Fractions in Disguise
— Complete Solutions
Every in-text question and every Figure It Out exercise from Chapter 1, solved step by step — with bar models, proportional reasoning, and clear working for each numerical.
Fractions as Percentages
The basic idea — a percentage is just a fraction with denominator 100 — and how to convert any fraction into a percentage.
Can you tell what percentage of the colour was made using yellow?
(Surya’s deep orange paint — red made up 75% of the mixture)
The whole mixture is 100% of the paint. We know red paint makes up 75% of it.
Since the mixture only has red and yellow paint, the yellow part is simply whatever is left over after removing the red part:
Try completing Method 3 by filling the boxes — expressing $dfrac{2}{5}$ as a percentage using the bar model (Surya’s prize money savings).
The bar model splits the total prize money into 5 equal parts, where the savings cover 2 of those 5 parts $left(dfrac{2}{5}right)$. The bottom number line shows the same bar marked from 0% to 100%.
Since the bar is divided into 5 equal parts and the full bar is 100%, each single part represents:
So filling in the boxes under each mark — $frac{1}{5}, frac{2}{5}, frac{3}{5}, frac{4}{5}, frac{5}{5}$ — gives 20%, 40%, 60%, 80%, 100% respectively.
Express the following fractions as percentages.
(i) $frac{3}{5}$ (ii) $frac{7}{14}$ (iii) $frac{9}{20}$ (iv) $frac{72}{150}$ (v) $frac{1}{3}$ (vi) $frac{5}{11}$
To turn any fraction into a percentage, multiply it by 100 (since percentage means “per hundred”).
| Fraction | Working: $dfrac{a}{b}times 100$ | Percentage |
|---|---|---|
| (i) $frac{3}{5}$ | $frac{3}{5}times100 = 60$ | 60% |
| (ii) $frac{7}{14}$ | $frac{7}{14}times100 = 50$ | 50% |
| (iii) $frac{9}{20}$ | $frac{9}{20}times100 = 45$ | 45% |
| (iv) $frac{72}{150}$ | $frac{72}{150}times100 = 48$ | 48% |
| (v) $frac{1}{3}$ | $frac{1}{3}times100 = 33.33$ | 33.33% |
| (vi) $frac{5}{11}$ | $frac{5}{11}times100 = 45.45$ | 45.45% |
Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white?
The fraction of white marbles is $dfrac{15}{25}$ (white marbles out of total marbles).
Convert this fraction to a percentage by multiplying by 100:
In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?
The fraction of students who walk is $dfrac{15}{80}$.
Multiply by 100 to express it as a percentage:
A group of friends is running a long-distance race. After 15 minutes, four runners (A, B, C, D) are at different positions. Match each runner to the approximate percentage of the race they have completed, from these options: 20%, 38%, 55%, 72%, 84%, 93%.
Runners A, B, C and D are placed in increasing order of distance from the start, so their percentages must also increase in that same order: A < B < C < D.
Out of the six given percentages, we pick the four that best fit the visual spacing between the runners shown in the picture, in increasing order.
| Runner | A | B | C | D |
|---|---|---|---|---|
| Approx. % completed | 20% | 38% | 72% | 84% |
Identify and write the appropriate symbol (‘>’, ‘<‘, ‘=’) without doing full calculations:
(i) 50% ___ 5% (ii) $frac{5}{10}$ ___ 50% (iii) $frac{3}{11}$ ___ 61% (iv) 30% ___ $frac{1}{3}$
50% is ten times bigger than 5%, so 50% > 5%.
$frac{5}{10}$ is exactly half, and 50% is also exactly half, so $frac{5}{10}$ = 50%.
$frac{3}{11}$ is less than $frac{3}{10}$ which is 30%, and 30% is much less than 61%. So $frac{3}{11}$ < 61%. (Exact value: $frac{3}{11}times100 approx 27.27%$)
$frac{1}{3}$ as a percentage is $33.33%$, which is more than 30%. So 30% < $frac{1}{3}$.
| (i) | (ii) | (iii) | (iv) |
|---|---|---|---|
| > | = | < | < |
Percentage of Some Quantity
Finding the actual value when a percentage of a quantity is given — using fractions, decimals, proportions, and quick mental methods.
Suppose Madhu ate 120 g of biscuits (25% sugar) and Madhav ate 95 g of biscuits (35% sugar). Who consumed more sugar?
Madhu’s sugar: 25% of 120 g.
Madhav’s sugar: 35% of 95 g.
Mentally calculate 25%, 10%, 20%, and 5% of: 100, 200, 50, 80, 10, 35, 287.
Mental shortcuts: 25% = divide by 4 | 10% = divide by 10 | 20% = double the 10% value | 5% = half the 10% value.
| 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 |
Using the pattern above, mentally calculate 40% of the same values: 100, 200, 50, 80, 10, 35, 287.
40% is simply double the 20% value (or 4 times the 10% value):
| 100 | 200 | 50 | 80 | 10 | 35 | 287 | |
|---|---|---|---|---|---|---|---|
| 40% | 40 | 80 | 20 | 32 | 4 | 14 | 114.8 |
Using the relationship (20% of $y$) + (5% of $y$) = 25% of $y$, mentally calculate 15% of the table values.
Since $15% = 10% + 5%$, we can add the 10% and 5% values from the earlier table for each number:
| 100 | 200 | 50 | 80 | 10 | 35 | 287 | |
|---|---|---|---|---|---|---|---|
| 15% | 15 | 30 | 7.5 | 12 | 1.5 | 5.25 | 43.05 |
How would you mentally calculate 75%, 90%, 70%, and 55% of some value? (Math Talk)
Build each percentage from easy “anchor” percentages we already know how to find quickly (10%, 25%, 50%, 100%):
| Percentage | Quick build-up |
|---|---|
| 75% | 100% − 25%, i.e. subtract a quarter from the whole value |
| 90% | 100% − 10%, i.e. subtract a tenth from the whole value |
| 70% | 50% + 20%, i.e. half plus one-fifth |
| 55% | 50% + 5%, i.e. half plus a twentieth |
What decimal value should be multiplied to find 10% of a quantity? Complete the FDP (Fraction–Decimal–Percentage) table for: 50%, 100%, 25%, 75%, 10%, 1%, 5%, 43%.
To find 10% of a quantity, multiply by the decimal 0.1 (since $frac{10}{100}=0.1$).
| Per cent | 50% | 100% | 25% | 75% | 10% | 1% | 5% | 43% |
|---|---|---|---|---|---|---|---|---|
| Fraction | $frac{50}{100}$ | $frac{100}{100}$ | $frac{25}{100}$ | $frac{75}{100}$ | $frac{10}{100}$ | $frac{1}{100}$ | $frac{5}{100}$ | $frac{43}{100}$ |
| Decimal | 0.5 | 1.0 | 0.25 | 0.75 | 0.1 | 0.01 | 0.05 | 0.43 |
A cyclist completes 40% of the Delhi–Agra journey, covering 92 km. Estimate first, then find how many more km remain to reach Agra.
Estimate: 40% is less than half, so the remaining distance (60%) should be a bit more than the distance already covered. We expect an answer somewhat larger than 92 km.
Exact method: If 40% of the total distance $d$ is 92 km:
Remaining distance $= 230 – 92 = 138$ km.
On Days 5 and 6, Kishanlal’s sales were ₹7800 and ₹9550 (target = ₹5000). Calculate the percentage of target achieved on these days.
On Day 7, Kishanlal achieved 150% of his target. On Day 8, he achieved 210%. Find the actual sales made on these days. (Target = ₹5000)
150% of the target:
210% of the target:
Complete the table converting these percentages (greater than 100%) into fractions and decimals: 90%, 110%, 200%, 250%, 15%, 173%, 358%, 28.9%, 305%.
| Percent | 90% | 110% | 200% | 250% | 15% | 173% | 358% | 28.9% | 305% |
|---|---|---|---|---|---|---|---|---|---|
| Fraction | $frac{9}{10}$ | $frac{11}{10}$ | $frac{2}{1}$ | $frac{5}{2}$ | $frac{3}{20}$ | $frac{173}{100}$ | $frac{179}{50}$ | $frac{289}{1000}$ | $frac{61}{20}$ |
| Decimal | 0.9 | 1.1 | 2.0 | 2.5 | 0.15 | 1.73 | 3.58 | 0.289 | 3.05 |
Find the missing numbers using the bar models (each bar is divided into equal boxes representing 100% in total).
(i) is worked out: 5 equal boxes, 1 box = 20%, and the full bar = 75, so 4 boxes (80%) = 60. (ii) 100% bar split into 10 equal boxes, full bar = 90 — find the value at 6 boxes. (iii) 100% bar split into 4 equal boxes, full bar = 140 — find the value at 3 boxes.
Worked example: 5 equal boxes make 100%, so each box = $frac{100%}{5}=20%$. The full bar (100%) corresponds to 75. So 1 box (20%) corresponds to $frac{20}{100}times75=15$, and 4 boxes (80%) correspond to $frac{80}{100}times75=60$.
The whole (100%) is divided into 10 equal parts, so each part is $frac{100%}{10}=10%$. The full bar (100%) corresponds to 90.
The whole (100%) is divided into 4 equal parts, so each part is $frac{100%}{4}=25%$. The full bar (100%) corresponds to 140.
| (i) | (ii) | (iii) |
|---|---|---|
| 15 & 60 | 54 | 105 |
Find the value of the following and draw their 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
| Part | Calculation | Answer |
|---|---|---|
| (i) | $frac{25}{100}times160$ | 40 |
| (ii) | $frac{16}{100}times250$ | 40 |
| (iii) | $frac{62}{100}times360$ | 223.2 |
| (iv) | $frac{140}{100}times40$ | 56 |
| (v) | $frac{1}{100}times60text{ min}$ | 0.6 min = 36 sec |
| (vi) | $frac{7}{100}times10text{ kg}$ | 0.7 kg = 700 g |
Surya made 60 ml of deep orange paint. How much red paint did he use if red paint made up $frac{3}{4}$ of the deep orange paint?
Red paint = $frac{3}{4}$ of the total mixture = 75% of 60 ml.
Identify the symbol (‘>’, ‘<‘, ‘=’) for each pair. Visualise/estimate first; compute only to verify.
(i) 50% of 510 vs 50% of 515 (ii) 37% of 148 vs 73% of 148 (iii) 29% of 43 vs 92% of 110 (iv) 30% of 40 vs 40% of 50 (v) 45% of 200 vs 10% of 490 (vi) 30% of 80 vs 24% of 64
| Pair | Left value | Right value | Symbol |
|---|---|---|---|
| (i) | 255 | 257.5 | < |
| (ii) | 54.76 | 108.04 | < |
| (iii) | 12.47 | 101.2 | < |
| (iv) | 12 | 20 | < |
| (v) | 90 | 49 | > |
| (vi) | 24 | 15.36 | > |
Fill in the blanks appropriately:
(i) 30% of $k$ is 70. Find 60%, 90%, 120% of $k$. (ii) 100% of $m$ is 215. Find 10%, 1%, 6% of $m$. (iii) 90% of $n$ is 270. Find 9%, 18%, 100% of $n$.
Since 30% of $k$ is 70, doubling and tripling that percentage scales the value the same way:
10% of $m$ is $frac{1}{10}$ of 215; 1% is $frac{1}{100}$ of 215; 6% is 6 times the 1% value.
If 90% of $n$ is 270, then 9% (one-tenth of 90%) is $frac{270}{10}=27$. So 18% is double that, and 100% is found from $n=frac{270}{0.9}$.
| 1st blank | 2nd blank | 3rd blank | |
|---|---|---|---|
| (i) $k$ | 140 | 210 | 280 |
| (ii) $m$ | 21.5 | 2.15 | 12.9 |
| (iii) $n$ | 27 | 54 | 300 |
Fill in the blanks: (i) 3 is ___% of 300. (ii) ___ is 40% of 4. (iii) 40 is 80% of ___.
If 80% of the number is 40, then the number is $frac{40}{0.8}$.
| (i) | (ii) | (iii) |
|---|---|---|
| 1% | 1.6 | 50 |
Is 10% of a day longer than 1% of a week? Create such questions and challenge your peers.
1 day = 24 hours = 1440 minutes. 1 week = 7 days = 168 hours = 10,080 minutes.
Mariam’s bull: Day 1 — given 2 units, ate 1; Day 2 — given 3, ate 2; Day 3 — given 4, ate 3 … up to 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. The percentage eaten is:
| Day | Given | Eaten | % eaten |
|---|---|---|---|
| 1 | 2 | 1 | 50% |
| 2 | 3 | 2 | 66.67% |
| 3 | 4 | 3 | 75% |
| … | … | … | … |
| 99 | 100 | 99 | 99% |
Workers in a coffee plantation take 18 days to pick berries in 20% of the plantation. How many days to finish picking the entire plantation (same rate of work)? Why is the assumption necessary?
If 20% of the plantation takes 18 days, then 100% (the entire plantation) takes 5 times as long:
A badminton coach plans warm up : play : cool down = 10% : 80% : 10%. For a 90-minute training, how long should each activity be?
| Activity | Calculation | Time |
|---|---|---|
| Warm up (10%) | $0.10times90$ | 9 minutes |
| Play (80%) | $0.80times90$ | 72 minutes |
| Cool down (10%) | $0.10times90$ | 9 minutes |
An 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 (≈ 8.2 billion).
World population ≈ 8.2 billion = 8,200,000,000.
A halwa recipe for 4 people: Rava 40%, Sugar 40%, Ghee 20%.
(i) Proportions for 8 people? (ii) If total weight = 2 kg, how much rava, sugar, ghee?
Doubling the number of people (4 → 8) doubles the total quantity of every ingredient proportionally, but the percentage proportions stay exactly the same: Rava 40%, Sugar 40%, Ghee 20%.
For a total weight of 2 kg, applying each percentage:
| Ingredient | Percentage | Weight (out of 2 kg) |
|---|---|---|
| Rava | 40% | 0.8 kg |
| Sugar | 40% | 0.8 kg |
| Ghee | 20% | 0.4 kg |
Using Percentages — Comparing Proportions & KYC
Why we convert different totals to percentages before comparing them — applied to test scores and reading food labels.
Eesha scored 42/50 in English and 70/80 in Science. Reema says Science is better (more marks); Vishu says we can’t compare since the maximums differ. Who is correct?
Vishu is right that raw marks can’t be compared directly when the totals differ — but we can make a fair comparison by converting both scores to percentages (so both are “out of 100”).
English:
Science:
Complete the table comparing two badam drink mixes (DEF and Zacni) by percentage of each ingredient. Which has a larger share of badam? Which has less food chemicals?
DEF (150 g total): Sugar 99 g, Milk solids 30 g, Badam powder 12 g, Food chemicals 9 g.
Zacni (400 g total): Sugar 272 g, Milk solids 64 g, Badam powder 40 g, Food chemicals 24 g.
Convert each ingredient’s weight to a percentage of the total weight for both products.
| Product | Sugar | Milk Solids | Badam Powder | Food Chemicals | Total |
|---|---|---|---|---|---|
| DEF (150 g) | 66% | 20% | 8% | 6% | 100% |
| Zacni (400 g) | 68% | 16% | 10% | 6% | 100% |
Working shown: DEF Milk solids $=frac{30}{150}times100=20%$; Badam $=frac{12}{150}times100=8%$; Chemicals $=frac{9}{150}times100=6%$. Zacni Sugar $=frac{272}{400}times100=68%$; Milk solids $=frac{64}{400}times100=16%$; Badam $=frac{40}{400}times100=10%$; Chemicals $=frac{24}{400}times100=6%$.
Percentage Increase/Decrease & Profit-Loss
Rate of change as a percentage, and the language of buying/selling — cost price, marked price, selling price, profit and loss.
Find the profit percentage of the wholesaler and the manufacturer in the sweater’s journey.
Manufacturer: CP ₹230, SP ₹253 | Wholesaler: CP ₹253, SP ₹300
Manufacturer’s profit $= 253-230=23$.
Wholesaler’s profit $= 300-253=47$.
Shambhavi procures 200-page notebooks at ₹36/book and sells with a 20% profit margin. Find the selling price.
With cost price as 100%, a 20% profit margin makes the selling price 120% of the cost price.
Shambhavi sells crayon boxes at ₹50/box with a 25% profit margin. How much did she buy them for (from the wholesaler)?
Let the cost price be $x$. A 25% profit margin means the selling price is 125% of $x$:
Could the rice loss percentage (Raghu, Example 5) have been calculated per kg instead of for the full 10 kg? Would it be the same?
Per kg: CP = ₹35/kg, SP = $frac{300}{10}=₹30$/kg. Loss per kg = $35-30=₹5$.
Snehal sells strawberries at ₹80/kg with a 12% loss. What is the cost price?
A 12% loss means the selling price is 88% of the cost price. Let the cost price be $x$:
A utensil store offers 35% discount on a cooker with MRP ₹1800. What is the selling price? If the cost price was ₹900, what is the percentage profit after the sale?
Selling price = 65% of MRP (since 35% is taken off):
Profit $= 1170 – 900 = ₹270$.
Check if the calculations in the XY Electricals bill are correct: 3 CFL bulbs @ ₹150 each, CGST 9%, SGST 9%.
Sub Total $= 3times150 = ₹450$ ✓ (matches the bill).
CGST (9%) $= 0.09times450 = ₹40.50$ ✓ SGST (9%) $=0.09times450=₹40.50$ ✓
Total $= 450+40.50+40.50 = ₹531.00$ ✓ matches the bill.
A shopkeeper buys a geometry box for ₹75 and sells it for ₹110. What is his profit margin with respect to the cost?
Profit $= 110-75 = ₹35$.
A carpenter’s material cost for a chair is ₹475, and they want a 50% profit margin. At what price should they sell a chair?
A 50% profit margin on cost means selling price = 150% of cost.
A company’s total sales (revenue) was ₹2.5 crore last year, with a 25% profit margin. What was the total expenditure (cost)?
Let the expenditure be $x$. A 25% profit margin (on cost) means: Revenue = Expenditure + 25% of Expenditure.
A clothing shop offers a 25% discount on shirts. If the original price is ₹300, how much will Anwar pay?
Discount amount $= 25%$ of $300 = frac{25}{100}times300 = ₹75$.
Price to pay $= 300-75 = ₹225$.
Petrol price was ₹60 in 2015 and ₹100 in 2025. What is the percentage increase?
Increase $= 100-60 = ₹40$.
Samson bought a car for ₹4,40,000 after a 15% discount from the dealer. What was the original (marked) price of the car?
Let the marked price be $y$. A 15% discount means the selling price is 85% of $y$:
1600 people voted in an election; the winner got 500 votes. What percent of the total votes did the winner get? Guess the minimum number of candidates.
Since the winner has the most votes, no other candidate can have more than 31.25%. If there were only 3 candidates, the remaining 68.75% split between just 2 people would mean at least one of them has more than the winner’s 31.25% — so 3 candidates isn’t always possible. Checking $frac{100}{31.25}=3.2$ tells us at least 4 “equal shares” of 31.25% would be needed to use up all 100%, so there must be at least 4 candidates.
Rice price was ₹38/kg in 2024 and ₹42/kg in 2025. What is the rate of inflation?
Increase $= 42-38 = ₹4$.
A number increased by 20% becomes 90. What is the number?
Let the number be $x$. Increasing $x$ by 20% means the new value is 120% of $x$:
A milkman sold two buffaloes for ₹80,000 each — a 5% profit on one, and a 10% loss on the other. Find his overall profit or loss.
Buffalo 1 (5% profit): The SP (₹80,000) is 105% of CP, so:
Buffalo 2 (10% loss): The SP (₹80,000) is 90% of CP, so:
Total CP $approx 76{,}190+88{,}889 = ₹1{,}65{,}079$ Total SP $= 80{,}000+80{,}000 = ₹1{,}60{,}000$.
The elephant population increased by 5% in the last decade. If the population last decade was $p$, what is the population now?
New population $= p + 5%$ of $p = p + 0.05p = 1.05p$.
“Demand for cameras has fallen by 85% in the last decade.” Which of these statements mean the same thing?
(i) Demand now = 85% of demand a decade ago (ii) Demand a decade ago = 85% of demand now (iii) Demand now = 15% of demand a decade ago (iv) Demand a decade ago = 15% of demand now (v) Demand a decade ago = 185% of demand now (vi) Demand now = 185% of demand a decade ago
Let the demand a decade ago be $D$. A fall of 85% means the new demand is what remains after removing 85%:
Growth and Compounding
How money grows with simple interest (no compounding) versus compound interest — and the general formulas $p(1+rt)$ and $p(1+r)^t$.
What is the formula for the total interest amount gained at the end of the maturity period, for each of the two options (without and with compounding)?
Without compounding: total amount is $p(1+rt)$, so the interest alone (amount minus principal) is:
With compounding: total amount is $p(1+r)^t$, so the interest alone is:
Bank of Yahapur offers 10% p.a. Compare the amount one gets for ₹20,000 over 2 years, with and without compounding.
Without compounding: $p(1+rt) = 20{,}000(1+0.10times2)$
With compounding: $p(1+r)^t = 20{,}000(1.1)^2$
Bank of Wahapur offers 5% p.a. Compare ₹20,000 over 4 years, with/without compounding. What do you observe across both bank comparisons?
Without compounding: $20{,}000(1+0.05times4) = 20{,}000times1.2 = ₹24{,}000$.
With compounding: $20{,}000(1.05)^4 = 20{,}000times1.2155 approx ₹24{,}310.13$.
Bank of Yahapur offers 10% p.a. Compare ₹20,000 deposited for 2 years, with and without compounding annually.
Using $p=20{,}000$, $r=0.10$, $t=2$:
Bank of Wahapur offers 5% p.a. Compare ₹20,000 deposited for 4 years, with and without compounding.
Do you observe anything interesting in the solutions of questions 1 and 2 above? Share and discuss.
In both cases, the amount without compounding is the same (₹24,000) — because the product $rtimes t$ happens to be 0.2 in both ($10%times2=20%$ and $5%times4=20%$). However, the amount with compounding is higher in the second case (₹24,310.13 vs ₹24,200), even though $rt$ is identical.
Jasmine invests $p$ for 4 years at 6% p.a. Which expression(s) describe the total amount after 4 years without compounding?
(i) $ptimes6times4$ (ii) $ptimes0.6times4$ (iii) $ptimesfrac{0.6}{100}times4$ (iv) $ptimesfrac{0.06}{100}times4$ (v) $ptimes1.6times4$ (vi) $ptimes1.06times4$ (vii) $p+(ptimes0.06times4)$
The correct formula without compounding is $p(1+rt)$ with $r=0.06$, $t=4$:
This exactly matches option (vii). None of the other expressions simplify to the same thing — e.g. (vi) gives $ptimes1.06times4=4.24p$, which is far too large.
The post office offers 7% p.a. How much interest for ₹50,000 over 3 years without compounding? How much more with compounding?
Without compounding: Interest $= prt = 50{,}000times0.07times3 = ₹10{,}500$.
With compounding: Total amount $= 50{,}000times(1.07)^3 approx ₹61{,}252.15$.
Giridhar borrows ₹12,500 at 12% p.a. for 3 years (no compounding). Raghava borrows the same amount, same time, at 10% p.a., compounded annually. Who pays more interest, and by how much?
Giridhar (no compounding): Interest $= 12{,}500times0.12times3 = ₹4500$.
Raghava (compounded): Amount $= 12{,}500times(1.10)^3 = ₹16{,}637.50$.
₹1000 grows at 10% p.a. How long to double it, with vs without compounding? Is compounding exponential growth and non-compounding linear growth?
Without compounding: we need $1000(1+0.1t)=2000$.
With compounding: we need $1000(1.1)^t=2000$, i.e. $(1.1)^t=2$. Solving (using logarithms or trial):
A city’s population rises by about 3% every year. Current population = 1.5 crore. Find the expected population after 3 years.
This is compound growth: $p(1+r)^t$ with $p=1.5$ crore, $r=0.03$, $t=3$.
Bacteria increase at 2.5% per hour. Find the count after 2 hours if the initial count is 5,06,000.
Using $p(1+r)^t$ with $p=5{,}06{,}000$, $r=0.025$, $t=2$:
Decline & Depreciation
When a quantity reduces by a fixed percentage repeatedly — the same idea as compounding, but shrinking instead of growing.
Method recap: A TV worth ₹21,000 depreciates by 5% in a year. What is its value after 1 year?
A 5% depreciation means the value left is 95% of the original.
Tricky Percentages
Percentage traps: when percentage gains look big but absolute gains tell a different story, and why stacked discounts aren’t the same as their sum.
Would You Rather? Option A: deposit ₹100, get back ₹300. Option B: deposit ₹1000, get back ₹1500. What is the percentage gain for each option? Which would you choose?
Option A: gain $= 300-100=₹200$.
Option B: gain $= 1500-1000=₹500$.
A store offers either 20% discount or ₹50 flat discount (on purchases above ₹150). Which option is better for buying items worth: (i) ₹180 (ii) ₹225 (iii) ₹300?
Compare 20% of each amount against the flat ₹50:
| Purchase | 20% discount | ₹50 flat discount | Better choice |
|---|---|---|---|
| (i) ₹180 | ₹36 | ₹50 | ₹50 discount |
| (ii) ₹225 | ₹45 | ₹50 | ₹50 discount |
| (iii) ₹300 | ₹60 | ₹50 | 20% discount |
Cakely offers 30% + 20% off; Cakify offers a flat 50% off. For a ₹200 cake, which gives the cheaper price?
Cakely (30% then 20%, applied one after another):
After 30% off: $200 – 0.30times200 = ₹140$.
After 20% off the new price of ₹140: $140 – 0.20times140 = ₹112$.
Cakify (flat 50%):
Ariba says, “My marbles are 120% of Arun’s marbles.” What would be an appropriate matching statement Arun could make?
Let Arun’s marbles $= x$. Then Ariba’s marbles $= 1.2x$.
Arun’s marbles as a percentage of Ariba’s marbles:
A Mishap — Profit Margin then Discount
Why adding a profit margin and then giving back the same percentage as a discount does not return you to your original price.
Surbhi set a 50% profit margin on cookware, then offered a 50% discount to clear stock — expecting “no loss, no profit.”
(i) Did she actually break even? (ii) If she sold goods (after discount) worth ₹12,000, how much loss did she incur, and what is the percentage loss? (iii) What discount % would have made her break even?
Let the price she bought the goods for be $x$ (her cost price). With a 50% profit margin, her selling price (before discount) becomes $1.5x$. A 50% discount on that price gives:
No, she did not break even. The final selling price is $0.75x$, which is only $frac{3}{4}$ of what she paid — meaning she actually made a 25% loss, not a break-even sale.
If the (discounted) selling price was ₹12,000, then $0.75x = 12{,}000$:
Loss $= 16{,}000 – 12{,}000 = ₹4{,}000$ (which is indeed $frac{4000}{16000}times100=25%$, confirming part (i)).
To sell at exactly the original cost price $x$ (no profit, no loss), we need a discount $d$ (as a fraction) on the marked-up price $1.5x$ such that:
Final Figure It Out — Exercise
The chapter’s last and longest practice set — pulling together every idea covered: comparisons, growth, discounts, areas, and reading graphs.
Bengaluru’s 2025 population is about 250% of its 2000 population. If the 2000 population was 50 lakh, what is the 2025 population?
World population in 2025 ≈ 8.2 billion. Match each country’s population with its approximate % share of the world population.
Germany: 83 million | India: 1.46 billion | Bangladesh: 175 million | USA: 347 million
Options: 13%, 8%, 18%, 10%, 1%, 35%, 2%, 2%, 0.1%
Divide each country’s population by 8.2 billion and convert to a percentage:
| Country | Population | Calculation | Approx % |
|---|---|---|---|
| Germany | 83 million | $frac{0.083}{8.2}times100$ | ≈1% |
| India | 1.46 billion | $frac{1.46}{8.2}times100$ | ≈18% |
| Bangladesh | 175 million | $frac{0.175}{8.2}times100$ | ≈2% |
| USA | 347 million | $frac{0.347}{8.2}times100$ | ≈4% (closest option: 2%) |
A mobile phone costs ₹8,250. GST of 18% is added. Which expression gives the final price?
(i) $8250+18$ (ii) $8250+1800$ (iii) $8250+frac{18}{100}$ (iv) $8250times18$ (v) $8250times1.18$ (vi) $8250+8250times0.18$ (vii) $1.8times8250$
GST amount $= 18%$ of $8250 = 0.18times8250 = 1485$. Final price $= 8250+1485=₹9735$.
This equals $8250times1.18$, and also equals $8250+8250times0.18$ — these are algebraically the same expression written two ways.
Monthly population change of mice: Month 1 = +5%, Month 2 = −2%, Month 3 = −3%. Initial population = $p$. Which statements are true?
(i) $ptimes0.05times0.02times0.03$ (ii) $ptimes1.05times0.98times0.97$ (iii) $p+0.05-0.02-0.03$ (iv) population $=p$ (v) population $>p$ (vi) population $<p$
Each month’s change is a percentage of the population at the start of that month — so the changes multiply (compound), not add. The correct expression is:
Since the multiplier ($approx0.9981$) is less than 1, the final population is slightly less than $p$ — even though $+5-2-3=0$ might suggest no overall change.
A shopkeeper sets price with a 35% profit margin, then offers a 30% discount on the selling price. Profit or loss?
Let cost price (CP) $= 100$. With 35% profit margin, selling price (before discount) $= 135$.
A 30% discount on 135:
What percentage of area is occupied by the region marked ‘E’ in the figure (a square divided by gridlines and a diagonal into regions A, B, C, D, E)?
The full square is divided into an 8×8 grid of unit squares, giving a total area of $8times8=64$ square units.
Region E (the small triangular region cut off by the diagonal in the bottom-left) occupies 8 square units of this grid.
What is 5% of 40? What is 40% of 5? What is 25% of 12? What is 12% of 25? What is 15% of 60? What is 60% of 15? What do you notice? Justify using algebra, comparing $x%$ of $y$ and $y%$ of $x$.
| Pair | First value | Second value |
|---|---|---|
| 5% of 40 vs 40% of 5 | 2 | 2 |
| 25% of 12 vs 12% of 25 | 3 | 3 |
| 15% of 60 vs 60% of 15 | 9 | 9 |
Observation: In every pair, both values are equal!
Algebraic proof:
40% of excursion students are Grade 8; rest are Grade 9. 60% of the Grade 8 students are girls.
(i) What % of all students are Grade 8 girls? (ii) If total students = 160, how many are Grade 8 girls?
Grade 8 girls are 60% of the 40% who are Grade 8 students — so multiply the two percentages (as decimals):
If total students $=160$:
A shopkeeper sells pencils such that the selling price of 3 pencils equals the cost price of 5 pencils. Profit or loss? What percentage?
Let the cost price of 1 pencil $= ₹1$ (a convenient unit). Then CP of 5 pencils $= ₹5$.
SP of 3 pencils $=$ CP of 5 pencils $= ₹5$, so SP of 1 pencil $= frac{5}{3} approx ₹1.67$.
Profit per pencil $= frac{5}{3} – 1 = frac{2}{3}$.
Bus fares increased by 3% last year and by 4% this year. What is the overall percentage price increase over the last 2 years?
Successive increases compound, so multiply the two growth factors:
If the length of a rectangle is increased by 10% and the area stays the same, by what percentage (exactly) does the breadth decrease?
Let original length $=L$, original breadth $=B$, original area $=Ltimes B$. New length $=1.1L$. For the area to stay the same:
Percentage decrease in breadth:
A 65 g chips packet contains: Potato 70%, Vegetable oil 24%, Salt 3%, Spices 3%. Find the weight of each ingredient.
| Ingredient | % | Calculation | Weight |
|---|---|---|---|
| Potato | 70% | $0.70times65$ | 45.5 g |
| Vegetable oil | 24% | $0.24times65$ | 15.6 g |
| Salt | 3% | $0.03times65$ | 1.95 g |
| Spices | 3% | $0.03times65$ | 1.95 g |
Three shops sell the same item at ₹100 each. 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 each shop, ranked cheapest to costliest. (ii) Percentage discount at each shop. (iii) Best shop if you need exactly 4 items?
Each deal: pay for some items, get one extra free. Effective price/item $=dfrac{text{money paid}}{text{items received}}$.
| Shop | Pay for | Items received | Effective price/item | Discount % |
|---|---|---|---|---|
| A | 1 item (₹100) | 2 | $frac{100}{2}=₹50$ | $frac{1}{2}times100=50%$ |
| B | 2 items (₹200) | 3 | $frac{200}{3}approx₹66.67$ | $frac{1}{3}times100approx33.33%$ |
| C | 3 items (₹300) | 4 | $frac{300}{4}=₹75$ | $frac{1}{4}times100=25%$ |
For exactly 4 items: Shop A needs 2 separate “buy 1 get 1” deals → pay for 2, get 4 → total ₹200. Shop C‘s deal exactly matches 4 items (buy 3, get 1 free) → total ₹300. Shop B‘s deal gives 3 items for ₹200, plus 1 more item at full price (₹100) → total ₹300.
In a room of 100 people, 99% are left-handed. How many left-handed people must leave to bring that percentage down to 98%?
Initially: 99 left-handed people, 1 right-handed person (total 100). Let $x$ left-handed people leave. The remaining left-handed count is $99-x$, and the remaining total people is $100-x$ (only left-handed people are leaving; the right-handed person stays).
We need the new percentage to be 98%:
Solving: $99-x = 0.98(100-x) = 98-0.98x$
Based on the “Ability to use computer by age and gender (2023)” bar graph, which statements are valid?
(i) Twenties are the most computer-literate age group. (ii) Women lag behind men across all age groups. (iii) There are more people in their twenties than teenagers. (iv) More than a quarter of people in their thirties can use computers. (v) Less than 1 in 10 aged 60+ can use computers. (vi) Half of people in their twenties can use computers.
True. The Twenties group has the highest bars for both Female (26%) and Male (37%), exceeding every other age group, including Teenage (24%, 29%).
True. In every single age category shown, the female percentage is lower than the male percentage (e.g. Twenties: 26% vs 37%; Forties: 7% vs 14%).
Cannot be determined / generally false assumption. The graph shows the percentage of each age group that can use a computer — it tells us nothing about how many actual people exist in each age group. We cannot conclude there are “more people in their twenties” just from these percentages.
False. In the Thirties group, Female = 14% and Male = 25%; even combined, this represents only those individually surveyed — neither group alone exceeds 25%, and we can’t simply add the two percentages since they refer to different populations (the question intends each separately, and Male thirties at 25% is right at the boundary, not clearly “more than a quarter” for the group as a whole).
True. For Seniors (60 and above), Female = 2% and Male = 4% — both are well under 10%.
False. In the Twenties group, Female = 26% and Male = 37% — neither is close to 50%, so “half” is not correct for this age group.