Percentage for RRB NTPC — Concepts, Shortcuts & Solved PYQs

beginner 18 min read

Concept

Percentage literally means "per hundred" — it is just a way of expressing a fraction with denominator 100. When you say 40%, you mean 40/100 = 2/5. That is the entire concept.

Here is why it matters beyond the definition: percentage is the universal language of comparison. Whether it is salary hikes, price changes, vote shares, or marks — everything gets converted to "out of 100" so you can compare apples to apples. This is also why nearly every chapter in Quant (profit-loss, interest, data interpretation, mixtures) secretly runs on percentage at its core.

Think of it this way — imagine you have a 10-litre bucket (the base) and you pour 3 litres of colour into it. The fraction coloured is 3/10. But your brain is trained to process comparisons on a scale of 100, so you scale up: 3/10 × 100 = 30%. That is all a percentage conversion is — a scaling operation to base 100.

The moment you stop thinking of percentage as a formula and start thinking of it as a fraction-in-disguise, the exam questions become significantly easier. Every percentage problem reduces to one of three operations:

  1. Finding a percentage of a number — "What is 35% of 240?"
  2. Finding what percentage one number is of another — "320 is what percent of 500?"
  3. Finding the original value given a percentage — "If 40% of X is 80, find X."

The chain percentage problems (successive increases/decreases) and the comparison problems (A is x% of B type) are just combinations of these three. Once you own the base three, you own the chapter.


Deep Dive

The Core Formula

Percentage = (Part / Whole) × 100

Keep this one line and derive everything else from it.

Fraction-Percentage Equivalents You Must Know

Do not solve 1/8 × 100 in the exam hall. Memorise these and read percentage questions at visual speed:

| Fraction | Percentage | |----------|------------| | 1/2 | 50% | | 1/3 | 33.33% | | 1/4 | 25% | | 1/5 | 20% | | 1/6 | 16.67% | | 1/7 | 14.28% | | 1/8 | 12.5% | | 1/9 | 11.11% | | 1/10 | 10% | | 1/11 | 9.09% | | 1/12 | 8.33% |

Once you have these, multiples are trivial: 3/8 = 3 × 12.5% = 37.5%.

Percentage Change Formula

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

If the result is positive — increase. Negative — decrease. The denominator is always the original (old) value, not the new one. This is the single most common error in this topic.

Successive/Chain Percentage Change

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

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

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

Example — 20% increase then 15% decrease: Net = 20 + (-15) + (20 × -15)/100 = 5 - 3 = 2% increase

This matches the PYQ in this set. You can verify: start with 100, after +20% get 120, after -15% of 120 get 120 - 18 = 102. Net change = 2%.

The formula saves you from working with actual numbers. Once you are comfortable with it, chain problems become two-step algebra.

Example — 15% increase then 20% decrease (population PYQ): Net = 15 + (-20) + (15 × -20)/100 = -5 - 3 = -8% (decrease)

The "Reverse" or Compensating Percentage Problem

Classic setup: "Price increases by 25%, by how much should consumption fall to keep expenditure constant?"

Expenditure = Price × Consumption = constant.

If price becomes 125% = 5/4 of original, consumption must become 4/5 = 80% of original. Reduction = 20%.

The shortcut: if price increases by x/y (as a fraction over original), consumption must decrease by x/(x+y) of original in percentage terms. For 25% increase (1/4 of original), reduction = 1/(1+4) = 1/5 = 20%.

This eliminates the need to set up algebra for every such question.

The Set-Theory Percentage (Overlap Problems)

When you have: "85% passed English, 90% passed Math, 10% failed both — find % passed in both."

The key identity: Passed in at least one = 100% - Failed in both = 90%

Then: P(E ∩ M) = P(E) + P(M) - P(E ∪ M) = 85 + 90 - 90 = 85%

Look — if you keep forgetting this formula, just remember the Venn diagram: the total overlap region is what you solve for. Draw two circles, fill in what you know, and the algebra is automatic.

Comparing Two Quantities via Percentage (A is x% of B type)

30% of A = 0.25% of B(30/100)A = (0.25/100)BA/B = 0.25/30 = 1/120

The trap here is unit confusion between 30% and 0.25% — both have /100, so they cancel each other out cleanly in the ratio. Do not drop either /100.


Memory Tricks & Shortcuts

patternFraction Flash — Read Percentages Without Division

Convert the percentage to its fraction equivalent immediately upon reading the question. "25% of 320" becomes "1/4 of 320 = 80" — no multiplication needed. "37.5% of 480" becomes "3/8 of 480 = 180." Standard method: 37.5 × 480 / 100 = ~45 seconds with long multiplication. Fraction method: 480/8 = 60, 60 × 3 = 180, under 10 seconds. Works for all fractions in the 1/2 to 1/12 family and their multiples.

patternNet Change Formula — Crack Successive % in One Line

For any two successive percentage changes a and b, net change = a + b + ab/100. Use signed values (negative for decrease). Chain problem: 20% up then 15% down → 20 - 15 + (20×-15)/100 = 5 - 3 = 2% increase. Standard method (pick a base, apply both changes, compute final %): 4 steps, ~40 seconds. Formula method: 1 line, ~10 seconds. The formula also works for three changes applied in two passes.

substitutionPrice-Consumption Inverse — The 1/(1+n) Rule

When price rises by 1/n (i.e., by 100/n %), the required consumption reduction is always 1/(n+1) of original, i.e., 100/(n+1) %. Price up 25% = 1/4 rise → reduce consumption by 1/(4+1) = 1/5 = 20%. Price up 20% = 1/5 rise → reduce by 1/6 = 16.67%. Standard algebra setup: ~50 seconds. Pattern recognition: ~8 seconds after two practice repetitions.

eliminationSet Theory Shortcut — Anchor on the Failure Side

In double-subject pass/fail problems, always start from: "Passed in at least one = 100% - Failed in both." This gives you P(A ∪ B) directly. Then plug into P(A ∩ B) = P(A) + P(B) - P(A ∪ B). Starting from the union side eliminates the confusion of "does 10% failed in both mean 90% passed everything?" — it does not. 90% passed at least one. That single anchor prevents the most common error in this question type. Saves one full re-read of the question (~20 seconds).

substitutionWeighted Average Percentage — Use Actual Counts, Not Averages

"60% are boys, 20% of boys and 25% of girls play sports — what % of total?" Do NOT average 20% and 25%. Instead, assign a base of 100, split (60 boys, 40 girls), compute parts separately (12 + 10 = 22), and read off the answer directly (22%). The trap of averaging gives (20+25)/2 = 22.5% which is wrong. Working with a base of 100 students: 3 arithmetic operations, ~15 seconds.


Fast-Solving Framework

Read the question and classify it in the first 5 seconds using this decision tree:

Is there a "of" relationship? → Core formula (Part/Whole × 100 or reverse). Use fraction equivalents where possible.

Are two changes applied in sequence? → Successive % formula: a + b + ab/100. Assign signs, compute in one line.

Is expenditure / total kept constant while components change? → Set up Original = New. Use the 1/(n+1) rule if one component changes by a unit fraction.

Are two sets (subjects, criteria) being compared with an overlap? → Venn diagram anchor: Union = 100 - Both_Fail, then solve for intersection.

Is a ratio A:B required from percentage equalities? → Cross-multiply directly after converting percentages to fractions. Both sides carry /100 — they cancel cleanly.

Does the question compare A's value to C's through B? → Chain multiplication: if B = kC and A = mB, then A = mkC. Express as percentage of C in one step.

If none of the above fits in 8 seconds, pick a clean base (usually 100 or the LCM of the given denominators) and work through it numerically. Never start with algebra when substitution resolves it in under 30 seconds.


Solved PYQs

Why this question: Direct percentage calculation — the most basic form. Tests whether you know the formula direction (marks ÷ total, not total ÷ marks).

Previous Year Questionपिछले वर्ष का प्रश्न
A student scored 320 marks out of 500. What percentage of marks did he score?
एक छात्र ने 500 में से 320 अंक हासिल किए। उसने कितने प्रतिशत अंक प्राप्त किए?
  1. 64%
  2. 62%
  3. 66%
  4. 68%
  1. 64%
  2. 62%
  3. 66%
  4. 68%
Solutionसमाधान
Percentage = (Marks obtained/Total marks) × 100 = (320/500) × 100 = 0.64 × 100 = 64%.
प्रतिशत = (प्राप्त अंक/कुल अंक) × 100 = (320/500) × 100 = 0.64 × 100 = 64%।

Solving path: (320/500) × 100. Notice 320/500 = 32/50 = 16/25. And 16/25 × 100 = 64. Done in under 10 seconds with the fraction simplification. No long division needed.


Why this question: A salary comparison through a third variable — extremely common in RRB NTPC. Tests your ability to chain two percentage changes without confusion about the base.

Previous Year Questionपिछले वर्ष का प्रश्न
If A's salary is 25% more than B's salary and B's salary is 20% less than C's salary, then A's salary is what percent of C's salary?
यदि A की सैलरी, B की सैलरी से 25% ज़्यादा है और B की सैलरी, C की सैलरी से 20% कम है, तो A की सैलरी, C की सैलरी का कितने प्रतिशत है?
  1. 100%
  2. 105%
  3. 110%
  4. 115%
  1. 100%
  2. 105%
  3. 110%
  4. 115%
Solutionसमाधान
Let C's salary = 100. B's salary = 100 - 20% = 80. A's salary = 80 + 25% of 80 = 80 + 20 = 100. Therefore, A's salary is 100% of C's salary.
मान लें C का वेतन = 100। B का वेतन = 100 - 20% = 80। A का वेतन = 80 + 80 का 25% = 80 + 20 = 100। इसलिए, A का वेतन C के वेतन का 100% है।

Solving path: Set C = 100. B is 20% less than C → B = 80. A is 25% more than B → A = 80 + 25% of 80 = 80 + 20 = 100. A/C = 100/100 = 100%. The key discipline: always fix the base at the variable being compared against, not at a convenient round number for a different variable.


Why this question: Successive percentage change — the net change formula in action. Appears regularly and trips up students who add the percentages directly.

Previous Year Questionपिछले वर्ष का प्रश्न
A number is increased by 20% and then decreased by 15%. What is the net percentage change?
किसी संख्या को पहले 20% बढ़ाया जाता है और फिर 15% घटाया जाता है। कुल मिलाकर कितने प्रतिशत का बदलाव हुआ?
  1. 2% increase
  2. 5% increase
  3. 2% decrease
  4. 5% decrease
  1. 2% की वृद्धि
  2. 5% की वृद्धि
  3. 2% की कमी
  4. 5% की कमी
Solutionसमाधान
Let the number be 100. After 20% increase = 120. After 15% decrease = 120 - 15% of 120 = 120 - 18 = 102. Net change = (102-100)/100 × 100% = 2% increase.
मान लें संख्या 100 है। 20% वृद्धि के बाद = 120। 15% कमी के बाद = 120 - 120 का 15% = 120 - 18 = 102। कुल परिवर्तन = (102-100)/100 × 100% = 2% वृद्धि।

Solving path: Use the formula: 20 + (-15) + (20 × -15)/100 = 5 - 3 = 2% increase. Or verify with base 100: 100 → 120 → 102. Net = +2%. The trap: 20% - 15% = 5% increase. Wrong. The base changes after the first operation.


Why this question: Population change with two successive operations, opposite in direction — the most tested version of chain percentage in RRB NTPC.

Previous Year Questionपिछले वर्ष का प्रश्न
The population of a town increases by 15% in the first year and decreases by 20% in the second year. What is the net percentage change in population?
एक शहर की जनसंख्या पहले साल 15% बढ़ती है और दूसरे साल 20% घट जाती है। जनसंख्या में कुल कितने प्रतिशत का बदलाव हुआ?
  1. 8% decrease
  2. 5% decrease
  3. 8% increase
  4. 5% increase
  1. 8% की कमी
  2. 5% की कमी
  3. 8% की वृद्धि
  4. 5% की वृद्धि
Solutionसमाधान
Let initial population = 100. After first year = 100 + 15% = 115. After second year = 115 - 20% of 115 = 115 - 23 = 92. Net change = (92-100)/100 × 100% = -8% (decrease).
मान लें प्रारंभिक जनसंख्या = 100। पहले वर्ष के बाद = 100 + 15% = 115। दूसरे वर्ष के बाद = 115 - 115 का 20% = 115 - 23 = 92। कुल परिवर्तन = (92-100)/100 × 100% = -8% (कमी)।

Solving path: Formula: 15 + (-20) + (15 × -20)/100 = -5 - 3 = -8%. Net result: 8% decrease. Base 100 verification: 100 → 115 → 115 - 23 = 92. Confirm: (92-100)/100 × 100 = -8%. The answer is decrease, not increase — many students flip the sign because the larger percentage change (20%) is a decrease.


Why this question: Set theory overlap in a percentage context — a distinct skill from straight calculation. The "failed in both" framing is a classic misdirection.

Previous Year Questionपिछले वर्ष का प्रश्न
In an exam, 85% students passed in English, 90% passed in Math, and 10% failed in both. What percentage of students passed in both subjects?
एक परीक्षा में 85% छात्र अंग्रेज़ी में पास हुए, 90% छात्र गणित में पास हुए, और 10% दोनों में फेल हुए। कितने प्रतिशत छात्र दोनों विषयों में पास हुए?
  1. 75%
  2. 80%
  3. 85%
  4. 90%
  1. 75%
  2. 80%
  3. 85%
  4. 90%
Solutionसमाधान
Students who failed in both = 10%. Students who passed in at least one = 90%. Using formula: P(E∪M) = P(E) + P(M) - P(E∩M). So, 90 = 85 + 90 - P(E∩M). Therefore, P(E∩M) = 85%.
दोनों में फेल छात्र = 10%। कम से कम एक में पास छात्र = 90%। सूत्र का उपयोग: P(E∪M) = P(E) + P(M) - P(E∩M)। अतः, 90 = 85 + 90 - P(E∩M)। इसलिए, P(E∩M) = 85%।

Solving path: Anchor at failures: 10% failed both → 90% passed at least one. So P(E ∪ M) = 90%. Now: P(E ∩ M) = P(E) + P(M) - P(E ∪ M) = 85 + 90 - 90 = 85%. The distractor answer 80% comes from incorrectly computing 90 - (100-85) - (100-90) = 90 - 15 - 10 = 65... or from just averaging. Neither is correct. Anchor on the union, not on intuition.


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