Percentage – A Complete Guide for Competitive Exam Aspirants
Introduction
Percentage is one of the most frequently tested topics in quantitative aptitude sections of examinations such as the JKSSB Social Forestry Worker exam, SSC, banking, railways, and various state‑level recruitment tests. Its ubiquity stems from the fact that percentages provide a uniform way to compare quantities, express changes, and evaluate performance irrespective of the absolute size of the numbers involved. Mastery of percentage concepts not only helps in solving direct questions but also forms the backbone of related topics like profit‑loss, simple and compound interest, mixture and alligation, data interpretation, and problems based on population growth.
In this article we will build a solid conceptual foundation, discuss the interplay between fractions, decimals and percentages, highlight key facts that often appear in shortcuts, work through a variety of illustrative examples, list exam‑focused pointers, provide a set of practice questions graded by difficulty, and finally answer common doubts in an FAQ section.
Concept Explanation
1. What is a Percentage?
The word “percent” originates from the Latin per centum, meaning “by the hundred”. A percentage expresses a number as a fraction of 100. Symbolically,
\[
x\% = \frac{x}{100}
\]
Thus, 25 % means 25 out of every 100, i.e., \(\frac{25}{100}=0.25\).
2. Converting Between Forms
| From | To | Formula | Example |
|---|---|---|---|
| Fraction → Percentage | Multiply by 100 and add % sign | \(\displaystyle \frac{a}{b}\times 100\%\) | \(\frac{3}{4}\times100 = 75\%\) |
| Decimal → Percentage | Multiply by 100 and add % sign | \(d \times 100\%\) | \(0.68 \times 100 = 68\%\) |
| Percentage → Fraction | Divide by 100 and simplify | \(\displaystyle \frac{p}{100}\) | \(40\% = \frac{40}{100} = \frac{2}{5}\) |
| Percentage → Decimal | Divide by 100 | \(\displaystyle \frac{p}{100}\) | \(85\% = 0.85\) |
These conversions are reversible and form the basis for solving most percentage‑based problems.
3. Basic Percentage Operations
- Finding p % of a number N:
\[
\text{Value} = \frac{p}{100}\times N \]
- What percentage is A of B?
\[
\% = \frac{A}{B}\times 100
\]
- Increase or Decrease by p %: – Increase: New value = \(N \left(1+\frac{p}{100}\right)\)
- Decrease: New value = \(N \left(1-\frac{p}{100}\right)\)
- Successive Percentage Changes: When a quantity undergoes two successive changes of \(p_1\%\) and \(p_2\%\), the net change is not simply \(p_1+p_2\%\). The combined effect is given by:
\[
\text{Net\%}= p_1 + p_2 + \frac{p_1 p_2}{100}
\]
(Use a minus sign for a decrease.)
- Percentage Point vs. Percent Change:
- A change from 20 % to 25 % is an increase of 5 percentage points, but the relative increase is \(\frac{25-20}{20}\times100 = 25\%\).
4. Important Derived Formulas
| Situation | Formula | Remarks |
|---|---|---|
| Profit % | \(\displaystyle \frac{\text{Profit}}{\text{Cost Price}}\times100\) | Profit = SP – CP |
| Loss % | \(\displaystyle \frac{\text{Loss}}{\text{Cost Price}}\times100\) | Loss = CP – SP |
| Discount % | \(\displaystyle \frac{\text{Marked Price} – \text{Selling Price}}{\text{Marked Price}}\times100\) | |
| Simple Interest (SI) | \(\displaystyle \frac{P \times R \times T}{100}\) | P = principal, R = rate % per annum, T = time (years) |
| Compound Interest (CI) for n years | \(\displaystyle P\left(1+\frac{R}{100}\right)^n – P\) | |
| Population after n years with r % growth | \(\displaystyle P_0\left(1+\frac{r}{100}\right)^n\) | |
| Percentage of a mixture (alligation) | Use rule of alligation to find ratio of components. |
Key Facts to Remember 1. 100 % = 1 (as a decimal): Multiplying any number by 1 leaves it unchanged; therefore, adding 100 % of a number doubles it.
- The commutative property of percentages: \(p\% \text{ of } q = q\% \text{ of } p\). This often simplifies calculations (e.g., 12 % of 250 = 250 % of 12).
- Base matters: Always identify the reference quantity (the “whole”) before applying a percentage. Mistakes arise when the base is changed mid‑problem. 4. Percentage increase/decrease is not symmetric: A 20 % increase followed by a 20 % decrease does not return to the original value; the net effect is a decrease of \(20 – 20 – \frac{20\times20}{100}= -4\%\). 5. When dealing with successive changes, use the net formula rather than applying each step separately unless the number of steps is small.
- In profit‑loss problems, if the cost price of two items is the same and one is sold at x % profit and the other at x % loss, the overall result is a loss of \(\frac{x^2}{100}\%\).
- For data interpretation, percentages often appear in pie charts, bar graphs, and tables; converting them to actual values requires knowledge of the total.
- Approximation tricks:
- 1 % of a number = shift decimal two places left.
- 10 % = shift decimal one place left.
- 25 % = quarter; 50 % = half; 75 % = three‑quarters.
- 33.33 % ≈ one‑third; 66.66 % ≈ two‑thirds. —
Illustrative Examples
Example 1 – Basic Computation
Question: What is 18 % of 250?
Solution:
\[
18\% \text{ of } 250 = \frac{18}{100}\times 250 = 0.18 \times 250 = 45
\]
Answer: 45
Example 2 – Finding What Percent One Number Is of Another
Question: 42 is what percent of 140?
Solution:
\[
\% = \frac{42}{140}\times 100 = 0.3 \times 100 = 30\%
\]
Answer: 30 %
Example 3 – Percentage Increase
Question: The price of a laptop increased from ₹32,000 to ₹40,000. Find the percentage increase.
Solution:
Increase = 40,000 – 32,000 = ₹8,000
\[
\% \text{ Increase} = \frac{8,000}{32,000}\times 100 = 0.25 \times 100 = 25\%
\]
Answer: 25 % increase
Example 4 – Successive Percentage Changes
Question: A shopkeeper marks up an item by 30 % and then offers a discount of 20 % on the marked price. What is the net profit or loss percent?
Solution:
Let cost price = ₹100 (for convenience).
Marked price after 30 % increase = \(100 \times (1+0.30) = 130\)
Selling price after 20 % discount = \(130 \times (1-0.20) = 130 \times 0.80 = 104\)
Net gain = 104 – 100 = ₹4
Profit % = \(\frac{4}{100}\times 100 = 4\%\)
Answer: 4 % profit
(Using the net formula: Net% = +30 –20 + (30×–20)/100 = 10 –6 = 4 %)
Example 5 – Population Growth
Question: The population of a town is 1,85,000. It grows at 6 % per annum. What will be the population after 2 years?
Solution:
\[
P = 1,85,000 \times \left(1+\frac{6}{100}\right)^2 = 1,85,000 \times (1.06)^2
\]
\((1.06)^2 = 1.1236\)
\[
P = 1,85,000 \times 1.1236 = 2,07,866 \text{ (approximately)}
\]
Answer: Approximately 2,07,866 —
Example 6 – Profit‑Loss with Equal Cost Prices
Question: Two articles are bought for the same cost price. One is sold at a profit of 15 % and the other at a loss of 15 %. Find the overall profit or loss percent.
Solution:
Let CP of each = ₹100.
Profit on first = 15 % of 100 = ₹15 → SP₁ = 115
Loss on second = 15 % of 100 = ₹15 → SP₂ = 85
Total CP = 200, Total SP = 115 + 85 = 200 Overall profit/loss = 0 → 0 %
Note: The individual percentages cancel only when the profit and loss percentages are equal. If they differ, a net loss appears (see formula in Key Facts).
Answer: No profit, no loss (0 %).
Example 7 – Data Interpretation from a Pie Chart
Question: In a pie chart showing the expenditure of a household, the sector for “Education” occupies 72°. If the total monthly expenditure is ₹45,000, how much is spent on education?
Solution:
Full circle = 360° corresponds to 100 % of expenditure.
Education’s fraction = \(\frac{72}{360} = \frac{1}{5} = 20\%\)
Amount spent = \(20\% \text{ of } 45,000 = \frac{20}{100}\times45,000 = 9,000\) Answer: ₹9,000
Exam‑Focused Points 1. Identify the base quickly – In word problems, the phrase “of” usually indicates the base (the whole).
- Use the 100‑base trick – Convert percentages to fractions with denominator 100; this often simplifies multiplication/division.
- Leverage symmetry – Remember \(p\% \text{ of } q = q\% \text{ of } p\). This can turn a tough multiplication into an easier one.
- Beware of successive changes – Apply the net formula; avoid doing each step separately unless the question explicitly asks for intermediate values.
- Profit‑Loss shortcuts –
- If SP = CP × (1 + p/100) → profit p %.
- If SP = CP × (1 – p/100) → loss p %.
- For two articles with same CP, net loss% = \(\frac{(p_1-p_2)^2}{100}\) when one is sold at p₁% profit and the other at p₂% loss.
- Interest formulas – Simple interest is linear in time; compound interest grows exponentially. Remember that for small rates and short periods, SI ≈ CI.
- Mixture/Alligation – When two ingredients are mixed to get a desired percentage, the ratio of quantities is inversely proportional to the differences from the target percentage.
- Approximation – In multiple‑choice questions, rounding to the nearest 5 % or 10 % can help eliminate options quickly.
- Units consistency – Ensure that when you compute percentage increase/decrease, the numerator and denominator are in the same units.
- Check for “percentage point” traps – Distinguish between absolute change in percent (%) and relative change (percent of original).
Practice Questions #### Level 1 (Basic)
- What is 12 % of 250?
- Express 0.375 as a percentage.
- If 45 is 15 % of a number, find the number.
- A shirt priced at ₹800 is sold at a discount of 12 %. What is the selling price? 5. The population of a village increased from 12,000 to 15,000. Find the percentage increase.
Level 2 (Intermediate)
- A number is increased by 20 % and then decreased by 20 %. What is the net percentage change?
- The price of a commodity rises from ₹150 to ₹180. By what percent must the new price be reduced to restore the original price?
- A trader marks his goods 40 % above cost price and allows a discount of 10 %. Find his profit percent.
- The simple interest on a sum for 3 years at 5 % per annum is ₹450. Find the principal.
- In an examination, a student scores 84 marks out of 120. What percentage did he obtain?
Level 3 (Advanced)
- The population of a town is 2,00,000. It grows at 8 % per annum for the first year and then at 12 % per annum for the second year. What is the population after two years?
- Two articles are sold for ₹1,200 each. One is sold at a profit of 20 % and the other at a loss of 20 %. Find the overall profit or loss percent.
- A mixture contains 30 % alcohol. How much pure alcohol must be added to 10 L of this mixture to make the alcohol concentration 50 %?
- The price of a commodity is reduced by 25 % and then increased by 25 %. What is the net change in price?
- A bank offers a nominal interest rate of 6 % per annum compounded semi‑annually. What is the effective annual rate?
Answers (for self‑check):
- 30
- 37.5 % 3. 300
- ₹704
- 25 %
- –4 % (a decrease of 4 %) 7. 16.67 %
- 26 % profit
- ₹3,000
- 70 %
- 2,33,280
- 4 % loss
- 20 L of pure alcohol
- –6.25 % (a decrease of 6.25 %)
- 6.09 % (approx.)
Frequently Asked Questions (FAQs)
Q1: Why is percentage always expressed out of 100?
A: Historically, fractions with denominator 100 were easy to handle in trade and taxation. Expressing a ratio as “per hundred” provides a common scale, making comparison intuitive irrespective of the actual sizes involved.
Q2: Can a percentage exceed 100 %? Give an example. A: Yes. A percentage over 100 indicates that the part is larger than the whole. For instance, if a company’s profit this year is ₹2 lakhs while last year’s profit was ₹1 lakhs, the profit has increased by 100 % (doubled). If this year’s profit is ₹3 lakhs, the increase is 200 % (tripled).
Q3: How do I avoid confusion between “percent” and “percentage points”?
A: Remember:
- Percent refers to a relative change (multiply or divide by 100).
- Percentage points refer to the simple arithmetic difference between two percentages.
Example: Interest rate rising from 5 % to 7 % is a 2 percentage point increase, but the relative increase is \(\frac{7-5}{5}\times100 = 40\%\).
Q4: Is there a quick way to compute 15 % of a number without a calculator?
A: Yes. 15 % = 10 % + 5 %. Compute 10 % by shifting the decimal one place left, then halve that value to get 5 %, and add them. For 240: 10 % = 24, 5 % = 12 → 15 % = 36.
Q5: In mixture problems, why do we use the rule of alligation?
A: Alligation provides a shortcut to find the ratio in which two or more ingredients at given concentrations must be mixed to achieve a desired concentration. It avoids setting up algebraic equations each time. The rule states:
\[
\frac{\text{Quantity of cheaper}}{\text{Quantity of dearer}} = \frac{\text{Dearer price} – \text{Mean price}}{\text{Mean price} – \text{Cheaper price}}
\]
The same principle works with percentages (treat % as “price”).
Q6: How does compound interest differ from simple interest in terms of percentage growth?
A: Simple interest adds the same absolute amount each period (fixed percentage of the original principal). Compound interest adds interest on the accumulated amount, so the effective percentage growth accelerates over time. For the same rate and time, CI > SI unless the time is one period (where they are equal).
Q7: What is the best strategy to tackle percentage questions in a timed exam? A:
- Read carefully and underline the base (the “of” quantity).
- Convert the given percentage to a fraction or decimal if it simplifies multiplication.
- Apply the appropriate formula (increase/decrease, profit/loss, interest, etc.).
- Check if successive changes are involved; use the net formula when possible.
- Estimate first to eliminate implausible options, then compute precisely if needed.
Q8: Are there any common pitfalls when dealing with percentages in data interpretation?
A: Yes.
- Forgetting that the percentages in a pie chart sum to 100 %; if they don’t, the chart is misleading.
- Assuming that a higher percentage always means a higher absolute value without checking the total.
- Misinterpreting stacked bar charts where each segment’s percentage is relative to the segment’s own total, not the overall total.
Q9: How can I quickly verify my answer for a percentage increase/decrease problem?
A: After computing the new value, check whether it lies logically between the original and the expected direction. For an increase, the new value must be greater than the original; for a decrease, it must be smaller. Additionally, plug the new value back into the percentage formula to see if you retrieve the given percent.
Q10: Is there any relation between percentage and ratio?
A: Yes. A percentage is just a ratio where the second term is fixed at 100. For example, the ratio 3:4 can be expressed as \(\frac{3}{4}\times100 = 75\%\). Conversely, any percentage p % can be written as the ratio p:100 (or its reduced form).
Closing Remarks Mastering percentages is less about memorizing endless formulas and more about understanding the underlying idea of “parts per hundred”. Once this concept is internalized, the same reasoning applies to profit‑loss, interest, mixture, population, and data‑interpretation questions that dominate competitive exams. Practice the problems above, internalize the shortcuts, and always verify the logic of each step. With consistent effort, percentage problems will become one of the most scoring sections in your preparation arsenal.
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End of Article.
