Figure It Out — Questions & Solutions

Given

Express 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 Find

Six exact percentage equivalents

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Multiply each fraction by 100: (i) $\frac{300}{5}$ = 60

   (ii) $\frac{700}{14}$ = 50

   (iii) $\frac{900}{20}$ = 45.

2. (iv) $\frac{7200}{150}$ = 48

   (v) $\frac{100}{3}$ = 33$\frac{1}{3}$

   (vi) $\frac{500}{11}$ = 45 $\frac{5}{11}$.

   Recurring decimals should not be replaced by an exact finite decimal.

Result

60%; 50%; 45%; 48%; 33$\frac{1}{3}$%; ($\frac{500}{11}$)%.

Given

15 of Nandini’s 25 marbles are white. Choose the percentage: 10%,15%,25%,60%,40%,none.

To Find

White-marble percentage

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. $\frac{15}{25}$ × 100 = $\frac{3}{5}$ × 100 = 60.

Result

60%, option (iv).

Given

15 of 80 students walk to school.

To Find

Percentage of students who walk to school

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. $\frac{15}{80}$ × 100 = $\frac{1500}{80}$ = 18.75.

Result

18.75%.

Given

Match the four runner positions A,B,C,D on the source start–finish line to the approximate options 55%,20%,38%,72%,84%,93%.

To Find

Best approximate option for each runner

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Compare each distance from Start with the whole Start–Finish distance.

2. A is between one-quarter and two-fifths, B just beyond half, C around three-quarters, and D close to the finish.

3. Among the supplied options, the nearest intended matches are A:38%, B:55%, C:72%, D:93%.

   These are approximate visual matches, not exact measurements of the drawing.

Result

A→38%; B→55%; C→72%; D→93%.

Given

Insert > , < or = : (i) 50% □ 5%; (ii) $\frac{5}{10}$ □ 50%; (iii) $\frac{3}{11}$ □ 61%; (iv) 30% □ $\frac{1}{3}$.

To Find

Four comparisons

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) 50 > 5 with the same percent unit.

   (ii) $\frac{5}{10}$ = $\frac{1}{2}$ = 50%.

2. (iii) $\frac{3}{11}$ < $\frac{1}{2}$ < 61%.

   (iv) 30% = $\frac{3}{10}$ < $\frac{1}{3}$ because 9 < 10 after cross-multiplication.

Result

(i) > ; (ii) = ; (iii) < ; (iv) < .

Given

Complete the equal-part bar models: (i) five parts, total 75, first four worth 60 (worked); (ii) ten parts, total 90, first six selected; (iii) four parts, total 140, first three selected (page 13).

To Find

Percentage per part and selected amounts

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) Each part is $\frac{100}{5}$ = 20%

   four parts give 80% of 75 = 60.

2. (ii) Each part is $\frac{100}{10}$ = 10%

   six parts give 60% of 90 = 54.

3. (iii) Each part is $\frac{100}{4}$ = 25%

   three parts give 75% of 140 = 105.

Result

(ii) 10% and 54; (iii) 25% and 105.

Given

Find and draw bar models for (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.

To Find

Six quantities and their bar proportions

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) Shade one of four equal parts of a 160 bar: $\frac{160}{4}$ = 40.

2. (ii) Shade 16 of 100 parts of a 250 bar: 0.16 × 250 = 40.

3. (iii) Shade 62 of 100 parts of a 360 bar: 0.62 × 360 = 223.2.

4. (iv) Use a full 40 bar and 40% of a second equal bar: 40+16 = 56.

5. (v) 1% of 60 minutes = 0.6 minute = 36 seconds.

6. (vi) 7% of 10 kg = 0.7 kg = 700 g.

Result

40; 40; 223.2; 56; 36 seconds; 0.7 kg.

Given

Red paint is $\frac{3}{4}$ of 60 ml of deep orange paint.

To Find

Volume of red paint

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. 60 ÷ 4 = 15 ml per quarter.

   Three quarters contain 3 × 15 = 45 ml.

Result

45 ml.

Given

Compare (i) 50% of 510 and 50% of 515; (ii) 37% and 73% of 148; (iii) 29% of 43 and 92% of 110; (iv) 30% of 40 and 40% of 50; (v) 45% of 200 and 10% of 490; (vi) 30% of 80 and 24% of 64.

To Find

Comparison symbol for each pair

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) 255 < 257.5

   (ii) same positive base with 37 < 73, so < .

2. (iii) 12.47 < 101.2.

   (iv) 12 < 20.

3. (v) 90 > 49.

   (vi) 24 > 15.36.

Result

(i) < ; (ii) < ; (iii) < ; (iv) < ; (v) > ; (vi) > .

Given

Complete: (i) 30% of k = 70; find 60%,90%,120%. (ii) 100% of m = 215; find 10%,1%,6%. (iii) 90% of n = 270; find 9%,18%,100%. (iv) Create two similar questions.

To Find

Missing proportional quantities and two illustrative learner-created questions

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) Double, triple and quadruple 70: 140,210,280.

2. (ii) Divide 215 by 10 and 100: 21.5,2.15

   six hundredths give 12.9.

3. (iii) One tenth of 270 is 27 at 9%

   double gives 54 at 18%

   n = $\frac{270}{0.9}$ = 300.

4. (iv) Example response: “25% of a is 30

   find 50% and 100%” gives 60 and 120.

   Another: “10% of b is 8

   find 5% and 20%” gives 4 and 16.

   These are examples of the requested learner-created questions, not additional textbook items.

Result

(i) 140,210,280; (ii) 21.5,2.15,12.9; (iii) 27,54,300; example responses supplied for (iv).

Given

Fill in: (i) 3 is □% of 300; (ii) □ is 40% of 4; (iii) 40 is 80% of □.

To Find

Three missing quantities

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) $\frac{3}{300}$ × 100 = 1.

   (ii) 0.4 × 4 = 1.6.

   (iii) Whole = $\frac{40}{0.8}$ = 50.

Result

(i) 1%; (ii) 1.6; (iii) 50.

Given

Is 10% of a day longer than 1% of a week? Create similar questions.

To Find

Comparison in the same time unit

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. 10% of 24 hours = 2.4 hours

   1% of 168 hours = 1.68 hours.

2. 2.4−1.68 = 0.72 hour = 43.2 minutes, so the first duration is longer.

3. Example learner-created comparison: 5% of a day is 1.2 hours, equal to 20% of 6 hours.

Result

Yes, by 43.2 minutes.

Given

On day n, for n = 1,…,99, Mariam gives n+1 units of fodder and the bull eats n units. Express the eaten proportions as percentages and describe the pattern.

To Find

Percentage eaten on every specified day

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. The base is fodder offered that day.

2. Day n: eaten percentage = 100$\frac{n}{\left(n+1\right)}$ = 100$\frac{-100}{\left(n+1\right)}$.

   This formula supplies all 99 values.

3. First values: day 1 = 50%, day 2 = 66$\frac{2}{3}$%, day 3 = 75%, day 4 = 80%

   day 99 = 99%.

4. As n grows the subtracted fraction decreases, so the eaten percentage increases but stays below 100%.

   One unit remains uneaten each day.

Result

Day n: (100$\frac{n}{\left(n+1\right)}$)%; rising from 50% to 99%.

Given

Workers pick 20% of a coffee plantation in 18 days. Find the time for the entire plantation if their rate is unchanged, and explain that assumption.

To Find

Total picking time

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. 100% is five times 20%, so total time = 5 × 18 = 90 days

   the remaining 80% takes 72 days.

2. This proportional scaling needs equal work per unit area and unchanged workforce/productivity.

   If crop density or working conditions differ, area alone does not determine time.

Result

90 days in total under the constant-rate assumption.

Given

Divide a 90-minute training session into warm-up, play and cool-down in proportions 10%,80%,10%.

To Find

Duration of each activity

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Warm-up = 0.1 × 90 = 9 minutes

   play = 0.8 × 90 = 72 minutes

   cool-down = 9 minutes.

2. 9+72+9 = 90 minutes.

Result

9,72,9 minutes respectively.

Given

About 90% of the world’s population is estimated to live in the Northern Hemisphere. Estimate the number using a world-population figure for the selected year.

To Find

Estimated Northern Hemisphere population

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. If the selected year’s worldwide estimate is P, use 0.9P.

   Record the year and source of P.

2. Using only the textbook’s supplied 2025 estimate of 8.2 billion from page 28 gives 0.9 × 8.2 = 7.38 billion.

   This is a 2025 example, not an asserted current-year population.

Result

0.9P; 7.38 billion when using the textbook’s 2025 figure.

Given

Halwa for four people has rava 40%, sugar 40%, ghee 20%. (i) What proportions for eight people? (ii) Find each mass when the total is 2 kg.

To Find

Scaled proportions and ingredient masses

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) Doubling all ingredient amounts leaves the proportions unchanged: 40%,40%,20%.

2. (ii) Rava = 0.4 × 2 = 0.8 kg

   sugar = 0.8 kg

   ghee = 0.2 × 2 = 0.4 kg.

   Total = 2 kg.

Result

Same proportions; 800 g rava, 800 g sugar, 400 g ghee.

Given

A geometry box costs ₹75 and sells for ₹110. Find the profit percentage with respect to cost.

To Find

Profit as a percentage of cost

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Profit = 110−75 = ₹35.

2. Profit percentage = $\frac{35}{75}$ × 100 = ($\frac{140}{3}$)%.

Result

46$\frac{2}{3}$% profit on cost.

Given

Materials for a chair cost ₹475; the carpenter wants a 50% profit margin. Find the selling price.

To Find

Selling price with the percentage base stated

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Using the cost-based convention of the preceding exercise: profit = 0.5 × 475 = ₹237.50, giving price ₹712.50.

2. If “margin” instead means profit as a share of sales, then 475 is 50% of sales and price is ₹950.

   The textbook has discussed both bases

   the wording here does not expressly name one.

Result

₹712.50 on cost; ₹950 if a 50% sales margin is intended. State the convention.

Given

A company had revenue ₹2.5 crore and profit margin 25%. Find total expenditure.

To Find

Expenditure under the stated margin convention

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Using the revenue-based company margin discussed on page 19, profit = 25% of 2.5 = ₹0.625 crore.

2. Expenditure = 2.5−0.625 = ₹1.875 crore.

3. If the percentage were instead on cost, expenditure would be $\frac{2.5}{1.25}$ = ₹2 crore.

   The revenue-based interpretation matches the surrounding discussion.

Result

₹1.875 crore using profit/revenue as the margin.

Given

A ₹300 shirt is discounted by 25%.

To Find

Price paid by Anwar

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Discount = 0.25 × 300 = ₹75

   payment = 300−75 = ₹225.

Result

₹225.

Given

Petrol increases from ₹60 in 2015 to ₹100 in 2025. Find the percentage increase.

To Find

Increase relative to the 2015 price

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Increase = ₹40.

   Divide by the original ₹60: $\frac{40}{60}$ × 100 = ($\frac{200}{3}$)%.

2. The listed 66.66% is a truncated approximation, not the exact value.

   Correct two-decimal rounding is 66.67%.

Result

66$\frac{2}{3}$% exactly; option (iv) is the intended approximate choice.

Given

Samson pays ₹440000 for a car after a 15% discount. Find its original price. The printed numbering restarts at Question 3 on this page.

To Find

Original car price

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. The paid amount is 85% of the original: 0.85P = 440000.

2. P = 440000 × $\frac{100}{85}$ = $\frac{8800000}{17}$ rupees.

Result

₹$\frac{8800000}{17}$, approximately ₹517647.06.

Given

1600 people vote; the winner receives 500 votes. Find the vote percentage and minimum possible candidate count.

To Find

Winner’s share and minimum candidates

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. $\frac{500}{1600}$ × 100 = 31.25%.

2. With only three candidates and no one exceeding the winner, at most 3 × 500 = 1500 votes can be accounted for.

   Thus at least four candidates are needed.

3. Four are possible: votes 500,400,400,300 sum to 1600 with a unique winner.

   This assumes all 1600 votes are valid candidate votes.

Result

31.25%; at least four candidates under the stated assumption.

Given

Rice rises from ₹38 per kg to ₹42 per kg. Find the inflation rate.

To Find

Percentage increase on ₹38

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Increase = ₹4.

   Rate = $\frac{4}{38}$ × 100 = ($\frac{200}{19}$)%.

Result

($\frac{200}{19}$)%, approximately 10.53%.

Given

A number becomes 90 after a 20% increase.

To Find

Original number

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. 1.2x = 90, so x = $\frac{90}{1.2}$ = 75.

Result

75.

Given

Two buffaloes sell for ₹80000 each, one at 5% profit and the other at 10% loss. Find the overall profit or loss.

To Find

Combined profit/loss

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. First cost = $\frac{80000}{1.05}$ = $\frac{1600000}{21}$ rupees.

   Second cost = $\frac{80000}{0.9}$ = $\frac{800000}{9}$ rupees.

2. Total cost = $\frac{10400000}{63}$ rupees

   total sales = 160000 = $\frac{10080000}{63}$ rupees.

3. Loss = $\frac{320000}{63}$ rupees.

   Loss percentage = $\frac{320000}{10400000}$ × 100 = ($\frac{40}{13}$)%.

Result

Loss ₹$\frac{320000}{63}$ ≈ ₹5079.37, or $\frac{40}{13}$% ≈3.08% on combined cost.

Given

An elephant population p increases by 5%. Which expression gives the new population?

To Find

New-population expression

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. New = p+0.05p = 1.05p.

Result

p × 1.05, option (iv).

Given

Which statement means demand has fallen by 85% from its value a decade ago?

To Find

Equivalent remaining-demand statement

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. The remaining share is 100%−85% = 15% of the old demand.

Result

Demand now is 15% of demand a decade ago, option (iii).

Given

Compare ₹20000 at 10% per year for two years with and without annual compounding.

To Find

Two final totals

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Without compounding: 20000(1+0.10 × 2) = ₹24000.

2. With annual compounding: 20000 × 1.${10}^2$ = ₹24200

   interest is ₹4200 instead of ₹4000.

Result

₹24000 without and ₹24200 with compounding; extra ₹200.

Given

Compare ₹20000 at 5% per year for four years with and without annual compounding.

To Find

Two final totals

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Without compounding: 20000(1+0.05 × 4) = ₹24000.

2. With compounding: 20000 × 1.${05}^4$ = ₹24310.125 exactly before monetary rounding.

3. At the end, rounding to paise gives ₹24310.13.

   The model assumes no intermediate rounding.

Result

₹24000 without; ₹24310.125 with compounding, or ₹24310.13 to paise.

Given

Compare the results of Questions 1 and 2: 10% for two years and 5% for four years on ₹20000.

To Find

The shared and differing growth patterns

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Both simple-interest models have rate × time = 0.20, so both give ₹24000.

2. Compounding gives different multipliers: 1.${10}^2$ = 1.21, whereas 1.${05}^4$ = 1.21550625.

3. The second compounded amount exceeds the first by ₹110.125, before rounding.

Result

Equal simple-interest totals; the four-year compounded total is larger.

Given

Jasmine invests p for four years at 6% per year without compounding. Select the expression for the total received.

To Find

Correct total-amount expression

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Interest = p × 0.06 × 4.

   Add back the principal: p+(p × 0.06 × 4) = 1.24p.

Result

Option (vii): p+(p × 0.06 × 4).

Given

Invest ₹50000 for three years at 7% per year. Find interest without compounding and the extra interest with annual compounding.

To Find

Simple interest and additional compounded interest

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Simple interest = 50000 × 0.07 × 3 = ₹10500.

2. Compounded total = 50000 × 1.${07}^3$ = ₹61252.15, so compound interest = ₹11252.15.

3. Extra interest = 11252.15−10500 = ₹752.15.

Result

Simple interest ₹10500; compounding earns ₹752.15 more.

Given

Giridhar borrows ₹12500 for three years at 12% simple interest. Raghava borrows the same sum for three years at 10% compounded annually. Who pays more?

To Find

Difference in interest paid

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Giridhar: 12500 × 0.12 × 3 = ₹4500.

2. Raghava: 12500(1.${10}^3$−1) = 12500 × 0.331 = ₹4137.50.

3. Difference = 4500−4137.50 = ₹362.50.

Result

Giridhar pays ₹362.50 more interest.

Given

How long does ₹1000 take to double at 10% per year with and without compounding? Is the growth exponential or linear?

To Find

Doubling time under the two models

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Simple interest adds ₹100 each year: 1000+100t = 2000 gives t = 10 years

   the growth is linear.

2. Annual compounding gives 1000 × 1.1ⁿ after n completed years.

   After seven years: ₹1948.7171

   after eight: ₹2143.58881.

3. Thus the first annual credit date at or above double is year 8

   no integer-year total is exactly ₹2000.

   This is exponential growth.

Result

10 years simple; at least double after 8 annual compounding periods.

Given

A city has 1.5 crore people and grows about 3% each year. Estimate its population after three years.

To Find

Three-year population estimate

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Multiply by 1.03 each year: 1.5 × 1.${03}^3$ = 1.6390905 crore.

2. This is 16390905 people under the model

   the input growth rate is approximate, so treat it as an estimate.

Result

Approximately 1.639 crore (about 16.39 million).

Given

Initially 506000 bacteria grow by 2.5% per hour. Estimate the count after two hours.

To Find

Two-hour model count

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. 506000 × 1.${025}^2$ = 506000 × 1.050625 = 531616.25.

2. The model can give a non-integer expectation

   an actual count is an integer.

   To the nearest bacterium, the estimate is 531616.

Result

Model value 531616.25; approximately 531616 bacteria.

Given

Using the source figures, Bengaluru’s 2025 population is 250% of its 2000 population of 50 lakh.

To Find

Population implied by the textbook figures

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. 250% = 2.5, so 2.5 × 50 = 125 lakh.

Result

125 lakh, or 1.25 crore.

Given

Use the textbook’s 2025 world population 8.2 billion to estimate shares for Germany 83 million, India 1.46 billion, Bangladesh 175 million and USA 347 million.

To Find

Approximate country shares

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Use world total 8200 million.

   Germany: $\frac{83}{8200}$ × 100≈1.01%.

   India: $\frac{1460}{8200}$ × 100≈17.80%.

2. Bangladesh: $\frac{175}{8200}$ × 100≈2.13%.

   USA: $\frac{347}{8200}$ × 100≈4.23%.

3. Nearest whole-percent matches are 1%,18%,2%,4%.

   The printed option bank contains no 4% option for the USA

   do not substitute an incorrect option.

Result

Germany ≈1%; India ≈18%; Bangladesh ≈2%; USA ≈4% (missing from printed choices).

Given

Add 18% GST to the stated ₹8250 phone price. Which expressions give the final amount?

To Find

Tax-inclusive price and matching expressions

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Final price = 8250+0.18 × 8250 = 8250 × 1.18.

2. GST = ₹1485

   total = ₹9735.

Result

Options (v) and (vi); ₹9735. This is the question’s stated tax rate.

Given

A mouse population p changes by +5%,−2%,−3% in successive months. Identify the correct multiplier and compare the final population with p.

To Find

Three-month model and net change

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Each percentage uses the previous month as base.

   Multiply 1.05 × 0.98 × 0.97 = 0.99813.

2. The model gives 0.99813p, a decrease of 0.187% for p > 0.

Result

Statements (ii) and (vi) are true; the final population is below p.

Given

A shopkeeper sets a price at 35% profit and then discounts that price by 30%. Does the sale make a profit or loss?

To Find

Final profit/loss relative to cost

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Using the cost-based profit convention, set cost C.

   Marked price = 1.35C

   discounted price = 0.70 × 1.35C = 0.945C.

2. Loss = C−0.945C = 0.055C, so loss percentage is 5.5%.

3. If “margin” is instead defined on sales, the initial price would be $\frac{C}{0.65}$ and the conclusion differs.

   The 5.5% answer explicitly uses profit on cost.

Result

5.5% loss under the cost-based convention.

Given

In the grid figure on page 29, find the percentage of the total square occupied by region E.

To Find

Area percentage of triangular region E

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. The outer square has 8 × 8 unit cells, area 64.

   The bottom-left 4 × 4 square is divided by a diagonal into D and E.

2. E is half of 16, so area(E) = 8.

   Its percentage is $\frac{8}{64}$ × 100 = 12.5.

Result

12.5%.

Given

Nonnegative numbers x and y.

To Prove

x% of y = y% of x

Reasoning

Multiplication is commutative.

Proof

1. 5% of 40 = 2 = 40% of 5.

2. 25% of 12 = 3 = 12% of 25.

3. 15% of 60 = 9 = 60% of 15.

4. x% of y = ($\frac{x}{100}$)y = $\frac{\mathrm{xy}}{100}$.

5. y% of x = ($\frac{y}{100}$)x = $\frac{\mathrm{yx}}{100}$.

6. xy = yx, so the expressions are equal.

Result

x% of y = y% of x.

Given

40% of excursion students are Grade 8, of whom 60% are girls. (i) Find the overall Grade 8 girl percentage. (ii) Apply it to a total of 160 students.

To Find

Nested percentage and consistency of the headcount

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) 0.60 × 0.40 = 0.24, so 24% of all excursion students are Grade 8 girls.

2. (ii) 0.24 × 160 = 38.4.

   A group cannot contain 38.4 students.

3. The source’s exact percentages and total are inconsistent as headcounts.

   Rounding to 38 would no longer satisfy both exact percentages, so it is not an exact solution.

Result

(i) 24%. (ii) The arithmetic gives 38.4; no exact whole-student solution exists with all the stated data.

Given

The selling price of three pencils equals the cost price of five. Find profit or loss percentage.

To Find

Profit percentage on pencil cost

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Let cost per pencil be c > 0.

   Selling price per pencil is 5$\frac{c}{3}$.

2. Profit per pencil = 5$\frac{c}{3}$−c = 2$\frac{c}{3}$

   divide by c and multiply by 100.

Result

66$\frac{2}{3}$% profit.

Given

Successive increases of 3% and 4%.

To Find

Overall percentage increase

Reasoning

Multiply the growth factors.

Solution

1. 1.03 × 1.04 = 1.0712.

2. 1.0712 − 1 = 0.0712.

3. 0.0712 × 100 = 7.12%.

Result

The increase is 7.12%, not 7%.

Given

Original length L > 0, breadth B > 0; new length 1.1L.

To Find

Required percentage decrease in breadth

Reasoning

Equal areas require 1.1L × new breadth = LB.

Solution

1. New breadth = $\frac{B}{1.1}$ = 10$\frac{B}{11}$.

2. Decrease = B − 10$\frac{B}{11}$ = $\frac{B}{11}$.

3. Relative decrease = $\frac{\left(\frac{B}{11}\right)}{B}$ × 100 = $\frac{100}{11}$%.

Result

Breadth decreases by exactly $\frac{100}{11}$% = 9 $\frac{1}{11}$%.

Given

A 65 g packet lists approximately 70% potato,24% vegetable oil,3% salt and 3% spices. Find ingredient masses.

To Find

Approximate mass of each ingredient

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. Potato = 0.70 × 65 = 45.5 g

   oil = 0.24 × 65 = 15.6 g.

2. Salt = 0.03 × 65 = 1.95 g

   spices = 1.95 g.

3. Check: 45.5+15.6+1.95+1.95 = 65 g.

Result

Approximately 45.5 g potato,15.6 g oil,1.95 g salt and 1.95 g spices.

Given

At ₹100 per item, compare A: buy 1 get 1 free; B: buy 2 get 1 free; C: buy 3 get 1 free. Find effective unit prices, discounts and the best shop for four items.

To Find

Unit prices, discount rates and four-item cost

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) A: ₹$\frac{100}{2}$ = ₹50 per item

   B: ₹$\frac{200}{3}$≈₹66.67

   C: ₹$\frac{300}{4}$ = ₹75.

   Order: A,B,C.

2. (ii) Free share of each bundle gives discounts $\frac{1}{2}$ = 50%,$\frac{1}{3}$ = 33$\frac{1}{3}$%,$\frac{1}{4}$ = 25%.

3. (iii) Assuming repeatable offers, A gives two bundles (four items) for ₹200.

   B gives a three-item bundle plus one paid item for ₹300.

   C gives four for ₹300.

Result

Choose A for four items: ₹200, assuming the advertised offer can be used twice.

Given

99 left-handed people and 1 right-handed person; only left-handed people leave.

To Find

Number of left-handed people who must leave

Reasoning

The unchanged right-handed count becomes 2% of the remaining total.

Solution

1. Remaining total = 1 ÷ 0.02 = 50.

2. Left-handed people remaining = 49.

3. 99 − 49 = 50

   check $\frac{49}{50}$ = 98%.

Result

50 left-handed people must leave.

Given

Using the source computer-ability graph, assess: (i) People in their twenties are the most computer-literate among all age groups. (ii) Women lag behind in the ability to use computers across 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 and above can use computers. (vi) Half of the people in their twenties can use computers. These are the source statements to evaluate, not conclusions.

To Find

Validity and limits of six interpretations

Reasoning

Percentage = part ÷ stated base × 100; use exact values before rounding.

Solution

1. (i) The twenties have the highest plotted rate within each gender (26% female,37% male).

   This supports the intended gender-wise comparison

   an exact combined-sex ranking also needs group weights.

2. (ii) Every plotted female rate is below the male rate in the same age group.

   This is a description of the displayed survey, not an inherent ability claim.

3. (iii) Population counts cannot be inferred from these within-group percentages.

4. (iv) False: thirties are 14% female and 25% male

   any weighted combined rate is at most 25%, not more than a quarter.

5. (v) Supported by the senior rates of 2% female and 4% male: any weighted combined rate is below 10%.

6. (vi) False: twenties rates 26% and 37% cannot combine to 50%.

   Do not add the two percentages.

Result

(ii) and (v) supported; (i) supported within each gender, with an aggregation caveat; (iii) not inferable; (iv) and (vi) false.