Episode 8 — Aptitude and Reasoning / 8.1 — Percentage
8.1.a Concepts and Formulas
1. What Does "Percentage" Mean?
The word percentage comes from the Latin per centum -- "for every hundred."
When we say "45%", we mean 45 out of every 100, or equivalently:
45% = 45/100 = 0.45
The symbol % is simply shorthand for "divided by 100."
2. Converting Between Fractions, Decimals, and Percentages
2.1 Fraction to Percentage
Multiply the fraction by 100 and attach the % sign.
Fraction --> Percentage: (Fraction) x 100 %
Example:
3/5 = (3/5) x 100 % = 60%
2.2 Percentage to Fraction
Divide by 100 and simplify.
Percentage --> Fraction: P% = P/100
Example:
75% = 75/100 = 3/4
2.3 Decimal to Percentage
Multiply the decimal by 100.
Decimal --> Percentage: Decimal x 100 %
Example:
0.375 = 0.375 x 100 % = 37.5%
2.4 Percentage to Decimal
Divide by 100 (move the decimal point two places left).
Percentage --> Decimal: P% = P / 100
Example:
12.5% = 12.5 / 100 = 0.125
2.5 Fraction to Decimal
Divide the numerator by the denominator.
3/8 = 3 / 8 = 0.375
2.6 Decimal to Fraction
Write the decimal over the appropriate power of 10, then simplify.
0.65 = 65/100 = 13/20
3. The Basic Percentage Formula
Percentage = (Part / Whole) x 100
This is the most fundamental formula. Everything else is derived from it.
Three rearrangements:
Part = (Percentage / 100) x Whole
Whole = (Part x 100) / Percentage
Percentage = (Part / Whole) x 100
Worked Example:
In a class of 80 students, 52 passed an exam. What percentage passed?
Percentage = (Part / Whole) x 100
= (52 / 80) x 100
= 0.65 x 100
= 65%
4. Finding a Percentage of a Number
To find P% of a number N:
P% of N = (P / 100) x N
Worked Example:
Find 35% of 240.
35% of 240 = (35/100) x 240
= 0.35 x 240
= 84
5. Finding the Whole When a Percentage Is Known
If P% of a number is V, then:
Whole = (V x 100) / P
Worked Example:
15% of a number is 45. Find the number.
Whole = (45 x 100) / 15
= 4500 / 15
= 300
Verification: 15% of 300 = 45. Correct.
6. Percentage Increase
When a value goes UP from an Old value to a New value:
Percentage Increase = [(New - Old) / Old] x 100
Key point: The denominator is always the original (old) value.
Worked Example:
A shirt's price increased from Rs 400 to Rs 480. Find the percentage increase.
% Increase = [(480 - 400) / 400] x 100
= [80 / 400] x 100
= 20%
7. Percentage Decrease
When a value goes DOWN from an Old value to a New value:
Percentage Decrease = [(Old - New) / Old] x 100
Worked Example:
A phone's price dropped from Rs 15,000 to Rs 12,750. Find the percentage decrease.
% Decrease = [(15000 - 12750) / 15000] x 100
= [2250 / 15000] x 100
= 15%
8. The Multiplier Method
This is the most efficient way to handle percentage changes.
8.1 For Increase
If a value increases by r%, the new value is:
New Value = Old Value x (1 + r/100)
The term (1 + r/100) is called the multiplier.
Example: Increase of 20% --> Multiplier = 1.20
If original = 500, new value = 500 x 1.20 = 600
8.2 For Decrease
If a value decreases by r%, the new value is:
New Value = Old Value x (1 - r/100)
Example: Decrease of 15% --> Multiplier = 0.85
If original = 500, new value = 500 x 0.85 = 425
8.3 Finding Original from New Value
Old Value = New Value / Multiplier
Worked Example:
After a 25% increase, a price becomes Rs 750. What was the original price?
Multiplier = 1 + 25/100 = 1.25
Original = 750 / 1.25 = Rs 600
9. Successive Percentage Changes
When two percentage changes occur one after another, the net effect is NOT simply their sum.
9.1 Formula for Two Successive Changes
If two successive changes are a% and b% (use negative sign for decrease):
Net % change = a + b + (ab / 100)
Worked Example:
A price increases by 20% and then decreases by 10%. What is the net percentage change?
a = +20, b = -10
Net change = 20 + (-10) + (20 x (-10)) / 100
= 20 - 10 - 200/100
= 20 - 10 - 2
= 8%
Net effect = 8% increase
9.2 Using Multipliers for Successive Changes
Alternatively, multiply the individual multipliers:
Net Multiplier = (1 + a/100) x (1 + b/100)
Same example:
Net Multiplier = 1.20 x 0.90 = 1.08
Net change = 1.08 - 1 = 0.08 = 8% increase
9.3 Three or More Successive Changes
For three successive changes a%, b%, and c%:
Net Multiplier = (1 + a/100) x (1 + b/100) x (1 + c/100)
Worked Example:
A population increases by 10%, then 20%, then decreases by 5%. If initial population is 10,000, find the final population.
Final = 10000 x 1.10 x 1.20 x 0.95
= 10000 x 1.254
= 12,540
10. Percentage Points vs Percentage Change
These are not the same thing. This is a very common source of confusion.
10.1 Percentage Points
The arithmetic difference between two percentages.
If interest rates go from 8% to 12%, the change is 4 percentage points.
10.2 Percentage Change
The relative change calculated using the standard formula.
If interest rates go from 8% to 12%:
% Change = [(12 - 8) / 8] x 100 = 50% increase
So the rate increased by 4 percentage points but the percentage increase is 50%.
11. Important Derived Results
11.1 If A is r% more than B, then B is less than A by:
B is less than A by = [r / (100 + r)] x 100 %
Worked Example:
If A's salary is 25% more than B's, by what percent is B's salary less than A's?
Required % = [25 / (100 + 25)] x 100
= [25 / 125] x 100
= 20%
B's salary is 20% less than A's.
Why this works: If B = 100, then A = 125. Difference = 25. Now the base is A = 125, not B.
(25 / 125) x 100 = 20%
11.2 If A is r% less than B, then B is more than A by:
B is more than A by = [r / (100 - r)] x 100 %
Worked Example:
If the sale price is 20% less than MRP, by what percent is MRP more than sale price?
Required % = [20 / (100 - 20)] x 100
= [20 / 80] x 100
= 25%
MRP is 25% more than the sale price.
12. Population / Depreciation (Growth & Decay)
12.1 Population Growth
Population after n years = P x (1 + r/100)^n
Where:
P = present population
r = annual growth rate (%)
n = number of years
12.2 Depreciation (Value Decay)
Value after n years = V x (1 - r/100)^n
Where:
V = present value
r = annual depreciation rate (%)
n = number of years
Worked Example:
A machine worth Rs 50,000 depreciates at 10% per annum. Find its value after 3 years.
Value = 50000 x (1 - 10/100)^3
= 50000 x (0.90)^3
= 50000 x 0.729
= Rs 36,450
13. Expenditure and Consumption Problems
A standard pattern: if the price of a commodity changes and the total expenditure is fixed, how does consumption change?
Expenditure = Price x Consumption
If price increases by r% and expenditure stays the same:
Reduction in consumption = [r / (100 + r)] x 100 %
If price decreases by r% and expenditure stays the same:
Increase in consumption = [r / (100 - r)] x 100 %
Worked Example:
Sugar price increases by 25%. By what percent must a family reduce consumption to keep expenditure the same?
Reduction = [25 / (100 + 25)] x 100
= [25 / 125] x 100
= 20%
14. Election / Voting Problems
In a two-candidate election:
Total valid votes = Total votes - Invalid votes
Winner's votes = P% of valid votes
Loser's votes = (100 - P)% of valid votes
Difference = Winner's votes - Loser's votes
Worked Example:
In an election, candidate A gets 60% of valid votes and wins by 4,800 votes. 20% of total votes were invalid. Find total votes.
Difference in vote share = 60% - 40% = 20% of valid votes
20% of valid votes = 4800
Valid votes = 4800 x 100 / 20 = 24,000
Valid votes = 80% of total votes (since 20% invalid)
Total votes = 24,000 x 100 / 80 = 30,000
15. Mixture Problems Involving Percentages
Problem type: A solution has a certain concentration (percentage) of a substance. Water or substance is added/removed.
Key principle: The amount of pure substance stays constant (unless substance is added/removed).
Worked Example:
40 litres of a mixture contains 10% alcohol. How much alcohol must be added so the concentration becomes 20%?
Initial alcohol = 10% of 40 = 4 litres
Let x litres of alcohol be added.
New alcohol = (4 + x) litres
New total = (40 + x) litres
(4 + x) / (40 + x) = 20/100 = 1/5
5(4 + x) = 40 + x
20 + 5x = 40 + x
4x = 20
x = 5 litres
Summary of All Formulas
1. Percentage = (Part / Whole) x 100
2. Part = (Percentage / 100) x Whole
3. Whole = (Part x 100) / Percentage
4. % Increase = [(New - Old) / Old] x 100
5. % Decrease = [(Old - New) / Old] x 100
6. New (after r% increase) = Old x (1 + r/100)
7. New (after r% decrease) = Old x (1 - r/100)
8. Successive changes (a,b) = a + b + ab/100
9. A is r% more than B,
B is less than A by = [r/(100+r)] x 100 %
10. A is r% less than B,
B is more than A by = [r/(100-r)] x 100 %
11. Population growth = P x (1 + r/100)^n
12. Depreciation = V x (1 - r/100)^n
13. Price up r%, reduce
consumption by = [r/(100+r)] x 100 %