Question 1

In a town, 25% of families own a phone and 15% own a car; 65% own neither. Exactly 2000 families own both a phone and a car. Consider the statements: (i) 10% own both, (ii) 35% own either a car or a phone, (iii) 40,000 families live in the town. Whi…

Accepted Answer

Introduction / Context: This is an inclusion–exclusion application coupled with converting counts to percentages. The ”neither” percent gives ”either” directly, and the given counts anchor the town’s population.Given Data / Assumptions:% with phone = 25.% with car = 15.% with neither = 65 ⇒ % with either = 35.Families owning both = 2000.Concept / Approach: Use inclusion–exclusion: P(phone ∪ car) = P(phone) + P(car) − P(both). Since 65% own neither, P(phone ∪ car) = 35%. Plug known phone and car percentages to get P(both). Then use the 2000 count to find total families.Step-by-Step Solution:Either = 100% − 65% = 35% ⇒ Statement (ii) is true.25% + 15% − both% = 35% ⇒ both% = 5% (not 10%) ⇒ Statement (i) is false.5% corresponds to 2000 families ⇒ Total families = 2000 / 0.05 = 40,000 ⇒ Statement (iii) is true.Verification / Alternative check: Check: 25 + 15 − 5 = 35% aligns with ”either,” consistent with 65% ”neither.”Why Other Options Are Wrong: Any option including (i) is invalid since both% is 5%, not 10%. ”Only II” omits the correct (iii).Common Pitfalls: Mixing ”either” with ”exactly one,” or forgetting that ”neither” immediately gives ”either.”Final Answer: II and III

Question 2

In a company, 60% of employees are men. Of the men, 40% earn more than ₹50000 per year. If 36% of all employees earn more than ₹50000 per year, what percentage of women earn less than ₹50000 per year?

Accepted Answer

Introduction / Context: Partition the population into men and women, then use the given proportion earning above ₹50000 to deduce the women’s share above that threshold. Finally, convert to the share earning below ₹50000.Given Data / Assumptions:Total employees = 100 units for convenience.Men = 60; Women = 40.Men above ₹50000 = 40% of 60 = 24.Total above ₹50000 = 36 ⇒ Women above ₹50000 = 12.Concept / Approach: The overall percentage above ₹50000 equals the sum of men-above and women-above (as percentages of total). Subtract to find women-above and convert to a percentage of the women subgroup. Then subtract from 100% to get women-below.Step-by-Step Solution:Women above ₹50000 = 36 − 24 = 12.Percentage of women above = 12 / 40 * 100% = 30%.Percentage of women below = 100% − 30% = 70%.Verification / Alternative check: Check totals above: men 24 + women 12 = 36% overall, matching the given.Why Other Options Are Wrong: 60% or 40% result from misplacing the base (total vs. women subgroup); 30% is women-above, not women-below.Common Pitfalls: Using total employees as the denominator when computing the within-women percentage.Final Answer: 70%

Question 3

720 sweets are shared equally among children such that each child receives a number of sweets equal to 20% of the total number of children. How many sweets does each child get?

Accepted Answer

Introduction / Context: The number of sweets per child depends on the number of children, and the problem links them: each child’s share is 20% of the headcount. This yields a quadratic relation in the number of children.Given Data / Assumptions:Total sweets = 720.If number of children = n, then sweets per child = 0.20 * n.All sweets are distributed exactly.Concept / Approach: Total sweets = (children) * (sweets per child) = n * (0.20n) = 0.20n^2. Set that equal to 720 and solve for n. Then compute each child’s share using 0.20n.Step-by-Step Solution:0.20n^2 = 720 ⇒ n^2 = 720 / 0.20 = 3600 ⇒ n = 60 (positive root).Sweets per child = 0.20 * n = 0.20 * 60 = 12.Verification / Alternative check: Total sweets check: 60 children * 12 sweets = 720, consistent.Why Other Options Are Wrong: 14, 11, 15, 10 do not satisfy the quadratic relationship when multiplied by the corresponding child count.Common Pitfalls: Treating 20% of children as 20% of sweets, or assuming a linear relation. The share per child depends on n, producing n^2.Final Answer: 12

Question 4

A pot initially contains pure alcohol. In each operation, 20% of the mixture is removed and replaced with water. This replacement is done three times in total. What percentage of alcohol remains after the third operation?

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Introduction / Context: Successive replacement problems rely on the fact that each operation removes a fixed fraction of whatever is currently in the vessel. If a fraction r is replaced each time, the remaining fraction of the original solute after k operations is (1 − r)^k.Given Data / Assumptions:Initially: 100% alcohol.Each time, 20% of the mixture is removed and replaced with water.Number of operations k = 3.No volume losses; perfect mixing after each operation.Concept / Approach: After one replacement, the fraction of original alcohol left is (1 − 0.20) = 0.80. After k operations, the fraction of alcohol remaining equals 0.80^k. Convert this fraction to a percentage for the final answer.Step-by-Step Solution:Remaining fraction after each operation = 1 − 0.20 = 0.80.After 3 operations: 0.80^3 = 0.512.Convert to percent: 0.512 * 100% = 51.2% alcohol remains.Verification / Alternative check: Equivalent formula: Remaining% = (initial%) * (1 − r)^k = 100% * 0.8^3 = 51.2%.Why Other Options Are Wrong: 49% and 38% are typical rounding or arithmetic slips; 29% corresponds to removing a larger fraction or more cycles.Common Pitfalls: Confusing ”removed fraction” with ”remaining fraction” or adding percentages instead of multiplying factors across steps.Final Answer: 51.2%

Question 5

If the price of petrol increases by 20%, by what percentage should consumption be reduced so that expenditure on petrol remains unchanged?

Accepted Answer

Introduction / Context: With expenditure E = price * quantity held constant, any increase in price must be offset by a proportional decrease in quantity. The relation is multiplicative, not additive.Given Data / Assumptions:Price increases by 20% ⇒ price multiplier = 1.20.Expenditure must remain the same.Concept / Approach: If original quantity is Q, new quantity Q′ must satisfy 1.20 * Q′ = Q, hence Q′ = Q / 1.20 = (5/6)Q. The reduction equals Q − Q′ = (1/6)Q, which as a percentage of Q is 16 2/3%.Step-by-Step Solution:Required factor for quantity = 1 / 1.20 = 5/6.Reduction fraction = 1 − 5/6 = 1/6.Convert to percent: (1/6)*100% = 16 2/3%.Verification / Alternative check: Example: Original spend = ₹120 at ₹12/L gives 10 L. New price ₹14.40/L; buying 8.333… L costs ₹120 again. Reduction is 1.666… L on 10 L = 16 2/3%.Why Other Options Are Wrong: 20% assumes additive logic; 15% or 8% under-adjust; 6 2/3% divides by the wrong base.Common Pitfalls: Adding −20% directly to quantity change instead of using the inverse multiplier.Final Answer: 16 2/3%

Question 6

In an organisation with 1240 employees, 25% are promoted. How many employees got promoted?

Accepted Answer

Introduction / Context: This is a direct percentage-of-a-number calculation. Find one quarter of 1240 (since 25% = 1/4).Given Data / Assumptions:Total employees = 1240.Promoted = 25% of total.Concept / Approach: 25% equals 1/4. Therefore, the number promoted is 1240 / 4. Alternatively, compute 0.25 * 1240.Step-by-Step Solution:1240 * 0.25 = 310.Verification / Alternative check: 1240 / 4 = 310, consistent.Why Other Options Are Wrong: 372 corresponds to 30%; 398 is 32.1%; 345 is ~27.8%; none match 25%.Common Pitfalls: Misplacing the decimal when applying percent to a four-digit number.Final Answer: 310

Question 7

The difference between 78% and 59% of a number is 323. What is 62% of that number?

Accepted Answer

Introduction / Context: Express percentage statements algebraically to solve for the unknown number, then apply the required percentage to find the requested value.Given Data / Assumptions:Let the number be N.0.78N − 0.59N = 323.Find 0.62N.Concept / Approach: Combine like terms on the left to isolate N, then multiply N by 0.62 to get 62% of the number.Step-by-Step Solution:(0.78 − 0.59)N = 0.19N = 323.N = 323 / 0.19 = 1700.62% of N = 0.62 * 1700 = 1054.Verification / Alternative check: 78% of 1700 = 1326; 59% = 1003; difference = 323, confirming N.Why Other Options Are Wrong: Values like 1159 or 1178 result from using 68% or 69% by mistake; 1037 reflects arithmetic slips.Common Pitfalls: Treating the 323 as 19% itself rather than as the result of 19% of N.Final Answer: 1054

Question 8

A pipe is cut into two pieces. The longer piece is 70% of the original length. By what percentage is the longer piece longer than the shorter piece?

Accepted Answer

Introduction / Context: The question asks for a relative comparison: how much longer the longer piece is compared to the shorter piece, expressed as a percentage of the shorter piece.Given Data / Assumptions:Total length = L.Longer piece = 70% of L = 0.70L.Shorter piece = 30% of L = 0.30L.Concept / Approach: Percentage by which A exceeds B is (A − B)/B * 100%. Substitute A = 0.70L and B = 0.30L and simplify; L cancels out.Step-by-Step Solution:Difference = 0.70L − 0.30L = 0.40L.Relative to shorter: (0.40L)/(0.30L) * 100% = (4/3)*100% = 133 1/3% = 400/3%.Verification / Alternative check: Let L = 100 units: longer 70, shorter 30. Excess = 40 over base 30 = 133 1/3%.Why Other Options Are Wrong: 40% uses the total length as the base; 140% and 133 1/3% are close but only 400/3% exactly represents 133 1/3% in fractional form as given in options.Common Pitfalls: Using the total length as denominator instead of the shorter piece, which changes the base and hence the percentage.Final Answer: 400/3%

Question 9

A brand's watch price is first marked 15% above its original price and then increased by an additional 10% due to higher demand. What is the overall percentage increase relative to the original price?

Accepted Answer

Introduction / Context: Successive percentage increases multiply, not add. We must compound the 15% and 10% increases to find the net change over the original price.Given Data / Assumptions:Initial price = P.After first increase: 1.15P.After second increase: 10% more on the new price ⇒ 1.10 * 1.15P.Concept / Approach: Net multiplier = 1.15 * 1.10 = 1.265. The overall increase is 1.265 − 1 = 0.265 = 26.5%.Step-by-Step Solution:Compute compounded factor: 1.15 * 1.10 = 1.265.Net percent increase = (1.265 − 1) * 100% = 26.5%.Verification / Alternative check: If P = ₹100, new price after steps = ₹126.50, yielding exactly 26.5% rise.Why Other Options Are Wrong: 25% and 27% are rounding guesses; 35% incorrectly adds 15% and 20% (or misreads the steps).Common Pitfalls: Adding 15% and 10% (which would give 25%) instead of compounding.Final Answer: 26.5%

Question 10

In an examination, 25% of students failed in Mathematics, 20% failed in Economics, and 5% failed in both. What percentage passed in both subjects?

Accepted Answer

Introduction / Context: Passing in both subjects is the complement of failing in at least one. Use inclusion–exclusion to find the percentage failing in at least one subject, then subtract from 100%.Given Data / Assumptions:% fail Mathematics = 25.% fail Economics = 20.% fail both = 5.Concept / Approach: % failing at least one = 25 + 20 − 5 = 40%. Therefore, % passing both = 100% − 40% = 60%.Step-by-Step Solution:Fail(at least one) = 25% + 20% − 5% = 40%.Pass(both) = 100% − 40% = 60%.Verification / Alternative check: A Venn diagram confirms the arithmetic: the intersection is not double-counted.Why Other Options Are Wrong: 45% and 39% mis-handle inclusion–exclusion; 62% is a rounding/guess.Common Pitfalls: Adding the single-subject failure rates without subtracting the overlap.Final Answer: 60%

Question 11

If the numerator of a fraction is increased by 200% and the denominator is increased by 150%, the resulting fraction is 9/35. What was the original fraction?

Accepted Answer

Introduction / Context: Percentage changes to numerator and denominator scale a fraction by separate multipliers. Translate the described changes into factors and solve for the original fraction.Given Data / Assumptions:Original fraction = x / y.Numerator +200% ⇒ new numerator = 3x.Denominator +150% ⇒ new denominator = 2.5y.New fraction = 9/35.Concept / Approach: Form the equation (3x)/(2.5y) = 9/35. Simplify the coefficient 3/2.5 to 6/5 or 3/(5/2). Solve for x/y and reduce to lowest terms.Step-by-Step Solution:(3x)/(2.5y) = 9/35 ⇒ (x/y) = (9/35) * (2.5/3).(2.5/3) = (5/6). So x/y = (9/35) * (5/6) = 45/210 = 3/14.Verification / Alternative check: Starting from 3/14: New value = (3*3)/(14*2.5) = 9/35 as required.Why Other Options Are Wrong: 3/10, 2/15, 2/7 do not transform to 9/35 under the stated multipliers.Common Pitfalls: Treating 200% increment as 2x instead of 3x, or 150% increment as 1.5y instead of 2.5y.Final Answer: 3/14

Question 12

Sahil scores: Hindi 45, Science 47, Mathematics 49, Social Science 45, and English 32 (each subject out of 50). What is his overall percentage?

Accepted Answer

Introduction / Context: Compute the total obtained marks over the total maximum marks and convert to a percentage. All subjects have equal maximum marks, simplifying the calculation.Given Data / Assumptions:Scores: 45, 47, 49, 45, 32.Maximum per subject = 50.Number of subjects = 5.Concept / Approach: Total obtained = sum of individual scores. Total maximum = 5 * 50 = 250. Percentage = (obtained / maximum) * 100%.Step-by-Step Solution:Total obtained = 45 + 47 + 49 + 45 + 32 = 218.Total maximum = 250.Percentage = (218 / 250) * 100% = 87.2%.Verification / Alternative check: 10% of 250 is 25; subtracting 7.8 from 100 gives 92.2? That route is error-prone; stick to direct fraction 218/250 = 0.872.Why Other Options Are Wrong: 75% or 78% underestimate; 56.4% mis-sums marks.Common Pitfalls: Arithmetic errors when summing or dividing by 250; forgetting each subject has the same maximum.Final Answer: 87.2%

Question 13

A jogger wants to complete a fixed course in one-quarter less time than usual (i.e., reduce time to 75% of normal). By what percentage must the jogger increase average speed to achieve this?

Accepted Answer

Introduction / Context: For a fixed distance, speed and time are inversely proportional: speed * time = constant. Reducing time requires increasing speed by the reciprocal factor.Given Data / Assumptions:Target time = 75% of usual time (reduce by 1/4).Distance is fixed; conditions unchanged.Concept / Approach: If original time is T and original speed is V, then V * T = constant. New time T′ = 0.75T ⇒ new speed V′ must satisfy V′ * 0.75T = V * T ⇒ V′ = (1 / 0.75) * V = (4/3) * V. Increase% = (V′ − V)/V * 100% = (1/3)*100% = 33 1/3%.Step-by-Step Solution:T′ = 0.75T ⇒ speed factor = 1 / 0.75 = 1.333…Increase% = (1.333… − 1) * 100% = 33 1/3%.Verification / Alternative check: Example: If usual speed is 12 km/h, new must be 16 km/h to cut time from 1 h to 45 min; increase is 4/12 = 33 1/3%.Why Other Options Are Wrong: 25% corresponds to reducing time to 80%; 50% is for halving time; 20% underestimates.Common Pitfalls: Subtracting percentages instead of using the inverse relationship for fixed distance.Final Answer: 33 1/3%

Question 14

A student's originally awarded marks were later reduced by 40% of the original award during re-evaluation, leaving a new score of 96. How many marks did the student lose due to re-evaluation?

Accepted Answer

Introduction / Context: The reduction is a percentage of the original marks, not of the revised marks. We first recover the original award, then compute the absolute loss.Given Data / Assumptions:Let original marks = M.Reduction = 40% of M.New score after reduction = 96.Concept / Approach: New score = M − 0.40M = 0.60M. Solve 0.60M = 96 for M, then subtract to find the loss amount (which is 0.40M).Step-by-Step Solution:0.60M = 96 ⇒ M = 96 / 0.60 = 160.Loss = 0.40M = 0.40 * 160 = 64.Verification / Alternative check: Check: Original 160, minus 64 equals 96, matching the revised score.Why Other Options Are Wrong: 56, 58, 60, 68 arise from computing 40% on the wrong base or arithmetic slips.Common Pitfalls: Treating ”reduced by 40% of original marks” as ”new score is 40% of original,” which would be incorrect here. The remainder is 60% of original.Final Answer: 64

Question 15

An alloy weighs 50 g and contains 80% gold. How many grams of pure gold must be added so that the new alloy is 90% gold?

Accepted Answer

Introduction / Context: Alloy-percentage adjustment problems conserve the existing quantity of gold while adding pure gold. Set up the target percentage equation based on the new total weight and new gold amount.Given Data / Assumptions:Initial alloy weight = 50 g.Initial gold = 80% of 50 = 40 g.Add x g of pure gold; no silver added or removed.Target gold% = 90% of (50 + x).Concept / Approach: Form equation: (40 + x) / (50 + x) = 0.90. Solve for x.Step-by-Step Solution:40 + x = 0.90(50 + x).40 + x = 45 + 0.90x ⇒ x − 0.90x = 45 − 40 ⇒ 0.10x = 5 ⇒ x = 50 g.Verification / Alternative check: New total = 100 g; new gold = 90 g ⇒ 90% gold, consistent.Why Other Options Are Wrong: 60 g would overshoot to 100 * (40+60)/110 ≈ 90.9%; 30 g or 40 g yield 70/80% respectively, not 90%.Common Pitfalls: Forgetting that only pure gold is added (the silver amount remains 10 g), so the denominator changes along with the numerator.Final Answer: 50g

Question 16

In a test, Rajesh scored 112, which is 32 more than the passing marks. Sonal scored 75% of the maximum and that was 70 more than the passing marks. What is the minimum passing percentage of the test?

Accepted Answer

Introduction / Context: Use two equations: one linking Rajesh’s score to the passing mark, and another linking Sonal’s percentage of maximum marks to the passing mark. Solve for passing marks and maximum, then compute the passing percentage.Given Data / Assumptions:Rajesh score = 112 = P + 32 ⇒ P = passing marks.Sonal score = 75% of maximum = 0.75M, and 0.75M − P = 70.All marks are on the same maximum M.Concept / Approach: From Rajesh: P = 112 − 32 = 80. Substitute into Sonal’s equation to get M, then compute passing% = (P/M) * 100%.Step-by-Step Solution:P = 112 − 32 = 80.0.75M − 80 = 70 ⇒ 0.75M = 150 ⇒ M = 200.Passing percentage = 80/200 * 100% = 40%.Verification / Alternative check: Sonal’s score 0.75 * 200 = 150, which is indeed 70 above the pass mark 80.Why Other Options Are Wrong: 35%, 45%, 48% arise from arithmetic errors or using Rajesh’s score as maximum.Common Pitfalls: Misinterpreting phrases ”more than the passing marks” vs. ”percent of maximum.”Final Answer: 40%

Question 17

Ram sells goods 25% cheaper than Shyam but 25% dearer than Hari. By what percentage are Hari's goods cheaper than Shyam's?

Accepted Answer

Introduction / Context: Express all prices relative to one base to compare Hari's and Shyam's prices directly. Using Ram as a bridge between them, set up proportions and simplify.Given Data / Assumptions:Ram = 25% cheaper than Shyam ⇒ Ram = 0.75 * Shyam.Ram = 25% dearer than Hari ⇒ Ram = 1.25 * Hari.All comparisons are price ratios; quality is identical.Concept / Approach: Set 0.75 * Shyam = 1.25 * Hari. Solve for the ratio Hari/Shyam, then convert the difference to a percentage with Shyam as base.Step-by-Step Solution:0.75S = 1.25H ⇒ S = (1.25/0.75)H = (5/3)H.Hari compared to Shyam: H / S = H / ((5/3)H) = 3/5.Thus Hari's goods are cheaper by (1 − 3/5) * 100% = 2/5 * 100% = 40%.Verification / Alternative check: Let Hari = ₹100 ⇒ Ram = ₹125 ⇒ Shyam = ₹125 / 0.75 = ₹166.67. Hari is 40% less than Shyam (difference ₹66.67 on base ₹166.67).Why Other Options Are Wrong: 25 and 33 1/3 misapply one step or ignore compounding through Ram; 50 overstates the gap.Common Pitfalls: Subtracting 25 and 25 to guess 0% or 50% without proper ratio handling.Final Answer: 40

Question 18

In a school, the ratio of boys to girls is 3 : 2. If 20% of the boys and 25% of the girls are scholarship holders, what percentage of the total students do not receive a scholarship?

Accepted Answer

Introduction / Context:This percentage-and-ratio question asks for the proportion of students who are not scholarship holders, given the gender ratio and the scholarship percentages for each group. The key is to use a convenient base matching the ratio, compute weighted adults (scholarship holders), then subtract from 100%.Given Data / Assumptions:Boys : Girls = 3 : 2.20% of boys get scholarships.25% of girls get scholarships.All percentages are of their respective groups.Concept / Approach:Work with a ratio base (e.g., 5k total students so that boys = 3k and girls = 2k). Compute scholarship holders in each group, sum them, convert to an overall percentage, and finally find the complement for non-scholarship students.Step-by-Step Solution:Assume total = 5k ⇒ boys = 3k, girls = 2k.Scholarship boys = 20% of 3k = 0.6k.Scholarship girls = 25% of 2k = 0.5k.Total scholarships = 0.6k + 0.5k = 1.1k.Scholarship % of all students = 1.1k / 5k * 100 = 22%.Non-scholarship % = 100% − 22% = 78%.Verification / Alternative check:Pick actual numbers (boys 300, girls 200). Scholarship holders = 20% of 300 (60) + 25% of 200 (50) = 110 of 500 total ⇒ 22% scholarships → 78% not on scholarship.Why Other Options Are Wrong:75% and 60% come from unweighted or incorrect averaging; 55% is far too low; 72% is a nearby distractor but not the computed complement of 22%.Common Pitfalls:Simply averaging 20% and 25% without weighting by the 3 : 2 ratio gives the wrong total scholarship rate.Final Answer:78%

Question 19

The price of ghee rises by 32%. A family reduces its consumption so that its total spending on ghee increases by only 10% compared to before. If the family earlier consumed 10 kg, what is the new consumption?

Accepted Answer

Introduction / Context:This problem ties together percentage change in price, adjustment in quantity, and overall expenditure. Expenditure = price * quantity, so multiple percentage changes combine multiplicatively, not additively.Given Data / Assumptions:Original consumption = 10 kg.Price increases by 32% ⇒ price factor = 1.32.Final expenditure rises by 10% ⇒ expenditure factor = 1.10.Concept / Approach:If E = P * Q, then new E / old E = (new P / old P) * (new Q / old Q). Hence 1.10 = 1.32 * (Q_new / Q_old). Solve for Q_new using Q_old = 10 kg.Step-by-Step Solution:1.10 = 1.32 * (Q_new / Q_old)Q_new / Q_old = 1.10 / 1.32 = 0.8333… = 5/6Q_new = (5/6) * 10 = 50/6 = 8 1/3 kgVerification / Alternative check:Original spend = 10 * P. New spend = 8 1/3 * 1.32P = (25/3)*1.32P ≈ 11P, which is 10% higher than 10P.Why Other Options Are Wrong:8 3/4 kg and 8 1/2 kg correspond to different expenditure targets; 9 kg would give too large a spending increase; 8 1/4 kg undershoots the required 10% rise.Common Pitfalls:Adding or subtracting percentages linearly (32 − 10), or assuming consumption falls by exactly 32% − 10% = 22%. Always use the multiplicative relation of E = P * Q.Final Answer:8 1/3 kg

Question 20

A family spends on wheat, meat, and vegetables in the ratio 12 : 17 : 3. If their prices increase by 20%, 30%, and 50% respectively (quantities unchanged), by what percentage does the total expense increase?

Accepted Answer

Introduction / Context:This is a weighted-percentage increase problem. Different items contribute differently to the total expense, so their price hikes must be combined using the original spending weights.Given Data / Assumptions:Spending ratio (wheat : meat : vegetables) = 12 : 17 : 3.Price increases: 20%, 30%, 50% respectively.Quantities consumed remain unchanged, so spending on each item scales by its price factor.Concept / Approach:Let original spends be 12x, 17x, and 3x (total 32x). After the increases, new spends are 12x*1.20, 17x*1.30, and 3x*1.50. Sum these to get the new total and compare with 32x.Step-by-Step Solution:Original total = 12x + 17x + 3x = 32xNew total = 12x*1.20 + 17x*1.30 + 3x*1.50= 14.4x + 22.1x + 4.5x = 41.0xIncrease factor = 41.0x / 32x = 1.28125Percentage increase = (1.28125 − 1) * 100 = 28.125% = 28 1/8%Verification / Alternative check:Compute weighted average of increases using weights 12, 17, and 3 over 32 total. Weighted increase = (12*20% + 17*30% + 3*50%) / 32 = (240 + 510 + 150) / 32% = 900/32% = 28.125%.Why Other Options Are Wrong:23 1/3%, 27 1/8%, and 25 1/7% result from incorrect weights or arithmetic; 30% is the largest single-item increase, not the weighted total.Common Pitfalls:Taking a simple average of 20%, 30%, 50% or using the largest increase directly instead of a weighted calculation.Final Answer:28 1/8%

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