1. Percentage

Basic Mathematics – Quick‑Revision Notes

(Tailored for JKSSB Social Forestry Worker Examination)

---

1. Percentage

Why it matters: Appears in profit‑loss, interest, mixture, and data‑interpretation questions.

Core Formulae

Concept	Formula	When to Use
Finding % of a number	\( \displaystyle \frac{x}{100}\times N \)	“What is 15 % of 240?”
Finding what % one number is of another	\( \displaystyle \frac{A}{B}\times100 \)	“What % is 30 of 120?”
Increase/Decrease by %	New value = \( N \times \left(1\pm\frac{p}{100}\right) \)	Price after 12 % hike, population after 8 % drop
Successive % changes	Overall change = \( (1+\frac{p_1}{100})(1+\frac{p_2}{100})-1 \)	Two‑step discount, compound growth
Percentage point vs. % change	A change from 20 % to 25 % is +5 percentage points (or +25 %)	Avoid confusion in data‑interpretation

Mnemonics

• “PER‑CENT = PER (OUT OF) CENT (100)” → Remember to always divide/multiply by 100.
• “INC‑DEC = 1 ± p/100” → For increase use +, for decrease use –.

Short‑cut Tricks

1. To find 1 % → divide the number by 100. Then scale up (e.g., 7 % = 1 % × 7).
2. To find 10 % → move decimal one place left. 5 % = half of 10 %.
3. To find 25 % → quarter the number (divide by 4).
4. To find 33⅓ % → one‑third of the number.
5. To find 66⅔ % → two‑thirds (or subtract 33⅓ %).

Typical Exam‑Style Problems

• If a forest nursery produces 850 saplings and 12 % die, how many survive?

Surviving = \(850 \times (1-0.12)=850\times0.88=748\). – The price of timber rose from ₹450 per quintal to ₹540. What is the percentage increase?

Increase = \(\frac{540-450}{450}\times100 = \frac{90}{450}\times100 =20\%\).

---

2. Average (Mean)

Why it matters: Used in estimating yields, workforce productivity, and analyzing test scores.

Core Formulae

Situation	Formula	Remarks
Simple average of n numbers	\( \displaystyle \bar{x}= \frac{\sum_{i=1}^{n} x_i}{n} \)	Direct sum divided by count
Weighted average	\( \displaystyle \bar{x}= \frac{\sum w_i x_i}{\sum w_i} \)	Weights = frequency, area, etc.
Average of two groups combined	\( \displaystyle \bar{x}_{comb}= \frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2} \)	Useful for merging data from two plantations
Change in average when a value is replaced	New avg = Old avg + \(\frac{\text{New value}-\text{Old value}}{n}\)	Quick update without re‑summing
Average of first n natural numbers	\( \displaystyle \frac{n+1}{2} \)	Derived from sum formula \(n(n+1)/2\)
Average of first n even numbers	\( n+1 \)	Because series is 2,4,…,2n
Average of first n odd numbers	\( n \)	Series 1,3,…,(2n‑1)

Mnemonics

• “AVG = SUM ÷ COUNT” – Visualize a scale balancing total weight against number of items.
• “WEIGHTED AVG = (WEIGHT×VALUE) SUM ÷ TOTAL WEIGHT” – Think of a seesaw where each weight pulls the balance.

Short‑cut Tricks

1. If all numbers increase by a constant k, the average also increases by k.
2. If all numbers are multiplied by a constant k, the average is multiplied by k.
3. To find the missing number when average is known:

Missing = \( (Desired\ Avg \times Total\ count) – Sum\ of\ known\ numbers \).

Typical Exam‑Style Problems

• The average height of 15 saplings is 120 cm. If one sapling of height 90 cm is removed, what is the new average?

Total height = 15 × 120 = 1800 cm.

After removal: new total = 1800 − 90 = 1710 cm; new count = 14.

New average = 1710 ÷ 14 ≈ 122.14 cm.

• In a plantation, 30 % of the area yields 8 qt/ha, 50 % yields 10 qt/ha, and the rest yields 6 qt/ha. Find the overall average yield.

Weighted avg = \(0.30×8 + 0.50×10 + 0.20×6 = 2.4 + 5 + 1.2 = 8.6\) qt/ha.

---

3. Time and Work

Why it matters: Determines manpower requirements for planting, weeding, harvesting, and maintenance tasks.

Core Concepts

• Work = Rate × Time
• Rate (work per unit time) = \( \frac{1}{\text{Time to complete whole job}} \) (if one person does the job alone).
• When multiple agents work together, their rates add.

Formulae Summary

Scenario	Formula	When to Use
One person completes work in D days	Rate = \( \frac{1}{D} \) (job/day)	Baseline
n identical workers, each takes D days alone	Combined rate = \( \frac{n}{D} \) → Time together = \( \frac{D}{n} \)	Same efficiency
Workers with different efficiencies (A takes a days, B takes b days)	Combined rate = \( \frac{1}{a} + \frac{1}{b} \) → Time = \( \frac{1}{\frac{1}{a}+\frac{1}{b}} \)	Different skill levels
Work done in t days by a worker with rate r	Work = \( r \times t \)	Partial completion
Pipe & Cistern (inlet fills, outlet empties)	Net rate = Σ(inlet rates) – Σ(outlet rates)	Analogous to work problems
Man‑Day concept	Total work = (Number of men) × (Number of days)	Useful when work is proportional to manpower

Mnemonics

• “RATE = 1⁄TIME” – Flip the time to get the rate.
• “ADD RATES, NOT TIMES” – When people work together, sum their rates.
• “MAN‑DAY = WORK UNIT” – Think of a man‑day as a packet of work.

Short‑cut Tricks

1. If A can do a job in a days and B in b days, together they finish in \( \displaystyle \frac{ab}{a+b} \) days. (Derived from harmonic mean).
2. If A is twice as good as B, then A’s time = \( \frac{1}{2} \) × B’s time.
3. To find time for a fraction of work: Multiply the full‑job time by the fraction.

E.g., if a job takes 20 days, 3/5 of it takes \(20×\frac{3}{5}=12\) days.

1. When a worker leaves after t days:

Work done = \( r × t \). Remaining work = 1 – (r×t). Then compute time for rest with remaining workers.

Typical Exam‑Style Problems – A labourer can plant 200 saplings in 5 days. How many labourers are needed to plant 1200 saplings in 3 days?

Rate of one labourer = \(200/5 = 40\) saplings/day.

Required total rate = \(1200/3 = 400\) saplings/day.

Number of labourers = \(400/40 = 10\).

• Two pumps fill a tank. Pump A fills it in 6 h, Pump B empties it in 9 h. If both are opened together, how long to fill the tank?

Fill rate = \(1/6\) tank/h, Empty rate = \(1/9\) tank/h.

Net rate = \(1/6 – 1/9 = (3-2)/18 = 1/18\) tank/h.

Time = \(18\) h.

---

4. Ratio and Proportion Why it matters: Used in mixing fertilizers, allocating land, comparing growth rates, and solving partnership problems.

Core Definitions

• Ratio \(a:b\) = \(\frac{a}{b}\) (compare two quantities of same kind).
• Proportion \(a:b = c:d\) means \(\frac{a}{b} = \frac{c}{d}\).
• Continued proportion: \(a:b = b:c\) → \(b^2 = ac\).

Formulae & Rules

Rule	Expression	Application
Componendo & Dividendo	If \(\frac{a}{b} = \frac{c}{d}\) then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\)	Simplifies complex ratio equations
Invertendo	\(\frac{a}{b} = \frac{c}{d} \Rightarrow \frac{b}{a} = \frac{d}{c}\)	Flip ratios
Alternendo	\(\frac{a}{b} = \frac{c}{d} \Rightarrow \frac{a}{c} = \frac{b}{d}\)	Swap means
Addendo	\(\frac{a}{b} = \frac{c}{d} \Rightarrow \frac{a+c}{b+d} = \frac{a}{b}\)	Useful for adding same ratio
Subtrahendo	\(\frac{a}{b} = \frac{c}{d} \Rightarrow \frac{a-c}{b-d} = \frac{a}{b}\)	Subtract same ratio
Mean Proportional	If \(a:x = x:b\) then \(x = \sqrt{ab}\)	Geometric mean
Third Proportional	If \(a:b = b:c\) then \(c = \frac{b^2}{a}\)	Extension of ratio
Fourth Proportional	If \(a:b = c:d\) then \(d = \frac{bc}{a}\)	Direct proportion

Mnemonics

• “C-D: Componendo‑Dividendo = (sum)/(difference)” – Remember the “+ / –” pattern.
• “I‑A‑S: Invertendo, Alternendo, Subtrahendo” – Helps recall the three transformation tricks.
• “MEAN = √(product)” – For mean proportional, think of geometric mean.

Short‑cut Tricks

1. To split a quantity Q in ratio p:q, compute:

• Part for p = \(Q × \frac{p}{p+q}\)
• Part for q = \(Q × \frac{q}{p+q}\)

1. If a:b = c:d, then \(ad = bc\) (cross‑multiplication). 3. To find the ratio of three numbers given pairwise ratios, align the common term. Example: \(a:b = 2:3\) and \(b:c = 4:5\) → make b equal: \(a:b = 8:12\), \(b:c = 12:15\) → \(a:b:c = 8:12:15\). 4. Inverse ratio – If speed ↑, time ↓ for same distance: speed ratio = inverse of time ratio.

Typical Exam‑Style Problems

• A fertilizer mix requires nitrogen, phosphorus, and potassium in the ratio 4:2:1. If 35 kg of phosphorus is used, how much nitrogen and potassium are needed?

Ratio N:P:K = 4:2:1 → total parts = 7.

Phosphorus corresponds to 2 parts → 1 part = 35/2 = 17.5 kg.

Nitrogen = 4×17.5 = 70 kg.

Potassium = 1×17.5 = 17.5 kg.

• The ratio of the number of trees planted by two teams A and B is 5:3. If team A planted 200 more trees than team B, how many trees did each plant?

Let A = 5x, B = 3x.

5x – 3x = 200 → 2x = 200 → x = 100. Hence A = 500 trees, B = 300 trees. —

5. Integrated Problem‑Solving Strategies

Step	Action	Example (Social Forestry Context)
1️⃣ Identify the concept	Look for keywords: percent, average, time‑work, ratio	“What percent of the saplings survived?” → Percentage
2️⃣ Extract given data	Write numbers, units, and what is unknown	850 saplings, 12 % died → find survivors
3️⃣ Choose the appropriate formula	Match to concept table	Surviving = Total × (1 − %/100)
4️⃣ Plug‑in & simplify	Perform arithmetic, cancel units	850 × 0.88 = 748
5️⃣ Check reasonableness	Does answer lie between logical bounds?	Survivors < total, >0 → OK
6️⃣ State answer with unit	Include “saplings”, “days”, “kg”, etc.	748 saplings

Quick‑Check List (keep handy on the back of your answer sheet):

• ☐ Percent → multiply/divide by 100.
• ☐ Average → sum ÷ count (or weighted).
• ☐ Time‑Work → rate = 1/time; add rates for combined work.
• ☐ Ratio → cross‑multiply; use part‑fraction method for splitting.

---

6. Key Highlights (One‑Liners for Last‑Minute Review)

• Percentage – “To find x % of N, move decimal two places left then multiply by x.”
• Average – “If every value increases by k, the average increases by k.”
• Time & Work – “Together time = \(\frac{ab}{a+b}\) for two workers.”
• Ratio – “To split Q in ratio a:b, Q×a/(a+b) and Q×b/(a+b).”
• Componendo‑Dividendo – “\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\).”
• Mean Proportional – “√(product) gives the middle term in a:b = b:c.”
• Work Unit – “Man‑day = 1 worker × 1 day; total work = men × days.”
• Percentage Point vs. % Change – “5 pp ≠ 5 %; the former is absolute, the latter relative.”
• Weighted Avg – “Think of each value pulling the average with its weight.”
• Inverse Ratio – “Speed ↑ ⇒ Time ↓ for same distance (ratio flips).”

---

7. Practice Problems (No Solutions – for Self‑Test)

1. A nursery produced 1 200 saplings. If 18 % were damaged during transport, how many saplings arrived intact?
2. The average yield of three plots is 9 qt/ha. Two plots give yields of 7 qt/ha and 11 qt/ha. Find the yield of the third plot. 3. Four workers can complete a forest‑clearing job in 6 days. How many days will 6 workers take, assuming equal efficiency?
3. The ratio of the area allocated to teak and eucalyptus in a plantation is 7:5. If the total area is 240 ha, find the area under each species.
4. A mixture of two fertilizers contains nitrogen and phosphorus in the ratio 3:2. If the mixture has 15 kg of nitrogen, how much phosphorus does it contain?
5. A pump can fill a water tank in 4 hours, while a leak can empty it in 6 hours. If both are opened simultaneously, how long will it take to fill the tank? 7. The average score of 12 students in a test is 68. If two students scoring 75 and 80 are added, what becomes the new average?
6. In a proportion, \( \frac{5}{x} = \frac{15}{45} \). Find x. 9. A team of 5 labourers can plant 350 saplings in 5 days. How many saplings can 8 labourers plant in 7 days?
7. The ratio of the ages of two forest officers is 4:3. Five years later, the sum of their ages will be 55 years. Find their present ages.

After attempting, verify your answers using the formulae and shortcuts discussed.

—

End of Revision Notes. Keep this sheet handy, revise the bullet points and tables daily, and you’ll be ready to tackle the mathematics section of the Social Forestry Worker exam with confidence. Good luck!