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Percentages for CUET General Aptitude

About 10 min to learn · 72 questions

Percentages turn up all over CUET's General Aptitude Test: in plain arithmetic, in word problems about marks, salaries and elections, and inside data tables. Almost every question uses the same few ideas: percent of a quantity, fraction–percent conversions, percentage change and its base, reverse percentages, percentage points, successive changes, price and consumption, and compound growth and depreciation. This lesson works through each with examples and the traps beside them.

Syllabus area
General Mental Ability, Numerical Ability
On this page · 9 sections
  1. 01The core idea
  2. 02Percent basics
  3. 03Percentage change
  4. 04Successive changes
  5. 05Growth and depreciation
  6. 06Exam shortcuts
  7. 07Common mistakes
  8. 08Quick revision
  9. 09Check your understanding

The core idea

Percent means out of every hundred, so a percentage is just a fraction with 100 underneath. Nearly every question in this chapter comes down to one habit: think in multipliers. A rise of 15% is ×1.15\times 1.15, a fall of 8% is ×0.92\times 0.92, and two changes in a row are two multipliers in a row.

The other habit is to find the base before you divide. The base is the value the percentage is taken of: the original value in a change, and the number after "of" in a share.

Percent basics

What does "percent" actually mean?

Percent means "out of every hundred", so 35% is 35100=0.35\tfrac{35}{100} = 0.35. To find a percentage of a quantity, change the percentage to a decimal or fraction and multiply: 35% of 480 is 0.35×480=1680.35 \times 480 = 168.

Watch out

A percentage above 100 means more than the whole: 150% of 60 is 1.5×60=901.5 \times 60 = 90. But "150% more than 60" is 60+90=15060 + 90 = 150. "Of" and "more than" give different answers, and both are usually among the options.

How do you convert between fractions and percentages?

To turn a fraction into a percentage, multiply by 100. To turn a percentage into a fraction, divide by 100 and simplify. So 35=60%\tfrac{3}{5} = 60\% and 45%=45100=92045\% = \tfrac{45}{100} = \tfrac{9}{20}.

FractionPercentageFractionPercentage
12\tfrac{1}{2}50%17\tfrac{1}{7}about 14.29%
13\tfrac{1}{3}3313%33\tfrac{1}{3}\%18\tfrac{1}{8}12.5%
14\tfrac{1}{4}25%19\tfrac{1}{9}1119%11\tfrac{1}{9}\%
15\tfrac{1}{5}20%110\tfrac{1}{10}10%
16\tfrac{1}{6}1623%16\tfrac{2}{3}\%111\tfrac{1}{11}9111%9\tfrac{1}{11}\%
112\tfrac{1}{12}813%8\tfrac{1}{3}\%116\tfrac{1}{16}6.25%

Exam shortcut

One conversion gives a family: from 18=12.5%\tfrac{1}{8} = 12.5\% you get 38=37.5%\tfrac{3}{8} = 37.5\%, 58=62.5%\tfrac{5}{8} = 62.5\% and 78=87.5%\tfrac{7}{8} = 87.5\%. For decimals, move the point two places right: 0.07=7%0.07 = 7\% and 0.325=32.5%0.325 = 32.5\%.

How do you find what percentage one number is of another?

Divide the part by the whole and multiply by 100. The number after the word "of" is always the base, so it goes underneath.

Example 1 · Part of a whole

A cricket academy has 75 trainees, of whom 27 are left-handed batters. What percentage are left-handed?

  1. The base is all 75 trainees.

  2. 2775×100=36%\tfrac{27}{75} \times 100 = 36\%.

Answer

36%

Watch out

Put both numbers in the same unit first. "What percent of 2 kg is 250 g?" is 2502000×100=12.5%\tfrac{250}{2000} \times 100 = 12.5\%, not 250 divided by 2.

How do you find the whole when a percentage of it is given?

If p%p\% of a number is NN, the number is N×100pN \times \tfrac{100}{p}. This is the most common reverse step in the chapter.

Example 2 · The whole from a part

24% of the seats in a hall are reserved, which is 132 seats. How many seats does the hall have?

  1. 132 is 24% of the whole.

  2. Whole: 132×10024=550132 \times \tfrac{100}{24} = 550.

Answer

550 seats

Pass-mark questions

A student needs 40% to pass, scores 170 and fails by 30 marks. The pass mark is 170+30=200170 + 30 = 200, which is 40% of the maximum, so the maximum is 200×10040=500200 \times \tfrac{100}{40} = 500. "Failed by" means add. "Passed by" means subtract.

Percentage change

How do you find percentage change?

Divide the change by the original value and multiply by 100. The base is always the original, whether the quantity went up or down. To increase a value by r%r\%, multiply by 1+r1001 + \tfrac{r}{100}; to decrease it, by 1−r1001 - \tfrac{r}{100}.

Percentage change

Percentage change=new−originaloriginal×100\text{Percentage change} = \frac{\text{new} - \text{original}}{\text{original}} \times 100

Example 3 · Percentage increase

A tea stall sold 250 cups a day last month and 290 this month. Find the percentage increase.

  1. Change: 290−250=40290 - 250 = 40.

  2. Increase: 40250×100=16%\tfrac{40}{250} \times 100 = 16\%.

Answer

16%

Dividing by the new value, 290, gives about 13.8%, the commonest wrong answer in the chapter.

How do you find the original value after a percentage increase or decrease?

Divide the final value by the multiplier: after a rise of r%r\%, original = final ÷(1+r100)\div (1 + \tfrac{r}{100}), and after a fall, final ÷(1−r100)\div (1 - \tfrac{r}{100}). Never subtract r%r\% of the final value.

Example 4 · Undoing a rise

After a 25% rise, a coaching fee is ₹7,500. What was it before?

  1. The multiplier is 1.25.

  2. Original: 7,500÷1.25=6,0007{,}500 \div 1.25 = 6{,}000.

  3. Check: 25% of 6,000 is 1,500, and 6,000+1,500=7,5006{,}000 + 1{,}500 = 7{,}500.

Answer

₹6,000

Subtracting 25% of 7,500 gives ₹5,625, which is wrong because the 25% was taken on the old fee.

What is the difference between percent and percentage points?

A change in percentage points is the plain difference between two percentages. A percent change divides that gap by the old percentage. An interest rate that moves from 5% to 6% has risen by 1 percentage point, but by 20%.

Example 5 · Points and percent

A school's pass rate rises from 60% to 75%. By how many percentage points did it rise, by what percentage, and by what percentage did the failure rate fall?

  1. Points: 75−60=1575 - 60 = 15.

  2. Percent: 1560×100=25%\tfrac{15}{60} \times 100 = 25\%.

  3. Failures fell from 40% to 25%: 1540×100=37.5%\tfrac{15}{40} \times 100 = 37.5\%.

Answer

15 points, a 25% rise, and a 37.5% fall in failures

If A is more than B by r%, how much less is B than A?

B is less than A by r100+r×100\tfrac{r}{100 + r} \times 100 percent. The two percentages differ because the base changes: "more than B" uses B as the base, and "less than A" uses A.

More than and less than

A is r% more than B⇒B is r100+r×100% less than A\text{A is } r\% \text{ more than B} \Rightarrow \text{B is } \frac{r}{100 + r} \times 100\% \text{ less than A}

If A is 25% more than B, B is 25125×100=20%\tfrac{25}{125} \times 100 = 20\% less than A. If A is 20% less than B, B is 2080×100=25%\tfrac{20}{80} \times 100 = 25\% more than A.

Successive changes

What is the fastest way to combine successive percentage changes?

Two successive changes of a%a\% and b%b\% give a net change of a+b+ab100a + b + \tfrac{ab}{100} percent, with a minus sign for a decrease. For three or more, multiply the multipliers. The order never matters.

ChangesNet effect
+10%, then +10%10+10+1=+21%10 + 10 + 1 = +21\%
+20%, then −10%20−10−2=+8%20 - 10 - 2 = +8\%
−15%, then −15%−15−15+2.25=−27.75%-15 - 15 + 2.25 = -27.75\%
+x%+x\%, then −x%-x\%−x2100-\tfrac{x^2}{100}, always a loss

Example 6 · Successive discounts

A jacket marked ₹2,400 has successive discounts of 25% and 12%. What does the customer pay, and what single discount is this equal to?

  1. Price paid: 2,400×0.75×0.88=1,5842{,}400 \times 0.75 \times 0.88 = 1{,}584.

  2. That is 0.75×0.88=0.660.75 \times 0.88 = 0.66 of the marked price, so the single discount is 34%.

  3. The formula agrees: −25−12+25×12100=−34-25 - 12 + \tfrac{25 \times 12}{100} = -34.

Answer

₹1,584, a single discount of 34%

Adding the two discounts gives 37%, a planted option.

Exam shortcut

For three changes, multiply: a rise of 5%, then 10%, then a fall of 20% gives 1.05×1.1×0.8=0.9241.05 \times 1.1 \times 0.8 = 0.924, a net fall of 7.6%.

How do price, consumption and expenditure affect each other?

Expenditure is price times consumption, so a change in price and a change in consumption combine like two successive changes. To keep spending the same after a price rise of r%r\%, consumption must fall by r100+r×100\tfrac{r}{100 + r} \times 100 percent.

  • Price up 20%, spending the same: cut consumption by 20120×100=1623%\tfrac{20}{120} \times 100 = 16\tfrac{2}{3}\%.
  • Price down 20%, spending the same: consumption can rise by 2080×100=25%\tfrac{20}{80} \times 100 = 25\%.
  • Price up 10% and consumption down 10%: spending changes by 10−10−1=−1%10 - 10 - 1 = -1\%.

Example 7 · Cheaper onions

A 25% fall in the price of onions lets a family buy 4 kg more for ₹360. Find the new and the original price per kg.

  1. The fall saves 25% of 360, which is ₹90.

  2. That ₹90 buys the extra 4 kg at the new price: 90÷4=22.5090 \div 4 = 22.50 per kg.

  3. The original price: 22.50÷0.75=3022.50 \div 0.75 = 30 per kg.

Answer

New price ₹22.50 per kg, original ₹30 per kg

Check: ₹360 buys 12 kg at ₹30 and 16 kg at ₹22.50.

How does area change when the sides of a shape change?

Area is a product of two lengths, so the same formula applies: length up a%a\% and breadth up b%b\% change the area by a+b+ab100a + b + \tfrac{ab}{100} percent. Volume multiplies three factors.

  • Both sides of a square up 10%: area up 21%.
  • Radius of a circle down 10%: area down 10+10−1=19%10 + 10 - 1 = 19\%.
  • Length up 30%, breadth down 30%: area down 9%.
  • Each edge of a cube up 10%: volume ×1.13=1.331\times 1.1^3 = 1.331, up 33.1%.

Growth and depreciation

How do you calculate population growth and depreciation?

Growth and depreciation are repeated percentage changes, each applied to the latest value. After nn years at r%r\% a year, the value is the original times (1+r100)n(1 + \tfrac{r}{100})^n for growth, or (1−r100)n(1 - \tfrac{r}{100})^n for depreciation.

Compound growth and depreciation

Value after n years=original×(1±r100)n\text{Value after } n \text{ years} = \text{original} \times \left(1 \pm \frac{r}{100}\right)^n

Example 8 · Growth and depreciation

A town of 75,000 grows at 4% a year. What is its population after 2 years? And a tractor bought for ₹5,00,000 loses 15% of its value a year. What is it worth after 2 years?

  1. Town: 75,000×1.04×1.04=75,000×1.0816=81,12075{,}000 \times 1.04 \times 1.04 = 75{,}000 \times 1.0816 = 81{,}120.

  2. Tractor: 5,00,000×0.85×0.85=5,00,000×0.7225=3,61,2505{,}00{,}000 \times 0.85 \times 0.85 = 5{,}00{,}000 \times 0.7225 = 3{,}61{,}250.

Answer

81,120 people; ₹3,61,250

Going backwards and finding the rate

  • Different rates each year: multiply the factors, so 5% then 8% is 1.05×1.08=1.1341.05 \times 1.08 = 1.134.
  • A value some years ago: divide by the factors. ₹43,200 after two years of 20% depreciation was 43,200÷0.64=67,50043{,}200 \div 0.64 = 67{,}500.
  • The rate from start and end over two years: 50,000 becoming 60,500 is a factor of 1.21=1.121.21 = 1.1^2, so 10% a year.
  • Powers worth knowing: 1.12=1.211.1^2 = 1.21, 1.13=1.3311.1^3 = 1.331, 0.92=0.810.9^2 = 0.81, 0.93=0.7290.9^3 = 0.729, 1.22=1.441.2^2 = 1.44, 0.82=0.640.8^2 = 0.64, 1.052=1.10251.05^2 = 1.1025.

How do percentages work in election and exam-score questions?

In a two-candidate election the winning margin, as a percentage, is the gap between the two shares: a winner with 62% of valid votes beats a loser with 38% by 24% of the valid votes. Elections often stack layers, voters on the list, votes cast and valid votes, so peel them off one factor at a time.

  • If 10% of listed voters did not vote and 4% of votes cast were invalid, valid votes = list ×0.9×0.96\times 0.9 \times 0.96. Work backwards by dividing by each factor.
  • Marks and percentages are different things: going from 60 to 72 out of 120 is 10 percentage points, from 50% to 60%, and a 20% rise in marks.

Exam shortcuts

  • Assume 100. When no actual number is given, start at 100 and follow the numbers.
  • Use multipliers. Up 12% is ×1.12\times 1.12. Down 12% is ×0.88\times 0.88.
  • Split awkward percentages. 17.5% of 640 is 10% (64) plus 5% (32) plus 2.5% (16), which is 112.
  • Swap the order. x%x\% of yy equals y%y\% of xx, so 16% of 25 is 25% of 16, which is 4.
  • Check by going forwards. After a reverse calculation, apply the change to your answer.

Common mistakes

  • The wrong base. Percentage change divides by the original, not the new value.
  • Reversing by subtracting. After a 20% rise, the original is final ÷1.2\div 1.2.
  • Adding successive changes. 10% then 20% is 32%, not 30%. A rise and fall of the same percentage always leave a loss.
  • Percent and percentage points. 8% to 10% is 2 points but a 25% rise.
  • Forgetting that the whole changes. New members raise the total as well as the part.
  • Simple instead of compound. 10% a year for 3 years is a 27.1% loss, not 30%.
  • "Of" and "more than". 200% of a number doubles it. 200% more than it triples it.
  • Mixed units. Convert hours to minutes or kilograms to grams first.
  • Answering the wrong quantity. If the question asks for the new salary, the old one is a wrong option.

Quick revision

Basics

  • p%p\% of NN = p100×N\tfrac{p}{100} \times N
  • Part as a percent: part ÷ whole × 100, the base after "of"
  • Whole from a part: N×100pN \times \tfrac{100}{p}

Change

  • Change ÷ original × 100
  • Up r%r\%: ×(1+r100)\times (1 + \tfrac{r}{100}). Reverse: divide
  • Points: plain difference of percentages
  • A r%r\% more than B: B is r100+r×100%\tfrac{r}{100 + r} \times 100\% less

Successive changes

  • a+b+ab100a + b + \tfrac{ab}{100}
  • +x+x then −x-x: a loss of x2100%\tfrac{x^2}{100}\%
  • Price up r%r\%, same spending: consumption down r100+r×100%\tfrac{r}{100 + r} \times 100\%

Growth

  • Original ×(1±r100)n\times (1 \pm \tfrac{r}{100})^n
  • Backwards: divide by the factors
  • 1.12=1.211.1^2 = 1.21, 0.92=0.810.9^2 = 0.81, 0.82=0.640.8^2 = 0.64

Profit, loss and discounts build on these ideas in profit, loss and discount, compound interest in simple and compound interest, and percentages on charts in data interpretation.

Next · 08Quick revision

Check your understanding

72 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Percentage basicsQ1 of 72Q1

Percentage basics

16 questions

Question 1

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The Nair family's monthly budget (total monthly income ₹60,000):

ItemAmount (₹)
Rent15,000
Food12,000
Education9,000
Transport4,800
Other7,200
Savings12,000

Using the Nair family's budget table, what percentage of the family's monthly income goes on rent?

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AnswerB · 25%

  1. Rent = ₹15,000; income = ₹60,000.

  2. Percentage = 1500060000×100=25%\frac{15000}{60000} \times 100 = 25\%.

Worth noting

The base is the income, ₹60,000. Dividing by the total spending of ₹48,000 (income minus savings) gives 31.25%, the tempting wrong answer.

Question 2

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A school has 640 students, and 35% of them travel by the school bus. How many students travel by the school bus?

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AnswerB · 224

  1. 35% of 640=35100×64035\% \text{ of } 640 = \frac{35}{100} \times 640.

  2. =0.35×640=224= 0.35 \times 640 = 224.

Worth noting

416 is the number who do NOT take the bus (65% of 640).

Question 3

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In a practice test, Tenzin got 7 out of every 8 questions right. What percentage of the questions did he get right?

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AnswerD · 87.5%

  1. Fraction right = 78\frac{7}{8}.

  2. 78×100=87.5%\frac{7}{8} \times 100 = 87.5\%.

Worth noting

Know by heart: 18=12.5%\frac{1}{8} = 12.5\%, so 78=7×12.5%=87.5%\frac{7}{8} = 7 \times 12.5\% = 87.5\%. 12.5% is the share he got wrong.

Question 4

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Of the 240 mangoes in a crate, 18 were found to be spoilt. What percentage of the mangoes were spoilt?

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AnswerB · 7.5%

  1. Percentage spoilt = 18240×100\frac{18}{240} \times 100.

  2. =1800240=7.5%= \frac{1800}{240} = 7.5\%.

Worth noting

0.75% comes from forgetting to multiply by 100 correctly; 18% treats the count as if it were already a percentage.

Question 5

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Which fraction is equal to 37.5%?

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AnswerC · 38\frac{3}{8}

  1. 37.5%=37.5100=375100037.5\% = \frac{37.5}{100} = \frac{375}{1000}.

  2. Divide top and bottom by 125: 3751000=38\frac{375}{1000} = \frac{3}{8}.

Worth noting

720=35%\frac{7}{20} = 35\% and 58=62.5%\frac{5}{8} = 62.5\%, which is 100%−37.5%100\% - 37.5\%.

Question 6

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Fatima saves 15% of her monthly stipend. Her savings come to ₹1,800 a month. What is her monthly stipend?

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AnswerD · ₹12,000

  1. 15% of the stipend = ₹1,800.

  2. Stipend = 180015×100=₹12,000\frac{1800}{15} \times 100 = ₹12{,}000.

Worth noting

₹270 is 15% of ₹1,800 — the percentage applied to the wrong number. ₹10,200 is what she spends, not her stipend.

Question 7

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Express 0.045 as a percentage.

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AnswerB · 4.5%

  1. To change a decimal to a percentage, multiply by 100.

  2. 0.045×100=4.5%0.045 \times 100 = 4.5\%.

Worth noting

Moving the decimal point one place instead of two gives 0.45%.

Question 8

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An old bus route carried 36,000 passengers a day. The new metro line that replaced it carries 250% of that number. How many passengers a day does the metro carry?

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AnswerC · 90,000

  1. 250%=2.5250\% = 2.5 times.

  2. 2.5×36000=90,0002.5 \times 36000 = 90{,}000.

Worth noting

"250% of" means 2.5 times. 1,26,000 is what you get if you read it as "250% more than" (36,000+90,000)(36{,}000 + 90{,}000).

Question 9

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The Nair family's monthly budget (total monthly income ₹60,000):

ItemAmount (₹)
Rent15,000
Food12,000
Education9,000
Transport4,800
Other7,200
Savings12,000

Using the Nair family's budget table, the family's spending on education is what percentage of its spending on food?

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AnswerC · 75%

  1. Education = ₹9,000; food = ₹12,000.

  2. "Education as a percentage of food" puts food in the denominator: 900012000×100=75%\frac{9000}{12000} \times 100 = 75\%.

Worth noting

The quantity after "of" is the base. Flipping the fraction gives 13313%133\frac{1}{3}\%; dividing by the income gives 15%.

Question 10

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In a college election with only two candidates, the winner got 58% of the valid votes and won by 4,800 votes. How many valid votes were cast?

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AnswerC · 30,000

  1. Loser's share = 100%−58%=42%100\% - 58\% = 42\%.

  2. Winning margin = 58%−42%=16%58\% - 42\% = 16\% of the valid votes.

  3. 16% of valid votes = 4,800, so valid votes = 480016×100=30,000\frac{4800}{16} \times 100 = 30{,}000.

Worth noting

The margin is the gap between the two shares (16%), not 58%−50%=8%58\% - 50\% = 8\%, which gives 60,000.

Question 11

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To pass an exam, a candidate needs 36% of the maximum marks. Gurpreet scored 198 marks and failed by 18 marks. What are the maximum marks?

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AnswerD · 600

  1. Pass mark = 198+18=216198 + 18 = 216.

  2. 36% of maximum = 216, so maximum = 21636×100=600\frac{216}{36} \times 100 = 600.

Worth noting

"Failed by 18" means he needed 18 MORE, so add. Using 198 alone gives 550; subtracting 18 gives 450.

Question 12

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Aarav's salary is 25% more than Joseph's. By what percentage is Joseph's salary less than Aarav's?

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AnswerA · 20%

  1. Let Joseph earn 100. Then Aarav earns 125.

  2. Joseph earns 25 less than Aarav; as a percentage of Aarav's salary: 25125×100=20%\frac{25}{125} \times 100 = 20\%.

Worth noting

The difference is the same 25, but the base changes to Aarav's 125. Shortcut: if A is r% more than B, B is r100+r×100%\frac{r}{100+r} \times 100\% less than A.

Question 13

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What is the value of 18% of 350 plus 35% of 180?

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AnswerC · 126

  1. 18% of 350=0.18×350=6318\% \text{ of } 350 = 0.18 \times 350 = 63.

  2. 35% of 180=0.35×180=6335\% \text{ of } 180 = 0.35 \times 180 = 63.

  3. Sum = 63+63=12663 + 63 = 126.

Worth noting

Both parts are equal because 18×350=35×180=630018 \times 350 = 35 \times 180 = 6300. A misplaced decimal point gives 12.6 or 1,260; stopping after one part gives 63.

Question 14

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In a class of 48 students, 37.5% are girls. Then 6 more girls join the class. What percentage of the class is now girls?

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AnswerC · 4449%44\frac{4}{9}\%

  1. Girls at first = 37.5% of 48=1837.5\% \text{ of } 48 = 18.

  2. After 6 join: girls = 24, class size = 48+6=5448 + 6 = 54.

  3. Percentage = 2454×100=49×100=4449%\frac{24}{54} \times 100 = \frac{4}{9} \times 100 = 44\frac{4}{9}\%.

Worth noting

The class grows too. Keeping 48 as the total gives 2448=50%\frac{24}{48} = 50\%, the tempting wrong answer.

Question 15

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Lalremsiami spends 27 minutes of her 3-hour study session on revision. What percentage of the session is revision?

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AnswerB · 15%

  1. Convert to the same unit: 3 hours = 180 minutes.

  2. Percentage = 27180×100=15%\frac{27}{180} \times 100 = 15\%.

Worth noting

Both quantities must be in the same unit. Treating 3 hours as 300 minutes gives 9%.

Question 16

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In a school exam, 70% of the students passed in Mathematics, 75% passed in English, and 10% failed in both subjects. If 990 students passed in both subjects, how many students took the exam?

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AnswerC · 1,800

  1. Failed in Mathematics = 30%; failed in English = 25%; failed in both = 10%.

  2. Failed in at least one subject = 30%+25%−10%=45%30\% + 25\% - 10\% = 45\%.

  3. Passed in both = 100%−45%=55%100\% - 45\% = 55\%.

  4. 55% of total = 990, so total = 99055×100=1,800\frac{990}{55} \times 100 = 1{,}800.

Worth noting

Ignoring the 10% who failed both gives passed-in-both = 70%+75%−100%=45%70\% + 75\% - 100\% = 45\% and a total of 2,200.

Percentage change

23 questions

Question 17

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A shop's monthly sales fell from ₹80,000 to ₹68,000. By what percentage did sales fall?

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AnswerB · 15%

  1. The fall itself is 80,000−68,000=12,00080{,}000 - 68{,}000 = 12{,}000.

  2. A percentage change is always measured against the value you started from, so divide by 80,000, not by 68,000.

  3. 12,00080,000=0.15\frac{12{,}000}{80{,}000} = 0.15, which is 15%.

Worth noting

Dividing by the new figure gives 12,00068,000≈17.6%\frac{12{,}000}{68{,}000} \approx 17.6\%. That number answers a different question: by what percentage the smaller figure would have to rise to get back.

Question 18

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The fare for a train journey rose from ₹450 to ₹540. By what percentage did the fare rise?

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AnswerB · 20%

  1. Rise = ₹540 − ₹450 = ₹90.

  2. Percentage rise = 90450×100=20%\frac{90}{450} \times 100 = 20\%.

Worth noting

The base is the OLD fare. Dividing by ₹540 gives 1623%16\frac{2}{3}\%. 120% is the new fare as a percentage of the old one.

Question 19

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The number of students living in a hostel fell from 1,250 to 1,100. By what percentage did the number fall?

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AnswerA · 12%

  1. Fall = 1,250−1,100=1501{,}250 - 1{,}100 = 150.

  2. Percentage fall = 1501250×100=12%\frac{150}{1250} \times 100 = 12\%.

Worth noting

Dividing by the new value 1,100 gives about 13.6%. 88% is how much of the original remains.

Question 20

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A factory made 360 bicycles a day. After new machines were installed, daily output increased by 35%. What is the new daily output?

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AnswerD · 486

  1. Increase = 35% of 360=12635\% \text{ of } 360 = 126.

  2. New output = 360+126=486360 + 126 = 486. (Or directly: 360×1.35=486360 \times 1.35 = 486.)

Worth noting

126 is only the increase. 395 comes from adding 35 bicycles instead of 35%.

Question 21

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A bank raised its fixed-deposit interest rate from 8% a year to 10% a year. By what percent (not percentage points) did the interest rate rise?

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AnswerD · 25%

  1. The rate rose by 10−8=210 - 8 = 2 percentage points.

  2. As a percent of the old rate: 28×100=25%\frac{2}{8} \times 100 = 25\%.

Worth noting

2 is the rise in percentage POINTS, not percent. Dividing by the new rate 10 gives 20%.

Question 22

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A party's vote share in a city rose from 32% to 40%. Which statement describes this change correctly?

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AnswerA · It rose by 8 percentage points, which is a 25% rise

  1. Difference of the two shares = 40−32=840 - 32 = 8 percentage points.

  2. Percent rise = 832×100=25%\frac{8}{32} \times 100 = 25\%.

Worth noting

Percentage points are a plain subtraction of two percentages; percent change divides that gap by the old value. 20% uses the new share, 40, as the base.

Question 23

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After a 20% rise, a gym's monthly fee is ₹1,440. What was the fee before the rise?

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AnswerB · ₹1,200

  1. New fee = 120% of old fee.

  2. Old fee = 14401.2=₹1,200\frac{1440}{1.2} = ₹1{,}200.

Worth noting

Taking 20% off ₹1,440 gives ₹1,152 — wrong, because the 20% was calculated on the OLD fee, not on ₹1,440.

Question 24

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A used scooter now costs ₹68,000, which is 15% less than its price last year. What was its price last year?

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AnswerC · ₹80,000

  1. Now = 85% of last year's price.

  2. Last year's price = 680000.85=₹80,000\frac{68000}{0.85} = ₹80{,}000.

Worth noting

Adding 15% of ₹68,000 gives ₹78,200 — wrong base. Check: 15% of ₹80,000 = ₹12,000, and ₹80,000 − ₹12,000 = ₹68,000.

Question 25

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During a slow year, a company cut every employee's salary by 5%. Meera earned ₹38,000 a month before the cut. What does she earn now?

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AnswerC · ₹36,100

  1. Cut = 5% of ₹38,000 = ₹1,900.

  2. New salary = ₹38,000 − ₹1,900 = ₹36,100.

Worth noting

₹1,900 is only the cut; ₹33,000 subtracts ₹5,000 instead of 5%.

Question 26

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The Nair family's monthly budget (total monthly income ₹60,000):

ItemAmount (₹)
Rent15,000
Food12,000
Education9,000
Transport4,800
Other7,200
Savings12,000

Using the Nair family's budget table: next month, food spending rises by 20% and transport spending rises by 25%. Income and every other item stay the same, and savings absorb the extra cost. By what percentage do the family's savings fall?

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AnswerC · 30%

  1. Extra food cost = 20% of ₹12,000 = ₹2,400.

  2. Extra transport cost = 25% of ₹4,800 = ₹1,200.

  3. Savings fall by ₹3,600, from ₹12,000 to ₹8,400.

  4. Percentage fall = 360012000×100=30%\frac{3600}{12000} \times 100 = 30\%.

Worth noting

The fall is measured on the old savings (₹12,000). Dividing by the income gives 6%; adding the two rates gives 45%.

Question 27

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Siddharth's marks in a test out of 80 rose from 56 to 70. By how many percentage points did his percentage score rise?

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AnswerB · 17.5

  1. Old percentage = 5680×100=70%\frac{56}{80} \times 100 = 70\%.

  2. New percentage = 7080×100=87.5%\frac{70}{80} \times 100 = 87.5\%.

  3. Rise = 87.5−70=17.587.5 - 70 = 17.5 percentage points.

Worth noting

25 is the percent increase in his marks (14÷56)(14 \div 56), not the rise in percentage points. 14 is the rise in raw marks.

Question 28

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After a 12% rise in membership, a library now has 1,344 members. How many new members did it gain?

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AnswerA · 144

  1. 1,344=112%1{,}344 = 112\% of the old membership.

  2. Old membership = 13441.12=1,200\frac{1344}{1.12} = 1{,}200.

  3. New members = 1,344−1,200=1441{,}344 - 1{,}200 = 144.

Worth noting

Taking 12% of 1,344 (about 161) uses the new figure as the base. 1,200 is the old membership, not the gain.

Question 29

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The price of petrol rose from ₹96 a litre to ₹104.64 a litre. By what percentage did the price rise?

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AnswerC · 9%

  1. Rise = ₹104.64 − ₹96 = ₹8.64.

  2. Percentage rise = 8.6496×100=9%\frac{8.64}{96} \times 100 = 9\%.

Worth noting

8.64% treats the rupee rise as a percentage. Dividing by the new price gives about 8.26%.

Question 30

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In a college union election, a party won 36% of 2,50,000 votes in 2025 and 42% of 3,00,000 votes in 2026. By what percentage did the number of votes it won increase?

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AnswerD · 40%

  1. 2025 votes = 36% of 2,50,000=90,00036\% \text{ of } 2{,}50{,}000 = 90{,}000.

  2. 2026 votes = 42% of 3,00,000=1,26,00042\% \text{ of } 3{,}00{,}000 = 1{,}26{,}000.

  3. Increase = 36,000, so percentage increase = 3600090000×100=40%\frac{36000}{90000} \times 100 = 40\%.

Worth noting

The share rose by 6 percentage points (a 1623%16\frac{2}{3}\% rise in share) and turnout rose 20%, but the question asks about the NUMBER of votes, which needs both.

Question 31

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A water tank lost 12.5% of its water through a leak and now holds 1,050 litres. How much water did it hold before the leak?

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AnswerD · 1,200 litres

  1. After the leak the tank holds 87.5% of the original.

  2. Original = 10500.875=1,200\frac{1050}{0.875} = 1{,}200 litres.

Worth noting

Adding 12.5% of 1,050 gives 1,181.25 litres — the 12.5% was of the ORIGINAL amount. Tip: 12.5%=1812.5\% = \frac{1}{8}, so 1,050 is 78\frac{7}{8} of the original.

Question 32

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Anjali scored 72% in her first term exam and 81% in her second term exam, both out of the same total. By what percent did her marks increase?

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AnswerC · 12.5%

  1. The rise is 81−72=981 - 72 = 9 percentage points.

  2. Percent increase in marks = 972×100=12.5%\frac{9}{72} \times 100 = 12.5\%.

Worth noting

9 is the change in percentage points, not percent. Dividing by 81 gives 11.1%.

Question 33

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An employee received a salary increase of ₹4,500, which was an 18% raise. What is the employee's new salary?

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AnswerD · ₹29,500

  1. 18% of the old salary = ₹4,500.

  2. Old salary = 450018×100=₹25,000\frac{4500}{18} \times 100 = ₹25{,}000.

  3. New salary = ₹25,000 + ₹4,500 = ₹29,500.

Worth noting

₹25,000 is the OLD salary — the question asks for the new one. ₹810 applies 18% to the raise itself.

Question 34

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A mobile plan's daily data was cut from 2 GB to 1.5 GB, but its price did not change. By what percentage did the price per GB rise?

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AnswerB · 3313%33\frac{1}{3}\%

  1. Let the price be ₹P. Old price per GB = P2\frac{P}{2}; new = P1.5\frac{P}{1.5}.

  2. Ratio new ÷ old = 21.5=43\frac{2}{1.5} = \frac{4}{3}.

  3. So the price per GB rose by 13\frac{1}{3}, i.e. 3313%33\frac{1}{3}\%.

Worth noting

The data fell by 25%, but the price per GB does not rise by 25%: a 25% fall needs a 2575×100=3313%\frac{25}{75} \times 100 = 33\frac{1}{3}\% rise to compensate.

Question 35

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The pass rate at a school rose from 80% to 92%, with the same number of students appearing both years. By what percentage did the number of students who failed fall?

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AnswerD · 60%

  1. Failures fell from 20% to 8% of the students.

  2. Fall = 12 percentage points; as a percent of the old failures: 1220×100=60%\frac{12}{20} \times 100 = 60\%.

Worth noting

12 is the change in percentage points; 15% is the percent rise in the number who PASSED (12÷80)(12 \div 80). The question is about those who failed.

Question 36

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The Nair family's monthly budget (total monthly income ₹60,000):

ItemAmount (₹)
Rent15,000
Food12,000
Education9,000
Transport4,800
Other7,200
Savings12,000

Using the Nair family's budget table: next year the family's income rises by 10%, while every spending item stays exactly the same. By what percentage do the family's savings rise?

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AnswerD · 50%

  1. Total spending = ₹60,000 − ₹12,000 = ₹48,000, and it does not change.

  2. New income = ₹60,000 × 1.1 = ₹66,000.

  3. New savings = ₹66,000 − ₹48,000 = ₹18,000, a rise of ₹6,000.

  4. Percentage rise = 600012000×100=50%\frac{6000}{12000} \times 100 = 50\%.

Worth noting

Savings do not rise by the same 10% as income: the whole ₹6,000 increase lands on savings. Dividing ₹6,000 by the new savings gives 3313%33\frac{1}{3}\%.

Question 37

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Priya spends 80% of her income and saves the rest. Her income then rises by 25% and her spending rises by 15%. By what percentage do her savings rise?

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AnswerD · 65%

  1. Let the old income be 100: spending 80, savings 20.

  2. New income = 125; new spending = 80×1.15=9280 \times 1.15 = 92.

  3. New savings = 125−92=33125 - 92 = 33, a rise of 13.

  4. Percentage rise = 1320×100=65%\frac{13}{20} \times 100 = 65\%.

Worth noting

Savings are a difference, so their percentage change is not 25−15=10%25 - 15 = 10\%. Taking 100 as the income makes every figure easy.

Question 38

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In a town election, 8% of the voters on the list did not vote, and 5% of the votes cast were invalid. There were only two candidates. The winner got 55% of the valid votes and won by 17,480 votes. How many voters were on the list?

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AnswerD · 2,00,000

  1. Margin = 55%−45%=10%55\% - 45\% = 10\% of valid votes, so valid votes = 17,480×10=1,74,80017{,}480 \times 10 = 1{,}74{,}800.

  2. Valid votes = 95% of votes cast, so votes cast = 1748000.95=1,84,000\frac{174800}{0.95} = 1{,}84{,}000.

  3. Votes cast = 92% of the list, so voters on the list = 1840000.92=2,00,000\frac{184000}{0.92} = 2{,}00{,}000.

Worth noting

Work backwards one layer at a time. Skipping the invalid votes gives 1,90,000; stopping early gives 1,74,800 or 1,84,000.

Question 39

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A company's profit was 20% of its sales. The next year its sales rose by 20% but its profit fell to 15% of sales. How did the company's profit change?

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AnswerA · It fell by 10%

  1. Let old sales = 100; old profit = 20.

  2. New sales = 120; new profit = 15% of 120=1815\% \text{ of } 120 = 18.

  3. Change = 18−2020×100=−10%\frac{18 - 20}{20} \times 100 = -10\%, a fall of 10%.

Worth noting

The margin fell by 5 percentage points (a 25% fall in the margin), but sales grew, so profit in rupees fell only 10%.

Successive percentage change

19 questions

Question 40

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A town's bus fare rises by 10% in January and by another 10% in July. By what percentage has the fare risen over the whole year?

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AnswerC · 21%

  1. Successive change formula: a+b+ab100a + b + \frac{ab}{100}.

  2. 10+10+10×10100=21%10 + 10 + \frac{10 \times 10}{100} = 21\%.

Worth noting

The second 10% is taken on the already-raised fare, so the rises do not simply add to 20%. 121% is the final fare as a percentage of the original.

Question 41

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A shopkeeper marks a shirt 25% above its original price and then gives a 20% discount on the marked price. How does the final price compare with the original price?

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AnswerA · It is equal to the original price

  1. Net change = 25+(−20)+25×(−20)10025 + (-20) + \frac{25 \times (-20)}{100}.

  2. =25−20−5=0%= 25 - 20 - 5 = 0\%, so the final price equals the original.

Worth noting

Check with 100: marked price 125, 20% off 125 is 25, final 100. Just subtracting 25−2025 - 20 gives the tempting "5% more".

Question 42

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A store offers two successive discounts of 20% and 10% on a sofa. What single discount is equivalent to these two?

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AnswerA · 28%

  1. Price paid = 0.8×0.9=0.720.8 \times 0.9 = 0.72 of the marked price.

  2. Single equivalent discount = 100%−72%=28%100\% - 72\% = 28\%.

Worth noting

The 10% is taken off the already-reduced price, so the discounts do not add to 30%. 72% is the price paid, not the discount.

Question 43

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The length of a rectangle is increased by 20% and its breadth by 10%. By what percentage does its area increase?

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AnswerC · 32%

  1. Area multiplies by 1.2×1.1=1.321.2 \times 1.1 = 1.32.

  2. So the area increases by 32%. (Formula: 20+10+20×10100=3220 + 10 + \frac{20 \times 10}{100} = 32.)

Worth noting

Adding the two changes gives 30%, missing the extra ab100=2%\frac{ab}{100} = 2\%.

Question 44

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Each side of a square garden is increased by 30%. By what percentage does the area of the garden increase?

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AnswerD · 69%

  1. Area multiplies by 1.3×1.3=1.691.3 \times 1.3 = 1.69.

  2. So the area increases by 69%. (Formula: 30+30+30×30100=6930 + 30 + \frac{30 \times 30}{100} = 69.)

Worth noting

Doubling 30% gives 60%, missing the 9% that comes from the increase on the increase.

Question 45

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The price of a book is increased by 20% and the new price is then reduced by 20%. Compared with the original price, the final price is

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AnswerB · 4% lower

  1. Take the original price as 100, which makes the arithmetic easy to follow.

  2. A 20% increase gives 100×1.2=120100 \times 1.2 = 120.

  3. The 20% reduction is applied to 120, not to 100: 120×0.8=96120 \times 0.8 = 96.

  4. So the final price is 96 against an original 100, a fall of 4%.

Worth noting

The two percentages are equal but they are taken on different amounts, which is why they do not cancel. An equal rise and fall always leaves you slightly worse off.

Question 46

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The price of cooking oil rises by 25%. By what percentage must a family cut its consumption of oil so that its spending on oil stays the same?

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AnswerA · 20%

  1. Spending = price × consumption. Price becomes 1.25 times, so consumption must become 11.25=0.8\frac{1}{1.25} = 0.8 times.

  2. Cut = 1−0.8=0.21 - 0.8 = 0.2, i.e. 20%. (Shortcut: 25100+25×100=20%\frac{25}{100 + 25} \times 100 = 20\%.)

Worth noting

Cutting consumption by the same 25% overshoots: 1.25×0.75=0.93751.25 \times 0.75 = 0.9375, so spending would fall.

Question 47

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A 20% fall in the price of tomatoes lets Deepa buy 3 kg more tomatoes for ₹240. What was the original price of tomatoes per kg?

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AnswerC · ₹20

  1. A 20% fall saves 20% of ₹240 = ₹48 on the old quantity.

  2. That ₹48 buys the extra 3 kg at the new price, so new price = 483=₹16\frac{48}{3} = ₹16 per kg.

  3. New price = 80% of original, so original = 160.8=₹20\frac{16}{0.8} = ₹20 per kg.

Worth noting

₹16 is the NEW price. Check: at ₹20, ₹240 buys 12 kg; at ₹16 it buys 15 kg — 3 kg more.

Question 48

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The price of petrol rises by 20%, and a family reduces its petrol consumption by 10%. By what percentage does the family's spending on petrol rise?

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AnswerB · 8%

  1. Spending = price × consumption, so it becomes 1.2×0.9=1.081.2 \times 0.9 = 1.08 times.

  2. Spending rises by 8%. (Formula: 20−10+20×(−10)100=820 - 10 + \frac{20 \times (-10)}{100} = 8.)

Worth noting

Simply subtracting gives 10%; the cross term ab100\frac{ab}{100} is −2 here, not +2 (which would give 12%).

Question 49

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A shop's electricity bill for May was 30% more than its bill for April. Its June bill was 5% more than the May bill and came to ₹2,730. What was the April bill?

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AnswerB · ₹2,000

  1. May bill = 27301.05=₹2,600\frac{2730}{1.05} = ₹2{,}600.

  2. April bill = 26001.3=₹2,000\frac{2600}{1.3} = ₹2{,}000.

Worth noting

Undo each change by dividing, in reverse order. Subtracting 35% from ₹2,730 gives ₹1,774.50; dividing by 1.35 gives about ₹2,022.22 — both wrong because successive rises multiply.

Question 50

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Aarav earns 20% more than Bhavna, and Bhavna earns 25% less than Chirag. How does Aarav's salary compare with Chirag's?

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AnswerA · Aarav earns 10% less than Chirag

  1. Let Chirag earn 100. Bhavna earns 25% less: 75.

  2. Aarav earns 20% more than Bhavna: 75×1.2=9075 \times 1.2 = 90.

  3. Aarav earns 10 less than Chirag's 100, i.e. 10% less.

Worth noting

The two percentages are on different bases (Bhavna's and Chirag's salaries), so +20−25=−520 - 25 = -5 is wrong.

Question 51

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In a quiz, Ayesha scored 90. Nikhil scored 20% less than Ayesha, and Farhan scored 20% more than Nikhil. What was Farhan's score?

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AnswerC · 86.4

  1. Nikhil = 90×0.8=7290 \times 0.8 = 72.

  2. Farhan = 72×1.2=86.472 \times 1.2 = 86.4.

Worth noting

A 20% fall followed by a 20% rise does not return to 90: the rise is on the smaller number. Net effect = −20+20−4=−4%-20 + 20 - 4 = -4\%, and 96% of 90=86.496\% \text{ of } 90 = 86.4.

Question 52

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The length of a rectangular plot is increased by 15% and its breadth is decreased by 20%. What happens to its area?

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AnswerA · It decreases by 8%

  1. Area multiplies by 1.15×0.8=0.921.15 \times 0.8 = 0.92.

  2. So the area decreases by 8%. (Formula: 15−20+15×(−20)100=−815 - 20 + \frac{15 \times (-20)}{100} = -8.)

Worth noting

15−20=−515 - 20 = -5 misses the cross term of −3.

Question 53

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A newspaper's number of subscribers fell by 20% in the first year, rose by 25% in the second year and rose by 10% in the third year. What was the overall percentage increase over the three years?

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AnswerB · 10%

  1. Overall multiplier = 0.8×1.25×1.10.8 \times 1.25 \times 1.1.

  2. 0.8×1.25=10.8 \times 1.25 = 1, so the multiplier is 1.1.

  3. Overall increase = 10%.

Worth noting

Adding the three rates (−20+25+10)(-20 + 25 + 10) gives 15%, which ignores that each change acts on a different base.

Question 54

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The price of sugar rises by 20%, but a family can increase its spending on sugar by only 8%. By what percentage must the family reduce its consumption of sugar?

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AnswerA · 10%

  1. Spending = price × consumption, so new consumption = 1.081.2\frac{1.08}{1.2} of the old.

  2. 1.081.2=0.9\frac{1.08}{1.2} = 0.9, so consumption must fall to 90%.

  3. Reduction = 10%.

Worth noting

20−8=12%20 - 8 = 12\% ignores that the cut applies at the higher price. 1623%16\frac{2}{3}\% would be the cut if spending could not rise at all.

Question 55

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The length of a cuboid-shaped box is increased by 10%, its breadth by 20%, and its height is decreased by 25%. What happens to its volume?

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AnswerA · It decreases by 1%

  1. Volume multiplies by 1.1×1.2×0.751.1 \times 1.2 \times 0.75.

  2. 1.1×1.2=1.321.1 \times 1.2 = 1.32, and 1.32×0.75=0.991.32 \times 0.75 = 0.99.

  3. So the volume decreases by 1%.

Worth noting

Adding the changes (10+20−25=5)(10 + 20 - 25 = 5) gives "increases by 5%", the trap. Multiply the factors instead.

Question 56

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The radius of a circular flower bed is reduced by 30%. By what percentage does its area decrease?

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AnswerD · 51%

  1. Area is proportional to r2r^2, so it multiplies by 0.7×0.7=0.490.7 \times 0.7 = 0.49.

  2. Decrease = 1−0.49=0.511 - 0.49 = 0.51, i.e. 51%.

Worth noting

49% is the area that REMAINS, not the decrease. Formula: −30−30+(−30)(−30)100=−51-30 - 30 + \frac{(-30)(-30)}{100} = -51.

Question 57

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Rohan's salary was cut by x%, and later the reduced salary was raised by x%. He now earns 4% less than his original salary. What is x?

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AnswerD · 20

  1. Net change for −x% then +x% is −x2100%-\frac{x^2}{100}\%.

  2. x2100=4\frac{x^2}{100} = 4, so x2=400x^2 = 400 and x=20x = 20.

  3. Check: 100 → 80 → 96, which is 4% less.

Worth noting

A cut and a rise by the same percentage always leave a net LOSS of x2100%\frac{x^2}{100}\%, whatever the order.

Question 58

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A seller cuts the price of a table fan by 10%, and a week later cuts the new price by another 10%. The fan now costs ₹1,215. What was its original price?

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AnswerD · ₹1,500

  1. After two 10% cuts the price is 0.9×0.9=0.810.9 \times 0.9 = 0.81 of the original.

  2. Original = 12150.81=₹1,500\frac{1215}{0.81} = ₹1{,}500.

Worth noting

Adding 20% (₹1,458) or adding 10% twice (₹1,470.15) to ₹1,215 uses the wrong base. Divide by the multiplier instead.

Population growth and depreciation

14 questions

Question 59

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The population of a town is 40,000 and grows by 10% every year. What will the population be after 2 years?

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AnswerD · 48,400

  1. After 1 year: 40000×1.1=44,00040000 \times 1.1 = 44{,}000.

  2. After 2 years: 44000×1.1=48,40044000 \times 1.1 = 48{,}400.

Worth noting

48,000 adds 10% of 40,000 twice (simple growth); the second year's growth is on 44,000. 8,400 is only the increase.

Question 60

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A car bought for ₹6,00,000 loses 20% of its value every year. What is its value after 2 years?

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AnswerC · ₹3,84,000

  1. Each year the value becomes 80% of the previous year's value.

  2. Value = 600000×0.8×0.8=600000×0.64=₹3,84,000600000 \times 0.8 \times 0.8 = 600000 \times 0.64 = ₹3{,}84{,}000.

Worth noting

₹3,60,000 subtracts 20% of the ORIGINAL price twice. ₹2,16,000 is the total loss in value, not the value.

Question 61

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A culture of 800 bacteria grows by 50% every hour. How many bacteria are there after 2 hours?

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AnswerD · 1,800

  1. After 1 hour: 800×1.5=1,200800 \times 1.5 = 1{,}200.

  2. After 2 hours: 1200×1.5=1,8001200 \times 1.5 = 1{,}800.

Worth noting

1,600 adds 50% of 800 twice; 1,200 stops after one hour; 1,000 is the increase only.

Question 62

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A machine is worth ₹50,000 now, and its value falls by 10% every year. What will it be worth after 3 years?

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AnswerB · ₹36,450

  1. Value = 50000×0.9350000 \times 0.9^3.

  2. 0.93=0.7290.9^3 = 0.729, so value = 50000×0.729=₹36,45050000 \times 0.729 = ₹36{,}450.

Worth noting

₹35,000 takes 10% of ₹50,000 off three times (simple depreciation). ₹40,500 is the value after only 2 years.

Question 63

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After growing by 8% in one year, a town's population is 27,000. What was the population a year earlier?

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AnswerB · 25,000

  1. 27,000=108%27{,}000 = 108\% of last year's population.

  2. Last year = 270001.08=25,000\frac{27000}{1.08} = 25{,}000.

Worth noting

Subtracting 8% of 27,000 gives 24,840 — the 8% was of the OLD population, so divide by 1.08.

Question 64

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A mobile phone loses 25% of its value every year. After 2 years it is worth ₹22,500. What was it worth when new?

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AnswerC · ₹40,000

  1. After 2 years the value is 0.75×0.75=0.56250.75 \times 0.75 = 0.5625 of the original.

  2. Original = 225000.5625=₹40,000\frac{22500}{0.5625} = ₹40{,}000.

Worth noting

Multiplying ₹22,500 by 1.25 twice gives ₹35,156.25 — the 25% was of the higher earlier value. ₹45,000 undoes a simple 50% loss.

Question 65

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A town has a population of 80,000. It increases by 10% in the first year and decreases by 5% in the second year. What is the population after 2 years?

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AnswerB · 83,600

  1. After year 1: 80000×1.1=88,00080000 \times 1.1 = 88{,}000.

  2. After year 2: 88000×0.95=83,60088000 \times 0.95 = 83{,}600.

Worth noting

84,000 uses 10−5=5%10 - 5 = 5\% on 80,000. The 5% fall is on 88,000, so it removes 4,400, not 4,000.

Question 66

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A scooter worth ₹80,000 loses 10% of its value every year. After how many whole years will its value first fall below ₹50,000?

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AnswerC · 5

  1. Values at the end of each year: ₹72,000, ₹64,800, ₹58,320, ₹52,488.

  2. Year 5: 52488×0.9=₹47,239.2052488 \times 0.9 = ₹47{,}239.20, the first value below ₹50,000.

  3. So it takes 5 years.

Worth noting

Simple depreciation (₹8,000 a year) would reach ₹48,000 in 4 years, but compound depreciation removes less each year.

Question 67

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A village's population grew from 62,500 to 72,900 in two years, at the same rate each year. What was the annual rate of growth?

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AnswerA · 8%

  1. Growth factor over 2 years = 7290062500=1.1664\frac{72900}{62500} = 1.1664.

  2. 1.1664=1.0821.1664 = 1.08^2, so the yearly factor is 1.08.

  3. Annual rate = 8%.

Worth noting

16.64% is the total growth over two years; halving it (8.32%) treats the growth as simple. Compound growth needs the square root of the factor.

Question 68

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Kavya's monthly salary is ₹40,000. It is raised by 10% at the end of every year. What will her monthly salary be after 3 raises?

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AnswerD · ₹53,240

  1. Salary = 40000×1.1340000 \times 1.1^3.

  2. 1.13=1.3311.1^3 = 1.331, so salary = 40000×1.331=₹53,24040000 \times 1.331 = ₹53{,}240.

Worth noting

₹52,000 adds 10% of ₹40,000 three times; each raise is on the already-raised salary.

Question 69

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A school's annual trip cost ₹1,44,000 last year and ₹1,62,000 this year. If the cost rises by the same percentage next year, what will it cost next year?

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AnswerB · ₹1,82,250

  1. Rise this year = ₹18,000, so rate = 18000144000×100=12.5%\frac{18000}{144000} \times 100 = 12.5\%.

  2. Next year = 162000×1.125=₹1,82,250162000 \times 1.125 = ₹1{,}82{,}250.

Worth noting

Adding the same ₹18,000 again gives ₹1,80,000 — the same PERCENTAGE rise on a bigger base is a bigger rupee rise.

Question 70

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A town's population rose by 20% in the first year, fell by 20% in the second year and rose by 10% in the third year. It is now 52,800. What was the population three years ago?

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AnswerC · 50,000

  1. Overall factor = 1.2×0.8×1.11.2 \times 0.8 \times 1.1.

  2. 1.2×0.8=0.961.2 \times 0.8 = 0.96, and 0.96×1.1=1.0560.96 \times 1.1 = 1.056.

  3. Original = 528001.056=50,000\frac{52800}{1.056} = 50{,}000.

Worth noting

A 20% rise and a 20% fall do not cancel (they leave 96%). Assuming they cancel gives 48,000.

Question 71

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A printing machine bought for ₹2,50,000 depreciates by 20% in the first year, 15% in the second year and 10% in the third year. What is its value after 3 years?

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AnswerC · ₹1,53,000

  1. After year 1: 250000×0.8=₹2,00,000250000 \times 0.8 = ₹2{,}00{,}000.

  2. After year 2: 200000×0.85=₹1,70,000200000 \times 0.85 = ₹1{,}70{,}000.

  3. After year 3: 170000×0.9=₹1,53,000170000 \times 0.9 = ₹1{,}53{,}000.

Worth noting

₹1,37,500 subtracts 45% of the original price; ₹97,000 is the total loss; ₹1,70,000 stops after two years.

Question 72

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A town of 40,000 people grows by 5% each year, but at the end of each year, after the growth, 1,000 people move away. What is the population after 2 years?

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AnswerB · 42,050

  1. Year 1: 40000×1.05=42,00040000 \times 1.05 = 42{,}000; minus 1,000=41,0001{,}000 = 41{,}000.

  2. Year 2: 41000×1.05=43,05041000 \times 1.05 = 43{,}050; minus 1,000=42,0501{,}000 = 42{,}050.

Worth noting

Growing 40,000 for two years first (44,100) and then subtracting 2,000 gives 42,100 — wrong, because the people who left in year 1 do not grow in year 2.

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