and rate.
a. Rs. 205
b. Rs. 175
c. Rs. 152
d. Rs. 108
View solution
Correct Option: (d)
Let the sum be Rs. P
Step1: We require the principal, to calculate simple interest. Hence to the find principal, use the formula of compound interest.
C.I. = P | 1 + | 50 | 2 | – P | = | 127P | ||||
3 × 100 | 216 |
We are given that, compound interest = Rs. 127
Therefore,
Principal = Rs. 216
Step 2: Calculate simple interest
S. I. = | P × R × N | = | 216 × 3 × 50 | = Rs. 108 |
100 | 100 × 3 |