Percentages

Percentages are the foundation of quantitative aptitude. Nearly every profit/loss, interest, and ratio problem uses percentage concepts.

Core Concepts

What is a Percentage?

A percentage is a fraction with denominator 100. The symbol % means “per hundred.”

25% = 25/100 = 1/4 = 0.25

Conversions Between Formats

Memorize these. They appear constantly in exams.

Percentage to Fraction Shortcut

To convert a percentage to a fraction, divide by 100 and simplify:

37.5% = 37.5/100 = 375/1000 = 3/8

Fraction to Percentage Shortcut

Multiply the fraction by 100:

3/8 × 100 = 37.5%

Key Formulas

Percentage Change

Percentage Change = (New - Old) / Old × 100

Example: Price increases from ₹200 to ₹250.

Change = (250 - 200) / 200 × 100 = 50/200 × 100 = 25% increase

Percentage of a Number

x% of y = (x/100) × y

Example: 15% of 240 = (15/100) × 240 = 36

Finding the Whole from a Percentage

If x% of a number is N, then the number = N × (100/x)

Example: 20% of what number is 60?

Number = 60 × (100/20) = 300

Successive Percentages

When two successive changes of a% and b% are applied, the net effect is:

Net % change = a + b + (a × b)/100

Note: Use negative values for decreases.

Example 1: Two Successive Increases

A price increases by 20% then by 10%.

Net change = 20 + 10 + (20 × 10)/100 = 20 + 10 + 2 = 32%

Example 2: Increase then Decrease

A value increases by 20% then decreases by 20%.

Net change = 20 + (-20) + (20 × (-20))/100 = 0 - 4 = -4%

Key insight: A% increase followed by A% decrease always results in a net decrease of A²/100 %.

Example 3: Two Successive Decreases

A price decreases by 10% then by 20%.

Net change = -10 + (-20) + ((-10) × (-20))/100 = -30 + 2 = -28%

Percentage Change in Product (Two Variables)

When two quantities multiply to give a product, and one changes by a% and the other by b%:

% change in product = a + b + (ab)/100

This is the same formula as successive percentages!

Example: If length increases by 20% and width decreases by 10%, area changes by:

20 + (-10) + (20 × (-10))/100 = 10 - 2 = 8% increase

Population/Depreciation Problems

Population Growth

P_future = P_present × (1 + r/100)^n

Where r = rate of growth, n = number of years.

Depreciation

Value_after = Value_initial × (1 - r/100)^n

Example: A car worth ₹5,00,000 depreciates at 20% per year. Value after 2 years?

= 5,00,000 × (1 - 20/100)²
= 5,00,000 × (0.8)²
= 5,00,000 × 0.64
= ₹3,20,000

Tricks & Shortcuts

Trick 1: x% of y = y% of x

This is extremely useful for mental math.

8% of 50 = 50% of 8 = 4

Trick 2: Finding 10%, 1%, 5%

• 10% of a number = move decimal one place left
• 1% of a number = move decimal two places left
• 5% = half of 10%

Example: 10% of 847 = 84.7, 1% of 847 = 8.47, 5% of 847 = 42.35

Trick 3: Percentage Increase/Decrease Table

Trick 4: Quick Percentage Breakdown

To find 17.5% of something:

17.5% = 10% + 5% + 2.5%
= 10% + 5% + half of 5%

Trick 5: If A is x% more than B, then B is less than A by:

B is less than A by = [x / (100 + x)] × 100 %

Example: If A is 25% more than B, then B is less than A by:

= [25 / (100 + 25)] × 100
= (25/125) × 100
= 20%

Trick 6: If A is x% less than B, then B is more than A by:

B is more than A by = [x / (100 - x)] × 100 %

Example: If A is 20% less than B, then B is more than A by:

= [20 / (100 - 20)] × 100
= (20/80) × 100
= 25%

Common Exam Patterns

Pattern 1: “What percentage is A of B?”

(A/B) × 100

Pattern 2: “A is what percent less/more than B?”

[(B - A)/B] × 100  → A is less than B by this %
[(A - B)/B] × 100  → A is more than B by this %

Pattern 3: Expenditure problems

Expenditure = Price × Consumption

If price increases by P% and we want expenditure unchanged:

Reduction in consumption = [P/(100+P)] × 100 %

Practice Questions

Q1: Successive Discounts

Two successive discounts of 20% and 10% are given. Find the equivalent single discount.

Solution: Using successive formula with a = -20, b = -10:

Net = -20 + (-10) + ((-20)(-10))/100 = -30 + 2 = -28%

Equivalent single discount = 28%

Q2: Percentage Error

A student multiplied a number by 3/5 instead of 5/3. What is the percentage error?

Solution:

Error = 5/3 - 3/5 = (25 - 9)/15 = 16/15
% Error = (16/15) / (5/3) × 100 = (16/15) × (3/5) × 100 = 48/75 × 100 = 64%

Q3: Election Problem

In an election, candidate A got 60% of votes and won by 4000 votes. Find total votes.

Solution:

A got 60%, B got 40%
Difference = 60% - 40% = 20% of total = 4000
Total = 4000 × 100/20 = 20,000 votes

Q4: Price and Consumption

If the price of sugar increases by 25%, by what percent should consumption decrease to keep expenditure the same?

Solution:

Decrease = [25/(100+25)] × 100 = (25/125) × 100 = 20%

Q5: Depreciation

A machine worth ₹80,000 depreciates by 10% annually. What is its value after 3 years?

Solution:

Value = 80,000 × (0.9)³ = 80,000 × 0.729 = ₹58,320

Q6: Population

A town’s population is 50,000. It increases by 10% in year 1, 20% in year 2, and decreases by 5% in year 3. Find the final population.

Solution:

Final = 50,000 × 1.1 × 1.2 × 0.95
= 50,000 × 1.254
= 62,700

Q7: Mixed Problem

A number is increased by 20% and then decreased by 20%. What is the net change?

Solution:

Net = 20 + (-20) + (20 × (-20))/100 = -4%
Net change = 4% decrease

Q8: If 30% of A = 40% of B, then A:B = ?

Solution:

30A/100 = 40B/100
30A = 40B
A/B = 40/30 = 4/3
A:B = 4:3

Summary Table