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Quantitative Aptitude · Percentages Simple and Compound Interest

Percentage - Formulas, Key Points and Examples

Explanation

🔵 PERCENTAGE 🟢 1. BASIC PERCENTAGE Percentage expresses a quantity as a fraction of 100.
Percentage=PartWhole×100 \text{Percentage} = \frac{\text{Part}}{\text{Whole}}\times100
Part=Percentage×Whole100 \text{Part} = \frac{\text{Percentage}\times\text{Whole}}{100}
Whole=Part×100Percentage \text{Whole} = \frac{\text{Part}\times100}{\text{Percentage}}
🟡 KEY POINT Percentage always means "per 100". 🟢 2. FRACTION TO PERCENTAGE To convert a fraction into a percentage, multiply the fraction by 100.
Percentage=Fraction×100 \text{Percentage} = \text{Fraction}\times100
Important conversions:
12=50% \frac{1}{2}=50\%
13=3313% \frac{1}{3}=33\frac{1}{3}\%
14=25% \frac{1}{4}=25\%
15=20% \frac{1}{5}=20\%
18=12.5% \frac{1}{8}=12.5\%
110=10% \frac{1}{10}=10\%
120=5% \frac{1}{20}=5\%
125=4% \frac{1}{25}=4\%
150=2% \frac{1}{50}=2\%
1100=1% \frac{1}{100}=1\%
🟢 3. PERCENTAGE TO FRACTION To convert a percentage into a fraction, divide it by 100 and simplify.
Fraction=Percentage100 \text{Fraction} = \frac{\text{Percentage}}{100}
🟢 4. PERCENTAGE TO DECIMAL To convert a percentage into a decimal, divide it by 100.
Decimal=Percentage100 \text{Decimal} = \frac{\text{Percentage}}{100}
🟢 5. FINDING A PERCENTAGE OF A NUMBER To find x% of a number N:
x% of N=x100×N x\%\text{ of }N = \frac{x}{100}\times N
🟢 6. PERCENTAGE INCREASE Percentage increase measures how much a quantity has increased compared with its original value.
Increase=New Value−Original Value \text{Increase} = \text{New Value}-\text{Original Value}
Percentage Increase=IncreaseOriginal Value×100 \text{Percentage Increase} = \frac{\text{Increase}}{\text{Original Value}}\times100
🔴 IMPORTANT The denominator is always the original value. 🟢 7. PERCENTAGE DECREASE Percentage decrease measures how much a quantity has decreased compared with its original value.
Decrease=Original Value−New Value \text{Decrease} = \text{Original Value}-\text{New Value}
Percentage Decrease=DecreaseOriginal Value×100 \text{Percentage Decrease} = \frac{\text{Decrease}}{\text{Original Value}}\times100
🔴 IMPORTANT The denominator is always the original value. 🟢 8. NEW VALUE AFTER PERCENTAGE INCREASE If an original value is increased by x%:
New Value=Original Value(1+x100) \text{New Value} = \text{Original Value} \left(1+\frac{x}{100}\right)
or
New Value=Original Value×100+x100 \text{New Value} = \text{Original Value}\times\frac{100+x}{100}
🟢 9. NEW VALUE AFTER PERCENTAGE DECREASE If an original value is decreased by x%:
New Value=Original Value(1−x100) \text{New Value} = \text{Original Value} \left(1-\frac{x}{100}\right)
or
New Value=Original Value×100−x100 \text{New Value} = \text{Original Value}\times\frac{100-x}{100}
🟢 10. FINDING ORIGINAL VALUE AFTER AN INCREASE If the final value is obtained after an x% increase:
Original Value=Final Value×100100+x \text{Original Value} = \frac{\text{Final Value}\times100}{100+x}
🟢 11. FINDING ORIGINAL VALUE AFTER A DECREASE If the final value is obtained after an x% decrease:
Original Value=Final Value×100100−x \text{Original Value} = \frac{\text{Final Value}\times100}{100-x}
🟢 12. PERCENTAGE COMPARISON To find what percentage A is of B:
Percentage=AB×100 \text{Percentage} = \frac{A}{B}\times100
🟢 13. SUCCESSIVE PERCENTAGE INCREASE If a quantity is increased successively by a% and b%:
Net Increase=(a+b+ab100)% \text{Net Increase} = \left(a+b+\frac{ab}{100}\right)\%
🔴 IMPORTANT Successive percentage increases cannot be added directly. 🟢 14. SUCCESSIVE PERCENTAGE DECREASE If a quantity is decreased successively by a% and b%:
Net Decrease=(a+b−ab100)% \text{Net Decrease} = \left(a+b-\frac{ab}{100}\right)\%
🟢 15. INCREASE FOLLOWED BY DECREASE If a quantity is increased by a% and then decreased by b%:
Net Change=(a−b−ab100)% \text{Net Change} = \left(a-b-\frac{ab}{100}\right)\%
🟢 16. DECREASE FOLLOWED BY INCREASE If a quantity is decreased by a% and then increased by b%:
Net Change=(b−a−ab100)% \text{Net Change} = \left(b-a-\frac{ab}{100}\right)\%
🟢 17. EQUAL PERCENTAGE INCREASE AND DECREASE If a quantity is increased by x% and then decreased by x%:
Net Decrease=x2100% \text{Net Decrease} = \frac{x^2}{100}\%
🔴 IMPORTANT An equal percentage increase and decrease do not cancel each other. 🟢 18. ONE NUMBER IS x% MORE THAN ANOTHER If A is x% more than B:
A=B(1+x100) A = B\left(1+\frac{x}{100}\right)
or
A=B×100+x100 A = B\times\frac{100+x}{100}
🟢 19. ONE NUMBER IS x% LESS THAN ANOTHER If A is x% less than B:
A=B(1−x100) A = B\left(1-\frac{x}{100}\right)
or
A=B×100−x100 A = B\times\frac{100-x}{100}
🟢 20. REVERSE PERCENTAGE If A is x% more than B:
A=B(100+x100) A = B\left(\frac{100+x}{100}\right)
Therefore:
B=A(100100+x) B = A\left(\frac{100}{100+x}\right)
The percentage by which B is less than A is:
x100+x×100% \frac{x}{100+x}\times100\%
🟢 21. PERCENTAGE AND MARKS To find percentage from marks:
Percentage=Marks ObtainedMaximum Marks×100 \text{Percentage} = \frac{\text{Marks Obtained}} {\text{Maximum Marks}}\times100
To find marks from percentage:
Marks Obtained=Percentage×Maximum Marks100 \text{Marks Obtained} = \frac{\text{Percentage}\times\text{Maximum Marks}}{100}
To find maximum marks:
Maximum Marks=Marks Obtained×100Percentage \text{Maximum Marks} = \frac{\text{Marks Obtained}\times100}{\text{Percentage}}
🟢 22. PROFIT PERCENTAGE Profit is the difference between Selling Price and Cost Price.
Profit=SP−CP \text{Profit} = SP-CP
Profit %=ProfitCP×100 \text{Profit \%} = \frac{\text{Profit}}{CP}\times100
Therefore:
Profit %=SP−CPCP×100 \text{Profit \%} = \frac{SP-CP}{CP}\times100
🔴 IMPORTANT Profit percentage is calculated on Cost Price. 🟢 23. LOSS PERCENTAGE Loss is the difference between Cost Price and Selling Price.
Loss=CP−SP \text{Loss} = CP-SP
Loss %=LossCP×100 \text{Loss \%} = \frac{\text{Loss}}{CP}\times100
Therefore:
Loss %=CP−SPCP×100 \text{Loss \%} = \frac{CP-SP}{CP}\times100
🔴 IMPORTANT Loss percentage is calculated on Cost Price. 🟢 24. INCOME, EXPENDITURE AND SAVINGS Savings are the difference between income and expenditure.
Savings=Income−Expenditure \text{Savings} = \text{Income}-\text{Expenditure}
Savings %=SavingsIncome×100 \text{Savings \%} = \frac{\text{Savings}}{\text{Income}}\times100
🟢 25. POPULATION, SALARY AND PRODUCTION For an x% increase:
New Value=Old Value×100+x100 \text{New Value} = \text{Old Value}\times\frac{100+x}{100}
For an x% decrease:
New Value=Old Value×100−x100 \text{New Value} = \text{Old Value}\times\frac{100-x}{100}
This formula can be used for population, salary, price, production, sales, income and expenditure. 🟢 26. IMPORTANT PERCENTAGE SHORTCUTS
10% of N=N10 10\%\text{ of }N = \frac{N}{10}
5% of N=N20 5\%\text{ of }N = \frac{N}{20}
1% of N=N100 1\%\text{ of }N = \frac{N}{100}
25% of N=N4 25\%\text{ of }N = \frac{N}{4}
50% of N=N2 50\%\text{ of }N = \frac{N}{2}
75% of N=3N4 75\%\text{ of }N = \frac{3N}{4}

Example

🟠 EXAMPLE 1: FINDING A PERCENTAGE QUESTION: Find 25% of 480. 🟢 SOLUTION:
25% of 480=25100×480 25\% \text{ of } 480 = \frac{25}{100}\times480
=120 =120
✅ ANSWER:
120 120
🟠 EXAMPLE 2: FINDING WHAT PERCENTAGE ONE NUMBER IS OF ANOTHER QUESTION: 45 is what percentage of 180? 🟢 SOLUTION:
Percentage=45180×100 \text{Percentage} = \frac{45}{180}\times100
=25% =25\%
✅ ANSWER:
25% 25\%
🟠 EXAMPLE 3: CONVERTING FRACTION INTO PERCENTAGE QUESTION: Convert 35\frac{3}{5} into a percentage. 🟢 SOLUTION:
35×100 \frac{3}{5}\times100
=60% =60\%
✅ ANSWER:
60% 60\%
🟠 EXAMPLE 4: CONVERTING DECIMAL INTO PERCENTAGE QUESTION: Convert 0.375 into a percentage. 🟢 SOLUTION:
0.375×100 0.375\times100
=37.5% =37.5\%
✅ ANSWER:
37.5% 37.5\%
🟠 EXAMPLE 5: PERCENTAGE INCREASE QUESTION: The price of a book increases from ₹400 to ₹500. Find the percentage increase. 🟢 SOLUTION: Increase:
500−400=100 500-400=100
Percentage increase:
100400×100 \frac{100}{400}\times100
=25% =25\%
✅ ANSWER:
25% 25\%
🟠 EXAMPLE 6: PERCENTAGE DECREASE QUESTION: The price of a shirt decreases from ₹800 to ₹680. Find the percentage decrease. 🟢 SOLUTION: Decrease:
800−680=120 800-680=120
Percentage decrease:
120800×100 \frac{120}{800}\times100
=15% =15\%
✅ ANSWER:
15% 15\%
🟠 EXAMPLE 7: FINDING THE ORIGINAL NUMBER QUESTION: 30% of a number is 150. Find the number. 🟢 SOLUTION: Let the number be xx.
30% of x=150 30\%\text{ of }x=150
30100x=150 \frac{30}{100}x=150
x=150×10030 x=\frac{150\times100}{30}
x=500 x=500
✅ ANSWER:
500 500
🟠 EXAMPLE 8: FINDING MARKS QUESTION: A student scored 72% marks in an examination of 500 marks. Find the marks obtained. 🟢 SOLUTION:
72100×500 \frac{72}{100}\times500
=360 =360
✅ ANSWER:
360 marks 360\text{ marks}
🟠 EXAMPLE 9: PASS MARKS QUESTION: The pass percentage in an examination is 40%. If the maximum marks are 600, find the minimum marks required to pass. 🟢 SOLUTION:
40100×600 \frac{40}{100}\times600
=240 =240
✅ ANSWER:
240 marks 240\text{ marks}
🟠 EXAMPLE 10: SUCCESSIVE INCREASE QUESTION: The price of an article increases by 20% and then by 10%. Find the overall percentage increase. 🟢 SOLUTION: Using the successive percentage formula:
Net increase=a+b+ab100 \text{Net increase} = a+b+\frac{ab}{100}
Here:
a=20,b=10 a=20,\quad b=10
Therefore:
20+10+20×10100 20+10+\frac{20\times10}{100}
=30+2 =30+2
=32% =32\%
✅ ANSWER:
32% 32\%
🟠 EXAMPLE 11: SUCCESSIVE DECREASE QUESTION: The price of an article decreases by 20% and then by 10%. Find the overall percentage decrease. 🟢 SOLUTION: Using:
Net decrease=a+b−ab100 \text{Net decrease} = a+b-\frac{ab}{100}
=20+10−20×10100 =20+10-\frac{20\times10}{100}
=30−2 =30-2
=28% =28\%
✅ ANSWER:
28% 28\%
🟠 EXAMPLE 12: INCREASE FOLLOWED BY DECREASE QUESTION: The price of an article is increased by 20% and then decreased by 20%. Find the overall percentage change. 🟢 SOLUTION: Using:
Net change=a−b−ab100 \text{Net change} = a-b-\frac{ab}{100}
Here:
a=20,b=20 a=20,\quad b=20
20−20−20×20100 20-20-\frac{20\times20}{100}
=0−4 =0-4
=−4% =-4\%
The negative sign indicates a decrease. ✅ ANSWER:
4% decrease 4\%\text{ decrease}
🟠 EXAMPLE 13: POPULATION INCREASE QUESTION: The population of a town is 50,000. If it increases by 12%, find the new population. 🟢 SOLUTION: Increase:
12100×50000 \frac{12}{100}\times50000
=6000 =6000
New population:
50000+6000 50000+6000
=56000 =56000
✅ ANSWER:
56000 56000
🟠 EXAMPLE 14: POPULATION DECREASE QUESTION: The population of a town is 80,000. If it decreases by 15%, find the new population. 🟢 SOLUTION: Decrease:
15100×80000 \frac{15}{100}\times80000
=12000 =12000
New population:
80000−12000 80000-12000
=68000 =68000
✅ ANSWER:
68000 68000
🟠 EXAMPLE 15: EXPENDITURE AND SAVINGS QUESTION: A person earns ₹40,000 per month and spends 75% of his income. Find his monthly savings. 🟢 SOLUTION: Expenditure:
75100×40000 \frac{75}{100}\times40000
=30000 =30000
Savings:
40000−30000 40000-30000
=10000 =10000
Alternatively:
100%−75%=25% 100\%-75\%=25\%
25100×40000=10000 \frac{25}{100}\times40000=10000
✅ ANSWER:
₹10000 ₹10000
🟠 EXAMPLE 16: PERCENTAGE OF GIRLS QUESTION: In a class of 80 students, 35% are girls. Find the number of boys. 🟢 SOLUTION: Number of girls:
35100×80 \frac{35}{100}\times80
=28 =28
Number of boys:
80−28 80-28
=52 =52
✅ ANSWER:
52 boys 52\text{ boys}
🟠 EXAMPLE 17: ELECTION VOTES QUESTION: A candidate receives 55% of 20,000 votes. How many votes did the candidate receive? 🟢 SOLUTION:
55100×20000 \frac{55}{100}\times20000
=11000 =11000
✅ ANSWER:
11000 votes 11000\text{ votes}
🟠 EXAMPLE 18: INCOME INCREASE QUESTION: A person's salary is ₹25,000. If his salary is increased by 16%, find his new salary. 🟢 SOLUTION: Increase:
16100×25000 \frac{16}{100}\times25000
=4000 =4000
New salary:
25000+4000 25000+4000
=29000 =29000
✅ ANSWER:
₹29000 ₹29000
🟠 EXAMPLE 19: INCOME DECREASE QUESTION: A person's salary is ₹30,000. If his salary is reduced by 10%, find his new salary. 🟢 SOLUTION: Decrease:
10100×30000 \frac{10}{100}\times30000
=3000 =3000
New salary:
30000−3000 30000-3000
=27000 =27000
✅ ANSWER:
₹27000 ₹27000
🟠 EXAMPLE 20: PROFIT PERCENTAGE QUESTION: An article is bought for ₹800 and sold for ₹1,000. Find the profit percentage. 🟢 SOLUTION: Profit:
1000−800=200 1000-800=200
Profit percentage:
200800×100 \frac{200}{800}\times100
=25% =25\%
✅ ANSWER:
25% 25\%
🟠 EXAMPLE 21: LOSS PERCENTAGE QUESTION: An article is bought for ₹1,200 and sold for ₹1,020. Find the loss percentage. 🟢 SOLUTION: Loss:
1200−1020=180 1200-1020=180
Loss percentage:
1801200×100 \frac{180}{1200}\times100
=15% =15\%
✅ ANSWER:
15% 15\%
🟠 EXAMPLE 22: DISCOUNT QUESTION: A shirt marked at ₹2,000 is sold at a discount of 15%. Find the selling price. 🟢 SOLUTION: Discount:
15100×2000 \frac{15}{100}\times2000
=300 =300
Selling price:
2000−300 2000-300
=1700 =1700
✅ ANSWER:
₹1700 ₹1700
🟠 EXAMPLE 23: FINDING MARKED PRICE QUESTION: An article is sold for ₹1,700 after giving a discount of 15%. Find the marked price. 🟢 SOLUTION: After a 15% discount, the selling price is:
100%−15%=85% 100\%-15\%=85\%
Therefore:
85% of MP=1700 85\%\text{ of MP}=1700
85100×MP=1700 \frac{85}{100}\times MP=1700
MP=1700×10085 MP=\frac{1700\times100}{85}
MP=2000 MP=2000
✅ ANSWER:
₹2000 ₹2000
🟠 EXAMPLE 24: EXAMINATION RESULT QUESTION: A student needs 40% marks to pass an examination. He scores 220 marks and fails by 20 marks. Find the maximum marks. 🟢 SOLUTION: Passing marks:
220+20=240 220+20=240
Therefore:
40% of maximum marks=240 40\%\text{ of maximum marks}=240
Let maximum marks be xx.
40100x=240 \frac{40}{100}x=240
x=240×10040 x=\frac{240\times100}{40}
x=600 x=600
✅ ANSWER:
600 marks 600\text{ marks}
🟠 EXAMPLE 25: NUMBER INCREASED BY PERCENTAGE QUESTION: A number is increased by 25% and becomes 500. Find the original number. 🟢 SOLUTION: After a 25% increase:
100%+25%=125% 100\%+25\%=125\%
Therefore:
125% of original number=500 125\%\text{ of original number}=500
125100x=500 \frac{125}{100}x=500
x=500×100125 x=\frac{500\times100}{125}
x=400 x=400
✅ ANSWER:
400 400
🟠 EXAMPLE 26: NUMBER DECREASED BY PERCENTAGE QUESTION: A number is decreased by 20% and becomes 640. Find the original number. 🟢 SOLUTION: After a 20% decrease:
100%−20%=80% 100\%-20\%=80\%
Therefore:
80% of original number=640 80\%\text{ of original number}=640
80100x=640 \frac{80}{100}x=640
x=640×10080 x=\frac{640\times100}{80}
x=800 x=800
✅ ANSWER:
800 800
🟠 EXAMPLE 27: MORE THAN QUESTION: A is 25% more than B. If B is 400, find A. 🟢 SOLUTION: Increase:
25% of 400=25100×400 25\%\text{ of }400 = \frac{25}{100}\times400
=100 =100
Therefore:
A=400+100 A=400+100
A=500 A=500
✅ ANSWER:
500 500
🟠 EXAMPLE 28: LESS THAN QUESTION: A is 20% less than B. If B is 750, find A. 🟢 SOLUTION: Decrease:
20% of 750=20100×750 20\%\text{ of }750 = \frac{20}{100}\times750
=150 =150
Therefore:
A=750−150 A=750-150
A=600 A=600
✅ ANSWER:
600 600
🟠 EXAMPLE 29: REVERSE PERCENTAGE QUESTION: The price of an article is increased by 20%. By what percentage should the new price be reduced to get the original price? 🟢 SOLUTION: Assume the original price is:
100 100
After a 20% increase:
100+20=120 100+20=120
Required decrease:
120−100=20 120-100=20
Percentage decrease from the new price:
20120×100 \frac{20}{120}\times100
=1623% =16\frac{2}{3}\%
✅ ANSWER:
1623% decrease 16\frac{2}{3}\%\text{ decrease}
🟠 EXAMPLE 30: SUCCESSIVE PERCENTAGE CHANGE QUESTION: The population of a town increases by 10% in the first year and 20% in the second year. If the initial population is 50,000, find the population after two years. 🟢 SOLUTION: After the first year:
50000×110100 50000\times\frac{110}{100}
=55000 =55000
After the second year:
55000×120100 55000\times\frac{120}{100}
=66000 =66000
Overall increase:
66000−50000=16000 66000-50000=16000
Percentage increase:
1600050000×100 \frac{16000}{50000}\times100
=32% =32\%
✅ ANSWER:
66000 66000
32% increase 32\%\text{ increase}