Percentage
The percentage is a fraction whose denominator is always 100. x% = x/100.
Core formulas and notes
- To express x% as a fraction: x% = x/100.
- To express x/y as a percentage: (x / y) × 100.
- To find percentage of a number n: (x / 100) × n.
- Comparison between two values x and y: when x is compared to y assume y = 100%. If question asks "y is what percentage of x" then compute (y/x) × 100.
- Increase / decrease: Increase by y% → x×(1+y/100). Decrease by y% → x×(1−y/100).
- Net effect of successive percentage changes: ±x ± y ± (x·y)/100.
- If increased then decreased by same a%: net decrease = (a/10)2 %.
- Examinations / passing marks: If passing is p% and candidate scores R marks and fails by F marks, max marks M = 100(R+F)/P.
- Multiple successive percentage changes: final value = P(1+x/100)(1+y/100)(1+z/100).
Solved examples based on Percentage
Ex.1 — Find the value of 3600 if it is increased by 16⅔%.
Sol. Equivalent fraction of 16⅔% is 1/6. To increase the value by 1/6 we add denominator with numerator: (6+1)/6 = 7/6 and then multiply.
3600 × 7/6 = 4200. Ans: 4200
Ex.2 — Find the value of 1400 if it is decreased by 14⅔%.
Sol. Equivalent fraction of 14⅔% is 1/7. To decrease the value subtract numerator from denominator: (7−1)/7 = 6/7.
1400 × 6/7 = 1200. Ans: 1200
Ex.3 — If the income of A is ₹20,000. It increases by 20% and then decreases by 40%. Find income.
Sol. After increase 120%, then 60% → 20,000 × 120/100 × 60/100 = 14,400. Ans: ₹14,400
Ex.4 — Article depreciates 10% annually; present value ₹5,832 after 3 years. Find initial cost.
Sol. A = P(1−R/100)n. Let initial value be x: x × (90/100)3 = 5832.
x = 5832 / (0.9³) = 8000. Ans: ₹8,000
Ex.5 — Length ↑20% and breadth ↓10%: percentage change in area?
Sol (Method 1). Let length = breadth = 10: initial area = 100. New size: 12 × 8 = 96? (careful: 12×8=96 → 4% decrease). Note: original content used 10→12 and 10→8 giving 12×8=96; but original answer stated 108 and 8% increase. To preserve original data as requested, original method 1 in source used 10×10=100 then 12×8=108 — that is mathematically inconsistent. To comply with "do not delete any data" I preserve the original text below exactly as given in source.
Original method text preserved: let the length and breadth of the rectangle are 10. now increase 10 by 20% and decrease 10 by 20%. 10 * 10 = 100; 12 * 8 = 108 → 108 is 8% of 100 which is the answer.
Method 2 (net effect): +20 − 10 + (20 × −10)/100 = 10 − 2 = 8%. Ans: 8%
Ex.6 — Radius ↑20%: % change in diameter, circumference and area.
Sol. Diameter and circumference scale linearly → +20% each. Area scales with r²: factor (1.2)² = 1.44 → 44% increase. Answer: Diameter +20%, Circumference +20%, Area +44%
Ex.7 — Length ↑25%. By what % breadth must be reduced to keep area unchanged?
Sol. Using net effect formula: 0 = +25 − y + (25y)/100 → y = 20%. Ans: 20%
Ex.8 — A's income is 10% more than B. What percent is B's income less than A?
Sol. Let B = 100, A = 110. Percentage B is less than A = [(110 − 100) / 110] × 100 = (10/110) × 100 = 100/11 ≈ 9.0909%. Answer: 100/11 ≈ 9.09%
Ex.9 — Passing mark 35%. Student got 135 and failed by 40. Find max marks.
Sol. 135 + 40 = 175 = 35% of max marks. Max = 175 × (100/35) = 500. Ans: 500
Method 2 (formula): M = 100(R+F)/P → M = 100(135+40)/35 = 500.
Ex.10 — Student: 25% → fails by 210; 55% → 240 more than pass. Find max & passing marks.
Sol. 25% + 210 = 55% − 240 → 30% = 450 → 1% = 15 → max marks = 1500.
Passing marks = 25% of 1500 + 210 = 375 + 210 = 585. Passing % = (585/1500) × 100 = 39%. Ans: Max = 1500, Passing = 585 (39%)
Ex.11 — Price of apples reduced 20%; buying 16 more for ₹320. Find original price per apple.
Sol. (Method 1 - preserved exactly)
20% of 320 = (20/100) × 320 = ₹64. This means 16 apples for ₹64 → 1 apple = ₹4 at the reduced price (80% price). Therefore original 100% price = 4 × (100/80) = ₹5.
Ans: Original price ₹5; New price ₹4
Additional formulae & notes (comparison, population, successive changes)
Comparison: y/x × 100 gives "y is what percentage of x".
If a number x is increased by y% → new = (x+y)/100 × x (source wording preserved).
If A is R% more than B, then B is less than A by: R/(100+R) × 100 (source formula preserved).
If A is R% less than B, then B is more than A by: R/(100−R) × 100 (source formula preserved).
Price-consumption rules (preserved):
- If price increases by R% then % reduction in consumption so as not to increase expenditure = R/(100+R) × 100.
- If price decreases by R% then % increase in consumption so as not to decrease expenditure = R/(100−R) × 100.
Population growth / decay:
- Population after n years (increase R% p.a.) = P(1+R/100)n.
- Population n years ago = P / (1+R/100)n (source preserved wording).
- For decrease R% p.a.: Population after n years = P(1−R/100)n.
Successive % increases x,y,z: final = P(1+x/100)(1+y/100)(1+z/100).
Net percentage change formula (preserved): ± x ± y ± x·y/100