Theory & Concepts

SSC CGL Percentage Questions, Tricks & Formula PDF 2026

Prepare Percentage for SSC CGL 2026 with formulas, short tricks, solved examples, practice questions, PYQs, and free PDF notes for faster exam preparation.

SSC CGL Percentage PDF Notes 2026

20 min readDifficulty: Beginner to Intermediate

Percentage is the absolute foundation of arithmetic in SSC CGL. Whether you're solving Profit & Loss, SI & CI, or Data Interpretation, a strong grip on percentage shortcuts is non-negotiable.

Recommended Sequence

Learning Path

5 Essential Steps

Fraction-to-Percentage conversion

Net change (AB Formula)

Population & Increase logic

10 standard solved problems

Exam-day speed tricks

1. Fundamental formulas

Fundamental Formula

\text{Percentage} = \left( \frac{\text{Part}}{\text{Whole}} \right) \times 100

Quick Conversion Table (Fraction to Percentage):

18 Formulas

\frac{1}{1} = 100\%

\frac{1}{2} = 50\%

\frac{1}{3} = 33.33\%

\frac{1}{4} = 25\%

\frac{1}{5} = 20\%

\frac{1}{6} = 16.66\%

\frac{1}{7} = 14.28\%

\frac{1}{8} = 12.5\%

\frac{1}{9} = 11.11\%

\frac{1}{10} = 10\%

\frac{1}{11} = 9.09\%

\frac{1}{12} = 8.33\%

\frac{1}{13} = 7.69\%

\frac{1}{14} = 7.14\%

\frac{1}{15} = 6.66\%

\frac{1}{16} = 6.25\%

\frac{1}{20} = 5\%

\frac{1}{25} = 4\%

Concept 02: AB Formula

Successive Percentage Change (AB Formula)

When a quantity is changed successively by two percentage values \(A\%\) and \(B\%\), the effective overall percentage change is given by the master AB Formula:

Master Formula

\text{Net Change} = A + B + \frac{A \times B}{100}

Case 1: Two Increases

Both \(A\) and \(B\) are positive. Example: \(+10\%\) and \(+20\%\) gives \(10 + 20 + 2 = +32\%\).

Case 2: Increase & Decrease

Substitute \(-B\) for decrease. Example: \(+20\%\) and \(-10\%\) gives \(20 - 10 - 2 = +8\%\).

Case 3: Successive Discounts

Single Equivalent Discount = \(d_1 + d_2 - \frac{d_1 \times d_2}{100}\).

SSC CGL Geometry Shortcut (Area & Volume):If the radius of a circle or side of a square increases by \(r\%\), the 2D Area increases by \(\left( 2r + \frac{r^2}{100} \right)\).

Concept 03: Reciprocal Law

Price, Consumption & Expenditure Balance Rule

The fundamental relation governing market commodities is:\(\text{Expenditure} = \text{Price} \times \text{Consumption}\)

Price Increase by \(r\%\)

To keep Expenditure constant, Consumption must decrease by:

\text{\% Decrease} = \left( \frac{r}{100 + r} \right) \times 100

Price Decrease by \(r\%\)

To keep Expenditure constant, Consumption can increase by:

\text{\% Increase} = \left( \frac{r}{100 - r} \right) \times 100

Quick Exam Memory Shortcuts:

Price +20% so Cons -16.66%

Price +25% so Cons -20.00%

Price -20% so Cons +25.00%

Price -25% so Cons +33.33%

2. 20 Solved Examples

Question 01Standard Pattern

Rahul spends 50% of his monthly income on household items, 20% on buying clothes, 5% on medicines, and saves the remaining rs. 11,250. What is Rahul’s monthly income?

Ars. 38,200

Brs. 34,000

Crs. 41,600

Drs. 45,000

Correct

Solution

Ans: a) rs. 45,000

Let total units = \( 100 \).

Total expenditure = \( 50 + 20 + 5 = 75 \text{ units} \).

Remaining savings = \( 100 - 75 = 25 \text{ units} \).

Given: \( 25 \text{ units} = rs. 11,250 \).

1 unit = \( 11250 / 25 = 450 \).

Total income = \( 100 \times 450 = rs. 45,000 \).

Question 02Standard Pattern

A student scored 32% marks in science subjects out of 300. How much should he score in language papers out of 200 if he is to get overall 46% marks?

A72%

B67%

Correct

C66%

D60%

Solution

Ans: a) 67%

Total marks possible = \( 300 + 200 = 500 \).

Target score = \( 46\% \text{ of } 500 = 230 \).

Already scored in Science = \( 32\% \text{ of } 300 = 96 \).

Remaining marks needed = \( 230 - 96 = 134 \).

Required percentage in language = \( \frac{134}{200} \times 100 = 67\% \).

Question 03Standard Pattern

The ratio of the number of boys to girls in a school is 4 : 1. If 75% of boys and 70% of girls are scholarship holders, find the percent who do NOT get scholarship.

A50%

B28%

C25%

D26%

Correct

Solution

Ans: a) 26%

Assume total students = \( 500 \) (Boys = \( 400 \), Girls = \( 100 \)).

Non-scholarship boys = \( 25\% \text{ of } 400 = 100 \).

Non-scholarship girls = \( 30\% \text{ of } 100 = 30 \).

Total non-scholarship students = \( 100 + 30 = 130 \).

Required percentage = \( \frac{130}{500} \times 100 = 26\% \).

Question 04Standard Pattern

A number is increased by 20% and then again by 20%. By what percent should the result be reduced to get back the original number?

A30(5/9)%

Correct

B44%

C33.33%

D25%

Solution

Ans: a) 30(5/9)%

Let original number = \( 100 \).

Step 1: Increase by 20% \( \implies 120 \).

Step 2: Increase 120 by 20% \( \implies 120 + 24 = 144 \).

Difference to get back to original = \( 144 - 100 = 44 \).

Required reduction percentage = \( \frac{44}{144} \times 100 = 30\frac{5}{9}\% \).

Question 05Standard Pattern

Two students appeared for an examination. One secured 24 marks more than the other and his marks were 65% of the sum of their marks. What are their marks?

A52 & 28

Correct

B78 & 54

C48 & 24

D60 & 36

Solution

Ans: a) 52 & 28

Let higher marks = \( x \) and lower marks = \( x - 24 \).

Sum of their marks = \( x + x - 24 = 2x - 24 \).

Given equation: \( x = 0.65 \times (2x - 24) \).

\( x = 1.3x - 15.6 \implies 0.3x = 15.6 \implies x = 52 \).

Their marks are 52 and 28.

Question 06Standard Pattern

What is 15% of 34% of 2000?

A102

Correct

B110

C98

D105

Solution

Ans: a) 102

Calculation: \( \frac{15}{100} \times \frac{34}{100} \times 2000 \).

\( 0.15 \times 34 \times 20 = 102 \).

Question 07Standard Pattern

In an election, a candidate got 40% of total votes and lost to his rival by 280 votes. Find total votes polled.

A1400

Correct

B1200

C1500

D1000

Solution

Ans: a) 1400

Losing candidate = 40%, Winning candidate = 60%.

Difference = \( 60\% - 40\% = 20\% \).

Given: \( 20\% \text{ of Total} = 280 \implies \text{Total} = 280 \times 5 = 1400 \).

Question 08Standard Pattern

Price of sugar is increased by 20%. By what percent should a householder reduce consumption so as not to increase expenditure?

A16(2/3)%

Correct

B20%

C25%

D15%

Solution

Ans: a) 16(2/3)%

Reciprocal Formula: \( \% \text{ Decrease} = \left( \frac{r}{100+r} \right) \times 100 \).

\( \left( \frac{20}{120} \right) \times 100 = \frac{1}{6} \times 100 = 16\frac{2}{3}\% \).

Question 09Standard Pattern

A number is first increased by 10% and then decreased by 10%. What is the net change in the number?

A1% decrease

Correct

BNo change

C1% increase

D2% decrease

Solution

Ans: a) 1% decrease

AB Formula: \( A = +10, B = -10 \).

\( \text{Net Change} = 10 - 10 + \frac{10 \times (-10)}{100} = -1\% \).

The number decreases by 1%.

Question 10Standard Pattern

If 15% of A = 20% of B, then A : B is equal to:

A4 : 3

Correct

B3 : 4

C5 : 4

D2 : 1

Solution

Ans: a) 4 : 3

\( 0.15 A = 0.20 B \implies \frac{A}{B} = \frac{0.20}{0.15} = \frac{4}{3} \).

Ratio \( A : B = 4 : 3 \).

Question 11Standard Pattern

In an examination, 65% of the candidates passed in Mathematics, 55% passed in English, and 20% failed in both subjects. If 280 candidates passed in both subjects, find the total number of candidates.

A700

Correct

B800

C600

D750

Solution

Ans: a) 700

Let total candidates = \( 100\% \).

Percentage passing in at least one subject = \( 100\% - 20\% = 80\% \).

By Venn diagram formula: \( P(M \cup E) = P(M) + P(E) - P(M \cap E) \).

\( 80\% = 65\% + 55\% - P(M \cap E) \implies P(M \cap E) = 120\% - 80\% = 40\% \).

Given: \( 40\% = 280 \implies 1\% = 7 \).

Total candidates = \( 100 \times 7 = 700 \).

Question 12Standard Pattern

The ratio of expenditure and savings of a person is 5 : 3. If his income increases by 20% and expenditure increases by 12%, then by what percent do his savings increase?

A33.33%

Correct

B30%

C28%

D32%

Solution

Ans: a) 33.33%

Let Expenditure = \( 50 \) and Savings = \( 30 \).

Initial Income = \( 50 + 30 = 80 \).

New Income = \( 80 + 20\% \text{ of } 80 = 80 + 16 = 96 \).

New Expenditure = \( 50 + 12\% \text{ of } 50 = 50 + 6 = 56 \).

New Savings = \( 96 - 56 = 40 \).

Increase in Savings = \( 40 - 30 = 10 \).

Percentage increase = \( \frac{10}{30} \times 100 = 33.33\% \).

Question 13Standard Pattern

In a survey of 1000 families, 60% buy Brand A, 40% buy Brand B, and 20% buy both Brands. Find the percentage of families who buy NEITHER Brand A nor Brand B.

A20%

Correct

B30%

C15%

D25%

Solution

Ans: a) 20%

Families buying only Brand A = \( 60\% - 20\% = 40\% \).

Families buying only Brand B = \( 40\% - 20\% = 20\% \).

Families buying both = \( 20\% \).

Total families buying at least one brand = \( 40\% + 20\% + 20\% = 80\% \).

Percentage buying neither = \( 100\% - 80\% = 20\% \).

Question 14Standard Pattern

A shopkeeper marks an item 40% above the cost price and allows two successive discounts of 10% and 15%. What is his net profit or loss percentage?

A7.1% profit

Correct

B5.2% profit

C6.8% loss

D8.5% profit

Solution

Ans: a) 7.1% profit

Let Cost Price (CP) = \( 100 \).

Marked Price (MP) = \( 100 + 40\% \text{ of } 100 = 140 \).

After 1st Discount of 10%: Price = \( 140 \times 0.90 = 126 \).

After 2nd Discount of 15%: Selling Price (SP) = \( 126 \times 0.85 = 107.1 \).

Net Profit = \( 107.1 - 100 = 7.1 \).

Profit Percentage = \( 7.1\% \).

Question 15Standard Pattern

In an election between two candidates, 10% of voters did not cast their vote, and 10% of votes polled were found invalid. The winning candidate got 54% of valid votes and won by 1620 votes. Find total enrolled voters.

A25,000

Correct

B22,500

C20,000

D30,000

Solution

Ans: a) 25,000

Let total enrolled voters = \( 100x \).

Votes polled = \( 90x \) (10% did not vote).

Valid votes = \( 90x \times 0.90 = 81x \) (10% invalid).

Winner got \( 54\% \) of valid votes, Loser got \( 46\% \).

Difference between Winner and Loser = \( 54\% - 46\% = 8\% \) of valid votes.

Winning margin = \( 8\% \text{ of } 81x = 6.48x \).

Given: \( 6.48x = 1620 \implies x = 250 \).

Total enrolled voters = \( 100 \times 250 = 25,000 \).

Question 16Standard Pattern

Fresh fruit contains 68% water and dry fruit contains 20% water. How many kg of dry fruit can be obtained from 100 kg of fresh fruit?

A40 kg

Correct

B32 kg

C50 kg

D45 kg

Solution

Ans: a) 40 kg

Pulp content in fruit remains constant during drying.

In Fresh Fruit: Water = 68%, so Pulp = \( 100\% - 68\% = 32\% \).

Pulp weight in 100 kg fresh fruit = \( 32\% \text{ of } 100 = 32 \text{ kg} \).

In Dry Fruit: Water = 20%, so Pulp = \( 100\% - 20\% = 80\% \).

Let quantity of dry fruit = \( x \text{ kg} \).

Pulp in dry fruit = \( 80\% \text{ of } x = 32 \text{ kg} \).

\( 0.80x = 32 \implies x = 40 \text{ kg} \).

Question 17Standard Pattern

If the rate of income tax is increased by 19%, the net income is reduced by 1%. Find the original rate of income tax.

A5%

Correct

B4%

C6%

D4.5%

Solution

Ans: a) 5%

Increase in Income Tax = Decrease in Net Income.

\( 19\% \text{ of Tax} = 1\% \text{ of Net Income} \).

\( \frac{\text{Tax}}{\text{Net Income}} = \frac{1}{19} \).

Total Gross Income = \( \text{Net Income} + \text{Tax} = 19 + 1 = 20 \).

Original Tax Rate = \( \frac{\text{Tax}}{\text{Gross Income}} \times 100 = \frac{1}{20} \times 100 = 5\% \).

Question 18Standard Pattern

Due to a 25% reduction in the price of rice, a customer can buy 4 kg more rice for ₹600. What is the reduced price per kg of rice?

A₹37.50

Correct

B₹30.00

C₹40.00

D₹35.00

Solution

Ans: a) ₹37.50

Money saved due to 25% price drop = \( 25\% \text{ of } 600 = \text{₹}150 \).

With this saved ₹150, customer buys 4 kg extra rice.

Reduced price per kg = \( \frac{150}{4} = \text{₹}37.50 \).

Question 19Standard Pattern

The population of a town increases by 10% in the first year and decreases by 10% in the second year. If the current population is 99,000, what was the population 2 years ago?

A1,00,000

Correct

B1,05,000

C98,000

D1,10,000

Solution

Ans: a) 1,00,000

Let population 2 years ago = \( P \).

After Year 1 (+10%): Population = \( P \times 1.10 \).

After Year 2 (-10%): Population = \( P \times 1.10 \times 0.90 = 0.99P \).

Given: \( 0.99P = 99,000 \implies P = 1,00,000 \).

Question 20Standard Pattern

A 40-liter mixture of milk and water contains 10% water. How much water must be added to make the water content 20% in the new mixture?

A5 liters

Correct

B4 liters

C6 liters

D7.5 liters

Solution

Ans: a) 5 liters

Initial mixture = 40 liters.

Water in initial mixture = \( 10\% \text{ of } 40 = 4 \text{ liters} \).

Milk in initial mixture = \( 40 - 4 = 36 \text{ liters} \) (Milk volume remains constant).

In new mixture: Water = 20%, so Milk = 80%.

Let new total mixture volume = \( X \text{ liters} \).

\( 80\% \text{ of } X = 36 \text{ liters} \implies 0.80X = 36 \implies X = 45 \text{ liters} \).

Water to be added = \( 45 - 40 = 5 \text{ liters} \).