Question 1

A number is decreased to 72 1/2% (i.e., to 0.725 of itself) and the result is 87. Find the original number.

Accepted Answer

Introduction / Context: This is the mirror of “increase to” problems. “Decreased to 72 1/2%” means the final equals 72.5% of the original. We reverse the percent change to recover the initial value. Given Data / Assumptions: Final value = 87. Final = 72 1/2% of original = 0.725 of original. 0.725 = 29/40 for exact calculation. Concept / Approach: If original is x, then 0.725x = 87 ⇒ x = 87 / 0.725. Using fractions avoids rounding: x = 87 / (29/40) = 87 * 40 / 29. Step-by-Step Solution: Write 72 1/2% = 72.5% = 29/40.Let x be original ⇒ (29/40)x = 87.x = 87 * 40 / 29 = 3480 / 29 = 120. Verification / Alternative check: Check: 120 reduced to 72.5% ⇒ 120 * 0.725 = 87. Correct. Why Other Options Are Wrong: 110, 135, 58 are results of using 27/40 or 3/4 mistakenly, or of subtracting percentages incorrectly. Common Pitfalls: Interpreting “decreased by 72 1/2%.” The phrase here is “decreased to,” which means multiply by 0.725, not subtract 72.5 points. Final Answer: 120

Question 2

Two numbers are 30% and 37% less than a third number T. By what percent is the second number less than the first?

Accepted Answer

Introduction / Context: Comparing two numbers that are each reduced from the same base is a common percent-comparison task. The trick is to express both numbers in terms of the shared base, then compute the relative difference with the correct denominator (the “first” number as the reference). Given Data / Assumptions: Third number = T (baseline). First number = 30% less than T ⇒ 0.70T. Second number = 37% less than T ⇒ 0.63T. Concept / Approach: Percent by which the second is less than the first = (First − Second) / First * 100%. Substitute the expressions 0.70T and 0.63T and simplify; T cancels. Step-by-Step Solution: First = 0.70T; Second = 0.63T.Difference = 0.70T − 0.63T = 0.07T.Relative to First: (0.07T) / (0.70T) * 100% = (0.07 / 0.70) * 100% = 0.1 * 100% = 10%. Verification / Alternative check: Choose T = 100 arbitrarily: numbers become 70 and 63; difference 7 is 10% of 70. Check passes. Why Other Options Are Wrong: 3% or 4% come from dividing by T instead of the first; 7% uses the absolute gap relative to 100; 11% is a rounding guess. Common Pitfalls: Using the wrong base when converting a difference to a percent. The phrase “less than the first” means divide by the first number. Final Answer: 10%

Question 3

In a college election, a candidate secures 62% of the votes and wins by a majority of 144 votes. What is the total number of votes polled?

Accepted Answer

Introduction / Context: Majority in a two-candidate election equals the difference between the winner’s and loser’s shares of the total votes. When percentages are given, the majority is simply (winner% − loser%) of total. Use that to solve for the total number of votes. Given Data / Assumptions: Winner gets 62% of total votes V. Loser gets the remaining 38% (assuming only two candidates). Majority = 144 votes. Concept / Approach: Majority = (62% − 38%) of V = 24% of V. Therefore, 0.24V = 144. Solving yields V. This avoids separately computing each candidate’s votes first. Step-by-Step Solution: Let total votes be V.Majority share = (0.62 − 0.38) * V = 0.24V.Set 0.24V = 144 ⇒ V = 144 / 0.24 = 600. Verification / Alternative check: Winner votes = 0.62*600 = 372; loser votes = 228; difference = 372 − 228 = 144. Correct. Why Other Options Are Wrong: 800, 925, 1200 produce majority values of 192, 222, and 288 respectively, not 144. Common Pitfalls: Misreading 62% as the majority itself, or forgetting that majority is a percentage of total, not of the winner’s votes. Final Answer: 600

Question 4

The price of sugar rises by 20%, but expenditure is kept constant. What is the ratio of the reduction in consumption to the original consumption?

Accepted Answer

Introduction / Context: When expenditure (Price * Quantity) is held constant and price changes, quantity must adjust inversely. A 20% price increase means the buyer must reduce consumption so that total spend remains unchanged. Given Data / Assumptions: Initial price = P, initial quantity = Q, expenditure = P * Q. New price = 1.2P (20% higher). Expenditure must remain P * Q. Concept / Approach: Let new quantity be Q′. With constant expenditure: 1.2P * Q′ = P * Q ⇒ Q′ = Q / 1.2 = (5/6)Q. Reduction = Q − Q′ = Q − (5/6)Q = (1/6)Q. Hence, reduction : original = (1/6)Q : Q = 1 : 6. Step-by-Step Solution: Q′ = Q / 1.2 = (5/6)Q.Reduction = Q − Q′ = Q − (5/6)Q = (1/6)Q.Ratio (reduction : original) = (1/6)Q : Q = 1 : 6. Verification / Alternative check: Take P = 10, Q = 6 ⇒ spend = 60. New price = 12; to keep spend 60, Q′ = 5. Reduction 1 on original 6 ⇒ 1 : 6. Why Other Options Are Wrong: Ratios 1:3, 1:4, 1:5 correspond to price changes of 50%, 33 1/3%, and 25% respectively for constant spend, not 20%. Common Pitfalls: Subtracting 20 from 100 directly to claim a “20:100 = 1:5” ratio without using the inverse relationship required by constant expenditure. Final Answer: 1 : 6

Question 5

Water tax is increased by 20% while consumption drops by 20%. By what percent does the total expenditure change?

Accepted Answer

Introduction / Context: Total expenditure equals price multiplied by quantity. When both change, multiply the corresponding factors. A neat way is to convert each percent change to a multiplier and then combine them. Given Data / Assumptions: Price multiplier = 1 + 20% = 1.20. Quantity multiplier = 1 − 20% = 0.80. Initial expenditure is the reference (100%). Concept / Approach: New expenditure factor = 1.20 * 0.80 = 0.96. This equals 96% of the old expenditure, i.e., a 4% decrease. Always multiply the independent effects; never add the percentages. Step-by-Step Solution: Compute combined factor: 1.20 * 0.80 = 0.96.Translate to percent: 0.96 = 96% of original.Decrease = 100% − 96% = 4% decrease. Verification / Alternative check: Start with ₹100. After +20% price: effective price factor 1.2. With −20% quantity: factor 0.8. Spend = 100 * 1.2 * 0.8 = ₹96, 4% lower. Why Other Options Are Wrong: “No change” would require 1.20 * 0.80 = 1; “5% decrease” is a rounding guess; “4% increased” flips the sign. Common Pitfalls: Adding +20% and −20% to claim 0%, which ignores multiplicative effects. Percentage changes on different factors must be multiplied, not added. Final Answer: 4% decreased

Question 6

A 20% alcohol solution: 15 litres of this mixture receives 3 litres of water. What is the new percentage strength of alcohol?

Accepted Answer

Introduction / Context: Dilution problems preserve the amount of solute (alcohol) while changing the total volume by adding solvent (water). The resulting concentration is solute amount divided by the new total volume, expressed as a percentage. Given Data / Assumptions: Initial mixture = 15 L at 20% alcohol. Added water = 3 L. No alcohol is lost or gained during mixing. Concept / Approach: Compute the absolute amount of alcohol initially, then divide by the new total volume after adding water. Finally convert the fraction to a percentage. Step-by-Step Solution: Alcohol initially = 20% of 15 L = 0.20 * 15 = 3 L.New total volume = 15 + 3 = 18 L.New strength = alcohol / total = 3 / 18 = 1/6.Convert to percent: 1/6 ≈ 16 2/3%. Verification / Alternative check: If 16 2/3% of 18 L is alcohol, quantity = (1/6)*18 = 3 L, consistent with the conserved alcohol amount. Why Other Options Are Wrong: 12 1/2% corresponds to adding more water or miscounting initial alcohol; 24% is impossible since adding pure water cannot raise concentration; 16% is due to rounding down. Common Pitfalls: Treating percent changes additively or forgetting to first compute the absolute solute before re-percentage on the new volume. Final Answer: 16 2/3%

Question 7

A TV’s price is reduced by 30% and sales increase by 20%. What is the overall percentage change in revenue (price × quantity)?

Accepted Answer

Introduction / Context: Revenue is the product of price and quantity sold. When both factors change, multiply the respective percentage multipliers to get the combined effect, and then translate back into a percentage change. Given Data / Assumptions: Price change: −30% ⇒ multiplier 0.70. Quantity change: +20% ⇒ multiplier 1.20. Initial revenue serves as 100% reference. Concept / Approach: New revenue factor = 0.70 * 1.20 = 0.84. That is 84% of the original, hence a 16% decrease overall. Step-by-Step Solution: Compute product of multipliers: 0.70 × 1.20 = 0.84.Interpretation: revenue is 84% of original.Change = 100% − 84% = 16% decrease. Verification / Alternative check: If original price = ₹100 and sales = 100 units ⇒ revenue = ₹10,000. New price = ₹70; new sales = 120 units; new revenue = ₹8400 ⇒ 16% down. Why Other Options Are Wrong: A 10% change comes from adding or subtracting percentages instead of multiplying; “16% increased” flips the sign. Common Pitfalls: Adding −30% and +20% to get −10% (wrong). Effects on different factors must be multiplied. Final Answer: 16% decrease

Question 8

A town has 8000 people. The population increases by 10% in the first year and by 20% in the second year. What will be the population after 2 years?

Accepted Answer

Introduction / Context: Successive percentage changes compound multiplicatively. Apply each year’s growth factor to the running total rather than summing the percentages. Given Data / Assumptions: Initial population = 8000. Year 1 growth = +10% ⇒ factor 1.10. Year 2 growth = +20% ⇒ factor 1.20 applied to the result after year 1. Concept / Approach: Final population = 8000 * 1.10 * 1.20. Multiply the factors first or compute year by year; both yield the same result. This is standard compound percentage growth. Step-by-Step Solution: After Year 1: 8000 * 1.10 = 8800.After Year 2: 8800 * 1.20 = 10560. Verification / Alternative check: Combine factors: 1.10 * 1.20 = 1.32 ⇒ 8000 * 1.32 = 10560. Same answer. Why Other Options Are Wrong: 10400 assumes adding percentages once (10% of 8000 plus 20% of 8000); 10620 is a mis-multiplication; “None of these” is not applicable. Common Pitfalls: Adding 10% and 20% to get 30% and then taking 30% of 8000 (2400) to claim 10400; that ignores compounding. Final Answer: 10560

Question 9

A papaya tree grows by 20% every year. Two years later its height is 540 cm. What was its height when planted (2 years ago)?

Accepted Answer

Introduction / Context: This is reverse compound growth. If a quantity grows by a fixed percentage each year, its current value equals initial value times the compounded growth factor. To find the initial value, divide the current value by that factor. Given Data / Assumptions: Annual growth = 20% ⇒ factor 1.20 each year. Elapsed time = 2 years. Current height = 540 cm. Concept / Approach: If H₀ is initial height, then present height H = H₀ * (1.20)^2. Therefore, H₀ = H / (1.20^2). Step-by-Step Solution: Compute growth factor: (1.20)^2 = 1.44.Initial height H₀ = 540 / 1.44 = 375 cm. Verification / Alternative check: Grow 375 by 20% twice: Year 1 → 375 * 1.2 = 450; Year 2 → 450 * 1.2 = 540. Matches. Why Other Options Are Wrong: 324 and 432 reflect using 10% or 15% instead of 20% or mixing up square vs double counting. Common Pitfalls: Subtracting 2*20% from 540 or dividing by (1 + 2*0.2) instead of squaring the growth factor. Final Answer: 375 cms

Question 10

A machine depreciates by 10% per year. If its present value is Rs. 4000, what will be its value after 2 more years?

Accepted Answer

Introduction / Context: Depreciation compounded annually reduces value by a fixed fraction each year. After n years at rate r, value multiplies by (1 − r)^n. Apply this to the current value to get the future value after more depreciation. Given Data / Assumptions: Current value V = ₹4000. Annual depreciation rate = 10% ⇒ factor 0.90 per year. Time = 2 years. Concept / Approach: Future value after 2 years: V * (0.90)^2. Compute exactly to avoid rounding errors: 0.9^2 = 0.81. Step-by-Step Solution: Compute factor: (0.90)^2 = 0.81.Future value = 4000 * 0.81 = ₹3240. Verification / Alternative check: Year 1: 4000 → 3600. Year 2: 3600 → 3240. Same outcome. Why Other Options Are Wrong: ₹3200 and ₹3280 reflect linear changes instead of compounded; ₹3260 is a rounding guess; ₹3000 is too low. Common Pitfalls: Subtracting 20% once instead of compounding 10% twice, or subtracting 10 percentage points of the original value each year. Final Answer: Rs. 3240

Question 11

A town grows 4% annually, but emigration reduces the population by 1/2% each year. What is the overall percentage increase over 3 years?

Accepted Answer

Introduction / Context: When two opposing percentage effects apply annually, combine them multiplicatively each year. Here, net annual growth equals +4% − 0.5% = +3.5%. Over multiple years, compound the net factor, not the separate components. Given Data / Assumptions: Birth/immigration growth = +4% ⇒ factor 1.04. Emigration loss = −0.5% ⇒ factor 0.995. Net yearly factor = 1.04 * 0.995 = 1.035 (i.e., +3.5%). Time = 3 years. Concept / Approach: Overall multiplier after 3 years = (1.035)^3. Convert the multiplier to a percent increase by subtracting 1 and multiplying by 100. Step-by-Step Solution: Compute (1.035)^2 ≈ 1.071225.Multiply by 1.035 again: 1.071225 * 1.035 ≈ 1.108718.Overall increase ≈ 1.108718 − 1 = 0.108718 ≈ 10.87% ≈ 10.8% (to one decimal). Verification / Alternative check: Log-based approximation: 3 * 3.5% = 10.5% plus a small compounding addition (~0.35%) gives ~10.85%, consistent. Why Other Options Are Wrong: 9.8 and 10 neglect compounding; 10.5 uses 3 × 3.5% without compounding; 11.5 overshoots realistic compounding. Common Pitfalls: Adding annual percentages linearly and ignoring multi-year compounding. Final Answer: 10.8

Question 12

A man’s wages are cut by 50%. Then the reduced wages are increased by 50%. Overall, what percent loss does he suffer relative to his original wages?

Accepted Answer

Introduction / Context: Opposite percentage changes of the same magnitude do not cancel out because they apply to different bases. A 50% decrease followed by a 50% increase never returns to the starting value. Given Data / Assumptions: Initial wage = W. Decrease 50% ⇒ W becomes 0.5W. Increase 50% of the reduced wage ⇒ 0.5W * 1.5. Concept / Approach: Compute the final wage explicitly and compare to the original. The product 0.5 * 1.5 = 0.75 indicates the final is 75% of the original. Step-by-Step Solution: After decrease: W → 0.5W.After increase: 0.5W × 1.5 = 0.75W.Loss percent = (W − 0.75W)/W * 100% = 25%. Verification / Alternative check: If W = ₹100, new wage after cut = ₹50; after raise = ₹75. Loss = ₹25 ⇒ 25%. Why Other Options Are Wrong: 0% assumes symmetric cancellation; 0.25% and 2.5% are decimal-place errors. Common Pitfalls: Adding −50% and +50% to get 0%, forgetting that the 50% increase acts on the reduced base. Final Answer: 25 %

Question 13

P is six times as large as q (P = 6q). By what percentage is q less than P?

Accepted Answer

Introduction / Context: “q is less than P by what percent” requires using P as the base for the percentage comparison. With P = 6q, compute the relative shortfall of q from P and express it as a fraction of P. Given Data / Assumptions: P = 6q, with q > 0. Percent less than P is (P − q)/P * 100%. Concept / Approach: Substitute P = 6q; then the numerator becomes 5q and the denominator 6q. The q cancels, leaving a pure ratio that converts directly to a percentage. Step-by-Step Solution: (P − q)/P = (6q − q)/(6q) = 5q/(6q) = 5/6.Convert to percent: (5/6)*100% = 83 1/3%. Verification / Alternative check: Choose q = 1 ⇒ P = 6. q is 5 less than P; 5 is 5/6 of 6 ⇒ 83 1/3%. Why Other Options Are Wrong: 16 2/3% is the complement (q as a percent of P), not the deficit; 60% and 90% ignore the correct base. Common Pitfalls: Using q as the base and computing (P − q)/q, which would answer “how much greater is P than q,” a different question. Final Answer: 83 1/3

Question 14

A’s salary is 30% more than B’s. By what percentage is B’s salary less than A’s?

Accepted Answer

Introduction / Context: Percent “more than” and percent “less than” are asymmetric. If A is 30% more than B, then the percent by which B is less than A is not 30%; it must be computed using A as the base. Given Data / Assumptions: A = 1.30B. We seek (A − B)/A * 100%. Concept / Approach: Express A in terms of B, compute the difference relative to A. This yields a fraction independent of B that can be converted to a mixed percent. Step-by-Step Solution: A − B = 1.30B − B = 0.30B.(A − B)/A = 0.30B / 1.30B = 0.30/1.30 = 3/13.Convert 3/13 to percent: (3/13)*100% ≈ 23.0769% = 23 1/13%. Verification / Alternative check: Let B = ₹100 ⇒ A = ₹130. B is ₹30 less than A; 30/130 = 3/13 ≈ 23.08%. Why Other Options Are Wrong: 30% and 33 1/3% misapply symmetry; 25% is a round guess; 20% is unrelated. Common Pitfalls: Assuming “more than” and “less than” percentages are identical in magnitude. Always use the correct base for the comparison. Final Answer: 23 1/13

Question 15

A man spends 75% of his income. Then his income rises by 20% and he increases his expenditure by 10% (relative to the old expenditure). By what percent do his savings increase?

Accepted Answer

Introduction / Context: Savings = Income − Expenditure. When both income and expenditure change by different percentages, compute new income and new expenditure carefully (noting the base of each change), then compare savings before and after. Given Data / Assumptions: Initial income = I. Initial expenditure = 75% of I ⇒ 0.75I. Initial savings = I − 0.75I = 0.25I. New income = 1.20I (20% rise). New expenditure = 1.10 × (old expenditure) = 1.10 × 0.75I = 0.825I. Concept / Approach: Compute new savings and then the percentage increase relative to the old savings. Be careful: the 10% increase applies to the old expenditure, not to the new income. Step-by-Step Solution: Initial savings S₀ = 0.25I.New savings S₁ = 1.20I − 0.825I = 0.375I.Increase = S₁ − S₀ = 0.375I − 0.25I = 0.125I.Percent increase = (0.125I / 0.25I) * 100% = 50%. Verification / Alternative check: Let I = ₹100: initial savings ₹25. New income ₹120; new expenditure 1.10 × 75 = ₹82.5; new savings ₹37.5. Increase ₹12.5 on ₹25 = 50%. Why Other Options Are Wrong: 25% and 37 1/2% mis-handle the base or apply 10% to the wrong quantity; 10% is far too small. Common Pitfalls: Applying the 10% increase to the new income rather than the initial expenditure, or double-counting the 20% income rise as extra savings without adjusting expenditure. Final Answer: 50%

Question 16

The price of an article was reduced by 25%. By what percent must the reduced price be increased to restore the original price?

Accepted Answer

Introduction / Context: Reversing a percentage decrease requires a larger percentage increase because the base has changed. A 25% reduction multiplies the price by 0.75; to restore, divide by 0.75 or find the percentage that takes 0.75 back to 1.00. Given Data / Assumptions: Original price = P. Reduced price = 0.75P. We need a percentage x such that (0.75P) * (1 + x) = P. Concept / Approach: Solve 0.75 * (1 + x) = 1 ⇒ 1 + x = 1 / 0.75 = 4/3 ⇒ x = 1/3 = 33 1/3%. Step-by-Step Solution: Reduced factor: 1 − 25% = 0.75.Required multiplier: 1 / 0.75 = 1.333….Required increase = 33 1/3%. Verification / Alternative check: Take P = ₹300. After −25% ⇒ ₹225. Increase ₹225 by 33 1/3% ⇒ ₹225 × 4/3 = ₹300. Restored. Why Other Options Are Wrong: 25% would only raise ₹225 to ₹281.25; 11 1/9% and 9 1/11% are too small; 66 2/3% is for reversing a 40% cut. Common Pitfalls: Assuming symmetry (25% down then 25% up). Percent changes are base-dependent. Final Answer: 33 1/3%

Question 17

If increasing the numerator by 20% and decreasing the denominator by 10% yields a value of 16/21, what is the original fraction?

Accepted Answer

Introduction / Context: Transformations of fractions via percentage changes are best handled algebraically. Treat the original fraction as x/y, apply the specified changes to numerator and denominator, and equate to the given resultant value. Given Data / Assumptions: Original fraction = x/y with y ≠ 0. New fraction = (1.20x) / (0.90y) = 16/21. All changes are exact percentages. Concept / Approach: Simplify the multiplier (1.20/0.90) to 4/3. Then (4/3)*(x/y) = 16/21 ⇒ x/y = (16/21) * (3/4). Reduce to lowest terms for clarity. Step-by-Step Solution: (1.20x)/(0.90y) = (4/3)*(x/y).Set (4/3)*(x/y) = 16/21.x/y = (16/21)*(3/4) = 48/84 = 4/7. Verification / Alternative check: Start from 4/7: Increase numerator by 20% ⇒ 4.8; decrease denominator by 10% ⇒ 6.3; new value = 4.8/6.3 = 48/63 = 16/21. Yes. Why Other Options Are Wrong: 3/5 and 2/3 are common guesses from adding or subtracting percentages; 5/7 is close but incorrect after the transformation. Common Pitfalls: Adding 20% and 10% or treating them as changes on the entire fraction rather than separate numerator and denominator adjustments. Final Answer: 4/7

Question 18

A 40-litre mixture of milk and water contains 10% water. How many litres of water should be added so that water becomes 20% of the new mixture?

Accepted Answer

Introduction / Context: Mixture percentage problems rely on conserving the amount of each component while adjusting total volume. Initially, the amount of water is known from the given percentage. Adding pure water increases both the water amount and the total volume; set up an equation for the target percentage. Given Data / Assumptions: Total initial volume = 40 L. Initial water = 10% of 40 L = 4 L (so milk = 36 L). Let added water be w L; no milk is added or removed. Target: water to be 20% of final mixture. Concept / Approach: Final water amount = 4 + w. Final total volume = 40 + w. Target ratio: (4 + w)/(40 + w) = 0.20. Solve for w. Step-by-Step Solution: Set (4 + w)/(40 + w) = 1/5.Cross-multiply: 5(4 + w) = 40 + w ⇒ 20 + 5w = 40 + w.Rearrange: 4w = 20 ⇒ w = 5 L. Verification / Alternative check: New mixture = 45 L with water 9 L? Wait, check: 4 + 5 = 9; 9/45 = 1/5 = 20%. Correct. Why Other Options Are Wrong: 4 L gives 8/44 ≈ 18.18%; 6.5 L or 7.5 L overshoot 20%. Common Pitfalls: Using milk volume as the base or forgetting that added water increases the denominator as well as the numerator. Final Answer: 5 litres

Question 19

Boys and girls in a college are in the ratio 3 : 2. If 20% of the boys and 25% of the girls are adults, what percentage of the total students are not adults?

Accepted Answer

Introduction / Context: This problem blends ratios with basic percentage calculations. We are given the gender composition of a college and the fraction of adult students within each gender, and we must find the overall percentage of students who are not adults. This is typical of demographic or survey-style aptitude questions where careful aggregation is key. Given Data / Assumptions: Boys : Girls = 3 : 2. Adults among boys = 20% of boys. Adults among girls = 25% of girls. Assume a convenient total matching the ratio (e.g., 5k students) to simplify arithmetic; results are ratio-invariant. Concept / Approach: Use a ratio base to avoid cumbersome algebra. With 3 parts boys and 2 parts girls, compute adult counts in each group using the given percentages. Sum adult counts to get total adults. Then convert to a percentage of the whole. The non-adult percentage is 100% minus the adult percentage. Step-by-Step Solution: Let total students = 5k (boys = 3k, girls = 2k) Adult boys = 20% of 3k = 0.6k Adult girls = 25% of 2k = 0.5k Total adults = 0.6k + 0.5k = 1.1k Adult % = (1.1k / 5k) * 100 = 22% Non-adult % = 100% − 22% = 78% Verification / Alternative check: Pick actual numbers: if total is 500, boys = 300, girls = 200. Adults = 20% of 300 (60) + 25% of 200 (50) = 110 adults. Adult % = 110/500 = 22%; non-adults = 78%. Why Other Options Are Wrong: 58%, 67.5%, and 82.5% arise from mixing or averaging percentages incorrectly. 72% is an arbitrary nearby value but does not match the computed adult complement of 22%. Common Pitfalls: Directly averaging 20% and 25% without weighting by the 3 : 2 ratio, or subtracting the wrong quantity from 100%. Weighted aggregation is essential. Final Answer: 78%

Question 20

A class has students where 72% took Biology and 44% took Mathematics. Each student took at least one of these subjects, and 40 students took both. What was the total number of students in the class?

Accepted Answer

Introduction / Context: This is a classic inclusion–exclusion percentage problem. We are told the percentages of students taking Biology and Mathematics, that everyone took at least one, and that 40 students took both. The task is to recover the total class size from these data points. Given Data / Assumptions: B% = 72% of total N took Biology. M% = 44% of total N took Mathematics. Both = 40 students. Every student took Biology or Mathematics (or both), so union = 100% of N. Concept / Approach: Use inclusion–exclusion: Biology + Mathematics − Both = Union. In percentage terms, 0.72N + 0.44N − Both = N. Solve this linear equation for N using the given Both = 40. Step-by-Step Solution: 0.72N + 0.44N − Both = N 1.16N − Both = N ⇒ 1.16N − N = Both 0.16N = Both = 40 N = 40 / 0.16 = 250 Verification / Alternative check: Check counts: Biology = 0.72*250 = 180; Mathematics = 0.44*250 = 110; both = 40; union = 180 + 110 − 40 = 250, matching total N. Why Other Options Are Wrong: 200, 240, and 320 do not satisfy 0.16N = 40. 180 is a distractor resembling Biology-only count, not the total. Common Pitfalls: Adding 72% and 44% to get 116% and assuming that equals the total, without subtracting the overlap. Always manage the double-counted intersection via inclusion–exclusion. Final Answer: 250

Percentage Questions