Percentage

A percentage is a fraction whose denominator is always 100, so x% = x/100.

Percentage with solved examples

Core formulas and notes

  • Percentage as a fractionx% = x/100
  • Fraction as a percentage(x / y) × 100
  • x% of a number n(x / 100) × n
  • Comparison of two valuesThe value you compare against is taken as 100%. "y is what percentage of x" = (y / x) × 100.
  • Increase / decrease by y%Increase: x × (1 + y/100)  |  Decrease: x × (1 − y/100)
  • Net effect of two successive changes (x% and y%)±x ± y ± (x·y)/100
  • Increase then decrease by the same a%Net decrease = (a/10)2 %
  • Passing marksIf the pass mark is P%, a candidate scores R and fails by F, then maximum marks M = 100(R + F) / P.
  • Successive changes x%, y%, z%Final value = P(1 + x/100)(1 + y/100)(1 + z/100)

Solved examples based on Percentage

Ex.1Find the value of 3600 if it is increased by 16⅔%.

Sol. 16⅔% = 1/6. To increase a value by 1/6, add the numerator to the denominator: (6 + 1)/6 = 7/6, then multiply.

3600 × 7/6 = 4200. Ans: 4200

Ex.2Find the value of 1400 if it is decreased by 142⁄7%.

Sol. 142⁄7% = 1/7. To decrease a value by 1/7, subtract the numerator from the denominator: (7 − 1)/7 = 6/7, then multiply.

1400 × 6/7 = 1200. Ans: 1200

Ex.3A's income is ₹20,000. It increases by 20% and then decreases by 40%. Find the final income.

Sol. After the increase the income is 120%, then 60% of that remains.

20,000 × 120/100 × 60/100 = 14,400. Ans: ₹14,400

Ex.4An article depreciates 10% annually. Its value after 3 years is ₹5,832. Find the initial cost.

Sol. A = P(1 − R/100)n. Let the initial value be x.

x × (90/100)3 = 5832, so x = 5832 / 0.729 = 8000. Ans: ₹8,000

Ex.5Length of a rectangle increases by 20% and breadth decreases by 10%. Find the percentage change in area.

Method 1. Let length = breadth = 10, so the original area = 10 × 10 = 100.

New length = 10 + 20% = 12, new breadth = 10 − 10% = 9. New area = 12 × 9 = 108.

Change = 108 − 100 = 8, which is 8% of 100.

Method 2 (net effect). +20 − 10 + (20 × −10)/100 = 10 − 2 = 8%.

Ans: 8% increase

Ex.6The radius of a circle increases by 20%. Find the percentage change in diameter, circumference and area.

Sol. Diameter and circumference change in the same ratio as the radius, so both increase by 20%.

Area depends on r², so the factor is (1.2)2 = 1.44, an increase of 44%.

Ans: Diameter +20%, Circumference +20%, Area +44%

Ex.7The length of a rectangle increases by 25%. By what percentage must the breadth be reduced to keep the area unchanged?

Sol. Let the breadth decrease by y%. Using the net effect formula with net change 0:

0 = 25 − y − (25 × y)/100, so 25 = 1.25y and y = 20. Ans: 20%

Ex.8A's income is 10% more than B's. By what percentage is B's income less than A's?

Sol. Let B = 100, so A = 110.

Required % = (110 − 100)/110 × 100 = 100/11 ≈ 9.09%. Ans: 100/11 % ≈ 9.09%

Ex.9The passing mark is 35%. A student got 135 marks and failed by 40. Find the maximum marks.

Method 1. 135 + 40 = 175 = 35% of the maximum marks, so maximum = 175 × 100/35 = 500.

Method 2 (formula). M = 100(R + F)/P = 100(135 + 40)/35 = 500.

Ans: 500

Ex.10A student who scores 25% fails by 210 marks. One who scores 55% gets 240 more than the pass mark. Find the maximum marks and the passing marks.

Sol. 25% + 210 = 55% − 240, so 30% = 450 and 1% = 15. Maximum marks = 1500.

Passing marks = 25% of 1500 + 210 = 375 + 210 = 585. Passing % = 585/1500 × 100 = 39%.

Ans: Maximum = 1500, Passing = 585 (39%)

Ex.11The price of apples is reduced by 20%, so 16 more apples can be bought for ₹320. Find the original price per apple.

Sol. 20% of 320 = ₹64, which buys the 16 extra apples at the new price. So the new price = 64/16 = ₹4 per apple.

The new price is 80% of the original, so original price = 4 × 100/80 = ₹5.

Ans: Original ₹5, new ₹4

Additional formulae and notes

Comparison and successive changes

y is what percentage of x: (y/x) × 100.

If x is increased by y%: new value = x × (1 + y/100).

If A is R% more than B, B is less than A by R/(100 + R) × 100 %.

If A is R% less than B, B is more than A by R/(100 − R) × 100 %.

Net percentage change: ±x ± y ± (x·y)/100

Price and consumption
  • If the price increases by R%, consumption must fall by R/(100 + R) × 100 % to keep expenditure the same.
  • If the price decreases by R%, consumption must rise by R/(100 − R) × 100 % to keep expenditure the same.
Population growth and decay
  • Population after n years (increase R% p.a.) = P(1 + R/100)n
  • Population n years ago = P / (1 + R/100)n
  • Population after n years (decrease R% p.a.) = P(1 − R/100)n
  • Successive increases of x%, y%, z%: final = P(1 + x/100)(1 + y/100)(1 + z/100)