PERCENTAGE – QUICK REVISION NOTES
(Tailored for JKSSB Social Forestry Worker – Basic Mathematics)
1. What is Percentage?
- Definition – A percentage is a number or ratio expressed as a fraction of 100.
- Symbol: % (read “per cent”).
- Mathematically: \(x\% = \dfrac{x}{100}\).
Key Highlight – “Percent” means out of every hundred.
2. Basic Conversions
| From → To | Formula / Trick | Example |
|---|---|---|
| Fraction → % | Multiply by 100 and add % | \( \frac{3}{5} \times 100 = 60\% \) |
| Decimal → % | Shift decimal two places right (×100) | 0.42 → 42 % |
| % → Fraction | Write over 100 and simplify | 75 % = \( \frac{75}{100}= \frac{3}{4} \) |
| % → Decimal | Divide by 100 (shift decimal two places left) | 8.5 % → 0.085 |
| Mixed Number → % | Convert mixed to improper fraction, then ×100 | \(1\frac{2}{5}= \frac{7}{5}\) → \( \frac{7}{5}\times100=140\% \) |
Mnemonic for conversion:
“F‑D‑P” → Fraction → Decimal → Percent (just move the decimal two places each step).
3. Finding Percentage of a Quantity
\[
\text{Percentage of } Q = \frac{p}{100}\times Q
\]
- Example: What is 18 % of 250?
\(\frac{18}{100}\times250 = 45\).
Shortcut Trick (10 % method):
- Find 10 % (divide by 10), then scale.
10 % of 250 = 25 → 1 % = 2.5 → 18 % = 25 + (8×2.5) = 25 + 20 = 45.
4. Express One Quantity as a Percentage of Another
\[
\% = \frac{\text{Part}}{\text{Whole}}\times100\]
- Example: 30 is what percent of 120?
\(\frac{30}{120}\times100 = 25\%\).
Tip: Reduce the fraction first for quicker calculation.
5. Increase / Decrease by a Percentage
| Situation | Formula | Quick Trick |
|---|---|---|
| Increase by \(p\%\) | New value = Original \(\times\bigl(1+\frac{p}{100}\bigr)\) | Add \(p\%\) of original to original. |
| Decrease by \(p\%\) | New value = Original \(\times\bigl(1-\frac{p}{100}\bigr)\) | Subtract \(p\%\) of original from original. |
| Find original after increase/decrease | Original = Final \(\div\bigl(1\pm\frac{p}{100}\bigr)\) | Reverse the multiplier. |
- Example: Salary increased by 12 % → new salary = ₹ 22,400. Find original.
Original = \(22,400 ÷ 1.12 = ₹20,000\).
6. Profit, Loss & Discount (Percentage Based)
| Concept | Formula | Percentage Form |
|---|---|---|
| Profit % | \(\displaystyle \frac{\text{Profit}}{\text{Cost Price}}\times100\) | Profit % = \(\frac{SP-CP}{CP}\times100\) |
| Loss % | \(\displaystyle \frac{\text{Loss}}{\text{Cost Price}}\times100\) | Loss % = \(\frac{CP-SP}{CP}\times100\) |
| Discount % | \(\displaystyle \frac{\text{Discount}}{\text{Marked Price}}\times100\) | Discount % = \(\frac{MP-SP}{MP}\times100\) |
| Selling Price after discount | \(SP = MP \times\bigl(1-\frac{d}{100}\bigr)\) | – |
| Marked Price from SP & discount% | \(MP = \frac{SP}{1-\frac{d}{100}}\) | – |
Mnemonic: P‑L‑D → Profit, Loss, Discount – all follow the same “difference over base ×100” pattern.
7. Successive Percentage Changes
When two (or more) percentage changes happen one after another, the net change is not simply the sum.
\[
\text{Net % change} = a + b + \frac{ab}{100}
\]
- Increase then Increase: both \(a,b\) positive.
- Increase then Decrease: \(a\) positive, \(b\) negative (use sign).
Example: Price increased by 20 % then decreased by 10 %.
Net change = \(20 + (-10) + \frac{20\times(-10)}{100}=20-10-2=8\%\) increase.
Extended to three changes: Apply the formula iteratively or use:
\[
\text{Final multiplier}= (1+\frac{a}{100})(1+\frac{b}{100})(1+\frac{c}{100})-1
\]
8. Population Growth / Depreciation (Compound Percentage)
Formula (same as compound interest):
\[A = P\bigl(1+\frac{r}{100}\bigr)^{n}
\]
- Growth → \(r\) positive.
- Depreciation → \(r\) negative (or use \(1-\frac{r}{100}\)).
Shortcut for small \(n\): Use binomial approximation if \(r\) is small and \(n\) ≤ 2.
Example: A forest area of 500 ha grows at 4 % per annum. After 3 years?
\(A = 500(1.04)^3 ≈ 500×1.1249 ≈ 562.5\) ha.
9. Mixture & Alligation (Percentage Based)
When mixing two ingredients with different concentrations to get a desired concentration:
\[
\frac{\text{Quantity of cheaper}}{\text{Quantity of dearer}} = \frac{C_{dearer}-C_{mean}}{C_{mean}-C_{cheaper}}
\]
- All quantities expressed as percentages (or fractions).
Example: Mix 30 % alcohol solution with 50 % alcohol to get 40 % alcohol.
Ratio = \((50-40):(40-30)=10:10=1:1\). So equal parts.
Mnemonic: “C‑D‑M” → Cheaper, Dearer, Mean – place them in the formula as shown.
10. Important Percentage Tricks & Shortcuts
| Trick | When to Use | How |
|---|---|---|
| 10 % method | Quick mental calculation of any % | Find 10 % (÷10), then 1 % = 10 %÷10, scale up/down. |
| 5 % method | When % ends in 5 or 0 | Find 10 % then halve for 5 %; combine. |
| 25 % method | Quarter calculations | 25 % = ÷4; 75 % = 100 %‑25 % = ¾ of value. |
| Complementary % | Finding what’s left | If something is \(x\%\), the remainder is \((100-x)\%\). |
| Fraction‑% equivalents | Speed up conversion | Memorise common ones: \(\frac{1}{2}=50\%\), \(\frac{1}{4}=25\%\), \(\frac{1}{5}=20\%\), \(\frac{1}{8}=12.5\%\), \(\frac{1}{10}=10\%\), \(\frac{1}{3}=33.\overline{3}\%\), \(\frac{2}{3}=66.\overline{3}\%\). |
| Percentage point vs. percent | Avoid confusion | A change from 20 % to 25 % is a 5‑percentage‑point increase, but a 25 % relative increase (\(\frac{5}{20}\times100\)). |
| Reverse % | Finding original after % change | Original = New ÷ (1 ± %/100). Use + for decrease, – for increase. |
11. Common Exam‑Style Problems (with Quick Solutions)
| Problem Type | Sample Question | Shortcut Solution |
|---|---|---|
| Find % of a number | What is 37 % of 680? | 10 % = 68 → 30 % = 204 → 5 % = 34 → 2 % = 13.6 → Total = 204+34+13.6 = 251.6 |
| Express as % | 45 is what % of 180? | \(\frac{45}{180}=0.25\) → 25 % |
| Increase/Decrease | A tree’s height increased from 3 m to 3.9 m. % increase? | Increase = 0.9 m → \(\frac{0.9}{3}\times100 = 30 %\) |
| Successive change | Price of saplings rose 15 % then fell 10 %. Net change? | Net = \(15-10+\frac{15×(-10)}{100}=5-1.5=3.5 %\) increase |
| Profit/Loss | CP = ₹ 800, SP = ₹ 920. Profit %? | Profit = 120 → \(\frac{120}{800}\times100 = 15 %\) |
| Discount | MP = ₹ 1500, discount 12 %. SP? | Discount = 12 % of 1500 = 180 → SP = 1500‑180 = ₹1320 |
| Population | Village population 12,000 grows at 3 % p.a. After 2 years? | \(12000×(1.03)^2 ≈ 12000×1.0609 = 12,731\) |
| Mixture | How many litres of 20 % urea solution must be mixed with 40 % urea to get 30 % urea, if total volume is 100 L? | Using alligation: Ratio = (40‑30):(30‑20)=10:10=1:1 → 50 L each. |
| Reverse % | After a 20 % discount, a shirt costs ₹ 640. Original price? | Original = \(640 ÷ (1-0.20)=640÷0.80=₹800\) |
| Percentage point | Interest rate rose from 6 % to 9 %. Change in percentage points? | 9‑6 = 3 percentage points (relative increase = 50 %). |
12. Formula Sheet (One‑Page Reference)
| Topic | Formula |
|---|---|
| Basic % | \( \% = \frac{Part}{Whole} \times 100 \) |
| Part from % | \( Part = \frac{\%}{100} \times Whole \) |
| Whole from Part & % | \( Whole = \frac{Part \times 100}{\%} \) |
| Increase | New = Original × \((1+\frac{p}{100})\) |
| Decrease | New = Original × \((1-\frac{p}{100})\) |
| Profit % | \(\frac{SP-CP}{CP}\times100\) |
| Loss % | \(\frac{CP-SP}{CP}\times100\) |
| Discount % | \(\frac{MP-SP}{MP}\times100\) |
| Successive % (two steps) | Net % = \(a+b+\frac{ab}{100}\) |
| Compound growth/depreciation | \(A = P(1+\frac{r}{100})^{n}\) |
| Alligation | \(\frac{Q_{cheap}}{Q_{dear}} = \frac{C_{dear}-C_{mean}}{C_{mean}-C_{cheap}}\) |
| Reverse % (find original) | Original = New ÷ \((1\pm\frac{p}{100})\) |
| Percentage point | Change in pp = New % – Old % (no % sign) |
13. Quick‑Check Mnemonics
| Mnemonic | Meaning |
|---|---|
| “F‑D‑P” | Fraction → Decimal → Percent (move decimal two places each step). |
| “P‑L‑D” | Profit, Loss, Discount – all use (difference/base)×100. |
| “10‑5‑1” | For mental %: find 10 %, halve for 5 %, divide 10 % by 10 for 1 %. |
| “A + B + AB/100” | Successive % change formula (A and B can be + or –). |
| “C‑D‑M” | Alligation: Cheaper, Dearer, Mean – place in numerator/denominator as shown. |
| “Half‑Quarter‑Eighth” | Common fraction‑% equivalents: ½=50 %, ¼=25 %, ⅛=12.5 %. |
14. Exam‑Day Tips
- Read the question carefully – identify whether they ask for percentage, percentage point, or actual amount. 2. Convert everything to the same form (either all to % or all to fractions) before applying formulas.
- Use the 10 % method for speed; it reduces chances of arithmetic slip.
- Watch out for successive changes – never just add the percentages unless explicitly stated “simple sum”.
- Check units – if the problem gives area in hectares, volume in litres, or money in rupees, keep the unit throughout; only the numeric part is percent‑based.
- When in doubt, estimate – eliminate options that are far off (e.g., a 150 % increase on a value cannot be less than the original).
- Mark the “base” – the quantity that the percentage is taken of (cost price, original population, marked price, etc.). Mistaking the base is a common error.
15. Practice Set (Solve Quickly – Answers at End)
- What is 22 % of 750? 2. A tank holds 500 L of water. After a leak, it contains 420 L. What percent of water is lost?
- The price of a fertilizer bag increased from ₹ 180 to ₹ 207. Find the percentage increase.
- A shopkeeper allows a discount of 18 % on an item marked at ₹ 2500. What is the selling price?
- If a number is increased by 25 % and then decreased by 20 %, what is the net percent change?
- The population of a town is 80 000. It grows at 6 % per annum. What will be the population after 3 years? (Give nearest whole number.)
- Mix 30 % alcohol solution with 70 % alcohol solution to obtain 50 % alcohol. In what ratio should they be mixed?
- After a 15 % discount, a pair of shoes costs ₹ 1700. Find the original marked price.
- A farmer’s wheat yield increased from 1.2 tons/acre to 1.5 tons/acre. Compute the percentage increase.
- A sum of money becomes ₹ 13,310 after 3 years at 10 % per annum compound interest. Find the principal.
Answers:
- 165
- 16 %
- 15 %
- ₹ 2050
- 0 % (no net change) – actually 25 % up then 20 % down → net = 25‑20‑(25×20/100)=5‑5=0 %
- 95,112 (≈)
- 1:1 (equal parts)
- ₹ 2000
- 25 %
- ₹ 10,000
End of Revision Notes – Review the tables, mnemonics, and shortcuts repeatedly; they are the fastest route to scoring full marks in the percentage section of the JKSSB Social Forestry Worker Basic Mathematics paper. Good luck!
