Percentages Advanced
Key Concepts & Formulas
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | Successive Percentage Change | Net change = A + B + (A×B)/100 (use – for decrease). E.g. +20 % then –10 % → 20 – 10 – (20×10)/100 = 8 % net rise. |
| 2 | Product–constancy (Expenditure) | If price ↑ r %, consumption must ↓ 100r/(100+r) % to keep expenditure same. |
| 3 | Population/Depreciation | Pₙ = P₀(1 ± r/100)ⁿ; use + for growth, – for decay. |
| 4 | Percentage–to–Fraction jump | 37½ % = 3/8; 16⅔ % = 1/6; 6.25 % = 1/16 → speeds up mental work. |
| 5 | Net effect on Area | Both length & width ↑ r % → Area ↑ (2r + r²/100) %. |
| 6 | Faulty–balance & % profit | If shopkeeper sells x g but gives y g, % profit = [(x–y)/y]×100. |
| 7 | Election & votes | Minimum % of votes to win in 2–candidate contest = 50 % + 1 vote. |
10 Practice MCQs
-
After a 25 % hike, a salary becomes ₹6,000. What was the original salary? Answer: ₹4,800
Solution: 6,000 = 125 % → 1 % = 6,000/125 = 48 → 100 % = 48×100 = 4,800.
Shortcut: Original = New/(1 + r/100) = 6,000/1.25 = 4,800.
Tag: Single percentage change. -
A number is increased by 20 % and then decreased by 20 %. The net change is Answer: 4 % decrease
Solution: +20 –20 – (20×20)/100 = –4 %.
Shortcut: Square the change and put – sign: –(20)²/100 = –4 %.
Tag: Successive change. -
Price of sugar rises by 20 %. By what % must consumption fall so that expenditure remains constant? Answer: 16⅔ %
Solution: 100r/(100+r) = 100×20/120 = 50/3 %.
Tag: Product-constancy. -
In an election, the winner gets 64 % votes and wins by 1,080 votes. Total votes polled are Answer: 3,000
Solution: 64 % – 36 % = 28 % ≡ 1,080 → 1 % = 1,080/28 → 100 % = 1,080×100/28 = 3,000.
Tag: Election %. -
A rectangle’s length & breadth both increase by 10 %. Area increases by Answer: 21 %
Solution: 2r + r²/100 = 2×10 + 100/100 = 21 %.
Shortcut: (1.1)² – 1 = 0.21 → 21 %.
Tag: Area % change. -
A trader uses 900 g instead of 1 kg while selling at cost price. His profit % is Answer: 11¹/₉ %
Solution: [(1,000–900)/900]×100 = 100/9 %.
Tag: Faulty balance. -
A machine depreciates 10 % yearly. After 3 years its value is ₹36,450. Original price was Answer: ₹50,000
Solution: 36,450 = P(0.9)³ → P = 36,450/0.729 = 50,000.
Tag: Depreciation. -
If A’s income is 25 % more than B, then B’s income is less than A’s by Answer: 20 %
Solution: 25/(100+25)×100 = 20 %.
Shortcut: r/(100+r) rule.
Tag: Reverse % comparison. -
A man spends 30 % on rent, 20 % of the rest on food, and saves ₹1,120. His income is Answer: ₹2,000
Solution: After rent 70 %; food = 20 % of 70 % = 14 %; total spend 44 % → save 56 % = 1,120 → 1 % = 20 → 100 % = 2,000.
Tag: Multi-layer % expense. -
Population of a town was 40,000. It increases 5 % in first year and decreases 5 % in second year. The population after 2 years is Answer: 39,900
Solution: 40,000×1.05×0.95 = 39,900.
Tag: Successive change.
5 Previous Year Questions
[RRB NTPC 2021] A shirt is marked 40 % above cost price and sold after 10 % discount. Profit % is
Answer: 26 %
Solution: CP = 100 → MP = 140 → SP = 126 → 26 % profit.
Tag: Mark-up & discount.
[RRB Group-D 2019] Price of petrol increased from ₹80 to ₹92. Find % rise.
Answer: 15 %
Solution: (92–80)/80×100 = 15 %.
Tag: Basic % increase.
[RRB ALP 2018] A salary is first reduced by 20 % and then increased by 25 %. Net change is
Answer: No change
Solution: –20 +25 – (20×25)/100 = 0 %.
Tag: Successive change.
[RRB NTPC 2016] In a school, 72 % students are boys and the number of girls is 420. Total students?
Answer: 1,500
Solution: 28 % ≡ 420 → 100 % = 420/0.28 = 1,500.
Tag: Complementary %.
[RRB JE 2015] A number is increased by 10 %, then 10 % of new number is taken. What % of original is the result?
Answer: 11 %
Solution: 1.1N × 0.1 = 0.11N → 11 %.
Tag: Sequential % operation.
Speed Tricks & Shortcuts
| Situation | Shortcut | Example |
|---|---|---|
| Net % change after +r % then –r % | –(r²/100) % | +15 % then –15 % ⇒ –2.25 % |
| Consumption cut when price ↑ r % | ↓ 100r/(100+r) % | r = 25 % ⇒ ↓ 20 % |
| Reverse comparison (A is r % > B) | B is 100r/(100+r) % < A | r = 50 % ⇒ 33⅓ % less |
| Area of rectangle both sides ↑ r % | ↑ (2r + r²/100) % | r = 5 % ⇒ 10.25 % |
| Population n years at r % growth | Multiply by (1 + r/100)ⁿ | 10 % p.a. 3 y ⇒ 1.331× |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Adding % directly | Treating +20 % then –20 % as 0 % | Use successive formula: –4 % |
| Using wrong base in reverse % | “A is 25 % more than B” ⇒ B is 25 % less than A (wrong) | Base is new value: 25/125 = 20 % less |
| Forgetting to square in area change | Thinking 10 % each side ⇒ 10 % area rise | Must use 2r + r²/100 |
| Mixing up depreciation sign | Writing (1 + r/100) for decay | Always (1 – r/100) for depreciation |
Quick Revision Flashcards
| Front | Back |
|---|---|
| Formula for successive +A % & +B % | Net = A+B+AB/100 |
| 37½ % as fraction | 3/8 |
| Price ↑ 20 %; consumption ↓ __ % to keep budget same | 16⅔ % |
| If a value becomes 1.44×, % increase is | 44 % |
| Election: win margin 12 % of total votes; win % = | 56 % |
| Depreciation 10 % p.a. for 2 y; value is __ % of original | 81 % |
| Area both sides ↑ 20 %; area ↑ | 44 % |
| Faulty 800 g instead of 1 kg; profit % | 25 % |
| Reverse: A 20 % < B ⇒ B __ % > A | 25 % |
| Population 10 % growth 3 y; overall growth | 33.1 % |
BETA
