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Quantitative Aptitude · Percentages Simple and Compound Interest

Simple and Compound Interest - Formulas, Key Points and Examples

Explanation

🔵 SIMPLE AND COMPOUND INTEREST 🟢 1. BASIC TERMS Principal (P): The original amount of money borrowed, invested, or deposited is called the Principal. Rate of Interest (R): The percentage of interest charged or earned per year is called the Rate of Interest. Time (T): The period for which money is borrowed or invested is called Time. Simple Interest (SI): Interest calculated only on the original principal throughout the entire period is called Simple Interest. Compound Interest (CI): Interest calculated on the principal as well as the accumulated interest from previous periods is called Compound Interest. Amount (A): The total value obtained by adding interest to the principal is called the Amount.
Amount=Principal+Interest \text{Amount} = \text{Principal}+\text{Interest}
🟢 2. SIMPLE INTEREST Under Simple Interest, interest is calculated only on the original principal. Therefore, the interest remains the same for every equal period when the rate remains constant. 🟢 3. SIMPLE INTEREST FORMULA
SI=P×R×T100 SI = \frac{P\times R\times T}{100}
Where: P = Principal R = Rate of interest per annum T = Time in years 🟢 4. AMOUNT UNDER SIMPLE INTEREST
A=P+SI A = P+SI
Substituting the formula for SI:
A=P+P×R×T100 A = P+\frac{P\times R\times T}{100}
Therefore:
A=P(1+RT100) A = P\left(1+\frac{RT}{100}\right)
🟢 5. FINDING PRINCIPAL FROM SIMPLE INTEREST From:
SI=P×R×T100 SI = \frac{P\times R\times T}{100}
We get:
P=SI×100R×T P = \frac{SI\times100}{R\times T}
🟢 6. FINDING RATE FROM SIMPLE INTEREST
R=SI×100P×T R = \frac{SI\times100}{P\times T}
🟢 7. FINDING TIME FROM SIMPLE INTEREST
T=SI×100P×R T = \frac{SI\times100}{P\times R}
🟢 8. SIMPLE INTEREST WHEN TIME IS IN MONTHS If time is given in months:
T=Number of Months12 T = \frac{\text{Number of Months}}{12}
Therefore:
SI=P×R×Number of Months100×12 SI = \frac{P\times R\times\text{Number of Months}} {100\times12}
🟢 9. SIMPLE INTEREST WHEN TIME IS IN DAYS If time is given in days, normally:
T=Number of Days365 T = \frac{\text{Number of Days}}{365}
Therefore:
SI=P×R×Number of Days100×365 SI = \frac{P\times R\times\text{Number of Days}} {100\times365}
🟡 KEY POINT Use the convention specified in the question if a different number of days is mentioned. 🟢 10. COMPOUND INTEREST Under Compound Interest, interest earned during one period is added to the principal for the next period. Therefore, interest is earned on: Original Principal + Previously Accumulated Interest. 🟢 11. COMPOUND AMOUNT FORMULA When interest is compounded annually:
A=P(1+R100)T A = P\left(1+\frac{R}{100}\right)^T
Where: P = Principal R = Annual rate of interest T = Time in years 🟢 12. COMPOUND INTEREST FORMULA Since:
CI=A−P CI=A-P
Therefore:
CI=P(1+R100)T−P CI = P\left(1+\frac{R}{100}\right)^T-P
or
CI=P[(1+R100)T−1] CI = P\left[ \left(1+\frac{R}{100}\right)^T-1 \right]
🟢 13. COMPOUNDING HALF-YEARLY When interest is compounded half-yearly: The rate is divided by 2. The number of periods is multiplied by 2. Therefore:
A=P(1+R200)2T A = P\left(1+\frac{R}{200}\right)^{2T}
and
CI=P[(1+R200)2T−1] CI = P\left[ \left(1+\frac{R}{200}\right)^{2T}-1 \right]
🟢 14. COMPOUNDING QUARTERLY When interest is compounded quarterly: The rate is divided by 4. The number of periods is multiplied by 4. Therefore:
A=P(1+R400)4T A = P\left(1+\frac{R}{400}\right)^{4T}
and
CI=P[(1+R400)4T−1] CI = P\left[ \left(1+\frac{R}{400}\right)^{4T}-1 \right]
🟢 15. COMPOUNDING MONTHLY When interest is compounded monthly: The annual rate is divided by 12. The number of periods is multiplied by 12.
A=P(1+R1200)12T A = P\left(1+\frac{R}{1200}\right)^{12T}
🟢 16. COMPOUND INTEREST FOR TWO YEARS For two years with annual compounding:
A=P(1+R100)2 A = P\left(1+\frac{R}{100}\right)^2
Expanding:
A=P(1+2R100+R210000) A = P \left( 1+\frac{2R}{100} +\frac{R^2}{10000} \right)
Therefore:
CI=2PR100+PR210000 CI = \frac{2PR}{100} + \frac{PR^2}{10000}
🟢 17. DIFFERENCE BETWEEN CI AND SI FOR TWO YEARS For two years:
CI−SI=P(R100)2 CI-SI = P\left(\frac{R}{100}\right)^2
Therefore:
CI−SI=PR210000 CI-SI = \frac{PR^2}{10000}
🟢 18. DIFFERENCE BETWEEN CI AND SI FOR THREE YEARS For three years:
CI−SI=P[(1+R100)3−1−3R100] CI-SI = P \left[ \left(1+\frac{R}{100}\right)^3 - 1 - \frac{3R}{100} \right]
After simplification:
CI−SI=P(3R210000+R31000000) CI-SI = P \left( \frac{3R^2}{10000} + \frac{R^3}{1000000} \right)
🟢 19. COMPOUND INTEREST FOR DIFFERENT RATES If the rate changes each year:
A=P(1+R1100)(1+R2100)(1+R3100) A = P \left(1+\frac{R_1}{100}\right) \left(1+\frac{R_2}{100}\right) \left(1+\frac{R_3}{100}\right)
For different rates in different years, multiply the corresponding growth factors. 🟢 20. POPULATION AND COMPOUND GROWTH If a population increases by R% every year:
Final Population=Initial Population(1+R100)T \text{Final Population} = \text{Initial Population} \left(1+\frac{R}{100}\right)^T
🟢 21. DEPRECIATION If the value of an asset decreases by R% every year:
Final Value=Initial Value(1−R100)T \text{Final Value} = \text{Initial Value} \left(1-\frac{R}{100}\right)^T
🟢 22. COMPOUND GROWTH AND DEPRECIATION Growth uses:
(1+R100)T \left(1+\frac{R}{100}\right)^T
Depreciation uses:
(1−R100)T \left(1-\frac{R}{100}\right)^T
🟢 23. SIMPLE INTEREST VS COMPOUND INTEREST Simple Interest: Interest is calculated only on the original principal.
SI=PRT100 SI = \frac{PRT}{100}
Compound Interest: Interest is calculated on the principal plus accumulated interest.
A=P(1+R100)T A = P\left(1+\frac{R}{100}\right)^T
CI=A−P CI=A-P
🟡 KEY POINT For the same positive principal, rate and time greater than one compounding period, Compound Interest is generally greater than Simple Interest when compounded at the same annual rate. 🟢 24. AMOUNT RATIO IN COMPOUND INTEREST If two investments have the same rate and time:
A1A2=P1P2 \frac{A_1}{A_2} = \frac{P_1}{P_2}
🟢 25. EFFECTIVE RATE OF INTEREST If interest is compounded more than once per year, the effective annual rate is greater than the stated nominal annual rate. For n compounding periods per year:
Effective Rate=(1+R100n)n−1 \text{Effective Rate} = \left(1+\frac{R}{100n}\right)^n-1
As a percentage:
Effective Rate %=[(1+R100n)n−1]×100 \text{Effective Rate \%} = \left[ \left(1+\frac{R}{100n}\right)^n-1 \right]\times100
🟢 26. IMPORTANT SHORTCUTS If the rate is 10%:
1+10100=1.1 1+\frac{10}{100}=1.1
If the rate is 20%:
1+20100=1.2 1+\frac{20}{100}=1.2
If the rate is 25%:
1+25100=1.25 1+\frac{25}{100}=1.25
For depreciation of 10%:
1−10100=0.9 1-\frac{10}{100}=0.9
For depreciation of 20%:
1−20100=0.8 1-\frac{20}{100}=0.8
🔴 IMPORTANT EXAM POINTS 1. In Simple Interest, always use the original principal. 2. In Compound Interest, previous interest becomes part of the principal for the next period. 3. If compounding is half-yearly, divide the rate by 2 and multiply the time by 2. 4. If compounding is quarterly, divide the rate by 4 and multiply the time by 4. 5. If time is given in months for Simple Interest, convert months into years. 6. Amount is always Principal + Interest. 7. Compound Interest is Amount − Principal. 8. For percentage growth, use the plus sign. 9. For depreciation, use the minus sign. 10. For successive changes, calculate each change on the updated amount.

Example

🟠 EXAMPLE 1: FIND SIMPLE INTEREST QUESTION: Find the simple interest on ₹5,000 at 8% per annum for 3 years. 🟢 SOLUTION: Using:
SI=P×R×T100 SI=\frac{P\times R\times T}{100}
Here:
P=5000,R=8,T=3 P=5000,\quad R=8,\quad T=3
Therefore:
SI=5000×8×3100 SI=\frac{5000\times8\times3}{100}
SI=1200 SI=1200
✅ ANSWER:
₹1200 ₹1200
🟠 EXAMPLE 2: FIND THE AMOUNT QUESTION: Find the amount on ₹8,000 at 10% per annum simple interest for 2 years. 🟢 SOLUTION: First find the simple interest:
SI=8000×10×2100 SI=\frac{8000\times10\times2}{100}
SI=1600 SI=1600
Amount:
A=P+SI A=P+SI
A=8000+1600 A=8000+1600
A=9600 A=9600
✅ ANSWER:
₹9600 ₹9600
🟠 EXAMPLE 3: FIND PRINCIPAL QUESTION: The simple interest on a sum at 6% per annum for 5 years is ₹1,500. Find the principal. 🟢 SOLUTION: Using:
SI=P×R×T100 SI=\frac{P\times R\times T}{100}
1500=P×6×5100 1500=\frac{P\times6\times5}{100}
1500=30P100 1500=\frac{30P}{100}
150000=30P 150000=30P
P=5000 P=5000
✅ ANSWER:
₹5000 ₹5000
🟠 EXAMPLE 4: FIND RATE QUESTION: A sum of ₹4,000 earns a simple interest of ₹800 in 5 years. Find the rate of interest. 🟢 SOLUTION: Using:
R=SI×100P×T R=\frac{SI\times100}{P\times T}
R=800×1004000×5 R=\frac{800\times100}{4000\times5}
R=8000020000 R=\frac{80000}{20000}
R=4 R=4
✅ ANSWER:
4% 4\%
🟠 EXAMPLE 5: FIND TIME QUESTION: A sum of ₹6,000 earns ₹1,800 as simple interest at 10% per annum. Find the time. 🟢 SOLUTION: Using:
T=SI×100P×R T=\frac{SI\times100}{P\times R}
T=1800×1006000×10 T=\frac{1800\times100}{6000\times10}
T=18000060000 T=\frac{180000}{60000}
T=3 T=3
✅ ANSWER:
3 years 3\text{ years}
🟠 EXAMPLE 6: SIMPLE INTEREST FOR MONTHS QUESTION: Find the simple interest on ₹12,000 at 9% per annum for 8 months. 🟢 SOLUTION: Convert months into years:
T=812=23 T=\frac{8}{12}=\frac23
Using:
SI=P×R×T100 SI=\frac{P\times R\times T}{100}
SI=12000×9×23100 SI=\frac{12000\times9\times\frac23}{100}
SI=720 SI=720
✅ ANSWER:
₹720 ₹720
🟠 EXAMPLE 7: SIMPLE INTEREST FOR DAYS QUESTION: Find the simple interest on ₹10,000 at 12% per annum for 6 months. 🟢 SOLUTION:
T=612=12 T=\frac{6}{12}=\frac12
Therefore:
SI=10000×12×12100 SI=\frac{10000\times12\times\frac12}{100}
SI=600 SI=600
✅ ANSWER:
₹600 ₹600
🟠 EXAMPLE 8: DIFFERENCE BETWEEN AMOUNT AND PRINCIPAL QUESTION: A sum of ₹7,500 is invested at 8% simple interest for 4 years. Find the amount. 🟢 SOLUTION:
SI=7500×8×4100 SI=\frac{7500\times8\times4}{100}
SI=2400 SI=2400
Therefore:
A=P+SI A=P+SI
A=7500+2400 A=7500+2400
A=9900 A=9900
✅ ANSWER:
₹9900 ₹9900
🟠 EXAMPLE 9: COMPOUND INTEREST FOR 2 YEARS QUESTION: Find the compound interest on ₹10,000 at 10% per annum for 2 years. 🟢 SOLUTION: Using:
A=P(1+R100)T A=P\left(1+\frac{R}{100}\right)^T
A=10000(1+10100)2 A=10000\left(1+\frac{10}{100}\right)^2
A=10000(1.1)2 A=10000(1.1)^2
A=12100 A=12100
Compound interest:
CI=A−P CI=A-P
CI=12100−10000 CI=12100-10000
CI=2100 CI=2100
✅ ANSWER:
₹2100 ₹2100
🟠 EXAMPLE 10: COMPOUND AMOUNT QUESTION: Find the amount on ₹8,000 at 5% per annum compounded annually for 3 years. 🟢 SOLUTION:
A=P(1+R100)T A=P\left(1+\frac{R}{100}\right)^T
A=8000(1+5100)3 A=8000\left(1+\frac{5}{100}\right)^3
A=8000(1.05)3 A=8000(1.05)^3
A=9261 A=9261
✅ ANSWER:
₹9261 ₹9261
🟠 EXAMPLE 11: COMPOUND INTEREST QUESTION: Find the compound interest on ₹20,000 at 8% per annum for 2 years, compounded annually. 🟢 SOLUTION:
A=20000(1+8100)2 A=20000\left(1+\frac{8}{100}\right)^2
A=20000(1.08)2 A=20000(1.08)^2
A=23328 A=23328
Therefore:
CI=A−P CI=A-P
CI=23328−20000 CI=23328-20000
CI=3328 CI=3328
✅ ANSWER:
₹3328 ₹3328
🟠 EXAMPLE 12: COMPOUND INTEREST FOR 3 YEARS QUESTION: Find the compound interest on ₹5,000 at 10% per annum for 3 years. 🟢 SOLUTION:
A=5000(1+10100)3 A=5000\left(1+\frac{10}{100}\right)^3
A=5000(1.1)3 A=5000(1.1)^3
A=6655 A=6655
Therefore:
CI=6655−5000 CI=6655-5000
CI=1655 CI=1655
✅ ANSWER:
₹1655 ₹1655
🟠 EXAMPLE 13: COMPOUND INTEREST HALF-YEARLY QUESTION: Find the compound interest on ₹10,000 at 12% per annum for 1 year, compounded half-yearly. 🟢 SOLUTION: For half-yearly compounding:
R=122=6% R=\frac{12}{2}=6\%
and:
T=1×2=2 T=1\times2=2
Therefore:
A=10000(1+6100)2 A=10000\left(1+\frac{6}{100}\right)^2
A=10000(1.06)2 A=10000(1.06)^2
A=11236 A=11236
Hence:
CI=11236−10000 CI=11236-10000
CI=1236 CI=1236
✅ ANSWER:
₹1236 ₹1236
🟠 EXAMPLE 14: COMPOUND INTEREST QUARTERLY QUESTION: Find the amount on ₹16,000 at 8% per annum for 1 year, compounded quarterly. 🟢 SOLUTION: For quarterly compounding:
R=84=2% R=\frac{8}{4}=2\%
and:
T=1×4=4 T=1\times4=4
Using:
A=P(1+R100)T A=P\left(1+\frac{R}{100}\right)^T
A=16000(1+2100)4 A=16000\left(1+\frac{2}{100}\right)^4
A=16000(1.02)4 A=16000(1.02)^4
A≈17318.91 A\approx17318.91
✅ ANSWER:
₹17318.91 ₹17318.91
🟠 EXAMPLE 15: DIFFERENCE BETWEEN CI AND SI FOR 2 YEARS QUESTION: Find the difference between compound interest and simple interest on ₹10,000 at 10% per annum for 2 years. 🟢 SOLUTION: For 2 years:
Difference=P(R100)2 \text{Difference}=P\left(\frac{R}{100}\right)^2
=10000(10100)2 =10000\left(\frac{10}{100}\right)^2
=10000(0.1)2 =10000(0.1)^2
=100 =100
✅ ANSWER:
₹100 ₹100
🟠 EXAMPLE 16: DIFFERENCE BETWEEN CI AND SI FOR 3 YEARS QUESTION: Find the difference between compound interest and simple interest on ₹20,000 at 10% per annum for 3 years. 🟢 SOLUTION: Simple interest:
SI=20000×10×3100 SI=\frac{20000\times10\times3}{100}
SI=6000 SI=6000
Compound amount:
A=20000(1.1)3 A=20000(1.1)^3
A=26620 A=26620
Compound interest:
CI=26620−20000 CI=26620-20000
CI=6620 CI=6620
Difference:
CI−SI=6620−6000 CI-SI=6620-6000
=620 =620
✅ ANSWER:
₹620 ₹620
🟠 EXAMPLE 17: FIND PRINCIPAL USING COMPOUND AMOUNT QUESTION: The amount after 2 years at 10% per annum compound interest is ₹12,100. Find the principal. 🟢 SOLUTION: Using:
A=P(1+R100)T A=P\left(1+\frac{R}{100}\right)^T
12100=P(1.1)2 12100=P(1.1)^2
12100=1.21P 12100=1.21P
P=121001.21 P=\frac{12100}{1.21}
P=10000 P=10000
✅ ANSWER:
₹10000 ₹10000
🟠 EXAMPLE 18: FIND COMPOUND RATE QUESTION: A sum of ₹10,000 becomes ₹12,100 in 2 years when compounded annually. Find the rate of interest. 🟢 SOLUTION: Using:
A=P(1+R100)T A=P\left(1+\frac{R}{100}\right)^T
12100=10000(1+R100)2 12100=10000\left(1+\frac{R}{100}\right)^2
1.21=(1+R100)2 1.21=\left(1+\frac{R}{100}\right)^2
Taking square root:
1.1=1+R100 1.1=1+\frac{R}{100}
R100=0.1 \frac{R}{100}=0.1
R=10 R=10
✅ ANSWER:
10% 10\%
🟠 EXAMPLE 19: COMPOUND INTEREST AND ANNUAL GROWTH QUESTION: A population increases by 10% every year. If the present population is 20,000, find the population after 2 years. 🟢 SOLUTION: This follows compound growth.
A=P(1+R100)T A=P\left(1+\frac{R}{100}\right)^T
A=20000(1.1)2 A=20000(1.1)^2
A=24200 A=24200
✅ ANSWER:
24200 24200
🟠 EXAMPLE 20: DEPRECIATION QUESTION: A machine worth ₹50,000 depreciates at 10% per annum. Find its value after 2 years. 🟢 SOLUTION: For depreciation:
A=P(1−R100)T A=P\left(1-\frac{R}{100}\right)^T
A=50000(1−0.1)2 A=50000(1-0.1)^2
A=50000(0.9)2 A=50000(0.9)^2
A=40500 A=40500
✅ ANSWER:
₹40500 ₹40500
🟠 EXAMPLE 21: DEPRECIATION FOR 3 YEARS QUESTION: A car costs ₹8,00,000 and depreciates at 15% per annum. Find its value after 3 years. 🟢 SOLUTION:
A=P(1−R100)T A=P\left(1-\frac{R}{100}\right)^T
A=800000(1−0.15)3 A=800000(1-0.15)^3
A=800000(0.85)3 A=800000(0.85)^3
A=800000(0.614125) A=800000(0.614125)
A=491300 A=491300
✅ ANSWER:
₹491300 ₹491300
🟠 EXAMPLE 22: COMPOUND INTEREST WITH HALF-YEARLY COMPOUNDING QUESTION: Find the compound interest on ₹25,000 at 10% per annum for 1 year, compounded half-yearly. 🟢 SOLUTION: Rate for each half-year:
R=102=5% R=\frac{10}{2}=5\%
Number of periods:
T=1×2=2 T=1\times2=2
Therefore:
A=25000(1+5100)2 A=25000\left(1+\frac{5}{100}\right)^2
A=25000(1.05)2 A=25000(1.05)^2
A=27562.50 A=27562.50
Compound interest:
CI=27562.50−25000 CI=27562.50-25000
CI=2562.50 CI=2562.50
✅ ANSWER:
₹2562.50 ₹2562.50
🟠 EXAMPLE 23: SIMPLE INTEREST VS COMPOUND INTEREST QUESTION: Find the simple interest and compound interest on ₹10,000 at 10% per annum for 2 years. 🟢 SOLUTION: Simple interest:
SI=10000×10×2100 SI=\frac{10000\times10\times2}{100}
SI=2000 SI=2000
Compound amount:
A=10000(1.1)2 A=10000(1.1)^2
A=12100 A=12100
Compound interest:
CI=12100−10000 CI=12100-10000
CI=2100 CI=2100
Therefore:
CI−SI=2100−2000 CI-SI=2100-2000
=100 =100
✅ ANSWER:
SI=₹2000 SI=₹2000
CI=₹2100 CI=₹2100
Difference=₹100 \text{Difference}=₹100
🟠 EXAMPLE 24: FIND AMOUNT USING SIMPLE INTEREST QUESTION: A sum of ₹15,000 is invested at 7% per annum simple interest for 4 years. Find the amount. 🟢 SOLUTION:
SI=15000×7×4100 SI=\frac{15000\times7\times4}{100}
SI=4200 SI=4200
Amount:
A=P+SI A=P+SI
A=15000+4200 A=15000+4200
A=19200 A=19200
✅ ANSWER:
₹19200 ₹19200
🟠 EXAMPLE 25: FIND TIME USING SIMPLE INTEREST QUESTION: At what time will ₹8,000 earn ₹2,400 as simple interest at 10% per annum? 🟢 SOLUTION:
T=SI×100P×R T=\frac{SI\times100}{P\times R}
T=2400×1008000×10 T=\frac{2400\times100}{8000\times10}
T=24000080000 T=\frac{240000}{80000}
T=3 T=3
✅ ANSWER:
3 years 3\text{ years}