Question 1

If the side of a square is increased by 30%, by what percentage does its area increase?

Accepted Answer

Introduction / Context:Area of a square scales with the square of its side. When a linear dimension changes by a certain percent, the area changes by the square of the scale factor. This checks understanding of geometric scaling and percentage growth. Given Data / Assumptions:Side increases by 30% ⇒ side factor = 1.30. Concept / Approach:If side increases by factor k, the area increases by factor k^2. The percentage increase in area is (k^2 − 1) * 100%. With k = 1.3, compute 1.3^2 and translate to a percent change relative to the original area. Step-by-Step Solution: Side factor = 1.30 Area factor = (1.30)^2 = 1.69 Area increase % = (1.69 − 1) * 100 = 0.69 * 100 = 69% Verification / Alternative check:Assume initial side s = 10 units. Initial area = 100. New side = 13; new area = 169. Increase = 69 over 100 ⇒ 69%. Why Other Options Are Wrong:30% treats area like a linear dimension. 60% is close but incorrect; 9% is 0.3^2 without referencing the total; 75% has no basis here. Common Pitfalls:Adding 30% twice (to get 60%) or squaring 30 (instead of 1.3). Always convert percent changes to scale factors before squaring for area. Final Answer:69%

Question 2

The radius of a circle is increased by 1%. What is the resulting percentage increase in its area?

Accepted Answer

Introduction / Context:Area of a circle is proportional to r^2. A small percentage change in radius causes approximately double that percentage change in area, with a slight adjustment due to compounding. This problem asks for the exact percentage using scale factors. Given Data / Assumptions:Radius increase = 1% ⇒ radius factor = 1.01. Concept / Approach:Area factor = (radius factor)^2. Thus new area = old area * (1.01)^2. The percentage increase is [(1.01)^2 − 1] * 100%. Multiplying out gives 1.0201, which equates to a 2.01% increase, slightly more than a simple doubling because of compounding. Step-by-Step Solution: Radius factor = 1.01 Area factor = (1.01)^2 = 1.0201 Increase % = (1.0201 − 1) * 100 = 2.01% Verification / Alternative check:For small p%, area increase ≈ 2p% for a circle. Here 2 * 1% = 2% is the approximation; exact value 2.01% confirms the small compounding adjustment. Why Other Options Are Wrong:1% and 1.1% treat area as linear. 2% ignores compounding. 2.1% overshoots the exact squared factor. Common Pitfalls:Forgetting that area scales with r^2, or ignoring the compounding effect of the square when percentages are not infinitesimally small. Final Answer:2.01%

Question 3

A square is transformed into a rectangle: Its length and breadth become 30% and 20% more than the original square’s side, respectively. By what percentage does the new rectangle’s area exceed the original square’s area?

Accepted Answer

Introduction / Context:We compare the area of a square with that of a rectangle formed by increasing the side in two perpendicular directions by different percentages. This is a standard compound-change area problem using multiplicative factors. Given Data / Assumptions:Square side = s. Rectangle length = 1.30s. Rectangle breadth = 1.20s. Concept / Approach:Original area = s^2. New area = (1.30s) * (1.20s) = 1.56 * s^2. The area therefore increases by a factor of 1.56 relative to the original, which corresponds to a 56% increase. Step-by-Step Solution: Original area = s * s = s^2 New area = (1.30s) * (1.20s) = 1.56 s^2 Increase % = (1.56 − 1) * 100 = 56% Verification / Alternative check:Assume s = 10 units; old area = 100. New area = 13 * 12 = 156. Excess = 56 over 100 ⇒ 56%. Why Other Options Are Wrong:36% and 50% mis-handle multiplicative compounding. 20% is far too small. 60% rounds up incorrectly. Common Pitfalls:Adding 30% + 20% and stopping at 50% without compounding. Always multiply the separate scale factors for perpendicular dimensions. Final Answer:56%

Question 4

A rectangle’s length increases by 60%. By what percentage must its width decrease to keep the area unchanged?

Accepted Answer

Introduction / Context:Area of a rectangle is length * width. If one dimension increases, the other must adjust inversely to maintain constant area. This inverse-percentage relation is a common aptitude theme involving scale factors and reciprocal adjustments. Given Data / Assumptions:Length factor = 1.60; area must remain the same. Concept / Approach:If A = L * W remains constant and L is multiplied by 1.6, then W must be multiplied by 1/1.6 to offset this change. Compute 1/1.6 and convert the resulting factor to a percentage decrease relative to the original width. Step-by-Step Solution: Required width factor = 1 / 1.6 = 0.625 Decrease = 1 − 0.625 = 0.375 Percentage decrease = 0.375 * 100 = 37.5% Verification / Alternative check:Let original width be 100 units. New width = 62.5. New length = 160 (if original was 100). New area = 160 * 62.5 = 10,000, matching the original 100 * 100. Why Other Options Are Wrong:60% or 75% ignore reciprocal scaling. “None of these” does not apply because 37.5% is exact. 25% is an under-correction. Common Pitfalls:Subtracting 60% from 100% to get 40% and misinterpreting as remaining width rather than the correct reciprocal. Always use the inverse factor for constant-product constraints. Final Answer:37.5%

Question 5

Coffee price rises by 15%, and a homemaker reduces consumption by 15%. How does her total expenditure on coffee change?

Accepted Answer

Introduction / Context:Total expenditure equals price * quantity. When both change by percentages, the new expenditure is the product of the two scale factors. This question highlights why successive equal and opposite percentage changes do not cancel exactly. Given Data / Assumptions:Price factor = 1.15 (increase 15%), quantity factor = 0.85 (decrease 15%). Concept / Approach:New expenditure factor = 1.15 * 0.85. Multiply these to see if the result is above 1 (increase), equal to 1 (unchanged), or below 1 (decrease). Translate the factor to a percentage change relative to the original expenditure. Step-by-Step Solution: Expenditure factor = 1.15 * 0.85 = 0.9775 Change % = (0.9775 − 1) * 100 = −2.25% Hence, expenditure decreases by 2.25%. Verification / Alternative check:Consider ₹100 original expenditure for easy comparison. New becomes ₹97.75, confirming a 2.25% drop. Why Other Options Are Wrong:“Remains unchanged” is incorrect because 1.15 and 0.85 do not multiply to 1. The 4% values come from approximations or additive reasoning, not accurate multiplication. Common Pitfalls:Assuming +15% and −15% cancel. They do not, because the bases differ: the reduction is applied after the increase, making the effect multiplicative. Final Answer:Decreased by 2.25%

Question 6

A vocational course has seats that increase by 15% annually. If there were 800 seats in 1992, how many seats will there be in 1994 (after two annual increases)?

Accepted Answer

Introduction / Context:This compounding growth problem involves applying the same percentage increase year over year. It mirrors practical scenarios like enrollment growth, population increases, or asset appreciation over discrete periods. Given Data / Assumptions:Initial seats (1992) = 800. Annual increase = 15% each year. We need the count in 1994, meaning two compounding steps (1992 → 1993 → 1994). Concept / Approach:Use the compound factor (1 + r)^n where r = 0.15 and n = 2 years. The 1994 seats = 800 * 1.15 * 1.15. Compute sequentially to avoid rounding errors. Step-by-Step Solution: 1993 seats = 800 * 1.15 = 920 1994 seats = 920 * 1.15 = 1058 Verification / Alternative check:Direct compounding: 800 * (1.15)^2 = 800 * 1.3225 = 1058 (exact integer due to the given numbers). Why Other Options Are Wrong:920 is after one year, not two. 1040 and 1060 are near misses from rough approximations. 1178 overshoots, reflecting a larger rate or extra year. Common Pitfalls:Applying the increase only once, or adding 15% + 15% to get 30% and computing 800 * 1.30 = 1040 (approximate but not correct under compounding). Final Answer:1058

Question 7

A 21% reduction in wheat price lets a buyer purchase 10.5 kg more for ₹100. What is the reduced price per kg?

Accepted Answer

Introduction / Context:Price–quantity inverse problems are common in percentage aptitude. When price falls, the same money buys more quantity. Here, a reduction of 21% increases the quantity purchasable with ₹100 by 10.5 kg. We must find the new (reduced) price per kg. Given Data / Assumptions: Original price = p (₹/kg). Reduced price = 0.79p (21% reduction). Budget = ₹100. Quantity increase = 10.5 kg. Concept / Approach:Original quantity = 100 / p. New quantity = 100 / (0.79p). The difference equals 10.5. Set up and solve for p, then compute reduced price 0.79p. Algebra is straightforward but handle decimals carefully. Step-by-Step Solution: 100/(0.79p) − 100/p = 10.5 100 * (1/(0.79p) − 1/p) = 10.5 = 100 * ((p − 0.79p)/(0.79p^2)) = 100 * (0.21p)/(0.79p^2) = 21/(0.79p) So, 21/(0.79p) = 10.5 ⇒ 21 = 10.5 * 0.79p ⇒ p ≈ 21 / 8.295 ≈ 2.53 Reduced price = 0.79p ≈ 0.79 * 2.53 ≈ ₹2.00 Verification / Alternative check:At ₹2/kg reduced price, ₹100 buys 50 kg. Original price ≈ ₹2.53/kg ⇒ original quantity ≈ 39.5 kg. Difference ≈ 10.5 kg, consistent. Why Other Options Are Wrong:₹2.25 and ₹2.50 are near but not exact. ₹30 is irrelevant. ₹1.90 undercuts the implied figures from the difference. Common Pitfalls:Sign errors when forming the difference, or treating 21% of ₹100 instead of 21% of price. Always apply the reduction to price, not to the budget. Final Answer:Rs. 2

Question 8

For a sphere with radius 10 cm, the numerical value of surface area is what percent of the numerical value of its volume?

Accepted Answer

Introduction / Context:This question compares the numerical values of a sphere’s surface area and volume for a given radius. Although area and volume have different units, aptitude problems often request the percentage ratio of their numerical magnitudes for a fixed measurement. Given Data / Assumptions:Radius r = 10 cm. Surface area S = 4πr^2. Volume V = (4/3)πr^3. Concept / Approach:Compute S/V * 100%. The π factors cancel. The power of r reduces to 1 in the denominator, leaving a simple arithmetic fraction multiplied by 100. Substitute r = 10 to obtain a clean percentage. Step-by-Step Solution: S = 4πr^2 = 4π * 100 = 400π V = (4/3)πr^3 = (4/3)π * 1000 = 4000π/3 S/V = (400π) / (4000π/3) = (400 * 3) / 4000 = 1200/4000 = 0.3 Percentage = 0.3 * 100 = 30% Verification / Alternative check:Approximate numerically: S ≈ 1256.6; V ≈ 4188.8; S/V ≈ 0.3 ⇒ 30%. Why Other Options Are Wrong:Values like 24%, 26.5%, or 45% do not match the exact ratio. 33.3% is a common distractor (one-third) but incorrect here. Common Pitfalls:Forgetting the factor 1/3 in volume, or failing to cancel π. Misplacing powers of r can also lead to wrong ratios. Final Answer:30%

Question 9

Income relations across three people A, B, and C: 5% of A equals 15% of B, and 10% of B equals 20% of C. If C earns ₹2000, what is A + B + C (in ₹)?

Accepted Answer

Introduction / Context:This chained-percentage relation converts to simple ratios among A, B, and C. Once ratios are established, use the given absolute income of C to scale and find the sum of all three incomes. Given Data / Assumptions:5% of A = 15% of B; 10% of B = 20% of C; C = ₹2000. Concept / Approach:From 5% of A = 15% of B, we get A/B = 15/5 = 3 ⇒ A = 3B. From 10% of B = 20% of C, we get B/C = 20/10 = 2 ⇒ B = 2C. Substitute C to find B, then A, and finally sum them. Step-by-Step Solution: B = 2C = 2 * 2000 = ₹4000 A = 3B = 3 * 4000 = ₹12000 Total = A + B + C = 12000 + 4000 + 2000 = ₹18000 Verification / Alternative check:Check the original relations: 5% of A = ₹600 equals 15% of B = ₹600; 10% of B = ₹400 equals 20% of C = ₹400. Consistent. Why Other Options Are Wrong:6000, 14000, 16000, and 20000 do not satisfy the ratio chain when C = 2000. Common Pitfalls:Inverting ratios, or mixing which percentage applies to which person. Convert each relation to a clean proportion before scaling. Final Answer:Rs. 18000

Question 10

A city has population 300,000, including 180,000 males. Overall, 50% are literate. If 70% of males are literate, how many literate females are there?

Accepted Answer

Introduction / Context:This problem uses aggregated literacy data to infer female literacy count. The trick is to compute total literates from the overall percentage, subtract literate males, and the remainder must be literate females. Given Data / Assumptions: Total population = 300,000. Males = 180,000 ⇒ Females = 120,000. Overall literates = 50% of total. Male literates = 70% of males. Concept / Approach:Compute literates: total literates first, then male literates. Subtract to find female literates. No need for percentages of females directly once male and total figures are known. Step-by-Step Solution: Total literates = 0.50 * 300,000 = 150,000 Male literates = 0.70 * 180,000 = 126,000 Female literates = 150,000 − 126,000 = 24,000 Verification / Alternative check:Female literacy rate = 24,000 / 120,000 = 20%. Check totals: 126,000 + 24,000 = 150,000, matches overall literates. Why Other Options Are Wrong:30,000; 54,000; 60,000; 42,000 do not satisfy the balancing condition with the total literate count and the given male figure. Common Pitfalls:Applying 50% separately to each gender or miscomputing the female population as 140,000. Keep track of group sizes and totals carefully. Final Answer:24000

Question 11

If the price of mangoes decreases by 40%, a buyer can get 32 more mangoes for ₹120 than before. What was the original price per mango (in ₹)?

Accepted Answer

Introduction / Context:Here, price falls and quantity bought with a fixed budget rises. We are told by how much the quantity changes, enabling us to back-calculate the original price. This is a standard inverse proportion question combining percentages and unit price logic. Given Data / Assumptions:Original price = p (₹/mango). New price after 40% decrease = 0.60p. Budget = ₹120. Quantity increase = 32 mangoes. Concept / Approach:Quantity with original price = 120/p. Quantity with reduced price = 120/(0.60p) = 200/p. Difference = 200/p − 120/p = 80/p. Set 80/p = 32 and solve for p. This avoids messy decimals and yields a clean value. Step-by-Step Solution: 80/p = 32 ⇒ p = 80 / 32 = 2.5 Therefore, original price = ₹2.50 per mango Verification / Alternative check:At original ₹2.50, 120 buys 48 mangoes. New price = 0.6 * 2.5 = ₹1.50, so 120 buys 80 mangoes. Increase = 32, matching the condition. Why Other Options Are Wrong:₹2.00 and ₹2.25 do not give a 32-mango increase. ₹3.00 and ₹3.50 are too high and produce smaller increases for a ₹120 budget. Common Pitfalls:Reducing the budget by 40% instead of the price, or mixing up which quantity grows when price falls. Always apply the percentage change to price only. Final Answer:₹ 2.50

Question 12

A student multiplied a number by 2/5 instead of the intended 5/2. What is the percentage error in the result relative to the correct value?

Accepted Answer

Introduction / Context:Miscalculations that invert a factor are common. The correct multiplier is 5/2, but the student used 2/5. We measure percentage error as the deviation from the correct result divided by the correct result, times 100%. Given Data / Assumptions:Correct factor = 5/2 = 2.5. Actual factor = 2/5 = 0.4. Concept / Approach:Let the true answer be T = 2.5N. The student’s answer A = 0.4N. Percentage error = |T − A| / T * 100%. Since factors scale the same base N, N cancels, leaving a comparison of the factors only. Step-by-Step Solution: Error magnitude = 2.5 − 0.4 = 2.1 Percentage error = (2.1 / 2.5) * 100 = 84% Verification / Alternative check:Using any N (say N = 10): Correct result = 25; student result = 4; error = 21; 21/25 * 100 = 84%. Why Other Options Are Wrong:82%, 86%, and 89% are plausible distractors from rough rounding. 80% corresponds to a different pair of factors, not this specific inversion. Common Pitfalls:Using the wrong base for percentage error (e.g., dividing by the wrong result). Always divide by the correct result when computing error percentage. Final Answer:84%

Question 13

Anirudh’s aggregate percentage from four tests: He scores 85%, 70%, 65%, and 60% in tests with maximum marks 100, 120, 140, and 160 respectively. What is his overall percentage?

Accepted Answer

Introduction / Context:Weighted averages are vital when different tests have different maximum marks. You must convert each percentage to actual marks scored, sum them, and divide by the sum of maximum marks to get the true aggregate percentage. Given Data / Assumptions:Percentages and corresponding maxima: (85% of 100), (70% of 120), (65% of 140), (60% of 160). Concept / Approach:Aggregate % = (Total scored / Total maximum) * 100. Compute each score from its percentage and add. Then divide by the overall maximum (sum of all test maxima). Avoid averaging percentages directly, which would be incorrect here because weights differ. Step-by-Step Solution: Scores: 0.85*100 = 85; 0.70*120 = 84; 0.65*140 = 91; 0.60*160 = 96 Total scored = 85 + 84 + 91 + 96 = 356 Total maximum = 100 + 120 + 140 + 160 = 520 Aggregate % = (356 / 520) * 100 ≈ 68.4615% = 68 6/13% Verification / Alternative check:Quick decimal check: 356/520 = 0.684615..., and 6/13 ≈ 0.461538..., confirming the mixed fraction representation. Why Other Options Are Wrong:Other fractions correspond to different totals or unweighted averaging. The only exact value with the given data is 68 6/13%. Common Pitfalls:Averaging 85, 70, 65, and 60 to get a simple mean (70%), which ignores varying test weights. Final Answer:68 6/13%

Question 14

In an exam, 950 boys and 250 girls appeared. If 90% of the boys and 60% of the girls passed, what percentage of all candidates failed?

Accepted Answer

Introduction / Context:This question tests aggregation of pass/fail percentages across groups of different sizes. Compute the total passes, subtract from total candidates, and express the failures as a percentage of the total. Given Data / Assumptions:Appeared: boys = 950, girls = 250. Pass rates: 90% of boys, 60% of girls. Concept / Approach:Calculate passes separately by group, sum to get total passes, subtract from total candidates to get failures, and finally convert to a percentage of total appeared. Step-by-Step Solution: Boys passed = 0.90 * 950 = 855 Girls passed = 0.60 * 250 = 150 Total passed = 855 + 150 = 1005 Total candidates = 950 + 250 = 1200 Failed = 1200 − 1005 = 195 Failure % = (195 / 1200) * 100 = 16.25% Verification / Alternative check:Passed % = 1005/1200 * 100 = 83.75%; hence failed % = 100 − 83.75 = 16.25%. Why Other Options Are Wrong:16.50% and 16.75% are close distractors, but exact arithmetic yields 16.25%. Common Pitfalls:Rounding early or using the average of 90% and 60% without accounting for different group sizes. Final Answer:16.25%

Question 15

A town had 1,50,000 people three years ago. If its population rose successively by 2%, 2.5%, and 4% in the last three years, what is the present population?

Accepted Answer

Introduction / Context:Successive percentage growth requires compounding, not addition. This population problem asks for the current value after three different annual growth rates are applied in sequence. Given Data / Assumptions:Initial population (3 years ago) = 150,000. Yearly growth factors: 1.02, then 1.025, then 1.04. Concept / Approach:Present population = 150,000 * 1.02 * 1.025 * 1.04. Multiply in order (multiplication is commutative) and avoid premature rounding to maintain accuracy. The final should be an integer in this crafted example. Step-by-Step Solution: After 1st year: 150,000 * 1.02 = 153,000 After 2nd year: 153,000 * 1.025 = 156,825 After 3rd year: 156,825 * 1.04 = 163,098 Verification / Alternative check:Combined factor: 1.02 * 1.025 * 1.04 = 1.08732; 150,000 * 1.08732 = 163,098. Why Other Options Are Wrong:Other values are near but reflect rounding or additive methods (2 + 2.5 + 4 = 8.5%), which is incorrect for successive growth. Common Pitfalls:Adding percentages directly or rounding after each step too aggressively. Use factors and keep precision to the end. Final Answer:163098

Question 16

In a survey, 80% owned a car and 60% owned a mobile phone. If 55% owned both, what percentage owned a car or a mobile phone or both?

Accepted Answer

Introduction / Context:An inclusion–exclusion application in percentage form: find the union of two ownership sets (car, phone) given each set’s size and their intersection. This is a standard survey analysis question. Given Data / Assumptions:Car = 80%, Phone = 60%, Both = 55%. We assume all percentages are of the same surveyed population. Concept / Approach:Union % = Car % + Phone % − Both %. Subtract the intersection once to correct for double counting of those who own both items. Step-by-Step Solution: Union = 80 + 60 − 55 = 85% Verification / Alternative check:If 85% own at least one, then 15% own neither. This is plausible and consistent with the given overlap of 55%. Why Other Options Are Wrong:65% and 80% underestimate by failing to add both sets properly. 97.5% is unrelated to inclusion–exclusion here. 90% overcounts the overlap. Common Pitfalls:Adding without subtracting the intersection or subtracting the intersection twice. Always use the formula A + B − A∩B. Final Answer:85%

Question 17

In an examination of 2000 candidates, every candidate took Physics or Mathematics or both. If 65.8% took Physics and 59.2% took Mathematics, how many took both subjects?

Accepted Answer

Introduction / Context:Again, an inclusion–exclusion principle question, but this time we must convert the union (100%, since every candidate took at least one) into an absolute count for the intersection using given percentages and total candidates. Given Data / Assumptions:Total candidates = 2000. Physics = 65.8%. Mathematics = 59.2%. Everyone took at least one. Concept / Approach:Physics + Mathematics − Both = 100% of the candidates. Therefore, Both % = Physics % + Mathematics % − 100%. Multiply by total to get headcount. Step-by-Step Solution: Both % = 65.8 + 59.2 − 100 = 25.0% Both count = 25% of 2000 = 0.25 * 2000 = 500 Verification / Alternative check:Union check: 65.8 + 59.2 − 25 = 100%. Numbers balance perfectly. Why Other Options Are Wrong:750, 250, 125, and 600 correspond to incorrect percentage conversions or misuse of inclusion–exclusion. Common Pitfalls:Forgetting that the union equals 100% here, or miscomputing percentages of the total population. Final Answer:500

Question 18

Two numbers X and Y are respectively 20% and 28% less than a third number Z. By what percentage is Y less than X?

Accepted Answer

Introduction / Context:This is a comparative percentage question with a common reference. Converting all numbers to the same base (Z) allows a direct comparison between X and Y, and then we compute the relative percentage difference with X as the base since we are asked “Y less than X.” Given Data / Assumptions:Let Z = 100 units for convenience. Then X = 80 (20% less), Y = 72 (28% less). Concept / Approach:Required percent = (X − Y) / X * 100. Using the normalized values avoids carrying Z through algebra and leads to quick computation. Step-by-Step Solution: X = 80, Y = 72 Difference = 80 − 72 = 8 Percentage less = (8/80) * 100 = 10% Verification / Alternative check:Proportional reasoning: Y/X = 72/80 = 0.9, so Y is 90% of X, meaning it is 10% less. Why Other Options Are Wrong:8% and 9% are near but not exact; 12% overshoots; 7.5% has no basis in the given reductions. Common Pitfalls:Comparing both directly to Z and then subtracting the percentages (28% − 20% = 8%) without converting to the same base for the final comparison. The reference for “less than X” must be X. Final Answer:10%

Question 19

Population growth with dual annual increases: A city's population rises at 5% per annum due to natural growth, and there is an additional 5% increase each year due to influx of job seekers (total compounded annually). What is the overall percentage …

Accepted Answer

Introduction / Context:This question tests compound percentage growth when two independent 5% increases act each year. Since both act annually, they combine to a single yearly multiplier and compound over multiple years.Given Data / Assumptions:Natural growth rate per year = 5%.Additional influx per year = 5%.Effective yearly increase = 5% + 5% = 10% (applied once per year).Time = 3 years; compound effect applies.Concept / Approach:Use compound growth: New population factor after n years = (1 + r)^n. Here r = 10% = 0.10 and n = 3.Step-by-Step Solution:Yearly factor = 1 + 0.10 = 1.10.3-year factor = 1.10^3 = 1.331.Overall percentage increase = (1.331 − 1) * 100% = 33.1%.Verification / Alternative check:Apply year by year: after Year 1: 1.10; Year 2: 1.21; Year 3: 1.331. This confirms 33.1% total increase.Why Other Options Are Wrong:33% and 33.24% are rounding guesses; exact compounding gives 33.1%.34% overstates the growth.32% ignores compounding.Common Pitfalls:Adding three times 10% (giving 30%) or mixing annual and total increases without compounding.Final Answer:33.1%

Question 20

Price–consumption adjustment: If the price of petrol rises by 20%, by what percentage must consumption be reduced so that total expenditure on petrol remains unchanged?

Accepted Answer

Introduction / Context:A classic inverse-percentage problem: when price goes up, to keep spending the same, consumption must drop in the exact inverse proportion.Given Data / Assumptions:Price increase r = 20%.Expenditure = Price * Quantity is to remain constant.We seek percentage decrease in quantity.Concept / Approach:When price increases by r%, quantity must be multiplied by 1/(1 + r). The percentage decrease needed is r/(100 + r) * 100%.Step-by-Step Solution:Required decrease = r / (100 + r) * 100%.Plug r = 20: decrease = 20 / 120 * 100%.Decrease = 1/6 * 100% = 16 2/3%.Verification / Alternative check:Assume initial price = 100 and quantity = 1 (expenditure = 100). New price = 120. To keep expenditure 100, quantity = 100/120 = 5/6, i.e., a reduction of 1/6 = 16 2/3%.Why Other Options Are Wrong:6 2/3%, 8%, 15%, 12.5% do not satisfy constant-expenditure arithmetic with a 20% price rise.Common Pitfalls:Subtracting 20% directly from quantity instead of using the inverse formula; that would under-compensate.Final Answer:16 2/3%

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