Percentage for RRB Group D — Concepts, Shortcuts & Solved PYQs

beginner 18 min read

Concept

Percentage literally means "per hundred" — it is a ratio where the denominator is always 100. When you say "30% of passengers carry luggage," you mean 30 out of every 100 passengers carry luggage. That is all percentage ever is: a fraction with 100 in the denominator, expressed with the % symbol.

Here is the analogy that makes this click permanently. Imagine a train with exactly 100 seats. Whatever fraction of seats are occupied, that fraction's numerator is the percentage. 75 seats occupied = 75%. This mental model handles every percentage question — increase, decrease, comparison — because you are always rescaling to a base of 100.

Three things every percentage question asks you to find, in some form:

  1. The percentage itself — what fraction of the base is the given quantity?
  2. The part — given the base and the percentage, find the actual quantity.
  3. The base — given the part and the percentage, find what it is a percentage of.

The formula connecting all three:

Percentage=PartBase×100\text{Percentage} = \frac{\text{Part}}{\text{Base}} \times 100

Rearranged:

That third rearrangement is the one most students forget, and it is the one RRB Group D loves to test. "There were 280 accidents this year, which is 70% of last year — find last year." That is Base = Part × (100/Percentage).

Percentage questions in RRB Group D fall into four buckets: finding the base from a reduced/increased value, successive percentage change, set overlap (two conditions, find both), and direct fraction-to-percent conversion. Learn to spot which bucket a question falls into, and you have already done half the work.


Deep Dive

Fraction-to-Percent Conversion Table

Stop computing these in the exam hall. These are fixed values — burn them in.

| 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% | | 3/4 | 75% | | 2/3 | 66.67% | | 3/5 | 60% |

The reverse table matters just as much: 20% = 1/5, 25% = 1/4, and so on. When a question says "decreased by 20%," the new value is 4/5 of the original. Work in fractions when they are clean — it is faster than working in decimals.

Finding the Base (Reverse Percentage)

This is the most tested form. Pattern: "After a change of X%, the result is Y — find the original."

Original=New Value(100±Change%)×100\text{Original} = \frac{\text{New Value}}{(100 \pm \text{Change\%})} \times 100

Use + if the value increased to reach Y, use if it decreased to reach Y.

Example: Accidents decreased by 30%, result is 280. Original=28070×100=400\text{Original} = \frac{280}{70} \times 100 = 400

Successive Percentage Change

When two percentage changes happen one after the other (like two years of growth), do NOT add them. They compound.

Net Change%=a+b+ab100\text{Net Change\%} = a + b + \frac{ab}{100}

where a and b are the two percentage changes (use negative sign for decreases).

Example: +15% then −10%: Net=15+(10)+15×(10)100=51.5=+3.5%\text{Net} = 15 + (-10) + \frac{15 \times (-10)}{100} = 5 - 1.5 = +3.5\%

This formula saves you from starting with a base of 100 and computing step by step — though that method also works if you prefer it.

Percentage Change in a Product (Revenue-type questions)

Revenue = Price × Passengers. If Price changes by p% and Passengers change by q%, the net revenue change is:

Net Revenue Change%=p+q+pq100\text{Net Revenue Change\%} = p + q + \frac{pq}{100}

Same formula as successive change. Example: Price +25%, Passengers −20%: Net=25+(20)+25×(20)100=55=0%\text{Net} = 25 + (-20) + \frac{25 \times (-20)}{100} = 5 - 5 = 0\%

Zero change. That is the answer to one of the PYQs below.

Set Overlap (Two-Condition Problems)

When you are told what percentage passed/failed in two subjects:

Passed in Both=Pass%A+Pass%BPass% in at least one\text{Passed in Both} = \text{Pass\%}_A + \text{Pass\%}_B - \text{Pass\% in at least one}

"Pass% in at least one" = 100% − "Fail% in both."

So the working formula is:

Both=Pass%A+Pass%B(100%Fail in both%)\text{Both} = \text{Pass\%}_A + \text{Pass\%}_B - (100\% - \text{Fail in both\%})

Example: 72% passed Math, 68% passed GK, 15% failed both.

This is the inclusion-exclusion principle applied to percentages. The formula looks different from what you see in textbooks, but the logic is the same: total = A + B − (A and B).

Efficiency as Percentage

Some RRB Group D questions link percentage to work. If helper's efficiency is 80% of maintainer's, then helper does 0.8 units of work per day compared to maintainer's 1 unit per day. Since Time = Work/Rate:

Helper’s Time=Maintainer’s Time0.8=Maintainer’s Time×54\text{Helper's Time} = \frac{\text{Maintainer's Time}}{0.8} = \text{Maintainer's Time} \times \frac{5}{4}

Note: Higher efficiency → less time. So if efficiency is 80% (less than 100%), the helper takes more time, not less.


Memory Tricks & Shortcuts

patternThe Multiplier Method

Instead of computing "increase by 25%" as adding 25% of the value, multiply directly by 1.25. For a decrease of 16%, multiply by 0.84. Chain two changes by multiplying two factors.

Example: +25% then −16% on base 100: 100 × 1.25 × 0.84 = 100 × 1.05 = 105 → 5% increase.

Standard method: compute 25% of 100 = 25, add to get 125, then compute 16% of 125 = 20, subtract to get 105. That is 5 steps. Multiplier method: one multiplication line. Time saved: ~20 seconds per question.

patternab/100 Net Change Formula

For any two successive percentage changes a% and b%, net change = a + b + ab/100. Use negative values for decreases.

Example: +15% and −10% → 15 − 10 + (15)(−10)/100 = 5 − 1.5 = 3.5% net increase.

Standard method: assume base 100, apply first change, then second change, compute difference. That is 4 arithmetic steps. This formula: 3 arithmetic operations. When a and b are both multiples of 5, ab/100 is always a whole or half number — zero mental load.

patternReverse Percentage via Fraction

When a value decreases by X% to reach Y, the original = Y × (100/(100−X)). When X is 20%, 25%, or 30%, convert to fractions immediately.

  • Decreased by 20% → multiply by 5/4
  • Decreased by 25% → multiply by 4/3
  • Decreased by 30% → multiply by 10/7

Example: decreased by 30% → 280. Original = 280 × 10/7 = 400.

Standard method: set up equation 0.7x = 280, solve x = 280/0.7 = 400 (involves decimal division). Fraction method: 280 × 10/7 = 40 × 10 = 400 in two mental steps. Saves ~15 seconds.

patternSet Overlap: Add and Subtract

For two-subject pass/fail overlap questions, one line does it all: Both = Pass_A + Pass_B − (100 − Fail_both).

Mentally: add the two pass percentages, subtract the "passed at least one" number.

Example: 72 + 68 = 140. Passed at least one = 85. Both = 140 − 85 = 55.

Standard method: draw a Venn diagram, label regions, set up equations. That is 4−5 steps. This formula: 3 additions/subtractions. Saves 30−40 seconds in exam conditions.

substitutionEfficiency-Time Inverse

If efficiency is given as a percentage of another worker's efficiency, flip it to get the time ratio. Efficiency 80% → Time ratio = 100/80 = 5/4.

So if the reference worker takes 15 days: helper takes 15 × 5/4 = 75/4 = 18.75 days.

Standard method: compute per-day work for each, set up equation, solve. Three separate steps. Substitution: one fraction multiplication. Note the direction — lower efficiency always means more days, so if your answer is less than the reference time, you have the fraction inverted.


Fast-Solving Framework

Read the question once and classify it into one of these four types:

Type 1 — Find the Base: Keywords: "after a change of X%, the value became Y." Use Original = Y × (100 / (100 ± X)). Determine sign: increased to reach Y → use +, decreased to reach Y → use −.

Type 2 — Successive Change: Two percentage changes applied one after another. Use net% = a + b + ab/100. If it is a product (revenue = price × quantity), same formula applies.

Type 3 — Set Overlap: Two conditions (passed in A, passed in B, failed in both). Use Both = A + B − (100 − Fail_both).

Type 4 — Direct Calculation: "X% of Y" or "what % is X of Y." Plug into Part = (P/100) × Base or P% = (Part/Base) × 100.

Decision rule: if the question mentions "original" or "last year" and gives you the current value → Type 1. If two changes happen sequentially → Type 2. If there are two groups with overlap → Type 3. Everything else → Type 4.

Do not start calculations before classifying. That classification step takes five seconds and prevents the most common errors.


Solved PYQs

Why this question: The classic reverse-percentage trap — students calculate 30% of 280 and subtract instead of finding the base.

Previous Year Questionपिछले वर्ष का प्रश्न
The number of railway accidents decreased by 30% this year compared to last year. If there were 280 accidents this year, how many accidents were there last year?
इस साल रेलवे दुर्घटनाओं की संख्या पिछले साल की तुलना में 30% कम हो गई। अगर इस साल 280 दुर्घटनाएं हुईं, तो पिछले साल कितनी दुर्घटनाएं हुई थीं?
  1. 364
  2. 400
  3. 420
  4. 450
  1. 364
  2. 400
  3. 420
  4. 450
Solutionसमाधान
Let last year's accidents = x. This year's accidents = x - 30% of x = 0.70x = 280. Therefore, x = 280 ÷ 0.70 = 400.
मान लें पिछले वर्ष की दुर्घटनाएं = x। इस वर्ष की दुर्घटनाएं = x - x का 30% = 0.70x = 280। अतः, x = 280 ÷ 0.70 = 400।

Solving path: Decreased by 30% means current value = 70% of original. So original = 280 ÷ 0.70. Faster with fractions: 280 × (10/7) = 400. Check: 30% of 400 = 120, and 400 − 120 = 280. Confirmed.


Why this question: Revenue = Price × Quantity is tested in disguise. Many students see +25% and −20% and guess "5% increase" by adding the two.

Previous Year Questionपिछले वर्ष का प्रश्न
If the price of railway tickets is increased by 25% and the number of passengers decreases by 20%, what is the percentage change in total revenue?
यदि रेलवे टिकट की कीमत 25% बढ़ा दी जाए और यात्रियों की संख्या 20% घट जाए, तो कुल राजस्व में कितने प्रतिशत बदलाव होगा?
  1. 0% change
  2. 5% decrease
  3. 10% increase
  4. 5% increase
  1. 0% बदलाव
  2. 5% की कमी
  3. 10% की वृद्धि
  4. 5% की वृद्धि
Solutionसमाधान
Let original price = 100, passengers = 100. New price = 125, new passengers = 80. Original revenue = 100×100 = 10000. New revenue = 125×80 = 10000. No change in revenue.
मान लें मूल कीमत = 100, यात्री = 100। नई कीमत = 125, नए यात्री = 80। मूल आय = 100×100 = 10000। नई आय = 125×80 = 10000। आय में कोई बदलाव नहीं।

Solving path: Apply the product change formula: 25 + (−20) + (25)(−20)/100 = 5 − 5 = 0%. Net change is zero. Alternatively, 100 × 100 = 10,000 original revenue; 125 × 80 = 10,000 new revenue. Same result, zero change.


Why this question: Set overlap is tested with railway-exam framing. The Venn diagram approach wastes 60+ seconds.

Previous Year Questionपिछले वर्ष का प्रश्न
In a railway examination, 72% candidates passed in Mathematics, 68% passed in General Knowledge, and 15% failed in both subjects. What percentage passed in both subjects?
एक रेलवे परीक्षा में 72% उम्मीदवार Mathematics में पास हुए, 68% General Knowledge में पास हुए, और 15% दोनों विषयों में फेल हुए। दोनों विषयों में कितने प्रतिशत उम्मीदवार पास हुए?
  1. 55%
  2. 60%
  3. 65%
  4. 70%
  1. 55%
  2. 60%
  3. 65%
  4. 70%
Solutionसमाधान
Failed in both = 15%, so passed in at least one = 85%. Using formula: Passed in both = (72 + 68) - 85 = 140 - 85 = 55%.
दोनों में असफल = 15%, अतः कम से कम एक में सफल = 85%। सूत्र का उपयोग: दोनों में सफल = (72 + 68) - 85 = 140 - 85 = 55%।

Solving path: Failed in both = 15%, so passed in at least one = 85%. Both = 72 + 68 − 85 = 140 − 85 = 55%.


Why this question: Two-year successive change — the classic mistake is 25 − 16 = 9% increase.

Previous Year Questionपिछले वर्ष का प्रश्न
A railway station's passenger traffic increased by 25% in the first year and decreased by 16% in the second year. What is the overall percentage change over two years?
एक रेलवे स्टेशन का यात्री यातायात पहले साल 25% बढ़ा और दूसरे साल 16% घट गया। दो सालों में कुल मिलाकर कितने प्रतिशत बदलाव हुआ?
  1. 5% increase
  2. 9% decrease
  3. 5% decrease
  4. 9% increase
  1. 5% की वृद्धि
  2. 9% की कमी
  3. 5% की कमी
  4. 9% की वृद्धि
Solutionसमाधान
Let initial traffic = 100. After first year: 125. After second year: 125 - (16% of 125) = 125 - 20 = 105. Overall change = (105-100)/100 × 100 = 5% increase.
मान लें प्रारंभिक यातायात = 100। पहले वर्ष के बाद: 125। दूसरे वर्ष के बाद: 125 - (125 का 16%) = 125 - 20 = 105। कुल परिवर्तन = (105-100)/100 × 100 = 5% वृद्धि।

Solving path: Net% = 25 + (−16) + (25)(−16)/100 = 9 − 4 = 5% increase. Or: 100 → 125 → 125 × 0.84 = 105. Net = 5% increase.


Why this question: Salary successive change with mixed signs. Direct application of the ab/100 formula.

Previous Year Questionपिछले वर्ष का प्रश्न
If the salary of a railway employee is increased by 15% and then decreased by 10%, what is the net percentage change in his salary?
यदि एक रेलवे कर्मचारी की सैलरी पहले 15% बढ़ाई जाती है और फिर 10% घटाई जाती है, तो उसकी सैलरी में कुल कितने प्रतिशत का बदलाव होगा?
  1. 3.5% increase
  2. 3.5% decrease
  3. 5% increase
  4. 2% decrease
  1. 3.5% की बढ़ोतरी
  2. 3.5% की कमी
  3. 5% की बढ़ोतरी
  4. 2% की कमी
Solutionसमाधान
Let original salary be 100. After 15% increase: 115. After 10% decrease: 115 - (10% of 115) = 115 - 11.5 = 103.5. Net change = 3.5% increase.
मान लें मूल वेतन 100 है। 15% वृद्धि के बाद: 115। 10% कमी के बाद: 115 - (115 का 10%) = 115 - 11.5 = 103.5। शुद्ध परिवर्तन = 3.5% वृद्धि।

Solving path: Net% = 15 + (−10) + (15)(−10)/100 = 5 − 1.5 = 3.5% increase.


Why this question: Direct percentage of total — tests whether you remember that percentages must add to 100.

Previous Year Questionपिछले वर्ष का प्रश्न
A railway station has 1500 passengers daily. If 36% are men, 44% are women, and the rest are children, how many children visit the station daily?
एक रेलवे स्टेशन पर रोज़ाना 1500 यात्री आते हैं। यदि 36% पुरुष हैं, 44% महिलाएं हैं, और बाकी बच्चे हैं, तो रोज़ाना कितने बच्चे स्टेशन पर आते हैं?
  1. 280
  2. 300
  3. 320
  4. 350
  1. 280
  2. 300
  3. 320
  4. 350
Solutionसमाधान
Men = 36%, Women = 44%, so Children = 100% - (36% + 44%) = 20%. Number of children = 20% of 1500 = 0.20 × 1500 = 300.
पुरुष = 36%, महिलाएं = 44%, अतः बच्चे = 100% - (36% + 44%) = 20%। बच्चों की संख्या = 1500 का 20% = 0.20 × 1500 = 300।

Solving path: Children% = 100 − 36 − 44 = 20%. Number of children = 20% of 1500 = 0.20 × 1500 = 300.


Why this question: Efficiency-time inverse relationship — lower efficiency means more time, not less.

Previous Year Questionपिछले वर्ष का प्रश्न
A railway helper's efficiency is 80% of a track maintainer's efficiency. If the track maintainer can complete a job in 15 days, in how many days can the helper complete the same job?
एक रेलवे हेल्पर की कार्यक्षमता एक ट्रैक मेंटेनर की कार्यक्षमता की 80% है। यदि ट्रैक मेंटेनर किसी काम को 15 दिनों में पूरा कर सकता है, तो हेल्पर उसी काम को कितने दिनों में पूरा करेगा?
  1. 18.75 days
  2. 19.5 days
  3. 20 days
  4. 21 days
  1. 18.75 days
  2. 19.5 days
  3. 20 days
  4. 21 days
Solutionसमाधान
Helper's efficiency = 80% of maintainer's efficiency. If maintainer takes 15 days, helper will take 15 ÷ 0.80 = 18.75 days. Time is inversely proportional to efficiency.
सहायक की दक्षता = रखरखावकर्ता की दक्षता का 80%। यदि रखरखावकर्ता 15 दिन लेता है, तो सहायक 15 ÷ 0.80 = 18.75 दिन लेगा। समय दक्षता के व्युत्क्रमानुपाती है।

Solving path: Helper does 80% of maintainer's work per day. Time = Work / Rate. Helper's time = Maintainer's time / 0.80 = 15 / 0.80 = 15 × (5/4) = 18.75 days. Sanity check: helper is slower, so 18.75 > 15. Correct direction.


Why this question: Reverse percentage on a percentage itself — "improved by 40%, result is 84%."

Previous Year Questionपिछले वर्ष का प्रश्न
A railway line's efficiency improved by 40% after maintenance. If the improved efficiency is 84%, what was the original efficiency?
मेंटेनेंस के बाद एक रेलवे लाइन की कार्यक्षमता 40% बढ़ गई। यदि बढ़ी हुई कार्यक्षमता 84% है, तो पहले की कार्यक्षमता क्या थी?
  1. 60%
  2. 65%
  3. 70%
  4. 75%
  1. 60%
  2. 65%
  3. 70%
  4. 75%
Solutionसमाधान
Let original efficiency = x%. After 40% improvement: x + 40% of x = 1.4x = 84%. Therefore, x = 84 ÷ 1.4 = 60%.
मान लें मूल दक्षता = x%। 40% सुधार के बाद: x + x का 40% = 1.4x = 84%। अतः, x = 84 ÷ 1.4 = 60%।

Solving path: Let original efficiency = x. After 40% improvement: 1.4x = 84. So x = 84 / 1.4 = 60%. In fractions: 84 × (10/14) = 84 × (5/7) = 60.


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