Percentages
Percentages for CGPSC CSAT: fraction–percent table, percentage change, successive change, percentage points, population and price problems with shortcuts.
Prerequisites
- Comfortable with fractions and decimals (Class VIII level): converting \( \frac{3}{8} \) to \( 0.375 \) without a pen.
- Multiplication tables up to 20 — they are the engine behind every shortcut on this page.
- This is the first topic of Basic Numeracy; Ratio and Proportion and Averages build directly on it.
The Hook: “Up 20%, Down 20% — Back to Square One?”
At a paddy procurement (धान खरीदी) centre in Mahasamund, two farmers are arguing over a notice board. One says, “Last season the society bought 20% more paddy than the season before, and this season 20% less — so we are back to where we started.” The other shakes his head.
Take a hypothetical 1,000 quintals. Up 20% gives 1,200. Down 20% of 1,200 is 240, leaving 960 — a 4% fall, not zero. The second 20% was taken of a bigger base.
The question this page answers: how do you track a quantity through one, two or several percentage changes — quickly, and without falling into the “base” trap that CSAT options are built around?
Why it matters for CGPSC: percentages (प्रतिशत) are the most reused tool in Prelims Paper II. They show up directly (price, population, marks questions) and hidden inside Data Interpretation tables, profit and loss, and growth comparisons. The reasoning-and-mathematics part of Mains Paper IV (GS-II) uses the same ideas, where you must also show the steps.
What a Percentage Is
“Per cent” means “per hundred”. A percentage is just a fraction with denominator 100:
$$\boxed{x\% = \frac{x}{100}} \qquad\qquad \boxed{x\% \text{ of } N = \frac{x}{100}\times N}$$Three conversions you must do mentally:
| Convert | Rule | Example |
|---|---|---|
| Fraction → % | Multiply by 100 | \( \frac{3}{8} = \frac{300}{8}\% = 37.5\% \) |
| % → fraction | Divide by 100, simplify | \( 35\% = \frac{35}{100} = \frac{7}{20} \) |
| % → decimal | Shift decimal two places left | \( 4.5\% = 0.045 \) |
The Fraction–Percentage Table
Memorise this table. It turns a “35 seconds with long division” question into a “5 seconds” one.
| Fraction | Percentage | Fraction | Percentage |
|---|---|---|---|
| \( \frac{1}{2} \) | 50% | \( \frac{1}{11} \) | \( 9\frac{1}{11}\% \approx 9.09\% \) |
| \( \frac{1}{3} \) | \( 33\frac{1}{3}\% \approx 33.33\% \) | \( \frac{1}{12} \) | \( 8\frac{1}{3}\% \approx 8.33\% \) |
| \( \frac{1}{4} \) | 25% | \( \frac{1}{15} \) | \( 6\frac{2}{3}\% \approx 6.67\% \) |
| \( \frac{1}{5} \) | 20% | \( \frac{1}{16} \) | 6.25% |
| \( \frac{1}{6} \) | \( 16\frac{2}{3}\% \approx 16.67\% \) | \( \frac{1}{20} \) | 5% |
| \( \frac{1}{7} \) | \( 14\frac{2}{7}\% \approx 14.29\% \) | \( \frac{1}{25} \) | 4% |
| \( \frac{1}{8} \) | 12.5% | \( \frac{2}{3} \) | \( 66\frac{2}{3}\% \approx 66.67\% \) |
| \( \frac{1}{9} \) | \( 11\frac{1}{9}\% \approx 11.11\% \) | \( \frac{3}{4} \) | 75% |
| \( \frac{1}{10} \) | 10% | \( \frac{5}{6} \) | \( 83\frac{1}{3}\% \approx 83.33\% \) |
Build the multiples, do not memorise them separately: \( \frac{1}{8} = 12.5\% \Rightarrow \frac{3}{8} = 37.5\%,\ \frac{5}{8} = 62.5\%,\ \frac{7}{8} = 87.5\% \).
Find \( 16\frac{2}{3}\% \) of 774.
\( 16\frac{2}{3}\% = \frac{1}{6} \), so the answer is \( 774 \div 6 = 129 \). No multiplication by 16.67 needed.
Percentage Change
$$\boxed{\text{Percentage change} = \frac{\text{New value} - \text{Old value}}{\text{Old value}}\times 100}$$The old (original) value is always the base. A positive answer is an increase; a negative answer is a decrease.
The multiplying factor
Change by \( r\% \) means multiply by a factor:
| Change | Multiplying factor | Example |
|---|---|---|
| Increase by \( r\% \) | \( 1 + \frac{r}{100} \) | +15% → \( \times 1.15 \) |
| Decrease by \( r\% \) | \( 1 - \frac{r}{100} \) | −20% → \( \times 0.80 \) |
Multiplying factors make multi-step questions mechanical: just multiply the factors in order.
“More than” and “less than” are not symmetric
If A is \( r\% \) more than B, then B is less than A by
$$\boxed{\frac{r}{100 + r}\times 100\ \%}$$and if A is \( r\% \) less than B, then B is more than A by \( \frac{r}{100 - r}\times 100\% \).
A’s monthly income is 25% more than B’s. By what percent is B’s income less than A’s?
Let B = 100, so A = 125. B is less by 25 on a base of 125: \( \frac{25}{125}\times 100 = 20\% \).
Shortcut: \( \frac{r}{100 + r}\times 100 \) with \( r = 25 \) gives the same 20%.
Successive Percentage Change
Two successive changes of \( a\% \) and \( b\% \) (use a negative sign for a decrease) equal a single change of
$$\boxed{a + b + \frac{ab}{100}\ \%}$$| Changes | Net effect | Calculation |
|---|---|---|
| +20%, then −20% | −4% | \( 20 - 20 + \frac{(20)(-20)}{100} = -4 \) |
| +10%, then +10% | +21% | \( 10 + 10 + \frac{100}{100} = 21 \) |
| +10%, then +20% | +32% | \( 10 + 20 + \frac{200}{100} = 32 \) |
| −10%, then −10% | −19% | \( -10 - 10 + \frac{100}{100} = -19 \) |
Three changes? Combine the first two into one, then combine that result with the third.
Price and consumption (expenditure stays fixed)
Expenditure = Price \( \times \) Consumption. If the price of an item rises by \( r\% \) and the household budget for it must not change, consumption must fall by
$$\boxed{\frac{r}{100 + r}\times 100\ \%}$$If the price falls by \( r\% \), consumption can rise by \( \frac{r}{100 - r}\times 100\% \) for the same spending.
The price of edible oil rises by 25%. By how much should a family in Durg cut its consumption so that its monthly spend on oil stays the same?
\( \frac{25}{125}\times 100 = 20\% \). Check: price factor \( 1.25 \), consumption factor \( 0.80 \), and \( 1.25 \times 0.80 = 1 \).
Percent vs Percentage Point
This is the single most common conceptual trap in CSAT and DI statements.
- Percentage point = simple difference between two percentages.
- Percent change = that difference divided by the old percentage, times 100.
In a hypothetical block, the literacy rate rises from 60% to 75%.
- Rise in percentage points: \( 75 - 60 = 15 \) percentage points.
- Rise in percent: \( \frac{15}{60}\times 100 = 25\% \).
Both statements are true; “the literacy rate rose by 15%” is false.
Population and Price Growth
When a quantity changes by the same rate every year, apply the factor once per year (this is compound growth):
$$\boxed{P_n = P\left(1 + \frac{r}{100}\right)^n} \qquad\qquad \boxed{P_n = P\left(1 - \frac{r}{100}\right)^n}$$The first is for growth (population, prices); the second for depreciation (the value of a tractor or a two-wheeler). To go backwards (population \( n \) years ago), divide instead: \( P_{\text{earlier}} = \frac{P}{\left(1 + \frac{r}{100}\right)^n} \).
A village near Rajnandgaon has 8,000 people and grows at 5% a year. What will its population be after 2 years?
\( 8000 \times 1.05 \times 1.05 = 8400 \times 1.05 = 8820 \).
Quick check with the successive-change formula: \( 5 + 5 + \frac{25}{100} = 10.25\% \), and 10.25% of 8,000 is 820.
A town of 50,000 grows 10% in one year and shrinks 10% the next. Population after 2 years?
Net change \( = -\frac{10^2}{100} = -1\% \), so \( 50000 \times 0.99 = 49500 \).
Marks and Elections: “Fails by” Questions
Turn the words into an equation in the maximum marks \( M \):
In a recruitment test, a candidate who scores 30% fails by 15 marks; another who scores 40% gets 25 marks more than the pass mark. Find the maximum marks and the pass mark.
Pass mark \( = 0.30M + 15 = 0.40M - 25 \Rightarrow 0.10M = 40 \Rightarrow M = 400 \).
Pass mark \( = 120 + 15 = 135 \). Check: \( 160 - 25 = 135 \).
Shortcut: the gap between the two scores is \( 10\% \) of \( M \) and equals \( 15 + 25 = 40 \) marks.
Memory Tricks
- “Than” marks the base. In “A is 25% more than B”, B is the base. Whatever follows “than” or “of” goes in the denominator.
- Successive change: “a plus b plus a-b by hundred”. Say it as a chant: \( a + b + \frac{ab}{100} \).
- The 25–20 pair (and 50–33⅓, 100–50). 25% up needs 20% down to come back; 50% up needs \( 33\frac{1}{3}\% \) down; 100% up needs 50% down. Same pairs for “more than / less than” and “price up / consumption down”.
- 1/7 = 0.142857… The same cyclic digits 142857 run through every seventh: \( \frac{1}{7} \approx 14.29\% \), \( \frac{2}{7} \approx 28.57\% \), \( \frac{3}{7} \approx 42.86\% \).
- Equal up-down = loss of \( \frac{x^2}{100} \): “square it, shift two places”.
Don’t Confuse
| This | vs | That |
|---|---|---|
| Rate rose from 60% to 75% = 15 percentage points | Same change = 25% increase | |
| A is 25% more than B (base B) | B is 20% less than A (base A) | |
| +20% then −20% = −4% | +20% then +20% = +44% | |
| Growth over 2 years at 10% = 21% (compound) | 10% × 2 = 20% (simple — wrong for population) |
Common Mistakes
Practice
Level 1: Foundation
Answer each without writing more than one line.
- Express \( \frac{3}{8} \) as a percentage.
- Find 12.5% of 640.
- Find 35% of 240.
- A price goes from ₹250 to ₹300. Find the percentage increase.
- A town grows 10% then falls 10%. What is the net change?
Answers
- 37.5% (three times 12.5%).
- \( \frac{1}{8}\times 640 = 80 \).
- \( 0.35 \times 240 = 84 \) (or 10% = 24, so 35% = 72 + 12 = 84).
- \( \frac{50}{250}\times 100 = 20\% \).
- A net decrease of 1% (\( \frac{10^2}{100} \)).
Level 2: Prelims MCQs
All figures below are hypothetical practice values, not real statistics.
Q1. The price of a quintal of a farm commodity at a mandi in Bemetara rises from ₹2,000 to ₹2,300. The percentage increase is:
(a) 13% (b) 15% (c) 17.5% (d) 30%
Answer: (b) 15%
Increase = ₹300 on an original price of ₹2,000. \( \frac{300}{2000}\times 100 = 15\% \).
Option (a) is \( \frac{300}{2300}\times 100 \approx 13\% \) — the wrong-base trap (dividing by the new price).
Q2. A cooperative society’s paddy procurement rose by 20% in one season and fell by 20% in the next. Compared with the first season, procurement in the third season is:
(a) unchanged (b) 4% higher (c) 4% lower (d) 2% lower
Answer: (c) 4% lower
Successive change: \( 20 + (-20) + \frac{(20)(-20)}{100} = -4\% \).
With numbers: 100 → 120 → \( 120 \times 0.8 = 96 \), i.e. 4% lower.
Q3. The price of edible oil increases by 25%. By what percentage must a household reduce its consumption so that its expenditure on oil remains unchanged?
(a) 20% (b) 25% (c) 30% (d) \( 33\frac{1}{3}\% \)
Answer: (a) 20%
Required cut \( = \frac{r}{100 + r}\times 100 = \frac{25}{125}\times 100 = 20\% \).
Check: new price factor 1.25, new consumption factor 0.8, product = 1, so spending is unchanged. Option (b) is the “same percentage” trap.
Q4. The literacy rate of a hypothetical development block rose from 60% to 75% over a decade. Consider the following statements:
- The literacy rate increased by 15 percentage points.
- The literacy rate increased by 25%.
- The literacy rate increased by 15%.
Which of the statements given above is/are correct?
(a) 1 only (b) 1 and 2 only (c) 2 and 3 only (d) 1, 2 and 3
Answer: (b) 1 and 2 only
Q5. In a recruitment test, a candidate who scores 30% marks fails by 15 marks, while another who scores 40% marks gets 25 marks more than the minimum pass marks. The maximum marks of the test are:
(a) 300 (b) 350 (c) 400 (d) 450
Answer: (c) 400
The two candidates differ by \( 40\% - 30\% = 10\% \) of the maximum marks, and in marks by \( 15 + 25 = 40 \). So \( 10\% \) of \( M = 40 \Rightarrow M = 400 \).
Pass mark \( = 30\% \text{ of } 400 + 15 = 135 \); check: \( 160 - 25 = 135 \).
Level 3: Speed Drill — Solve in 60 Seconds
Prelims Paper II rewards speed as much as accuracy, and the reasoning-and-mathematics part (Part 2) of Mains Paper IV (GS-II) tests the same arithmetic with steps shown. Time yourself: each item below should take under 60 seconds using the shortcuts on this page. If one takes longer, note which shortcut you missed.
- Find \( 16\frac{2}{3}\% \) of 774.
- A salary rises by 10% and then by 20%. Net percentage rise?
- A’s income is 25% more than B’s. By what percent is B’s income less than A’s?
- A village of 8,000 people grows at 5% per year. Population after 2 years?
- Find 37.5% of 480.
- A two-wheeler worth ₹80,000 depreciates 10% a year. Its value after 2 years? (hypothetical)
Answers with the shortcut used
- 129 — \( 16\frac{2}{3}\% = \frac{1}{6} \); \( 774 \div 6 = 129 \).
- 32% — \( 10 + 20 + \frac{10\times 20}{100} \).
- 20% — \( \frac{25}{125}\times 100 \).
- 8,820 — net \( 10.25\% \); \( 8000 + 820 \).
- 180 — \( 37.5\% = \frac{3}{8} \); \( 480 \div 8 \times 3 \).
- ₹64,800 — net \( -10 - 10 + 1 = -19\% \); \( 80000 \times 0.81 = 64800 \).
Quick Revision Table
| Idea | Formula / Fact |
|---|---|
| x% of N | \( \frac{x}{100}\times N \); x% of y = y% of x |
| Percentage change | \( \frac{\text{New} - \text{Old}}{\text{Old}}\times 100 \) |
| Successive change | \( a + b + \frac{ab}{100} \) |
| Equal rise and fall of x% | Net loss \( \frac{x^2}{100}\% \) |
| A is r% more than B | B is \( \frac{r}{100+r}\times 100\% \) less than A |
| Price up r%, same spend | Cut consumption \( \frac{r}{100+r}\times 100\% \) |
| Compound growth | \( P\left(1 + \frac{r}{100}\right)^n \) |
| Percentage point | Simple difference of two percentages |
Key Takeaways
- The original value is the base — “than” and “of” point to it.
- Know the fraction–percentage table cold; most CSAT percentage questions are fraction questions in disguise.
- Successive changes multiply, they do not add: \( a + b + \frac{ab}{100} \).
- An equal rise and fall always ends in a net loss of \( \frac{x^2}{100}\% \).
- Percentage points compare two percentages by subtraction; percent change divides by the old value.
What’s Next
- Ratio and Proportion — percentages are ratios with denominator 100; the next page generalises the idea.
- Averages — why you cannot simply average two percentages.
- Data Interpretation — percentage change and percentage points are the core of every DI set.
- General Mental Ability and Logical Reasoning — the other half of the CSAT section.
- Basic Numeracy — Quick Revision — every formula from this chapter on one page.