Percentage for SSC CGL — Concept, Shortcuts, and Solved PYQs

beginner 18 min read

Concept

Percentage is one of those topics where you already know the mechanics — what trips SSC CGL candidates is not the formula, it is recognizing which version of the formula applies in which disguise.

The word "percent" means "per hundred." When you say 40%, you mean 40 parts out of every 100. That is it. Every percentage problem is just a fraction with 100 in the denominator, written in a more readable way.

Here is the analogy that locks it in: think of 100 as your reference village of exactly 100 people. Any percentage question is asking you how many of those 100 people satisfy some condition. If 25% have phones, 25 people out of 100 have phones. If you add 20 more people to the village, you have to re-calculate the fraction against the new total — that is where percentage change questions come from.

The three core operations you need:

1. Converting between forms

2. Finding a percentage of a number

P% of X = (P/100) × X

This is the bedrock. Everything else builds on it.

3. Finding what percentage one number is of another

(Part / Whole) × 100 = Percentage

In CGL, these three operations get combined, chained, or disguised inside word problems about salaries, populations, prices, and mixtures. Your job is to strip away the story and identify which operation is being chained.

One important mindset shift: always ask "percentage of what?" before writing anything. The base (denominator) changes the entire answer. 20% of 500 and 20% of 600 give completely different numbers. In successive percentage change problems, the base keeps shifting — that is the trap.

Deep Dive

The Percentage Change Formula

This is the most tested single formula in SSC CGL percentage problems:

Percentage Change = [(New Value - Original Value) / Original Value] × 100

Positive result = increase. Negative result = decrease. Never put the new value in the denominator — it is always the original.

Successive Percentage Changes

When a value is changed by a% first and then by b%, the net effect is NOT (a + b)%. The correct net percentage change is:

Net Change = a + b + (ab/100)%

Where a and b carry their signs (positive for increase, negative for decrease).

Example: 20% increase followed by 15% decrease.

Net = 20 + (-15) + [20 × (-15)/100] = 20 - 15 - 3 = 2% increase.

Standard method using 100 as base: 100 → 120 → 102. Net = 2% increase. Same answer, but the formula is faster once you internalize the sign rule.

The Consumption-Expenditure Inverse Formula

Look — this is a dedicated exam trap. When price increases by r% and you want expenditure to stay constant, the reduction in consumption required is:

Reduction = [r / (100 + r)] × 100%

When price decreases by r% and you want expenditure to stay constant, the increase in consumption required is:

Increase = [r / (100 - r)] × 100%

Do not derive this in the exam. Memorize the template.

Example: Price rises 25%. Consumption reduction = 25/125 × 100 = 20%.

Percentage in Mixtures

The standard mixture question gives you a solution of X% concentration and asks how much pure substance to add to reach Y% concentration.

Setup: Let the mixture volume be V and concentration be c₁%. You add x ml of pure substance (100% concentration) to reach c₂%.

(c₁/100 × V + x) / (V + x) = c₂/100

Solve for x. Always set up this equation from first principles — do not try to memorize a sub-formula, because the question can vary (adding water instead of pure substance, etc.).

Salary / Population Comparisons

"A is 25% more than B" means A = B × (1 + 25/100) = 1.25B.

"A is 25% less than B" means A = B × (1 - 25/100) = 0.75B.

Here is where candidates make the classic reversal error: "If A is 25% more than B, by what percentage is B less than A?"

B is less than A by: (A - B)/A × 100 = (0.25B/1.25B) × 100 = 20%

Not 25%. The base has shifted from B to A.

Successive Multipliers for Population / Growth

For a quantity growing at r% per year, after n years:

Final = Initial × (1 + r/100)ⁿ

For two years with the same rate, expand the square rather than computing step-by-step:

Final = Initial × (1 + r/100)²

For small r, the expansion gives: Initial × [1 + 2r/100 + r²/10000]

In the exam, two-year growth is almost always faster to compute step-by-step: Year 1 result × multiplier again. Three or more years, use the compound formula.

Markup and Discount (Profit-and-Loss Bridge)

Selling Price = Marked Price × (1 - Discount%/100)

Profit% = (Selling Price - Cost Price)/Cost Price × 100

When a question gives you markup and discount together, chain the multipliers:

SP = CP × (1 + Markup%/100) × (1 - Discount%/100)

Plug CP = 100, compute SP, and the difference is the profit or loss percent directly.

Memory Tricks & Shortcuts

patternThe Fraction-to-Percent Table

Memorize these cold: 1/8 = 12.5%, 1/6 = 16.67%, 1/5 = 20%, 1/4 = 25%, 1/3 = 33.33%, 3/8 = 37.5%, 1/2 = 50%, 5/8 = 62.5%, 2/3 = 66.67%, 3/4 = 75%, 5/6 = 83.33%, 7/8 = 87.5%.

When a question says "40% of 250 = 20% of x", convert: 40% = 2/5, so (2/5 × 250) = (1/5 × x). Reading fractions instead of decimals takes 3 steps vs 5 steps in the standard approach. Standard method: 45s. This: 15s.

patternThe Net-Change Formula for Successive Percentages

For two successive changes a% and b% (with signs): Net = a + b + ab/100.

Micro-example: 20% increase then 15% decrease. 20 + (-15) + (20×(-15)/100) = 5 - 3 = 2% increase. You skip the intermediate calculation entirely. Standard method (build 100 → 120 → 102): 4 steps. Formula: 1 step. Speed gain: standard 30s vs shortcut 8s.

patternInverse Base Formula for Price-Consumption Problems

If price rises by r%, consumption drop needed = r/(100+r) × 100. If price falls by r%, consumption rise needed = r/(100−r) × 100.

Micro-example: Price rises 25% → drop = 25/125 × 100 = 20%. If you set this up from scratch with variables, it costs 6 algebraic steps. The template collapses it to one division. Standard: 40s. Template: 8s.

substitutionMarkup-Discount Chain Multiplier

Set CP = 100 always. Then SP = 100 × (1 + m/100) × (1 − d/100). The profit or loss percent is simply (SP − 100).

Micro-example: Markup 20%, discount 10%. SP = 100 × 1.2 × 0.9 = 108. Profit = 8%. No separate formula for profit%, no ratio setup. Standard method requires computing MP, then SP, then (SP−CP)/CP: 5 steps. Chain method: 2 steps. Time: 45s → 12s.

estimationTwo-Year Population — Step-by-Step Over Formula

For exactly two years, avoid expanding (1 + r/100)² algebraically. Just multiply twice: Year 1 = Initial × (1 + r/100), Year 2 = Year 1 × (1 + r/100). For 5% growth on 80,000: 80,000 × 1.05 = 84,000; 84,000 × 1.05 = 88,200. Faster than squaring because the numbers stay clean. Formula expansion: 4 arithmetic operations. Step-by-step: 2 multiplications of similar difficulty. Speed: roughly equal, but step-by-step has lower error rate.

Fast-Solving Framework

Read the question and locate the base first — every percentage is "of" something. Write it down explicitly before touching numbers.

Decision tree:

  1. Single percentage of a number? → Direct: (P/100) × X. Use fraction equivalents from memory if P is 12.5, 16.67, 33.33, 66.67.

  2. Comparison ("A is X% more/less than B")? → Write A = B × (1 ± X/100). If the question then asks "B is what % of A", the base has shifted — recompute from B/A × 100.

  3. Two successive percentage changes? → Use net-change formula: a + b + ab/100 with signs. Verify sign of ab term (both positive = positive, one negative = negative).

  4. Price-consumption / salary-savings inverse problem? → Use r/(100 ± r) × 100 template. Sign: +r in denominator if original variable increased, -r if decreased.

  5. Mixture / concentration problem? → Set up the single equation: (existing amount + added) / (existing volume + added) = target %. Solve for the unknown.

  6. Markup + discount? → Set CP = 100, chain multipliers, read profit/loss off SP directly.

If you cannot slot the question into one of these six types within 10 seconds, take CP = 100 or the given base value as your starting number and work forwards numerically. Do not get stuck in algebra.

Solved PYQs

Why this question: Tests the most basic salary comparison — confirms you know which is the base.

Previous Year Questionपिछले वर्ष का प्रश्न
If A's salary is 25% more than B's salary and B's salary is ₹4800, what is A's salary?
यदि A की सैलरी, B की सैलरी से 25% ज़्यादा है और B की सैलरी ₹4800 है, तो A की सैलरी कितनी है?
  1. ₹5800
  2. ₹6000
  3. ₹6200
  4. ₹6400
  1. ₹5800
  2. ₹6000
  3. ₹6200
  4. ₹6400
Solutionसमाधान
A's salary = B's salary + 25% of B's salary = 4800 + (25/100) × 4800 = 4800 + 1200 = ₹6000.
A की सैलरी = B की सैलरी + B की सैलरी का 25% = 4800 + (25/100) × 4800 = 4800 + 1200 = ₹6000।

Solving path: B = ₹4800. A = 4800 + 25% of 4800 = 4800 + 1200 = ₹6000. The base is B's salary, not A's. One-line calculation.


Why this question: Mixture-concentration problems appear regularly and catch candidates who set up the wrong denominator.

Previous Year Questionपिछले वर्ष का प्रश्न
A mixture contains 20% alcohol. How much pure alcohol should be added to 500 ml of this mixture to make it 40% alcohol?
एक मिश्रण में 20% अल्कोहल है। इस मिश्रण के 500 ml में कितना शुद्ध अल्कोहल मिलाया जाए कि अल्कोहल की मात्रा 40% हो जाए?
  1. 100 ml
  2. 125 ml
  3. 150 ml
  4. 166.67 ml
  1. 100 ml
  2. 125 ml
  3. 150 ml
  4. 166.67 ml
Solutionसमाधान
Initial alcohol = 20% of 500 = 100 ml. Let x ml pure alcohol be added. Then (100 + x)/(500 + x) = 40/100. Solving: 100 + x = 0.4(500 + x), so x = 166.67 ml.
प्रारंभिक एल्कोहल = 500 का 20% = 100 ml। माना x ml शुद्ध एल्कोहल मिलाया जाता है। तब (100 + x)/(500 + x) = 40/100। हल करने पर: x = 166.67 ml।

Solving path: Existing alcohol = 20% of 500 = 100 ml. Let x ml of pure alcohol be added. Equation: (100 + x)/(500 + x) = 40/100. Cross-multiply: 250(100 + x) = 100(500 + x) — wait, simplify as 100 + x = 0.4(500 + x)100 + x = 200 + 0.4x0.6x = 100x = 166.67 ml. Confirm: (100 + 166.67)/(500 + 166.67) = 266.67/666.67 = 40%. Correct.


Why this question: The "P% of A = Q% of B" format is a staple in CGL and can be solved in one step with fraction recognition.

Previous Year Questionपिछले वर्ष का प्रश्न
If 40% of 250 is equal to 20% of x, find the value of x:
यदि 250 का 40%, x के 20% के बराबर है, तो x का मान ज्ञात कीजिए:
  1. 400
  2. 500
  3. 600
  4. 750
  1. 400
  2. 500
  3. 600
  4. 750
Solutionसमाधान
40% of 250 = (40/100) × 250 = 100. So 20% of x = 100, which gives x = 100 × 100/20 = 500.
250 का 40% = (40/100) × 250 = 100। अतः x का 20% = 100, जिससे x = 100 × 100/20 = 500।

Solving path: 40% of 250 = 100. So 20% of x = 100, meaning x/5 = 100, so x = 500. Fraction approach: 40% = 2/5, so 2/5 × 250 = 100. Then 20% = 1/5, so x/5 = 100, x = 500. Under 10 seconds.


Why this question: Two-year growth at a fixed rate — confirms correct compounding (not simple addition of 5% + 5%).

Previous Year Questionपिछले वर्ष का प्रश्न
The population of a city increases by 5% every year. If the current population is 80,000, what will be the population after 2 years?
एक शहर की आबादी हर साल 5% बढ़ती है। यदि अभी की आबादी 80,000 है, तो 2 साल बाद आबादी कितनी होगी?
  1. 88,200
  2. 88,000
  3. 90,000
  4. 92,000
  1. 88,200
  2. 88,000
  3. 90,000
  4. 92,000
Solutionसमाधान
After 1 year = 80,000 × 1.05 = 84,000. After 2 years = 84,000 × 1.05 = 88,200.
1 साल बाद = 80,000 × 1.05 = 84,000। 2 साल बाद = 84,000 × 1.05 = 88,200।

Solving path: Year 1: 80,000 × 1.05 = 84,000. Year 2: 84,000 × 1.05 = 88,200. The wrong answer (88,000) would result from simple-interest-style thinking: 80,000 + 2 × 4,000 = 88,000. Compounding adds the extra 200.


Why this question: Successive percentage change — the most frequently tested percentage sub-type in SSC CGL.

Previous Year Questionपिछले वर्ष का प्रश्न
A number is increased by 20% and then decreased by 15%. What is the net percentage change?
एक संख्या को पहले 20% बढ़ाया जाता है, फिर 15% घटाया जाता है। कुल प्रतिशत बदलाव क्या होगा?
  1. 2% increase
  2. 2% decrease
  3. 5% increase
  4. 5% decrease
  1. 2% की बढ़ोतरी
  2. 2% की कमी
  3. 5% की बढ़ोतरी
  4. 5% की कमी
Solutionसमाधान
Let original number be 100. After 20% increase = 120. After 15% decrease = 120 × 85/100 = 102. Net change = 102 - 100 = 2, so 2% increase.
माना मूल संख्या 100 है। 20% बढ़ने के बाद = 120। 15% कम होने के बाद = 120 × 85/100 = 102। शुद्ध परिवर्तन = 102 - 100 = 2, अतः 2% वृद्धि।

Solving path: Use net-change formula: a = +20, b = -15. Net = 20 + (-15) + (20 × (-15)/100) = 5 - 3 = 2% increase. Or numerically: 100 → 120 → 120 × 0.85 = 102. Net change = +2%.


Why this question: Classic inverse percentage — tests whether you recognize the base has changed from original price to new price.

Previous Year Questionपिछले वर्ष का प्रश्न
If the price of sugar increases by 25%, by what percentage should consumption be reduced to keep expenditure same?
यदि चीनी का दाम 25% बढ़ जाए, तो खर्च को उतना ही रखने के लिए खपत में कितने प्रतिशत की कमी करनी चाहिए?
  1. 20%
  2. 25%
  3. 30%
  4. 33.33%
  1. 20%
  2. 25%
  3. 30%
  4. 33.33%
Solutionसमाधान
If price increases by 25%, new price = 125% of original. To maintain same expenditure, consumption should be reduced by (25/125) × 100 = 20%.
यदि मूल्य 25% बढ़ता है, तो नया मूल्य = मूल का 125%। समान खर्च बनाए रखने के लिए, उपभोग (25/125) × 100 = 20% कम करना चाहिए।

Solving path: Price rises 25%, so new price = 125. Expenditure = Price × Consumption. For expenditure to stay at 100 (old price × old consumption = 100 × 1 = 100), new consumption = 100/125 = 0.8. Reduction = 20%. Formula: 25/(100+25) × 100 = 25/125 × 100 = 20%.


Why this question: Markup plus discount — the chain multiplier approach saves time and avoids a common step-skipping error.

Previous Year Questionपिछले वर्ष का प्रश्न
A shopkeeper marks up his goods by 20% and then gives a discount of 10%. What is his net profit percentage?
एक दुकानदार अपने सामान पर 20% मार्कअप करता है और फिर 10% की छूट देता है। उसका कुल लाभ प्रतिशत कितना है?
  1. 8%
  2. 10%
  3. 12%
  4. 15%
  1. 8%
  2. 10%
  3. 12%
  4. 15%
Solutionसमाधान
If cost price is 100, marked price = 120. After 10% discount, selling price = 120 × 90/100 = 108. Net profit = 108 - 100 = 8, so profit percentage = 8%.
यदि क्रय मूल्य 100 है, तो अंकित मूल्य = 120। 10% छूट के बाद, विक्रय मूल्य = 120 × 90/100 = 108। शुद्ध लाभ = 108 - 100 = 8, अतः लाभ प्रतिशत = 8%।

Solving path: Set CP = 100. Marked price = 100 × 1.20 = 120. After 10% discount: SP = 120 × 0.90 = 108. Profit = 108 − 100 = 8%. So net profit = 8%. Do not add or subtract the percentages directly (20% − 10% ≠ 10%; the base shifts).

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