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Logical Reasoning · Logical Connectives, Syllogisms and Venn Diagrams

Logical Connectives - Concepts, 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)\%
If the result is positive, there is a net increase. If the result is negative, there is a net decrease. 🟢 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)\%
If the result is positive, there is a net increase. If the result is negative, there is a net decrease. 🟢 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

🔵 PERCENTAGE — SOLVED EXAMPLES 🟢 EXAMPLE 1: FINDING PERCENTAGE QUESTION: A student scored 360 marks out of 500. Find the percentage scored. SOLUTION:
Percentage=Marks ObtainedMaximum Marks×100 \text{Percentage} = \frac{\text{Marks Obtained}}{\text{Maximum Marks}}\times100
=360500×100 = \frac{360}{500}\times100
=72% =72\%
🔵 ANSWER:
72% 72\%
🟢 EXAMPLE 2: FINDING THE PART QUESTION: Find 25% of 640. SOLUTION:
Part=Percentage×Whole100 \text{Part} = \frac{\text{Percentage}\times\text{Whole}}{100}
=25×640100 = \frac{25\times640}{100}
=160 =160
🔵 ANSWER:
160 160
🟢 EXAMPLE 3: FRACTION TO PERCENTAGE QUESTION: Convert \frac{3}{8} into a percentage. SOLUTION:
Percentage=38×100 \text{Percentage} = \frac{3}{8}\times100
=37.5% =37.5\%
🔵 ANSWER:
37.5% 37.5\%
🟢 EXAMPLE 4: PERCENTAGE TO FRACTION QUESTION: Convert 45% into a fraction. SOLUTION:
Fraction=45100 \text{Fraction} = \frac{45}{100}
Simplifying:
=920 =\frac{9}{20}
🔵 ANSWER:
920 \frac{9}{20}
🟢 EXAMPLE 5: PERCENTAGE TO DECIMAL QUESTION: Convert 72% into a decimal. SOLUTION:
Decimal=72100 \text{Decimal} = \frac{72}{100}
=0.72 =0.72
🔵 ANSWER:
0.72 0.72
🟢 EXAMPLE 6: FINDING A PERCENTAGE OF A NUMBER QUESTION: Find 18% of 750. SOLUTION:
18% of 750=18100×750 18\%\text{ of }750 = \frac{18}{100}\times750
=135 =135
🔵 ANSWER:
135 135
🟢 EXAMPLE 7: PERCENTAGE INCREASE QUESTION: The price of a book increases from Rs. 400 to Rs. 500. Find the percentage increase. SOLUTION:
Increase=500−400 \text{Increase} = 500-400
=100 =100
Percentage Increase=100400×100 \text{Percentage Increase} = \frac{100}{400}\times100
=25% =25\%
🔵 ANSWER:
25% 25\%
🟢 EXAMPLE 8: PERCENTAGE DECREASE QUESTION: The price of a bag decreases from Rs. 800 to Rs. 680. Find the percentage decrease. SOLUTION:
Decrease=800−680 \text{Decrease} = 800-680
=120 =120
Percentage Decrease=120800×100 \text{Percentage Decrease} = \frac{120}{800}\times100
=15% =15\%
🔵 ANSWER:
15% 15\%
🟢 EXAMPLE 9: NEW VALUE AFTER INCREASE QUESTION: A salary of Rs. 25,000 is increased by 12%. Find the new salary. SOLUTION:
New Salary=25000(1+12100) \text{New Salary} = 25000\left(1+\frac{12}{100}\right)
=25000×112100 = 25000\times\frac{112}{100}
=28000 =28000
🔵 ANSWER:
Rs. 28,000 Rs.\ 28,000
🟢 EXAMPLE 10: NEW VALUE AFTER DECREASE QUESTION: A machine costs Rs. 50,000. Its price is reduced by 8%. Find the new price. SOLUTION:
New Price=50000(1−8100) \text{New Price} = 50000\left(1-\frac{8}{100}\right)
=50000×92100 = 50000\times\frac{92}{100}
=46000 =46000
🔵 ANSWER:
Rs. 46,000 Rs.\ 46,000
🟢 EXAMPLE 11: FINDING ORIGINAL VALUE AFTER INCREASE QUESTION: After a 20% increase, the price of an article is Rs. 1,440. Find its original price. SOLUTION:
Original Value=Final Value×100100+20 \text{Original Value} = \frac{\text{Final Value}\times100}{100+20}
=1440×100120 = \frac{1440\times100}{120}
=1200 =1200
🔵 ANSWER:
Rs. 1,200 Rs.\ 1,200
🟢 EXAMPLE 12: FINDING ORIGINAL VALUE AFTER DECREASE QUESTION: After a 15% decrease, the price of an article is Rs. 1,700. Find its original price. SOLUTION:
Original Value=Final Value×100100−15 \text{Original Value} = \frac{\text{Final Value}\times100}{100-15}
=1700×10085 = \frac{1700\times100}{85}
=2000 =2000
🔵 ANSWER:
Rs. 2,000 Rs.\ 2,000
🟢 EXAMPLE 13: PERCENTAGE COMPARISON QUESTION: 45 is what percentage of 180? SOLUTION:
Percentage=45180×100 \text{Percentage} = \frac{45}{180}\times100
=25% =25\%
🔵 ANSWER:
25% 25\%
🟢 EXAMPLE 14: SUCCESSIVE PERCENTAGE INCREASE QUESTION: A number is increased by 10% and then by 20%. Find the net percentage increase. SOLUTION:
Net Increase=(10+20+10×20100)% \text{Net Increase} = \left(10+20+\frac{10\times20}{100}\right)\%
=(30+2)% = (30+2)\%
=32% =32\%
🔵 ANSWER:
32% 32\%
🟢 EXAMPLE 15: SUCCESSIVE PERCENTAGE DECREASE QUESTION: The price of an article is decreased successively by 10% and 20%. Find the net percentage decrease. SOLUTION:
Net Decrease=(10+20−10×20100)% \text{Net Decrease} = \left(10+20-\frac{10\times20}{100}\right)\%
=(30−2)% = (30-2)\%
=28% =28\%
🔵 ANSWER:
28% 28\%
🟢 EXAMPLE 16: INCREASE FOLLOWED BY DECREASE QUESTION: A number is increased by 20% and then decreased by 10%. Find the net percentage change. SOLUTION:
Net Change=(20−10−20×10100)% \text{Net Change} = \left(20-10-\frac{20\times10}{100}\right)\%
=(20−10−2)% = (20-10-2)\%
=8% =8\%
🔵 ANSWER:
8% increase 8\%\text{ increase}
🟢 EXAMPLE 17: EQUAL PERCENTAGE INCREASE AND DECREASE QUESTION: The price of an article is increased by 20% and then decreased by 20%. Find the net percentage change. SOLUTION:
Net Decrease=x2100% \text{Net Decrease} = \frac{x^2}{100}\%
Here:
x=20 x=20
Therefore:
Net Decrease=202100% \text{Net Decrease} = \frac{20^2}{100}\%
=4% =4\%
🔵 ANSWER:
4% decrease 4\%\text{ decrease}
🟢 EXAMPLE 18: ONE NUMBER IS MORE THAN ANOTHER QUESTION: A is 25% more than B. If B = 800, find A. SOLUTION:
A=B(1+25100) A = B\left(1+\frac{25}{100}\right)
=800×125100 = 800\times\frac{125}{100}
=1000 =1000
🔵 ANSWER:
A=1000 A=1000
🟢 EXAMPLE 19: ONE NUMBER IS LESS THAN ANOTHER QUESTION: A is 20% less than B. If B = 750, find A. SOLUTION:
A=B(1−20100) A = B\left(1-\frac{20}{100}\right)
=750×80100 = 750\times\frac{80}{100}
=600 =600
🔵 ANSWER:
A=600 A=600
🟢 EXAMPLE 20: REVERSE PERCENTAGE QUESTION: A is 25% more than B. By what percentage is B less than A? SOLUTION:
Percentage=x100+x×100% \text{Percentage} = \frac{x}{100+x}\times100\%
Here:
x=25 x=25
Therefore:
=25125×100% = \frac{25}{125}\times100\%
=20% =20\%
🔵 ANSWER:
20% 20\%
🟢 EXAMPLE 21: PERCENTAGE AND MARKS QUESTION: A student obtains 420 marks out of 600. Find the percentage. SOLUTION:
Percentage=420600×100 \text{Percentage} = \frac{420}{600}\times100
=70% =70\%
🔵 ANSWER:
70% 70\%
🟢 EXAMPLE 22: FINDING MARKS FROM PERCENTAGE QUESTION: A student has to score 65% in an examination of 800 marks. How many marks must the student score? SOLUTION:
Marks Obtained=65×800100 \text{Marks Obtained} = \frac{65\times800}{100}
=520 =520
🔵 ANSWER:
520 marks 520\text{ marks}
🟢 EXAMPLE 23: PROFIT PERCENTAGE QUESTION: An article is bought for Rs. 800 and sold for Rs. 960. Find the profit percentage. SOLUTION:
Profit=SP−CP \text{Profit} = SP-CP
=960−800 =960-800
=160 =160
Profit %=160800×100 \text{Profit \%} = \frac{160}{800}\times100
=20% =20\%
🔵 ANSWER:
20% 20\%
🟢 EXAMPLE 24: LOSS PERCENTAGE QUESTION: An article is bought for Rs. 1,200 and sold for Rs. 1,020. Find the loss percentage. SOLUTION:
Loss=CP−SP \text{Loss} = CP-SP
=1200−1020 =1200-1020
=180 =180
Loss %=1801200×100 \text{Loss \%} = \frac{180}{1200}\times100
=15% =15\%
🔵 ANSWER:
15% 15\%
🟢 EXAMPLE 25: INCOME, EXPENDITURE AND SAVINGS QUESTION: A person's monthly income is Rs. 40,000 and expenditure is Rs. 32,000. Find the savings percentage. SOLUTION:
Savings=40000−32000 \text{Savings} = 40000-32000
=8000 =8000
Savings %=800040000×100 \text{Savings \%} = \frac{8000}{40000}\times100
=20% =20\%
🔵 ANSWER:
20% 20\%
🟢 EXAMPLE 26: POPULATION INCREASE QUESTION: The population of a town is 80,000. If it increases by 15%, find the new population. SOLUTION:
New Population=80000×100+15100 \text{New Population} = 80000\times\frac{100+15}{100}
=80000×115100 = 80000\times\frac{115}{100}
=92000 =92000
🔵 ANSWER:
92,000 92,000
🟢 EXAMPLE 27: POPULATION DECREASE QUESTION: The population of a village is 50,000. If it decreases by 12%, find the new population. SOLUTION:
New Population=50000×100−12100 \text{New Population} = 50000\times\frac{100-12}{100}
=50000×88100 = 50000\times\frac{88}{100}
=44000 =44000
🔵 ANSWER:
44,000 44,000
🟢 EXAMPLE 28: PERCENTAGE SHORTCUT QUESTION: Find 25% of 480 using a shortcut. SOLUTION:
25% of N=N4 25\%\text{ of }N = \frac{N}{4}
Therefore:
25% of 480=4804 25\%\text{ of }480 = \frac{480}{4}
=120 =120
🔵 ANSWER:
120 120
🟢 EXAMPLE 29: PERCENTAGE SHORTCUT QUESTION: Find 75% of 640 using a shortcut. SOLUTION:
75% of N=3N4 75\%\text{ of }N = \frac{3N}{4}
Therefore:
75% of 640=3×6404 75\%\text{ of }640 = \frac{3\times640}{4}
=480 =480
🔵 ANSWER:
480 480
🟢 EXAMPLE 30: MIXED PERCENTAGE PROBLEM QUESTION: A number is increased by 25% and the resulting number is 500. Find the original number. SOLUTION:
Original Value=Final Value×100100+25 \text{Original Value} = \frac{\text{Final Value}\times100}{100+25}
=500×100125 = \frac{500\times100}{125}
=400 =400
Verification:
400×125100=500 400\times\frac{125}{100}=500
🔵 ANSWER:
400 400