Key Concepts & Formulae at a Glance
- Per cent (from Latin per centum, "by the hundred") means "out of 100." A percentage is simply a fraction with denominator 100: \(x\%=\dfrac{x}{100}\).
- Fraction → percentage: multiply the fraction by 100, e.g. \(\dfrac34\times100=75\), so \(\dfrac34=75\%\).
- Percentage → fraction/decimal: \(y\%=\dfrac{y}{100}\); e.g. \(31\%=\dfrac{31}{100}=0.31\).
- Percentage of a quantity: \(y\%\) of \(z\) is \(\dfrac{y}{100}\times z\).
- Percentages can exceed 100% — this simply means the value is more than the whole/base being compared to.
- Percentage increase/decrease: \(\dfrac{\text{amount of change}}{\text{original amount (base)}}\times100\).
- Profit/loss percentage is conventionally calculated with respect to the cost price (CP) unless stated otherwise: \(\%\text{profit/loss}=\dfrac{|SP-CP|}{CP}\times100\).
- Discount: a percentage reduction from the marked price (MP); successive discounts (like 30%+20%) compound and are not the same as their sum (30%+20% ≠ 50% off).
- Simple (non-compounding) growth over \(t\) terms at rate \(r\): final amount \(=p(1+rt)\) — the base stays fixed every term (linear growth).
- Compounding over \(t\) terms at rate \(r\): final amount \(=p(1+r)^t\) — the base grows each term (exponential growth), always giving a larger result than simple growth for \(t \gt 1\).
1.1 Fractions as Percentages
The symbol % is read as per cent, from the Latin per centum ("by the hundred" / "out of hundred"). So 25 per cent (25%) means 25 out of every 100 — 25 people out of 100, ₹25 out of ₹100, or 25 marks out of 100. If 50% of some quantity \(s\) is meant, it means \(50\times\frac{1}{100}\times s = \frac{50}{100}\times s = \frac12 s\). Thus, percentages are simply fractions where the denominator is 100: \(20\%=\frac{20}{100}=\frac{2}{10}=\frac15\), and \(33\%=\frac{33}{100}\).
Expressing Fractions as Percentages
Example 1 — Surya's paint. Surya mixes red and yellow paint to make a deep orange colour for a sunset painting. Red paint makes up \(\frac34\) of the mixture. \(\frac34\) is 3 out of every 4, which is 6 out of every 8 (equivalent fraction), which is 30 out of every 40, which is 75 out of every 100. This means 75% is red.
Two systematic methods for converting \(\frac34\) to a percentage (finding the equivalent fraction with denominator 100):
| Method 1 (scale the fraction) | Method 2 (solve for x) |
| \(\dfrac34=\dfrac{3\times25}{4\times25}=\dfrac{75}{100}=75\%\) | \(\dfrac34=\dfrac{x}{100}\). Multiplying both sides by 100: \(x=\dfrac34\times100=75\). |
Bar model showing the equivalence between 3/4 (red) and 75%, with the remaining 1/4 (yellow) as 25%.
MTMath Talk — Can you tell what percentage of the colour was made using yellow?
Since red is \(\frac34=75\%\) of the mixture, and the whole mixture is 100%, yellow makes up the rest: \(100\%-75\%=25\%\). This also follows directly since yellow is \(\frac14\) of the mixture, and \(\frac14\times100=25\).
Example 2 — Surya's prize money. Surya wants to save \(\frac25\) of his prize money to buy a new canvas.
| Method 1 | Method 2 |
| \(\dfrac25=\dfrac{20}{50}=\dfrac{40}{100}=40\%\) | \(\dfrac25=\dfrac{x}{100} \Rightarrow x=\dfrac25\times100=40\) |
MTMath Talk — Method 3: Complete the bar model relating savings (as a fraction, marked in fifths) to the percentage scale below it. Isn't it the same as finding \(\frac25\)th of 100?
Yes — Method 3 is exactly the same idea as Methods 1 and 2, viewed as a bar model. The "savings for canvas" portion spans 2 of the 5 equal sections (since it's \(\frac25\) of the total). Each fifth of the bar corresponds to \(\frac{100}{5}=20\%\) on the percentage scale below. So the 2 shaded sections correspond to \(2\times20\%=40\%\) — matching the boxes to fill in as 20%, 40%, 60%, 80%, 100% under the tick marks \(\frac15,\frac25,\frac35,\frac45,\frac55\).
Several problems in mathematics can be approached and solved in different ways — while the method you come up with may be dear to you, it can be enriching to see how others thought about it. In short: a fraction is of a unit, while a percentage is per 100. To express a fraction as a percentage, just multiply the fraction by 100.
MTMath Talk — Example 3: Given a percentage, can you express it as a fraction? For example, express 24% as a fraction.
Since a percentage is already a fraction, \(24\%=\dfrac{24}{100}\). Other equivalent forms: \(\dfrac{24}{100}=\dfrac{12}{50}=\dfrac{6}{25}=\dfrac{48}{200}\).
In general, a percentage \(z\%\) can be expressed by any fraction equivalent to \(\dfrac{z}{100}\).
Figure it Out — Set 1
Five questions from pages 3–4 of the textbook.
1Express the following fractions as percentages: (i) \(\frac35\) (ii) \(\frac{7}{14}\) (iii) \(\frac{9}{20}\) (iv) \(\frac{72}{150}\) (v) \(\frac13\) (vi) \(\frac{5}{11}\)
Multiplying each fraction by 100:
(i) \(\dfrac35\times100=60\%\)
(ii) \(\dfrac{7}{14}\times100=50\%\)
(iii) \(\dfrac{9}{20}\times100=45\%\)
(iv) \(\dfrac{72}{150}\times100=48\%\)
(v) \(\dfrac13\times100=33.\overline{3}\%\)
(vi) \(\dfrac{5}{11}\times100\approx45.45\%\)
2Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white? (i) 10% (ii) 15% (iii) 25% (iv) 60% (v) 40% (vi) None of these
\(\dfrac{15}{25}\times100=60\%\).
3In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?
\(\dfrac{15}{80}\times100=18.75\%\).
4A group of friends is participating in a long-distance run. The positions of A, B, C, D after 15 minutes are shown on a race track. Match each runner's approximate position to a percentage of the race completed, from: 55%, 20%, 38%, 72%, 84%, 93%.
5Identify and write the appropriate symbol '>', '<', '=' (without calculating where possible): (i) 50% ___ 5% (ii) \(\frac{5}{10}\) ___ 50% (iii) \(\frac{3}{11}\) ___ 61% (iv) 30% ___ \(\frac13\)
(i) 50% is ten times 5%, so \(50\%\gt5\%\).
(ii) \(\frac{5}{10}=\frac12=50\%\), so \(\frac{5}{10}=50\%\).
(iii) \(\frac{3}{11}\) is a little more than a quarter (\(\frac{3}{11}\approx27.3\%\)), which is far less than 61%, so \(\frac{3}{11}\lt61\%\).
(iv) \(\frac13\approx33.3\%\), which is more than 30%, so \(30\%\lt\frac13\).
Why 100? & Percentages Around Us
If percentages are just a particular type of fraction, why do we need them at all? Consider: a biscuit factory has sugar making up \(\frac{9}{34}\) of Variety 1 and \(\frac{13}{45}\) of Variety 2 — it isn't obvious at a glance which is sweeter. But presented as percentages (26.47% vs. 28.88%), the comparison becomes immediate. If we want a common denominator, why choose 100 specifically, rather than 10, 50, 1000, or 43? In principle any denominator could be chosen, but 100 has real advantages: since our number system is base 10, numbers like 10, 100, and 1000 fit easily with decimals (e.g. \(31\%=\frac{31}{100}=0.31\)), making conversions between fractions, decimals, and percentages quick and intuitive. The number 100 is also large enough to give useful detail, yet simple enough to grasp mentally — "per 10" would be too coarse for many purposes, while statistics sometimes uses "per 1000" or "per lakh" for even finer detail.
The idea of "per hundred" is very old: it appears as early as the 4th century BCE in Kautilya's Arthashastra, describing monthly interest rates "per cent." Around the same time, the Romans used taxes of \(\frac{1}{20}\) and \(\frac{1}{100}\) in trade and auctions, and 15th-century Italian manuscripts used expressions like "xx p cento" (20%), "x p cento" (10%), and "vii p cento" (7%).
Percentages show up everywhere: the human body is, on average, about 60% water by weight; ice cream is about 30–50% air by volume; roughly 99.86% of the Solar System's mass is in the Sun; and an estimated 52% of the world's agricultural land is degraded.
1.2 Percentage of Some Quantity
Example 1 — Madhu and Madhav's biscuits. Madhu's biscuits had 25% sugar, Madhav's had 35% sugar. If both ate 100 g of biscuits, Madhav clearly ate more sugar (35 g vs. 25 g). But suppose Madhu ate 120 g and Madhav ate 95 g — who ate more sugar now?
Madhu ate 120 g of biscuits with 25% sugar. Since 25% sugar means 5 g sugar per 20 g biscuits, that's 30 g sugar per 120 g biscuits — or via the proportional relationship \(\frac{25}{100}=\frac{s}{120}\), giving \(s=\frac{25}{100}\times120=30\) g.
Madhav ate 95 g of biscuits with 35% sugar. Since 100 g has 35 g sugar, 1 g has \(\frac{35}{100}\) g sugar, so 95 g has \(\frac{35}{100}\times95=33.25\) g sugar.
Madhav ate more sugar (33.25 g vs. 30 g). More generally, \(y\%\) of some value \(z\) is \(\dfrac{y}{100}\times z\).
Free-hand Computations
We just calculated 25% of 120. Is 25% the same as \(\frac14\)? Yes, since \(25\) is \(\frac14\) of 100 (\(\frac{25}{100}=\frac14\)). So 25% of 40 is the same as \(\frac14\times40=10\).
MTMath Talk — Mentally calculate 25%, 10%, 20%, and 5% of the values 100, 200, 50, 80, 10, 35, 287. What relationships do you notice?
| 100 | 200 | 50 | 80 | 10 | 35 | 287 | |
|---|---|---|---|---|---|---|---|
| 25% | 25 | 50 | 12.5 | 20 | 2.5 | 8.75 | 71.75 |
| 10% | 10 | 20 | 5 | 8 | 1 | 3.5 | 28.7 |
| 20% | 20 | 40 | 10 | 16 | 2 | 7 | 57.4 |
| 5% | 5 | 10 | 2.5 | 4 | 0.5 | 1.75 | 14.35 |
20% of a value is always exactly double 10% of the same value, since 20 parts out of 100 is twice 10 parts out of 100. Also, (20% of \(y\)) + (5% of \(y\)) = 25% of \(y\), since \(\left(\frac{20}{100}\times y\right)+\left(\frac{5}{100}\times y\right)=\left(\frac{25}{100}\times y\right)\) for every \(y\).
MTMath Talk — Using the 20%=2×10% relationship, mentally calculate 40% of each value in the table above. Then find 15% using 10%+5%.
40% = 2×20%, so simply double each 20%-row value: 40, 80, 20, 32, 4, 14, 114.8 (for 100, 200, 50, 80, 10, 35, 287 respectively).
15% = 10%+5%, so add the 10%-row and 5%-row values: 15, 30, 7.5, 12, 1.5, 5.25, 43.05.
MTMath Talk — How would you mentally calculate 75%, 90%, 70%, and 55% of some value?
75% = \(\frac34\), so take a quarter and multiply by 3, or subtract 25% from the whole (100%−25%).
90% = 100%−10%, so subtract one-tenth of the value from the value itself.
70% = 7×10%, so find 10% and multiply by 7 (or 50%+20%).
55% = 50%+5%, so take half the value, then add one-twentieth of it.
The FDP Trio — Fractions, Decimals, and Percentages
Example 2. We find 50% of a value by multiplying by \(\frac12\). Will multiplying by the decimal 0.5 also work? Yes, since \(\frac12=0.5\): \(50\%=\frac{50}{100}=\frac12=\frac{0.5}{1}=0.5\). E.g. 50% of 24 = 12, and \(0.5\times24=12\) too.
MTMath Talk — What decimal value multiplies a quantity to give 10% of it? Complete the fraction/decimal table for 50%, 100%, 25%, 75%, 10%, 1%, 5%, 43%.
To find 10% of a quantity, multiply by the decimal 0.1 (since \(10\%=\frac{10}{100}=0.1\)).
| Per cent | 50% | 100% | 25% | 75% | 10% | 1% | 5% | 43% |
|---|---|---|---|---|---|---|---|---|
| Fraction | 50/100 | 100/100 | 25/100 | 75/100 | 10/100 | 1/100 | 5/100 | 43/100 |
| Decimal | 0.5 | 1.0 | 0.25 | 0.75 | 0.1 | 0.01 | 0.05 | 0.43 |
Activity: How Close Can You Get? Pair up. Each person picks a number (\(a\) and \(b\), with \(a\lt b\)), then both estimate the percentage equivalent to \(\frac{a}{b}\) within 5 seconds — whoever's estimate is closer wins the round. Play 10 rounds. (This is a practice activity with no fixed numeric answer — its value is in building quick estimation skill.)
Example 3 — Zubin's test. Maximum marks are 75; an A grade needs 80% or above. Minimum marks needed:
| Fraction multiplication | \(\dfrac{80}{100}\times75=\dfrac45\times75=60\) |
| Decimal multiplication | \(0.8\times75=60\) |
| Proportional reasoning | Out of 100, the minimum is 80; out of 75, it's \(\dfrac{75\times80}{100}=60\) |
Example 4 — Millet kanji. Millet to water ratio for boiling is 2:7. What percentage is millet, and how much millet is needed for 500 mL?
The ratio of millet to the total mixture is \(2:9\) (since \(2+7=9\)). In one unit of mixture, millet occupies \(\frac29\) units. Estimating first: half of 9 is 4.5, so \(\frac29\) is clearly less than 50%; half of 4.5 is 2.25, so \(\frac29\) is less than 25% too. Since 10% of 9 is 0.9 and 20% of 9 is 1.8, \(\frac29\) (=2) sits between 20–25%.
Exact percentage of millet: \(\frac29\times100\approx22.22\%\). Percentage of water: \(100-22.22=77.78\%\). Since a mixture with 22.22% millet means 100 mL has 22.22 mL millet, 500 mL will have \(5\times22.22=111.1\) mL of millet.
Example 5 — Cyclist Delhi to Agra. A cyclist completes 40% of the journey, covering 92 km. How much further to Agra?
| Method 1 | 40% is 92 km, so 20% is 46 km. This makes 60% = 92+46 = 138 km. |
| Method 2 | \(40:92::60:r \Rightarrow \dfrac{40}{92}=\dfrac{60}{r} \Rightarrow r=60\times\dfrac{92}{40}=138\) |
| Method 3 | \(\dfrac{40}{100}=\dfrac{92}{d} \Rightarrow d=92\times\dfrac{100}{40}=230\). Remaining \(=230-92=138\) km. |
| Method 4 | If \(x\) is remaining, total \(=x+92\), and \(\frac{40}{100}(x+92)=92 \Rightarrow x+92=230 \Rightarrow x=138\) |
Percentages Greater than 100
Can a percentage exceed 100? Example 6 — Kishanlal's garment shop. Kishanlal's daily sales target is at least ₹5000. Sales on Days 1–2 were ₹2000 and ₹3500 — that's \(\frac{2000}{5000}\times100=40\%\) and \(\frac{3500}{5000}\times100=70\%\) of target (equivalently, 60% and 30% short of target).
MTMath Talk — On Days 3 and 4, Kishanlal made ₹5000 and ₹6000. What percentage of target are these? On Days 5 and 6, he made ₹7800 and ₹9550 — find these percentages too. On Day 7 he achieved 150% of target and on Day 8, 210% — find the actual sales made those days.
Day 3: \(\frac{5000}{5000}\times100=100\%\) (exactly met target). Day 4: \(\frac{6000}{5000}\times100=120\%\) — since ₹1000 is 20% of ₹5000, ₹6000 (=5000+1000) is 100%+20%=120% of target.
Day 5: \(\frac{7800}{5000}\times100=156\%\). Day 6: \(\frac{9550}{5000}\times100=191\%\).
Day 7 (150% of target): sales \(=\frac{150}{100}\times5000=7500\). Day 8 (210% of target): sales \(=\frac{210}{100}\times5000=10{,}500\).
MTMath Talk — Complete the fraction/decimal table for 90%, 110%, 200%, 250%, 15%, 173%, 358%, 28.9%, 305%.
| Percent | 90% | 110% | 200% | 250% | 15% | 173% | 358% | 28.9% | 305% |
|---|---|---|---|---|---|---|---|---|---|
| Fraction | 9/10 | 11/10 | 2/1 | 5/2 | 3/20 | 173/100 | 179/50 | 289/1000 | 61/20 |
| Decimal | 0.9 | 1.1 | 2.0 | 2.5 | 0.15 | 1.73 | 3.58 | 0.289 | 3.05 |
Example 7 — Farmer's wheat harvest. Last year's harvest was 260 kg; this year's is 650 kg. This year's harvest as a percentage of last year's: \(\frac{650}{260}\times100=250\%\) — meaning this year's harvest is 2.5 times last year's.
Figure it Out — Set 2
Twelve questions from pages 12–14 of the textbook. Estimate first before computing.
1Find the missing numbers in the bar-model pairs (percentage bar vs. value bar), for totals 75, 90, and 140. The first has been worked out.
The method: each pair shows the same bar split into equal sections, once labelled with percentages (100% total) and once with the actual value (the given total). Match the percentage mark to the corresponding value mark using \(\dfrac{\%}{100}=\dfrac{\text{part}}{\text{total}}\), exactly as in Example 5's cyclist problem.
2Find the value of the following and draw bar models: (i) 25% of 160 (ii) 16% of 250 (iii) 62% of 360 (iv) 140% of 40 (v) 1% of 1 hour (vi) 7% of 10 kg
(i) \(\dfrac{25}{100}\times160=40\)
(ii) \(\dfrac{16}{100}\times250=40\)
(iii) \(\dfrac{62}{100}\times360=223.2\)
(iv) \(\dfrac{140}{100}\times40=56\)
(v) 1 hour = 60 min, so \(\dfrac{1}{100}\times60=0.6\) min = 36 seconds
(vi) \(\dfrac{7}{100}\times10=0.7\) kg
Bar models for (i) 25% of 160 = 40, and (iv) 140% of 40 = 56 — note how the second bar extends past the "100%" mark since the percentage exceeds 100.
3Surya made 60 mL of deep orange paint. How much red paint did he use if red made up \(\frac34\) of the mixture?
\(\dfrac34\times60=45\) mL.
4Write '>', '<', or '=' (visualise/estimate first): (i) 50% of 510 ⬜ 50% of 515 (ii) 37% of 148 ⬜ 73% of 148 (iii) 29% of 43 ⬜ 92% of 110 (iv) 30% of 40 ⬜ 40% of 50 (v) 45% of 200 ⬜ 10% of 490 (vi) 30% of 80 ⬜ 24% of 64
(i) Same percentage, but 515>510, so \(\lt\).
(ii) Same base (148), and 37%<73%, so \(\lt\).
(iii) \(29\%\text{ of }43\approx12.5\), \(92\%\text{ of }110\approx101.2\), so \(\lt\).
(iv) \(30\%\text{ of }40=12\), \(40\%\text{ of }50=20\), so \(\lt\).
(v) \(45\%\text{ of }200=90\), \(10\%\text{ of }490=49\), so \(\gt\).
(vi) \(30\%\text{ of }80=24\), \(24\%\text{ of }64=15.36\), so \(\gt\).
5Fill in the blanks: (i) 30% of \(k\) is 70, so 60% of \(k\) is ___, 90% of \(k\) is ___, 120% of \(k\) is ___. (ii) 100% of \(m\) is 215, so 10% of \(m\) is ___, 1% of \(m\) is ___, 6% of \(m\) is ___. (iii) 90% of \(n\) is 270, so 9% of \(n\) is ___, 18% of \(n\) is ___, 100% of \(n\) is ___.
(i) Since 60%, 90%, 120% are 2×, 3×, 4× of 30%: \(60\%\text{ of }k=2\times70=140\); \(90\%\text{ of }k=3\times70=210\); \(120\%\text{ of }k=4\times70=280\).
(ii) 10% of \(m\) is \(\frac{215}{10}=21.5\); 1% of \(m\) is \(\frac{215}{100}=2.15\); 6% of \(m\) is \(6\times2.15=12.9\).
(iii) Since 90% of \(n\)=270, \(n=300\). So 9% of \(n\)=27, 18% of \(n\)=54, and 100% of \(n\)=300.
6Fill in the blanks: (i) 3 is ___ % of 300. (ii) ___ is 40% of 4. (iii) 40 is 80% of ___.
(i) \(\frac{3}{300}\times100=1\%\).
(ii) \(\frac{40}{100}\times4=1.6\).
(iii) \(40=\frac{80}{100}\times x \Rightarrow x=50\).
7Is 10% of a day longer than 1% of a week? Create such questions and challenge your peers.
10% of a day: \(\frac{10}{100}\times24\times60=144\) minutes = 2.4 hours. 1% of a week: \(\frac{1}{100}\times7\times24\times60=100.8\) minutes = 1.68 hours.
8Math Talk — Mariam's bull: Day 1 given 2 units of fodder, ate 1; Day 2 given 3, ate 2; ... Day 99 given 100, ate 99. Represent these as percentages. What do you observe?
On Day \(n\), the bull is given \(n+1\) units and eats \(n\) units, so the percentage eaten is \(\frac{n}{n+1}\times100\). Day 1: \(\frac12\times100=50\%\). Day 2: \(\frac23\times100\approx66.67\%\). Day 3: \(\frac34\times100=75\%\). … Day 99: \(\frac{99}{100}\times100=99\%\).
9Workers take 18 days to pick coffee berries in 20% of a plantation. How many days for the entire plantation, assuming the rate of work stays the same? Why is this assumption necessary?
\(20\%:18\text{ days}::100\%:x \Rightarrow x=18\times\dfrac{100}{20}=90\) days.
The "same rate of work" assumption is necessary because without it, the workers could speed up, slow down, or take breaks, and the simple proportional scaling (5× the area needing 5× the time) would no longer hold.
10A badminton coach plans warm up : play : cool down as 10% : 80% : 10% of a 90-minute session. How long is each activity?
Warm up = \(\frac{10}{100}\times90=9\) min. Play = \(\frac{80}{100}\times90=72\) min. Cool down = \(\frac{10}{100}\times90=9\) min.
11An estimated 90% of the world's population lives in the Northern Hemisphere. Find the approximate number of people living there, based on this year's world population.
Using a current world population estimate of about 8.2 billion: \(\frac{90}{100}\times8{,}200{,}000{,}000\approx7{,}380{,}000{,}000\) (7.38 billion).
12A halwa recipe for 4 people uses Rava: 40%, Sugar: 40%, Ghee: 20%. (i) What is the proportion of each for 8 people? (ii) If the total weight of ingredients is 2 kg, how much rava, sugar, and ghee are present?
(i) The percentages (proportions) stay exactly the same regardless of how many people it serves — Rava: 40%, Sugar: 40%, Ghee: 20% — only the total quantity made doubles.
(ii) Rava \(=\frac{40}{100}\times2=0.8\) kg; Sugar \(=\frac{40}{100}\times2=0.8\) kg; Ghee \(=\frac{20}{100}\times2=0.4\) kg.
1.3 Using Percentages
To Compare Proportions
Example 1 — Eesha's test scores. Eesha scored 42/50 in English and 70/80 in Science. Since the maximum marks differ, we convert both to percentages: English \(=\frac{42}{50}\times100=84\%\); Science \(=\frac{70}{80}\times100=87.5\%\). Science is the better relative score (87.5% > 84%), even though the raw marks lost (8 in English vs. 10 in Science) might suggest otherwise.
Know Your Contents (KYC)
Example 2 — Badam drink mix. Madhu and Madhav compare two badam (almond) drink-mix products by their ingredient labels:
DEF (total 150 g): Sugar 99 g, Milk Solids 30 g, Badam Powder 12 g, Food Chemicals 9 g.
Zacni (total 400 g): Sugar 272 g, Milk Solids 64 g, Badam Powder 40 g, Food Chemicals 24 g.
DEF's sugar content as a percentage of total weight: \(\frac{99}{150}\times100=66\%\).
MTMath Talk — Complete the table by calculating percentages for both products, and check they add up to 100%. Which product has a larger share of badam? Which uses a smaller proportion of food chemicals?
| Sugar | Milk Solids | Badam Powder | Food Chemicals | |
|---|---|---|---|---|
| DEF | 66% | 20% | 8% | 6% |
| Zacni | 68% | 16% | 10% | 6% |
Both rows sum to exactly 100% (66+20+8+6=100 and 68+16+10+6=100), confirming the ingredients account for the entire weight.
Zacni has the larger share of badam powder (10% vs. DEF's 8%). Both products use the same proportion of food chemicals (6% each) — a tie.
Percentage Increase or Decrease
Percentages are often used to describe rate of change: \(\text{Percentage increase/decrease}=\dfrac{\text{amount of change}}{\text{original amount (base)}}\times100\).
1. Tomato prices. Price 3 years ago was ₹30, now ₹42. Increase = ₹12. Percentage increase \(=\frac{12}{30}\times100=40\%\).
2. Theatre footfall. Average footfall was 160 before COVID, now 100. Decrease = 60. Percentage decrease \(=\frac{60}{160}\times100=37.5\%\).
MTMath Talk — Example 3: Do these mean the same thing? (i) The population of a state in 1991 is 165% of that in 1961. (ii) The population increased by 65% from 1961 to 1991.
Yes — let \(p\) = population in 1961, \(q\) = population in 1991.
Statement (i): \(q=165\%\text{ of }p=\frac{165}{100}p=1.65p\).
Statement (ii): \(q=p+65\%\text{ of }p=p+0.65p=1.65p\).
Both give exactly \(q=1.65p\) — the 1991 population is 1.65 times the 1961 population either way.
Profit and Loss
The marked price (MP) is what the shopkeeper quotes (sometimes the MRP). The selling price (SP) is what the customer actually pays (often after a discount). The cost price (CP) is what the shopkeeper paid to acquire the item. A sweater's journey: Manufacturing (CP ₹230, MP ₹255, SP ₹253) → Wholesale (CP ₹253, MP ₹310, SP ₹300) → Retail (CP ₹300, MP ₹480, SP ₹430). Kishanlal (retailer) buys at ₹300, marks it at ₹480, and after bargaining sells at ₹430 — a profit of ₹430−₹300=₹130. (If SP were less than CP, it would be a loss.)
MTMath Talk — Example 4: Find the percentage profit Kishanlal made. Also find the profit percentage of the wholesaler and the manufacturer.
Considering CP as 100% (the standard convention): profit percentage \(=\dfrac{\text{profit}}{CP}\times100\).
Kishanlal (retailer): profit \(=430-300=130\); \(\%\text{profit}=\frac{130}{300}\times100\approx43.3\%\).
Wholesaler: bought at CP ₹253, sold at SP ₹300; profit \(=47\); \(\%\text{profit}=\frac{47}{253}\times100\approx18.58\%\).
Manufacturer: bought at CP ₹230, sold at SP ₹253; profit \(=23\); \(\%\text{profit}=\frac{23}{230}\times100=10\%\).
MTMath Talk — Shambhavi buys 200-page notebooks at ₹36 each and sells with a 20% profit margin — find the selling price. She sells crayon boxes at ₹50 per box with a 25% profit margin — find her buying price.
Notebooks: \(SP=CP\times\left(1+\frac{20}{100}\right)=36\times1.2=\)₹43.20.
Crayons: \(SP=CP\times1.25=50 \Rightarrow CP=\frac{50}{1.25}=\)₹40.
Example 5 — Raghu's rice. He bought rice at ₹35/kg (so 10 kg cost ₹350), and cleared old stock selling 10 kg for ₹300 — a loss of ₹50. Percentage loss \(=\frac{50}{350}\times100\approx14.28\%\).
MTMath Talk — Could we have calculated the loss percentage per kg instead? Would it be the same?
Yes — per kg, CP = ₹35, SP = ₹300÷10 = ₹30, loss = ₹5 per kg. Percentage loss \(=\frac{5}{35}\times100\approx14.28\%\) — exactly the same, since scaling both the loss amount and the base (CP) by the same factor (dividing by 10) leaves their ratio, and hence the percentage, unchanged.
Example 6 — Shyamala's damaged vase. Bought at ₹2650/piece; sells the damaged one at an 18% loss.
| Method 1 | SP is 18% less than CP, i.e. 82% of CP: \(0.82\times2650=\)₹2173. |
| Method 2 | Loss amount \(=\frac{18}{100}\times2650=477\). SP \(=2650-477=\)₹2173. |
MTMath Talk — Snehal sells strawberries at ₹80/kg with a 12% loss. What is the cost price?
SP is 88% of CP (since 100%−12%=88%): \(88\%\text{ of }CP=80 \Rightarrow CP=\dfrac{80}{0.88}\approx\)₹90.91/kg.
MTMath Talk — A utensil store offers a 35% discount on a cooker with MRP ₹1800. What is the selling price? If the cost price was ₹900, what is the percentage profit made?
Selling price \(=65\%\text{ of }1800=0.65\times1800=\)₹1170.
Profit \(=1170-900=270\). Percentage profit \(=\frac{270}{900}\times100=30\%\).
Gross vs. net profit. If Kishanlal's total sales last month were ₹80,000, and the goods sold cost him ₹48,000, the difference (₹32,000) is gross profit. After deducting other expenses (transport, salaries, electricity, etc.) of ₹8,000, the remaining ₹24,000 is net profit. (This chapter uses "profit" to mean gross profit, unless stated otherwise.)
Manisha's fertiliser. She buys 50 kg bags at ₹500/bag, sells at ₹750/bag — profit ₹250. With respect to CP: \(\frac{250}{500}\times100=50\%\). With respect to SP (revenue): \(\frac{250}{750}\times100\approx33.33\%\). Profit percentage based on revenue answers "how much profit compared to what I earned?", while profit percentage based on cost answers "how much profit compared to what I invested?" — these are genuinely different questions with different numeric answers.
Suppose in a month Manisha made sales of ₹1,50,000, with goods costing ₹1,00,000 (gross profit ₹50,000) and monthly expenses of ₹5,000 (net profit ₹45,000). Net profit as a percentage of revenue: \(\frac{45{,}000}{1{,}50{,}000}\times100=30\%\).
Taxes
Tax rates — like GST (Goods and Services Tax) or Income Tax — are also specified as percentages. GST is part of the amount paid, and this portion goes to the government.
MTMath Talk — Check if the calculations are correct in this bill: 3 CFL Bulbs at ₹150 each, Sub Total ₹450, CGST 9%, SGST 9%, Total.
Sub Total: \(3\times150=450\) ✓.
CGST (9% of 450): \(\frac{9}{100}\times450=40.5\) ✓ matches the ₹40.50 shown.
SGST (9% of 450): also \(40.5\) ✓ matches the ₹40.50 shown.
Total: \(450+40.5+40.5=531\) ✓ matches the ₹531.00 shown.
Figure it Out — Set 3
Twelve questions from pages 19–20 of the textbook.
1A shopkeeper buys a geometry box for ₹75 and sells it for ₹110. What is his profit margin with respect to the cost?
Profit \(=110-75=35\). \(\%\text{profit}=\frac{35}{75}\times100\approx46.67\%\).
2A carpenter's chair costs ₹475 in materials. For a 50% profit margin, what price should the chair sell for?
\(SP=475\times1.5=\)₹712.50.
3A company's total sales (revenue) last year was ₹2.5 crore, with a healthy profit margin of 25%. What was the total expenditure (costs) last year?
Following this chapter's convention that "profit margin" means profit with respect to cost (as in the Kishanlal/Manisha examples): if cost is \(x\), then \(x\times1.25=2.5\text{ crore} \Rightarrow x=\frac{2{,}50{,}00{,}000}{1.25}=2{,}00{,}00{,}000\).
4A clothing shop offers a 25% discount on all shirts. If the original price of a shirt is ₹300, how much will Anwar pay?
\(300\times(1-0.25)=300\times0.75=\)₹225.
5Petrol was ₹60 in 2015 and ₹100 in 2025. What is the percentage increase? (i) 50% (ii) 40% (iii) 60% (iv) 66.66% (v) 140% (vi) 160.66%
Increase \(=100-60=40\). \(\%\text{increase}=\frac{40}{60}\times100\approx66.67\%\).
6Samson bought a car for ₹4,40,000 after a 15% discount. What was the original price?
SP is 85% of the original price MP: \(0.85\times MP=4{,}40{,}000 \Rightarrow MP=\dfrac{4{,}40{,}000}{0.85}\approx\)₹5{,}17{,}647.06.
71600 people voted; the winner got 500 votes. What percent of total votes did the winner get? Guess the minimum number of candidates.
Winner's percentage: \(\frac{500}{1600}\times100=31.25\%\).
The remaining \(1600-500=1100\) votes went to other candidates. For the winner to genuinely win, every other individual candidate must have received fewer votes than 500. With just 1 opponent, that opponent alone would get all 1100 votes — far more than 500, so the winner couldn't have won. With 2 opponents, the maximum either can have (while both staying under 500) is \(2\times499=998\), which is still less than 1100 — still impossible. With 3 opponents, the maximum staying under 500 each is \(3\times499=1497\ge1100\) — now it's possible (e.g. votes split as 400+400+300=1100). So at least 3 opponents are needed.
8Rice was ₹38/kg in 2024, ₹42/kg in 2025. What is the rate of inflation?
Increase \(=42-38=4\). Inflation rate \(=\frac{4}{38}\times100\approx10.53\%\).
9A number increased by 20% becomes 90. What is the number?
\(x\times1.2=90 \Rightarrow x=75\).
10A milkman sold two buffaloes for ₹80,000 each — 5% profit on one, 10% loss on the other. Find his overall profit or loss.
Buffalo 1 (5% profit): \(CP_1\times1.05=80{,}000 \Rightarrow CP_1=\dfrac{80{,}000}{1.05}\approx\)₹76{,}190.48.
Buffalo 2 (10% loss): \(CP_2\times0.9=80{,}000 \Rightarrow CP_2=\dfrac{80{,}000}{0.9}\approx\)₹88{,}888.89.
Total CP \(\approx1{,}65{,}079.37\); Total SP \(=1{,}60{,}000\). Net result \(=1{,}60{,}000-1{,}65{,}079.37\approx-5{,}079.37\) — a loss.
Overall loss percentage \(=\frac{5{,}079.37}{1{,}65{,}079.37}\times100\approx3.08\%\).
11An elephant population increased by 5% in the last decade. If last decade's population is \(p\), the population now is: (i) \(p\times0.5\) (ii) \(p\times0.05\) (iii) \(p\times1.5\) (iv) \(p\times1.05\) (v) \(p+1.50\)
A 5% increase means the new value is the original plus 5% of it: \(p+0.05p=p(1+0.05)=1.05p\).
12Which statement(s) mean the same as "camera demand has fallen by 85% in the last decade"? (i) demand now is 85% of a decade ago (ii) demand a decade ago was 85% of now (iii) demand now is 15% of a decade ago (iv) demand a decade ago was 15% of now (v) demand a decade ago was 185% of now (vi) demand now is 185% of a decade ago
Let \(D_0\)=demand a decade ago, \(D_1\)=demand now. "Fallen by 85%" means \(D_1=D_0-0.85D_0=0.15D_0\), i.e. demand now is 15% of demand a decade ago — this matches statement (iii) exactly. All the other statements describe a different relationship (e.g. (i) would mean only a 15% fall, not 85%).
Growth and Compounding
Interest is extra money paid by banks/post offices on deposits (or by borrowers on loans). "p.a." (per annum) means "for every year." If a bank pays 10% p.a. on a ₹6000 deposit, after 1 year you get \(6000+(0.10\times6000)=6000+600=\)₹6600 — equivalently, the deposit becomes 110% of itself: \(6000\times1.1=6600\). In general, Amount after 1 year = Principal + (rate × Principal).
Example 7 — ₹6000 deposit for 3 years. Two possibilities:
Option 1 (no compounding): interest is paid out each year, and the principal stays ₹6000 throughout. Each year earns ₹600 interest (10% of 6000); over 3 years that's ₹1800 interest, plus the ₹6000 principal returned = ₹7800 total.
Option 2 (compounding): interest is added back to the principal each year. Year 1: \(6000\times1.1=6600\). Year 2: \(6600\times1.1=7260\). Year 3: \(7260\times1.1=7986\). Total received = ₹7986 — more than without compounding.
MTMath Talk — Example 8: What percent is the total amount received, with respect to the amount deposited, in both options?
Without compounding: \(\frac{7800}{6000}\times100=130\%\). This is \(6000\times(1+0.1+0.1+0.1)=6000\times1.3\) — a percentage gain of 30% over 3 years.
With compounding: \(\frac{7986}{6000}\times100=133.1\%\). This is \(6000\times1.1\times1.1\times1.1=6000\times1.331\) — a percentage gain of 33.1% over 3 years.
In general, for principal \(p\), rate \(r\), and \(t\) years:
| No compounding | Interest per term stays \(p\times r\), so after \(t\) terms: total \(=p+(p\times r\times t)=p(1+rt)\). |
| With compounding | Each term's principal grows by factor \((1+r)\): total \(=p\times(1+r)\times(1+r)\times\dots\times(1+r)\) (\(t\) times) \(=p(1+r)^t\). |
Figure it Out — Set 4
Nine questions from pages 22–24 of the textbook.
1Bank of Yahapur offers 10% p.a. Compare the amounts from depositing ₹20,000 for 2 years, with and without compounding.
Without compounding: \(20{,}000\times(1+0.1\times2)=20{,}000\times1.2=\)₹24{,}000.
With compounding: \(20{,}000\times(1.1)^2=20{,}000\times1.21=\)₹24{,}200.
2Bank of Wahapur offers 5% p.a. Compare depositing ₹20,000 for 4 years, with and without compounding.
Without compounding: \(20{,}000\times(1+0.05\times4)=20{,}000\times1.2=\)₹24{,}000.
With compounding: \(20{,}000\times(1.05)^4=20{,}000\times1.21550625=\)₹24{,}310.13 (rounded).
3Math Talk — Do you observe anything interesting comparing the solutions to Questions 1 and 2?
Both scenarios have the same total simple-interest rate (\(rt=10\%\times2=20\%=5\%\times4\)), so without compounding, both give exactly the same result (₹24,000). But with compounding, Bank of Wahapur's 4 smaller, more frequent compounding periods (₹24,310.13) beat Bank of Yahapur's 2 larger, less frequent periods (₹24,200) — even though the "total" rate (\(rt\)) is identical in both cases.
MTMath Talk — Example 9: General formula for ₹6000 at 10% p.a. for 't' years — and the formula for the total interest amount (not the total amount).
Without compounding: total amount \(=p(1+rt)\), so total interest \(=p(1+rt)-p=prt\).
With compounding: total amount \(=p(1+r)^t\), so total interest \(=p(1+r)^t-p=p\left[(1+r)^t-1\right]\).
4Jasmine invests amount 'p' for 4 years at 6% p.a. Which expression(s) give the total amount after 4 years without compounding? (i) \(p\times6\times4\) (ii) \(p\times0.6\times4\) (iii) \(p\times\frac{0.6}{100}\times4\) (iv) \(p\times\frac{0.06}{100}\times4\) (v) \(p\times1.6\times4\) (vi) \(p\times1.06\times4\) (vii) \(p+(p\times0.06\times4)\)
Without compounding, total \(=p(1+rt)=p(1+0.06\times4)=p\times1.24\). Checking option (vii): \(p+(p\times0.06\times4)=p(1+0.24)=1.24p\) ✓ — matches exactly. None of the other options simplify to \(1.24p\).
5The post office offers 7% p.a. How much interest for ₹50,000 over 3 years, without compounding? How much more if compounded?
Without compounding: interest \(=50{,}000\times0.07\times3=\)₹10{,}500.
With compounding: amount \(=50{,}000\times(1.07)^3=50{,}000\times1.225043=\)₹61{,}252.15; interest \(=61{,}252.15-50{,}000=\)₹11{,}252.15.
Extra from compounding \(=11{,}252.15-10{,}500=\)₹752.15.
6Giridhar borrows ₹12,500 at 12% p.a. for 3 years, no compounding. Raghava borrows the same for the same period at 10% p.a., compounded annually. Who pays more interest, and by how much?
Giridhar (simple): interest \(=12{,}500\times0.12\times3=\)₹4{,}500.
Raghava (compound): amount \(=12{,}500\times(1.1)^3=12{,}500\times1.331=\)₹16{,}637.50; interest \(=16{,}637.50-12{,}500=\)₹4{,}137.50.
7Math Talk — ₹1000 grows at 10% p.a. How long to double, with vs. without compounding? Is compounding exponential growth and non-compounding linear growth?
Without compounding (linear): \(p(1+rt)=2p \Rightarrow rt=1 \Rightarrow t=\frac{1}{0.1}=10\) years exactly.
With compounding (exponential): \(p(1+r)^t=2p \Rightarrow (1.1)^t=2 \Rightarrow t=\frac{\log2}{\log1.1}\approx7.27\) years — noticeably faster.
Yes — non-compounding grows the same fixed amount every year (additive/linear growth), while compounding multiplies by the same factor every year (multiplicative/exponential growth), so compounding always reaches any target (like doubling) sooner.
8A city's population rises by about 3% every year. If the current population is 1.5 crore, what is the expected population after 3 years?
\(1.5\text{ crore}\times(1.03)^3=1.5\times1.092727=1.6390905\) crore.
9Bacteria increase at 2.5% per hour. Find the count after 2 hours if the initial count is 5,06,000.
\(5{,}06{,}000\times(1.025)^2=5{,}06{,}000\times1.050625=5{,}31{,}616.25\).
Decline (Depreciation)
Many items lose value over time due to use and age — called depreciation.
Example 10 — TV depreciation. Bought at ₹21,000; depreciates 5% after 1 year.
| Method 1 | Reduction \(=5\%\text{ of }21{,}000=0.05\times21{,}000=1050\). Current value \(=21{,}000-1050=\)₹19{,}950. |
| Method 2 | Value after 1 year is 95% of current value: \(0.95\times21{,}000=\)₹19{,}950. |
Example 11 — Village population decline. Reducing by 10% every decade; current population 1250. Expected population after 3 decades:
\(1250\times0.9\times0.9\times0.9=1250\times0.729=911.25\).
(Working decade-by-decade with rounding at each step: 1250→1125→1013 (rounding 1112.5)→912 — very close to the exact 911.25.) Rounding off, the expected population after 3 decades is around 910.
Tricky Percentages
MTMath Talk — Would You Rather? Option A: deposit ₹100, get back ₹300. Option B: deposit ₹1000, get back ₹1500. What percentage gain does each give? Which would you choose (once only), and why?
Option A: gain \(=300-100=200\); \(\%\text{gain}=\frac{200}{100}\times100=200\%\).
Option B: gain \(=1500-1000=500\); \(\%\text{gain}=\frac{500}{1000}\times100=50\%\).
Option A gives a far higher percentage return (200% vs. 50%), but Option B gives more absolute profit (₹500 vs. ₹200). Which is "better" depends on what you're optimising for: if you have exactly ₹1000 available and want the most cash back, Option B wins (₹500 profit); if you only have ₹100 to spare, or you value the rate of return itself, Option A is far more efficient per rupee invested.
While comparing percentages, we must be mindful that we're comparing fractions or proportions, not absolute values — a club that "grew by 80%" isn't necessarily bigger than one that "grew by 100%," since the starting sizes may differ wildly (e.g. 80% growth on a large base can add more members than 100% growth on a tiny base).
MTMath Talk — A store offers either a 20% discount or a flat ₹50 off (for purchases above ₹150). Which is better for buying items worth (i) ₹180 (ii) ₹225 (iii) ₹300?
(i) ₹180: 20% discount \(=\)₹36 vs. flat ₹50 — flat ₹50 is better (saves more).
(ii) ₹225: 20% discount \(=\)₹45 vs. flat ₹50 — flat ₹50 is still better.
(iii) ₹300: 20% discount \(=\)₹60 vs. flat ₹50 — the 20% discount is better now.
The break-even point is exactly ₹250 (where 20% of the purchase equals ₹50) — below ₹250, the flat discount wins; above it, the percentage discount wins.
Example 12 — Cakely vs. Cakify. Cakely offers "30%+20% off"; Cakify offers a flat 50% off. Mathematically 30%+20%=50%, but successive discounts compound rather than simply add. For a ₹200 cake: Cakely's 30% off first gives ₹200−₹60=₹140, then 20% off ₹140 gives ₹140−₹28=₹112. Cakify's flat 50% off gives ₹200−₹100=₹100. Cakify is actually cheaper (₹100 vs. ₹112), even though "30%+20%" sounds identical to "50%."
A Mishap
Example 13 — Surbhi's cookware. She marks up all products by a 50% profit margin, then later offers a 50% discount to clear stock, expecting to break even. Modelling the cost as \(x\): selling price with margin \(=1.5x\); after a 50% discount, worth \(=0.75x\).
(i) Since \(0.75x=\frac34x\) is less than \(x\), she did make a loss — specifically a 25% loss, not a break-even.
(ii) If she sold goods (originally) worth ₹12,000 after the discount: \(0.75x=12{,}000 \Rightarrow x=16{,}000\). She lost ₹16,000−₹12,000=₹4,000.
(iii) For true break-even (selling at exactly the cost price \(x\)), the discount \(d\) (as a fraction) on the marked-up price \(1.5x\) must satisfy \(1.5x-d(1.5x)=x \Rightarrow d=\frac13\approx0.333\). So the discount should have been 33.33%, not 50%.
TTTry This — Ariba says, "My marbles are 120% of Arun's." What could Arun say, comparing his marbles to Ariba's?
Let Arun have \(a\) marbles; Ariba has \(1.2a\). Arun's marbles as a percentage of Ariba's: \(\dfrac{a}{1.2a}\times100=\dfrac{100}{1.2}\approx83.33\%\).
Figure it Out — Final Set
Fifteen questions from pages 27–30 of the textbook.
1Bengaluru's population in 2025 is about 250% of its 2000 population. If the 2000 population was 50 lakhs, what is the 2025 population?
\(50\text{ lakh}\times2.5=125\text{ lakh}=1.25\text{ crore}\).
2World population in 2025 is about 8.2 billion. Match each country's population to its approximate percentage share: Germany (83 million), India (1.46 billion), Bangladesh (175 million), USA (347 million) — options: 13%, 8%, 18%, 10%, 1%, 35%, 2%, 2%, 0.1%.
Computing each share against 8.2 billion:
Germany: \(\frac{83\text{ million}}{8200\text{ million}}\times100\approx1.01\%\) → matches 1%.
India: \(\frac{1460}{8200}\times100\approx17.8\%\) → closest match is 18%.
Bangladesh: \(\frac{175}{8200}\times100\approx2.13\%\) → matches 2%.
USA: \(\frac{347}{8200}\times100\approx4.23\%\).
3A mobile phone costs ₹8,250. GST of 18% is added. Which gives the final price? (i) 8250+18 (ii) 8250+1800 (iii) \(8250+\frac{18}{100}\) (iv) \(8250\times18\) (v) \(8250\times1.18\) (vi) \(8250+8250\times0.18\) (vii) \(1.8\times8250\)
Final price \(=8250\times1.18=8250+8250\times0.18=9735\). Both options (v) and (vi) give this correctly (they're algebraically identical: \(8250\times1.18=8250\times(1+0.18)=8250+8250\times0.18\)).
4Monthly mice-population changes: Month 1: +5%, Month 2: −2%, Month 3: −3%. Initial population \(p\). Which statement(s) are true? (i) \(p\times0.05\times0.02\times0.03\) (ii) \(p\times1.05\times0.98\times0.97\) (iii) \(p+0.05-0.02-0.03\) (iv) population unchanged (v) more than \(p\) (vi) less than \(p\)
Each month's change compounds onto the previous month's population, so the correct expression is \(p\times1.05\times0.98\times0.97\) — statement (ii). (Statement (i) and (iii) both misrepresent how percentage changes combine.)
Computing the combined factor: \(1.05\times0.98\times0.97\approx0.9981\), which is less than 1 — so the population after 3 months is less than \(p\), despite starting with an increase.
5A shopkeeper sets a price with a 35% profit margin, then offers a 30% discount on the selling price. Profit or loss?
Let CP \(=x\). Marked-up SP \(=1.35x\). After a 30% discount: \(0.7\times1.35x=0.945x\).
Since \(0.945x\lt x\), this is a loss of \(1-0.945=0.055\), i.e. 5.5%.
6What percentage of area is occupied by region 'E' in the dot-grid figure (regions A, B, C, D, E)?
7Math Talk — What is 5% of 40? 40% of 5? 25% of 12? 12% of 25? 15% of 60? 60% of 15? What do you notice? Make a general statement using algebra.
\(5\%\text{ of }40=2\), \(40\%\text{ of }5=2\) — equal. \(25\%\text{ of }12=3\), \(12\%\text{ of }25=3\) — equal. \(15\%\text{ of }60=9\), \(60\%\text{ of }15=9\) — equal.
In every case, \(x\%\text{ of }y = y\%\text{ of }x\). Algebraically: \(x\%\text{ of }y=\dfrac{x}{100}\times y=\dfrac{xy}{100}\), and \(y\%\text{ of }x=\dfrac{y}{100}\times x=\dfrac{xy}{100}\) — identical expressions, so they're always equal for any \(x,y\).
8An excursion has 40% Grade 8 students (rest Grade 9); 60% of Grade 8 students are girls. (i) What percentage of all students are Grade 8 girls? (ii) If 160 total students go, how many are Grade 8 girls?
(i) Grade 8 girls as a fraction of everyone \(=40\%\times60\%=\frac{40}{100}\times\frac{60}{100}=0.24=24\%\).
(ii) For 160 total students: \(24\%\text{ of }160=0.24\times160=38.4\), i.e. about 38 girls (rounding to a whole number of students).
9Try This — A shopkeeper sells pencils so that the selling price of 3 pencils equals the cost price of 5 pencils. Profit or loss? What percentage?
Let the cost of 1 pencil be \(c\). Selling price of 3 pencils \(=5c\) (given), so selling price of 1 pencil \(=\frac{5c}{3}\).
Profit per pencil \(=\frac{5c}{3}-c=\frac{2c}{3}\). Percentage profit \(=\dfrac{2c/3}{c}\times100=\dfrac{200}{3}\approx66.67\%\).
10Bus fares increased 3% last year, then 4% this year. What is the overall percentage increase?
Combined factor \(=1.03\times1.04=1.0712\), i.e. a 7.12% overall increase — not simply \(3\%+4\%=7\%\), because the second increase compounds on the already-increased fare.
11If a rectangle's length increases by 10% and the area is unchanged, by what exact percentage does the breadth decrease?
Let original length \(=L\), breadth \(=B\), area \(=LB\). New length \(=1.1L\). For area to stay the same: \(1.1L\times B'=LB \Rightarrow B'=\dfrac{B}{1.1}=\dfrac{10B}{11}\).
Decrease in breadth \(=B-\frac{10B}{11}=\frac{B}{11}\). Percentage decrease \(=\dfrac{B/11}{B}\times100=\dfrac{100}{11}\approx9.09\%\).
12A 65 g chips packet lists: Potato 70%, Vegetable oil 24%, Salt 3%, Spices 3%. Find the weight of each ingredient.
Potato \(=\frac{70}{100}\times65=45.5\) g. Vegetable oil \(=\frac{24}{100}\times65=15.6\) g. Salt \(=\frac{3}{100}\times65=1.95\) g. Spices \(=\frac{3}{100}\times65=1.95\) g.
Check: \(45.5+15.6+1.95+1.95=65\) g ✓.
13Three shops sell identical items at the same price with deals: Shop A "Buy 1 Get 1 Free," Shop B "Buy 2 Get 1 Free," Shop C "Buy 3 Get 1 Free." (i) Effective price per item at ₹100/item — rank cheapest to costliest. (ii) Percentage discount at each shop. (iii) For exactly 4 items, which shop is best?
(i) Shop A: pay for 1, get 2 → effective price \(=\frac{100}{2}=\)₹50/item. Shop B: pay for 2, get 3 → \(\frac{200}{3}\approx\)₹66.67/item. Shop C: pay for 3, get 4 → \(\frac{300}{4}=\)₹75/item. Cheapest to costliest: A < B < C.
(ii) Percentage discount \(=\dfrac{\text{free items}}{\text{total items received}}\times100\). Shop A: \(\frac{1}{2}\times100=50\%\). Shop B: \(\frac{1}{3}\times100\approx33.33\%\). Shop C: \(\frac{1}{4}\times100=25\%\).
(iii) For exactly 4 items: Shop A — buy 2 sets (pay for 2, get 4) = ₹200. Shop B — buy one "2 get 1" set (₹200, get 3) plus 1 more at full price (₹100) = ₹300 for 4 items. Shop C — buy one "3 get 1" set exactly = ₹300 for 4 items. Shop A is cheapest for exactly 4 items, at ₹200.
14Try This — In a room of 100 people, 99% are left-handed. How many left-handed people must leave to bring that down to 98%?
Let \(x\) left-handed people leave (only left-handed people leave; the room started with 99 left-handed and 1 right-handed person). Remaining: \((99-x)\) left-handed out of \((100-x)\) total, and we want this to equal 98%:
\(\dfrac{99-x}{100-x}=0.98 \Rightarrow 99-x=98-0.98x \Rightarrow 1=0.02x \Rightarrow x=50\).
Check: \(\frac{99-50}{100-50}=\frac{49}{50}=98\%\) ✓.
15Try This — Based on a bar graph of computer-use ability by age/gender (Children 4%/4%; Teenage 24%F/29%M; Twenties 26%F/37%M; Thirties 14%F/25%M; Forties 7%F/14%M; Fifties 4%F/9%M; Seniors 2%F/4%M), which statement(s) are valid? (i) Twenties are the most computer-literate age group (ii) women lag behind across age groups (iii) more people are in their twenties than teenagers (iv) more than a quarter of thirty-somethings can use computers (v) less than 1 in 10 aged 60+ can use computers (vi) half of twenty-somethings can use computers
(i) Valid. Twenties-Male at 37% is the single highest bar in the whole graph, and Twenties overall shows the highest usage rates among all age groups.
(ii) Mostly valid, with a caveat. In every age group, the male percentage is greater than or equal to the female percentage; for Children, the two are exactly equal (4% = 4%), so technically women don't "lag" in that one group — everywhere else, they clearly do.
(iii) Not valid — a common misreading. This bar graph shows the percentage of each age group able to use computers, not the actual number of people in each age group. We cannot conclude anything about relative population sizes ("more people in their twenties than teenagers") from ability percentages alone.
(iv) Not valid, assuming roughly equal numbers of men and women in their thirties: average usage \(\approx\frac{14+25}{2}=19.5\%\), which is less than a quarter (25%).
(v) Valid. Seniors: 2% (F) and 4% (M); average \(\approx3\%\), clearly less than 10%.
(vi) Not valid. Twenties: 26% (F) and 37% (M); average \(\approx31.5\%\), well short of 50%.
Puzzle Time — Peaceful Knights
Place 8 Knights So None Attacks Another
A knight moves in an "L-shape": two steps in one direction (vertical or horizontal) plus one step perpendicular to that. Place 8 knights on a standard 8×8 chessboard so that no knight can capture any other.
A simple, always-correct solution: place all 8 knights on squares of the same colour (for example, every square where the row and column indices add to an even number). A knight's move always lands on a square of the opposite colour to the one it started on — so if every knight starts on the same colour, none of them can ever reach a square occupied by another knight. One easy same-colour arrangement: place knights on both endpoints of every other row along the top two rows, e.g. the squares (row 1, columns 1,3,5,7) and (row 3, columns 1,3,5,7) — 8 knights total, all on light (or all on dark) squares, verified to have zero mutual attacks.
Frequently Asked Questions
Continue with Part 2, Chapter 2
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