Percentage — SSC CGL Quant Notes

Concepts and rules

Percent means 'per hundred': x% = x/100. So 35% = 35/100 = 7/20 = 0.35, and 3/8 = 3/8 × 100% = 37.5%.

Fraction table to know by heart: 1/2 = 50%, 1/3 = 33 1/3%, 1/4 = 25%, 1/5 = 20%, 1/6 = 16 2/3%, 1/7 = 14 2/7%, 1/8 = 12.5%.

More of the table: 1/9 = 11 1/9%, 1/11 = 9 1/11%, 1/12 = 8 1/3%, 1/16 = 6.25%, 1/20 = 5%, 1/25 = 4%, 1/40 = 2.5%.

x% of y = y% of x, because both equal x × y/100. So 16% of 25 = 25% of 16 = 4.

'A is what percent of B' = A/B × 100. The number after 'of' is the base (the 100%).

Percentage change = (new − old)/old × 100. The base is always the old (original) value, never the new one.

If A is r% more than B, then A = B × (100 + r)/100, and B is less than A by r/(100 + r) × 100%. If A is r% less than B, then B is more than A by r/(100 − r) × 100%.

Chained percentages multiply: if A is x% of B and B is y% of C, then A is (x × y/100)% of C.

Percentage points measure the plain difference between 2 percentages; a rise from 20% to 25% is 5 percentage points but a 25% relative rise.

Method and shortcuts

Finding the whole: if p% of N is K, then N = K × 100/p. Find 1% first if the numbers are awkward, then scale.

Replace percentages by fractions from the table: 37.5% of 480 = 3/8 × 480 = 180, which is faster than multiplying by 0.375.

For comparison questions, give the base a convenient value of 100 (or 100k) and build every other quantity from it.

Base-switch shortcut: r% more ⇒ r/(100 + r) less; r% less ⇒ r/(100 − r) more. 25% more ⇒ 20% less; 20% less ⇒ 25% more.

Shortcut limit: never add or subtract percentages that have different bases; convert each to the same base first.

Example 1: percent of a number

Question: Find 12.5% of 640.

Solution: 12.5% = 1/8, so the value is 640 ÷ 8 = 80.

Answer: 80

Example 2: switching the base

Question: Anil's salary is 25% more than Bharat's. By what percent is Bharat's salary less than Anil's?

Solution: Let Bharat = 100, so Anil = 125. The gap is 25 on a base of 125: 25/125 × 100 = 20%.

Answer: 20%

Example 3: from one percent to another

Question: 35% of a number is 84. What is 65% of the number?

Solution: 1% = 84 ÷ 35 = 2.4, so the number is 240. Then 65% = 65 × 2.4 = 156.

Answer: 156

Example 4: chained comparisons

Question: A is 20% more than B, and B is 25% less than C. A is what percent of C? (a) 90% (b) 95% (c) 85% (d) 105%

Solution: Let C = 100. B = 75. A = 75 × 120/100 = 90. So A is 90% of C.

Option (b) 95% comes from wrongly doing 100 + 20 − 25; percentages on different bases cannot be added.

Answer: (a) 90%

Example 5: percentage from an equation

Question: If 20% of (A + B) = 50% of (A − B), then B is what percent of A? (a) 30% (b) 233 1/3% (c) 42 6/7% (d) 60%

Solution: 0.2A + 0.2B = 0.5A − 0.5B, so 0.7B = 0.3A and B/A = 3/7.

3/7 × 100% = 300/7% = 42 6/7%. Option (b) is A as a percent of B, the reversed base.

Answer: (c) 42 6/7%

Example 6: base values from a rupee gap

Question: Asha's salary is 40% more than Bina's, and Chetan's salary is 20% less than Asha's. If Chetan earns ₹2,400 more than Bina, Bina's salary is: (a) ₹24,000 (b) ₹20,000 (c) ₹18,000 (d) ₹22,500

Solution: Let Bina = 100k. Asha = 140k. Chetan = 140k × 80/100 = 112k.

Chetan − Bina = 12k = ₹2,400, so k = 200 and Bina = 100 × 200 = ₹20,000. Check: Asha ₹28,000, Chetan ₹22,400.

Answer: (b) ₹20,000

Common mistakes

Taking the new value as the base for percentage change. Right: a rise from 80 to 100 is 20/80 = 25%, not 20/100 = 20%.

Assuming 'A is 25% more than B' means 'B is 25% less than A'. Right: B is 20% less than A.

Adding percentages of different bases, as in 20% more than B and 25% less than C. Right: build each from one base value.

Confusing percent with percentage points. Right: 10% to 12% is a 2 percentage point rise and a 20% relative rise.

Answering 'B as a percent of A' with A/B. Right: the quantity after 'of' goes in the denominator.

Practice questions

Q1. If 14 2/7% of a number is 36, the number is: (a) 245 (b) 252 (c) 260 (d) 216

Q2. The price of a pen is 60% of the price of a diary. The price of the diary is how much percent more than that of the pen? (a) 40% (b) 60% (c) 66 2/3% (d) 33 1/3%

Q3. If 40% of x = 25% of y, then x is what percent of y? (a) 160% (b) 37.5% (c) 75% (d) 62.5%

Answers

Q1. (b) 14 2/7% = 1/7, so the number is 36 × 7 = 252.

Q2. (c) Diary = 100, pen = 60; 40/60 × 100 = 66 2/3%.

Q3. (d) 0.4x = 0.25y gives x/y = 25/40 = 0.625 = 62.5%.

Quick revision

Base = the quantity after 'of' or the original value.

r% more ⇒ r/(100 + r) × 100% less; r% less ⇒ r/(100 − r) × 100% more.

Take the base as 100 or 100k and build the rest from it.

Know 1/6 = 16 2/3%, 1/7 = 14 2/7%, 1/8 = 12.5%, 1/9 = 11 1/9%, 1/12 = 8 1/3%.

Concepts and rules

Each change is a multiplying factor: an r% increase multiplies by (100 + r)/100 and an r% decrease by (100 − r)/100. So +25% is × 5/4 and −20% is × 4/5.

Net of 2 successive changes a% and b%: a + b + ab/100 percent, with a decrease entered as negative. The order of the changes does not matter.

An increase of r% followed by a decrease of r% (or the reverse) always gives a net loss of r²/100 percent: +10% then −10% gives −1%.

Restoring: after an r% decrease, you need an increase of r/(100 − r) × 100% to return; after an r% increase, a decrease of r/(100 + r) × 100%.

Expenditure = price × consumption. If price rises r% and expenditure must stay the same, consumption must fall by r/(100 + r) × 100%; if price falls r%, consumption can rise by r/(100 − r) × 100%.

Area = length × breadth works the same way: length +a% and breadth +b% change the area by a + b + ab/100 percent. A square's side +r% changes its area by 2r + r²/100 percent.

Growth at r% per year for n years: P × (1 + r/100)ⁿ. Depreciation: P × (1 − r/100)ⁿ. Different rates in different years: multiply the separate factors.

Going back in time: value n years ago = present value ÷ (growth factor)ⁿ. Dividing by 1.1 is not the same as subtracting 10%.

Method and shortcuts

Step 1: write each change as a fraction factor. Step 2: multiply the factors, cancelling as you go. Step 3: compare the product with 1 to read the net percent.

For 3 changes, use a + b + ab/100 on the first 2, then apply the formula again with the 3rd change.

Price-fall-more-quantity questions: the money saved by the fall, spent at the new price, buys the extra quantity. New price = (r% of the amount) ÷ extra quantity.

Use fractions: 10% = 1/10, 12.5% = 1/8, 20% = 1/5, 25% = 1/4, 33 1/3% = 1/3. A 1/n rise needs a 1/(n + 1) fall to undo it.

Shortcut limit: a + b + ab/100 is for multiplied quantities only; it does not apply when 2 percentages are of different totals that are then added.

Example 1: up and down by the same percent

Question: A price is increased by 20% and then decreased by 20%. Find the net change.

Solution: 20 − 20 + (20 × −20)/100 = −4. Check with 100: 100 → 120 → 96.

Answer: a 4% decrease

Example 2: population with changing rates

Question: A town of 50,000 people grows 10% in year 1, falls 10% in year 2 and grows 20% in year 3. Find the population after 3 years.

Solution: 50,000 × 11/10 × 9/10 × 6/5 = 55,000 × 9/10 × 6/5 = 49,500 × 6/5 = 59,400.

Answer: 59,400

Example 3: keeping expenditure fixed

Question: The price of sugar rises by 25%. By what percent must a family cut its consumption so that its spending on sugar does not change?

Solution: Price factor 5/4, so consumption factor must be 4/5, a fall of 1/5 = 20%. By formula: 25/125 × 100 = 20%.

Answer: 20%

Example 4: change in area

Question: The length of a rectangle is increased by 20% and its breadth decreased by 10%. Its area: (a) increases by 10% (b) decreases by 8% (c) increases by 2% (d) increases by 8%

Solution: 20 + (−10) + (20 × −10)/100 = 20 − 10 − 2 = 8, a positive value, so the area rises by 8%. Check: 6/5 × 9/10 = 54/50 = 1.08.

Answer: (d) increases by 8%

Example 5: working back through 3 changes

Question: A town's population rose 10% in the first year and 20% in the second, then fell 10% in the third. It is now 1,18,800. The population 3 years ago was: (a) 1,00,000 (b) 96,000 (c) 1,04,000 (d) 98,000

Solution: Combined factor = 11/10 × 6/5 × 9/10 = 594/500 = 1.188.

Original = 1,18,800 ÷ 1.188 = 1,00,000. Check forward: 1,00,000 → 1,10,000 → 1,32,000 → 1,18,800.

Answer: (a) 1,00,000

Example 6: price fall and extra quantity

Question: A 20% fall in the price of rice lets Meena buy 5 kg more for ₹800. The original price per kg was: (a) ₹32 (b) ₹36 (c) ₹40 (d) ₹45

Solution: The fall saves 20% of ₹800 = ₹160 on the old quantity, and this ₹160 buys the extra 5 kg at the new price. New price = 160 ÷ 5 = ₹32 per kg.

Original price = 32 ÷ (4/5) = ₹40 per kg. Check: ₹800 buys 20 kg before and 25 kg after. Option (a) ₹32 is the new price, a trap.

Answer: (c) ₹40

Common mistakes

Adding successive changes directly: +10% then +10% is +21%, not +20%.

Thinking +20% then −20% returns to the start. Right: it is a net 4% loss, since the second change acts on a larger base.

Going back by subtracting the percentage: the value before a 10% rise to 1,100 is 1,000, not 990.

Answering with the new price when the question asks for the original price, as in Example 6. Read which price is wanted.

Using 2r for a square's area change. Right: side +10% gives area +21%, i.e. 2r + r²/100.

Practice questions

Q1. A price is raised by 10%, then by 20%, and then reduced by 25%. The net change is: (a) 5% increase (b) 1% decrease (c) 1% increase (d) no change

Q2. Each side of a square is increased by 30%. Its area increases by: (a) 60% (b) 90% (c) 69% (d) 72%

Q3. A number is decreased by 20%. By what percent must the result be increased to get 8% more than the original number? (a) 28% (b) 30% (c) 32% (d) 35%

Answers

Q1. (b) 11/10 × 6/5 × 3/4 = 198/200 = 0.99, a 1% decrease.

Q2. (c) 30 + 30 + 900/100 = 69%.

Q3. (d) From 100 the number becomes 80; the target is 108; 28/80 × 100 = 35%.

Quick revision

Net of a% and b%: a + b + ab/100 (decrease negative).

+r% then −r% ⇒ loss of r²/100 %.

Price up r%, same spending ⇒ consumption down r/(100 + r) × 100%.

Back in time: divide by the factors; never subtract the percent.

Concepts and rules

Pass-mark questions: 2 scores given as percentages of the maximum marks, with marks above or below the pass mark. The gap in percent equals the gap in marks.

Elections: votes cast = valid + invalid. Candidates' percentages are usually of valid votes; the margin = winner − loser. Read carefully whether a percentage is of total voters, votes cast or valid votes.

With 2 candidates and all valid votes, if the winner gets w%, the margin is (2w − 100)% of the valid votes.

Income = expenditure + savings. If income and expenditure change by different percentages, work with actual amounts (base 100) to find the savings change.

Dry-mass rule: when water evaporates or is added, the non-water part stays the same. Fresh fruit with w% water has (100 − w)% solid.

Weighted percent of a group: overall % = (share of part 1 × its rate + share of part 2 × its rate)/100, e.g. literacy of a town from male and female rates.

Budget rule: if price rises a% and spending rises b%, consumption changes by the factor (100 + b)/(100 + a).

Method and shortcuts

Step 1: decide the base (maximum marks, valid votes, income, solid part). Step 2: express every given figure as a percent of that base. Step 3: equate a known number of units to its percent and scale.

Pass marks shortcut: maximum = (difference in marks)/(difference in percents) × 100, where 'marks' are signed (+ above pass, − below pass).

Mixture shortcut: the part that is not being added stays fixed. New total = fixed part ÷ (its new percent/100).

Savings shortcut: with income 100 and expenditure E, new savings = new income − new expenditure; never apply the income percent to savings directly.

Example 1: maximum and pass marks

Question: Ravi scores 30% and fails by 15 marks. Sita scores 40% and gets 25 marks more than the pass mark. Find the maximum marks and the pass mark.

Solution: The scores differ by 40% − 30% = 10% of the maximum, and by 15 + 25 = 40 marks. So 10% = 40 and the maximum is 400.

Pass mark = 30% of 400 + 15 = 120 + 15 = 135. Check: 40% of 400 = 160 = 135 + 25.

Answer: maximum 400, pass mark 135

Example 2: change in savings

Question: Kiran spends 75% of her income. Her income rises by 20% and her expenditure by 10%. By what percent do her savings rise?

Solution: Income 100, expenditure 75, savings 25. New income 120, new expenditure 82.5, new savings 37.5. Rise = 12.5/25 × 100 = 50%.

Answer: 50%

Example 3: fresh and dry fruit

Question: Fresh grapes contain 80% water and dry grapes 10% water. How much dry grape can be made from 45 kg of fresh grapes?

Solution: Solid part = 20% of 45 = 9 kg, and it does not change. In dry grapes it is 90% of the weight, so weight = 9 ÷ 0.9 = 10 kg.

Answer: 10 kg

Example 4: election with invalid votes

Question: In an election between 2 candidates, 20% of the votes cast were invalid. The winner got 60% of the valid votes and won by 1,200 votes. The total votes cast were: (a) 6,000 (b) 7,200 (c) 7,500 (d) 8,000

Solution: Margin = 60% − 40% = 20% of valid votes = 1,200, so valid votes = 6,000.

Valid votes are 80% of the votes cast, so votes cast = 6,000 ÷ 0.8 = 7,500. Option (a) 6,000 stops at the valid votes.

Answer: (c) 7,500

Example 5: weighted literacy

Question: In a town, 55% of the people are men. 40% of the men and 60% of the women are literate. If 25,500 people are illiterate, the population is: (a) 50,000 (b) 48,000 (c) 52,000 (d) 45,000

Solution: Women are 45%. Literate share = 55 × 40/100 + 45 × 60/100 = 22 + 27 = 49% of the population, so 51% are illiterate.

51% = 25,500, so 1% = 500 and the population is 50,000.

Answer: (a) 50,000

Example 6: incomes, savings and an expenditure gap

Question: Arjun's income is 25% more than Vikram's. Arjun saves 20% of his income and Vikram saves 10% of his. If Arjun spends ₹6,000 more than Vikram, Vikram's income is: (a) ₹48,000 (b) ₹75,000 (c) ₹54,000 (d) ₹60,000

Solution: Let Vikram's income = 100k, so Arjun's = 125k. Arjun spends 80% of 125k = 100k; Vikram spends 90% of 100k = 90k.

Gap = 10k = ₹6,000, so k = 600 and Vikram's income = ₹60,000. Check: Arjun ₹75,000 spends ₹60,000; Vikram spends ₹54,000.

Answer: (d) ₹60,000

Common mistakes

Taking the candidates' percentages on total votes when the question says 'valid votes'. Find valid votes first, then total.

In pass-mark questions, subtracting the marks when one student is below and one above. Right: the gap is 15 + 25 = 40, not 25 − 15.

Assuming savings change by the same percent as income. Right: compute new savings from new income minus new expenditure.

In water-content problems, keeping the water fixed. Right: the solid part is fixed; water is what changes.

Averaging rates without weights: (40 + 60)/2 = 50% literacy is wrong when men and women are not equal in number.

Practice questions

Q1. In an election between 2 candidates, the loser got 35% of the votes and lost by 2,400 votes. All votes were valid. The winner got: (a) 5,200 (b) 4,800 (c) 5,600 (d) 2,800

Q2. A 40 litre mixture of milk and water has 10% water. How much water must be added to make the water 20% of the new mixture? (a) 4 litres (b) 4.5 litres (c) 5 litres (d) 8 litres

Q3. The price of petrol rises by 25% but a man's spending on petrol rises by only 10%. His consumption falls by: (a) 15% (b) 12% (c) 10% (d) 14%

Answers

Q1. (a) Margin = 65% − 35% = 30% = 2,400, so the total is 8,000 and the winner got 65% of 8,000 = 5,200.

Q2. (c) Milk = 36 litres stays fixed and must be 80%, so the new total is 45 litres; add 45 − 40 = 5 litres.

Q3. (b) Consumption factor = 110/125 = 0.88, a fall of 12%.

Quick revision

Fix the unchanging quantity first: maximum marks, valid votes, solid part, income base.

Pass marks: difference of marks (signed) = difference of percents.

2 candidates, w% to the winner ⇒ margin (2w − 100)% of valid votes.

Savings = income − expenditure; compute it, never assume its percent.