Question 1

Class distribution of sweets: There are 40 students and 5 teachers. Each student receives sweets equal to 15% of the number of students, and each teacher receives sweets equal to 20% of the number of students. How many sweets were distributed in tot…

Accepted Answer

Introduction / Context: This problem is about interpreting percentage-based allocations relative to a fixed reference count (the number of students). Both students and teachers receive sweets calculated as percentages of the student count, not of recipients. Given Data / Assumptions: Total students = 40; total teachers = 5. Each student gets 15% of 40 sweets. Each teacher gets 20% of 40 sweets. All distributions use whole-number arithmetic derived from the percentages. Concept / Approach: Compute per-recipient sweets first, since the percentage base is the number of students. Then multiply by the number of recipients in each group and add the results to obtain the total sweets needed. Step-by-Step Solution: Per student = 0.15 * 40 = 6 sweetsPer teacher = 0.20 * 40 = 8 sweetsStudents total = 40 * 6 = 240Teachers total = 5 * 8 = 40Grand total = 240 + 40 = 280 sweets Verification / Alternative check: Check bases: both percentages reference 40 students, not 45 total people, so the above multiplication is correct. Why Other Options Are Wrong: 240 omits teacher sweets; 320 and 360 overshoot; 260 undercounts a subgroup. Common Pitfalls: Using the total people (45) as the percentage base or calculating different bases for students and teachers leads to incorrect totals. Final Answer: 280

Question 2

Moisture content and mass balance: Fresh grapes are 80% water and dry grapes are 10% water. If the weight of dry grapes is 500 kg, what would be the equivalent weight when fresh?

Accepted Answer

Introduction / Context: Moisture problems are solved via mass balance on solids because the solid portion remains constant when water content changes. Here we convert a dry batch to its fresh equivalent using solid fractions. Given Data / Assumptions: Fresh grapes: 80% water ⇒ 20% solids. Dry grapes: 10% water ⇒ 90% solids. Dry-grape batch weight = 500 kg. Solids are unchanged between states. Concept / Approach: Compute solid mass in the dry batch (90% of 500 kg). Then equate this to solid mass in the fresh state (20% of fresh weight). Solve for fresh weight by dividing by the fresh solid fraction. Step-by-Step Solution: Dry solids = 0.90 * 500 = 450 kgFresh solids fraction = 0.20Fresh weight = 450 / 0.20 = 2250 kg Verification / Alternative check: Reverse check: 20% of 2250 is 450 kg solids; converting to dry (90% solids) gives 450/0.90 = 500 kg. Consistent. Why Other Options Are Wrong: 2350, 2085, 2555, 2000 do not preserve the solids at 450 kg when converted back and forth between 20% and 90% solids states. Common Pitfalls: Comparing water weights directly instead of conserving solids; or mistakenly applying percent to the wrong base (total vs solids). Final Answer: 2250 kg

Question 3

Rice price reduced by 6.25% (i.e., 1/16): With ₹120, Vandana can buy 1 kg more than before. What is the reduced (new) price per kilogram?

Accepted Answer

Introduction / Context: Here, a small percent reduction in unit price leads to a unit increase in quantity for a fixed budget. Recognizing 6.25% as 1/16 simplifies the algebra significantly. Given Data / Assumptions: Old price = p (₹/kg); new price = p * (1 − 0.0625) = (15/16) * p. Budget = ₹120 (same before and after). Quantity increase due to the reduction = 1 kg. Concept / Approach: Original quantity = 120/p. New quantity = 120 / ((15/16)p) = 120 * 16 / (15p). The difference equals 1 kg. Solve the resulting linear equation for p and then compute the reduced price (15/16)p. Step-by-Step Solution: Increase = 120 * (1/((15/16)p) − 1/p) = 1= 120 * (16/(15p) − 1/p) = 120 * ((16/15 − 1)/p)= 120 * ((1/15)/p) = (120/15) * (1/p) = 8/p8/p = 1 ⇒ p = ₹8 (old price)New price = (15/16)*8 = ₹7.50 per kg Verification / Alternative check: Old qty = 120/8 = 15 kg. New price ₹7.50 ⇒ new qty = 120/7.50 = 16 kg. Gain = 1 kg, matches statement. Why Other Options Are Wrong: ₹9, ₹5.50, ₹19, ₹8 do not yield exactly a 1 kg increase on a ₹120 budget with a 6.25% drop. Common Pitfalls: Using 6.25 as 0.625 (tenfold error) or mixing old vs new prices when computing the quantity difference. Final Answer: ₹ 7.50 per kg

Question 4

Wheat price rises from ₹24/kg to ₹27/kg: By what percentage should a family reduce consumption to keep expenditure unchanged?

Accepted Answer

Introduction / Context: For a fixed budget, quantity purchased is inversely proportional to price. When price increases, quantity must decrease by the reciprocal factor to maintain constant expenditure. Given Data / Assumptions: Old price = ₹24/kg; new price = ₹27/kg. Expenditure remains constant. Concept / Approach: New quantity factor = old price / new price = 24/27 = 8/9. The percentage reduction equals 1 − (new factor) expressed in percent. Step-by-Step Solution: Quantity factor = 24/27 = 8/9 ≈ 0.888…Reduction = 1 − 8/9 = 1/9 ≈ 0.111…Percentage reduction ≈ 11.1% Verification / Alternative check: Suppose original buy = 9 kg (cost = 9*24 = 216). New price 27 ⇒ affordable qty = 216/27 = 8 kg, which is 1 kg less on a base of 9 kg = 11.11%. Why Other Options Are Wrong: 10.2%, 12.1%, 12.3%, 9.1% do not represent the exact 1/9 reduction required by the price ratio 24:27. Common Pitfalls: Comparing absolute rupee differences rather than ratios; percentage change must be computed on quantity, given a fixed expenditure constraint. Final Answer: 11.1%

Question 5

Commission scaling with hours and rate: A marketing manager earns ₹240 for a 40-hour week. In the final week, the rate increases by 50% and he works 60 hours. What commission does he earn that week?

Accepted Answer

Introduction / Context: This item connects hourly pay rate, hours worked, and proportional changes. A percentage increase in rate multiplies the base hourly rate while hours scale total earnings linearly. Given Data / Assumptions: Earnings at base = ₹240 for 40 hours ⇒ base rate = ₹240/40 = ₹6 per hour. Rate increases by 50% ⇒ new rate = 1.5 * 6 = ₹9 per hour. Final week hours = 60 hours. Concept / Approach: Total commission = (hourly rate) * (hours). Apply the rate hike first, then multiply by the longer work duration to get the final week’s commission. Step-by-Step Solution: Base rate = 240 / 40 = ₹6/hNew rate after 50% increase = 6 * 1.5 = ₹9/hCommission = 9 * 60 = ₹540 Verification / Alternative check: Relative scaling: earnings scale by both rate and time. Compared to base, factor = 1.5 (rate) * (60/40) = 1.5 * 1.5 = 2.25. Base week ₹240 * 2.25 = ₹540. Same result. Why Other Options Are Wrong: 500, 520, 600, 480 do not match the exact multiplication of rate and hours given by the scenario. Common Pitfalls: Adding 50% of weekly earnings and then scaling hours, instead of adjusting the hourly rate first and then multiplying by hours. Final Answer: 540

Question 6

Scoring percentage and total questions: A candidate answered 12 questions and scored full marks on them. If his overall score is 60% with all questions carrying equal marks, what is the total number of questions in the test?

Accepted Answer

Introduction / Context: Equal-mark questions allow us to map the number of correct answers directly to a percentage of the total. If a candidate gets full marks on a certain count of questions, that count represents the percentage of the total corresponding to the overall score. Given Data / Assumptions: The candidate attempted 12 and scored full marks on these 12. Overall percentage = 60% of the total possible marks. Each question carries the same marks; no negative marking is implied. Concept / Approach: Let total questions be T. Full marks on 12 questions equals 12/T of the total marks. Since his overall is 60%, we have 12/T = 0.60. Solve for T directly. Step-by-Step Solution: 12 / T = 0.60T = 12 / 0.60 = 20 Verification / Alternative check: If the test has 20 questions, 60% correct equals 12 questions. This aligns perfectly with “full marks in all of them.” Why Other Options Are Wrong: 36, 30, 25, and 24 do not satisfy 12/T = 0.60 with T as an integer. Common Pitfalls: Interpreting “attempted” as including partially correct marks; the phrase “secured full marks in all of them” implies 12 fully correct questions mapping directly to 60%. Final Answer: 20

Question 7

Monthly income from chained percentages: Pradeep spends 40% on food, then 50% of the remaining on clothes and conveyance, and finally saves one-third of the balance. If his annual savings are ₹19,200, what is his monthly income?

Accepted Answer

Introduction / Context: This problem chains multiple percentage operations on income: first on the whole, then on the remainder, then a fraction of the final remainder is saved. We reverse the chain using multipliers to recover the base monthly income from annual savings. Given Data / Assumptions: Food = 40% of income. Clothes & conveyance = 50% of the remainder. From the remaining amount after these expenses, one-third is saved. Annual savings = ₹19,200 ⇒ monthly savings = ₹1,600. Concept / Approach: Let monthly income be S. After food, balance = 0.60S. After clothes & conveyance, balance = 0.50 * 0.60S = 0.30S. Saving = one-third of this balance = (1/3) * 0.30S = 0.10S. Equate 0.10S to ₹1600 to find S. Step-by-Step Solution: Monthly saving = 0.10SAnnual savings ₹19,200 ⇒ monthly = 19,200 / 12 = ₹1,6000.10S = 1,600 ⇒ S = 1,600 / 0.10 = ₹16,000 Verification / Alternative check: Forward check: Food 40% ⇒ ₹6,400; remainder ₹9,600. Clothes & conveyance 50% of remainder ⇒ ₹4,800; new remainder ₹4,800. Savings = one-third ⇒ ₹1,600 per month, matching data. Why Other Options Are Wrong: 12k, 18k, 20k, 24k do not yield ₹1,600 monthly savings after applying the stated sequence of percentages. Common Pitfalls: Adding or subtracting percentages as if they were on the original base each time; the second expense is on the remaining amount, not on S. Final Answer: Rs. 16000

Question 8

Money transfers and fractions: Ashok gives 40% of his money to Jayant. Jayant gives one-fourth of what he received to Prakash. After paying ₹200 to a taxi from that amount, Prakash has ₹600 left. How much money did Ashok originally have?

Accepted Answer

Introduction / Context: This chained-transfer problem requires careful tracking of fractions of an original sum through multiple recipients and a fixed deduction, then reversing to obtain the original amount. Given Data / Assumptions: Ashok’s money = A. Jayant receives 40% of A ⇒ 0.40A. Prakash receives one-fourth of Jayant’s receipt ⇒ 0.25 * 0.40A = 0.10A. Prakash pays ₹200 taxi and still has ₹600 ⇒ amount received by Prakash = 800. Concept / Approach: Set 0.10A equal to ₹800 and solve for A. The fixed deduction happens after Prakash receives the amount, so we add it back to get his receipt before the deduction. Step-by-Step Solution: Prakash’s final = 600; taxi = 200 ⇒ received = 600 + 200 = ₹8000.10A = 800 ⇒ A = 800 / 0.10 = ₹8,000 Verification / Alternative check: Jayant gets 40% of 8000 = ₹3200. Prakash gets one-fourth: ₹800. After paying ₹200, Prakash has ₹600, consistent. Why Other Options Are Wrong: 1200, 4000, 2000 do not yield a 0.10A of ₹800; “Data inadequate” is incorrect since the chain uniquely determines A. Common Pitfalls: Subtracting the taxi fare from A or Jayant’s share directly; remember the deduction applies only to Prakash’s receipt. Final Answer: 8000

Question 9

Interpreting mixed accuracy data: Rajesh solved 80% questions correctly. The statement provides inconsistent partial counts (37 correct out of 41, and 5 correct out of another 8). What is the total number of questions?

Accepted Answer

Introduction / Context: Sometimes test statements contain conflicting partial data. Here we are told Rajesh solved 80% correctly and also given two overlapping fragments: “37 correct out of 41” and “5 correct out of 8” from the remainder. Interpreting these simultaneously produces incompatible totals. Given Data / Assumptions: Overall: 80% of total questions answered correctly. Fragment A: 37 correct out of a set of 41. Fragment B: 5 correct out of another set of 8. All questions carry equal marks; no negative marking is mentioned. Concept / Approach: If we combine the fragments as disjoint parts, total correct = 37 + 5 = 42. Then 42 is 80% of total ⇒ total = 52.5, which is impossible. Alternative readings also fail to yield a consistent integer total. Hence the data are insufficient or contradictory for a unique total. Step-by-Step Solution: Assume disjoint sets: correct = 37 + 5 = 42Total = 42 / 0.80 = 52.5 ⇒ not integral ⇒ inconsistencyOther interpretations (subset/superset) likewise do not produce a definitive integer total. Verification / Alternative check: Try treating 41 as the total attempted; then 37 correct contradicts the “80% of total” unless the total equals 46.25, also impossible. The statement remains ambiguous. Why Other Options Are Wrong: 75, 65, 60 provide arbitrary totals not supported by the given fragments and 80% condition. Common Pitfalls: Forcing a calculation by rounding or discarding parts of the data; instead, recognize when the information is inadequate to fix a unique value. Final Answer: Can't be determined

Question 10

Language distribution in a class of 60: 30% speak only English, 20% speak both Hindi and English, and the rest speak only Hindi. How many children can speak Hindi?

Accepted Answer

Introduction / Context: This is a basic set-partition problem with percentages. We partition the class into mutually exclusive groups: only English, both languages, and only Hindi, then count Hindi speakers (both + only Hindi). Given Data / Assumptions: Total children = 60. Only English = 30% of 60. Both Hindi and English = 20% of 60. Remaining are only Hindi. Concept / Approach: Compute counts for “only English” and “both,” subtract from total to get “only Hindi.” Hindi speakers = both + only Hindi. Each group is disjoint and covers the class. Step-by-Step Solution: Only English = 0.30 * 60 = 18Both = 0.20 * 60 = 12Only Hindi = 60 − 18 − 12 = 30Hindi speakers = both + only Hindi = 12 + 30 = 42 Verification / Alternative check: Sum of groups: 18 + 12 + 30 = 60, consistent partition. Why Other Options Are Wrong: 36, 30, 48, 40 miscount either the “both” group or the complements leading to incorrect totals of Hindi speakers. Common Pitfalls: Adding “both” to “only English” or double-counting those who speak both languages. Final Answer: 42

Question 11

Population split by gender and children: The male:female ratio is 7:8. Among males, 25% are children; among females, 20% are children. If adult females number 156,800, what is the total population?

Accepted Answer

Introduction / Context: This demographic problem uses ratios and within-group percentages. We are given adult females; using the female children percentage, we recover total females, then total males from the ratio, and finally the total population. Given Data / Assumptions: Males : Females = 7 : 8. Children among males = 25% ⇒ adults among males = 75%. Children among females = 20% ⇒ adults among females = 80%. Adult females = 156,800. Concept / Approach: Let females = 8k. Adult females = 80% of 8k = 6.4k. Set 6.4k = 156,800 to find k. Then males = 7k. Total population = males + females = 15k. Step-by-Step Solution: 6.4k = 156,800 ⇒ k = 156,800 / 6.4 = 24,500Females = 8k = 8 * 24,500 = 196,000Males = 7k = 171,500Total population = 196,000 + 171,500 = 367,500 Verification / Alternative check: Adult females check: 80% of 196,000 = 156,800, matching the given. Ratios align as 171,500 : 196,000 ≈ 7 : 8. Why Other Options Are Wrong: 245,000 and 312,000 ignore the 20% children among females; 196,000 and 171,500 are subgroup counts, not totals. Common Pitfalls: Applying the adult percentage to total population or confusing adult female count with total females. Final Answer: 367,500

Question 12

Year-on-year exam counts by state: In 1998, A:B:C = 3:5:6. In 1999, A and C both increase by 20%, B by 10%, and A:C becomes 1:2. What is the number of students in state A in 1999?

Accepted Answer

Introduction / Context: This ratio problem spans two years with percentage increases. However, without at least one absolute number, pure ratios do not determine unique counts, even if a ratio condition holds in the next year. Given Data / Assumptions: 1998 counts: A:B:C = 3x : 5x : 6x. 1999 increases: A by 20%, B by 10%, C by 20%. 1999 ratio A:C = 1:2. Concept / Approach: Compute 1999 A and C as 1.2 * 3x = 3.6x and 1.2 * 6x = 7.2x. The ratio 3.6x : 7.2x simplifies to 1 : 2 automatically for any x. Thus the A:C condition provides no new information to fix x, so absolute values remain undetermined. Step-by-Step Solution: 1999 A = 3x * 1.2 = 3.6x1999 C = 6x * 1.2 = 7.2xA:C = 3.6x : 7.2x = 1 : 2, true for all xNo absolute count (x) is given ⇒ cannot compute numeric A for 1999 Verification / Alternative check: Any x (e.g., x = 1,000) will satisfy the given ratio in 1999, producing different absolute values for A. Hence a unique solution is impossible. Why Other Options Are Wrong: 7200, 6000, 7500 are arbitrary choices that cannot be derived from the provided data alone. Common Pitfalls: Assuming x = 1000 or some default base; without an absolute reference, ratios do not yield unique totals. Final Answer: Data inadequate

Question 13

Allocations and remaining balance: After spending 40% on machinery, 25% on building, 15% on raw material, and 5% on furniture, Hari Lal has ₹1305 left. What was the initial amount?

Accepted Answer

Introduction / Context: Multiple fixed-percentage expenditures reduce the initial amount to a final balance. Sum the percentages to find the remaining fraction of the original and then recover the starting amount from the final balance. Given Data / Assumptions: Expenditures: 40% + 25% + 15% + 5% = 85% of the initial sum. Remaining = 15% of the initial sum. Final balance = ₹1305. Concept / Approach: If S is the initial amount, remaining = 0.15S. Set 0.15S = 1305 and solve for S. Ensure percentages are applied on the same base (the original) since these are independent allocations stated as parts of the whole. Step-by-Step Solution: Remaining fraction = 100% − 85% = 15% = 0.150.15S = 1305 ⇒ S = 1305 / 0.15 = 8700 Verification / Alternative check: Check: 15% of 8700 = 1305, so expenditures sum to 7395, consistent with the stated percentages totaling 85%. Why Other Options Are Wrong: 6500, 7225, 1390, 7800 do not yield ₹1305 as exactly 15%. Common Pitfalls: Sequentially applying percentages (compounding) when the problem describes allocations as fractions of the original total; here they are parts of one whole. Final Answer: Rs. 8700

Question 14

Targeting an overall 50%: A student scores 30% of 180 in Paper 1. What percent of 150 must he score in Paper 2 to achieve at least 50% overall?

Accepted Answer

Introduction / Context: The student needs a target overall percentage across two papers with different maxima. Compute the required total marks, subtract Paper 1 marks, and then express the remainder as a percentage of Paper 2’s maximum. Given Data / Assumptions: Paper 1 max = 180; scored = 30% of 180. Paper 2 max = 150; required overall ≥ 50% across both papers. No negative marks or scaling adjustments. Concept / Approach: Total maximum = 180 + 150 = 330. Target marks = 0.50 * 330 = 165. Paper 1 marks = 0.30 * 180 = 54. Needed in Paper 2 = 165 − 54. Convert this to a percentage of 150. Step-by-Step Solution: Paper 1 = 0.30 * 180 = 54Target overall = 0.50 * 330 = 165Required in Paper 2 = 165 − 54 = 111% on Paper 2 = 111 / 150 * 100 = 74% Verification / Alternative check: Total achieved with 74% on Paper 2: 54 + 0.74*150 = 54 + 111 = 165, exactly 50% overall. Why Other Options Are Wrong: 76%, 70%, 80%, 72% overshoot or undershoot the exact requirement for reaching 50% overall. Common Pitfalls: Taking 50% of each paper separately; the correct approach uses the combined maximum and total marks needed. Final Answer: 74%

Question 15

Pass threshold shortfall: To pass, 36% of the maximum marks is required. A student scored 113 and fell short by 85 marks. What are the maximum marks?

Accepted Answer

Introduction / Context: Here we relate a fixed passing percentage to a student’s mark and shortfall. The pass mark equals the student’s score plus the shortfall; compute the maximum from the pass percentage. Given Data / Assumptions: Pass mark = 36% of maximum marks. Student’s score = 113. Shortfall to pass = 85 ⇒ pass mark = 113 + 85 = 198. Concept / Approach: Let maximum marks be M. Then 0.36M = 198. Solve for M by dividing both sides by 0.36. Ensure correct base (maximum marks) for the pass percentage. Step-by-Step Solution: 0.36M = 198M = 198 / 0.36 = 550 Verification / Alternative check: 36% of 550 = 198. The student at 113 is short by 85, matching the given condition. Why Other Options Are Wrong: 500, 640, 1008, 480 do not yield a pass mark of 198 at 36%. Common Pitfalls: Using 36% of the student’s marks or adding 36 to the score. Passing percentage always applies to the maximum possible marks. Final Answer: 550

Question 16

Sequential spending and balance: From salary S, 10% is house rent; of the remainder, 15% is spent on children’s education; of the new balance, 10% is spent on clothes. If ₹1377 remains, what is S?

Accepted Answer

Introduction / Context: This is a chain-percentage problem on successive remainders. Each percentage is taken on the current balance, not on the original salary. Convert each step to a multiplier and multiply to get the final fraction of S that remains. Given Data / Assumptions: S = salary. Rent = 10% ⇒ balance = 0.90S. Education = 15% of balance ⇒ new balance = 0.85 * 0.90S. Clothes = 10% of balance ⇒ final balance = 0.90 * 0.85 * 0.90S. Final balance = ₹1377. Concept / Approach: Compute the net multiplier: 0.90 * 0.85 * 0.90 = 0.6885. Then 0.6885S = 1377 ⇒ solve S = 1377 / 0.6885. Use fraction reduction to avoid rounding errors. Step-by-Step Solution: Net factor = 0.90 * 0.85 * 0.90 = 0.68850.6885S = 1377 ⇒ S = 1377 / 0.6885Using 0.6885 = 6885/10000, S = 1377 * 10000 / 6885 = 2000 Verification / Alternative check: Forward: 2000 → 1800 → (−15%) 1530 → (−10%) 1377. Matches the given final amount. Why Other Options Are Wrong: 2040, 2100, 2200 do not reproduce ₹1377 after the specified deductions. Common Pitfalls: Adding percentages directly (10% + 15% + 10%) rather than compounding them on the successive balances. Final Answer: Rs. 2000

Question 17

Compensating for a 25% price increase: By what percentage should a family reduce the use of cooking oil so that expenditure does not increase?

Accepted Answer

Introduction / Context: To keep spending constant when prices rise, consumption must drop in inverse proportion to the price increase. This is a standard inverse relationship for fixed expenditure scenarios. Given Data / Assumptions: Price increase = 25% ⇒ new price factor = 1.25 times old price. Budget (expenditure) is fixed. Concept / Approach: If P becomes 1.25P and expenditure is fixed, quantity Q must scale by 1 / 1.25 = 0.80. That corresponds to a 20% reduction (since 1 − 0.80 = 0.20). Step-by-Step Solution: New quantity factor = 1 / 1.25 = 0.80Required reduction = 1 − 0.80 = 0.20 = 20% Verification / Alternative check: Example budget B at old price P buys Q = B/P. At new price 1.25P, same B buys B/(1.25P) = 0.80 * (B/P), i.e., 20% less quantity. Why Other Options Are Wrong: 15% and 10% are insufficient; 25% and 30% are excessive relative to the exact inverse factor 0.80. Common Pitfalls: Assuming the reduction equals the price increase; reductions for constant spend use the reciprocal, not a simple subtraction. Final Answer: 20%

Question 18

In a group of 85 children playing sports (badminton or table tennis or both), the total number of girls equals 70% of the number of boys. Among the boys, the count playing only badminton is 50% of all boys, and the total number of boys playing badmi…

Accepted Answer

Introduction / Context: This problem combines percentages with a two-set (badminton, table tennis) count. We translate the text into equations for boys and girls across three disjoint categories: only badminton, only table tennis, and both. Careful book-keeping is key to avoid double counting. Given Data / Assumptions: Total children = 85. Let boys = B, girls = G; G = 0.70 * B. Boys playing only badminton = 50% of B. Total boys playing badminton (only or both) = 60% of B. Only table tennis (boys + girls) = 40% of 85 = 34. Both games (boys + girls) = 12. Concept / Approach: Convert the gender relation to counts, then use category totals to solve for each sub-group. The equation G = 0.70B and total B + G = 85 determine B and G. Next, the badminton details determine how many boys are in ”both”, which in turn yields girls in ”both”. Finally, use the only-table-tennis count to finish the table and extract girls playing only badminton. Step-by-Step Solution: B + 0.70B = 85 ⇒ 1.70B = 85 ⇒ B = 50, G = 35.Boys only badminton = 0.50 * 50 = 25.Total boys playing badminton = 0.60 * 50 = 30 ⇒ boys both = 30 − 25 = 5.Both total = 12 ⇒ girls both = 12 − 5 = 7.Only table tennis total = 34. Let boys only TT = x ⇒ 25 (only B) + x + 5 (both) = 50 ⇒ x = 20; hence girls only TT = 34 − 20 = 14.Girls total 35 = (girls only B) + 14 + 7 ⇒ girls only B = 14. Verification / Alternative check: Sum across categories equals gender totals and overall total 85, confirming consistency. Why Other Options Are Wrong: 16 and 17 arise from arithmetic slips in distributing only-table-tennis or both; ”Data inadequate” is incorrect because the provided constraints uniquely determine the counts. Common Pitfalls: Forgetting that ”total boys playing badminton” includes those who also play table tennis, or misapplying the 40% to only one gender instead of all children. Final Answer: 14

Question 19

In an examination, 40% of students failed in Hindi and 50% failed in English. If 21% failed in both subjects, what percentage passed in Hindi?

Accepted Answer

Introduction / Context: The question asks only for the percentage who passed Hindi, independent of their English outcome. That depends solely on the failure rate in Hindi. Given Data / Assumptions: Failed Hindi = 40% of students. Failed English = 50% of students. Failed both = 21% of students (not actually needed to answer the asked part). Concept / Approach: Percentage passed in Hindi = 100% − percentage failed in Hindi. Overlaps with English do not affect the Hindi pass count, as ”passed Hindi” includes those who may have failed English. Step-by-Step Solution: Pass Hindi = 100% − 40% = 60%. Verification / Alternative check: Construct a Venn diagram: regardless of the English circle, the complement of the Hindi-fail region is 60%. Why Other Options Are Wrong: 40% and 31% relate to confusions with English failure or double-counting both; 55% uses arbitrary adjustments; 79% incorrectly adds/subtracts the overlap. Common Pitfalls: Overcomplicating with unnecessary inclusion–exclusion. The English results do not matter for ”passed Hindi”. Final Answer: 60%

Question 20

A student attempted 24 questions and scored full marks on each. If this equals 40% of the total test marks (all questions carry equal marks), how many questions were in the test?

Accepted Answer

Introduction / Context: With equal-mark questions, the number correct is proportional to total marks. If a perfect score on 24 questions equals 40% of overall marks, the total question count scales accordingly. Given Data / Assumptions: All questions carry equal marks. Marks from 24 correct = 40% of total marks. No negative marking or partial credit is implied. Concept / Approach: Let the total number of questions be N and each question be worth 1 unit for convenience. Then the student’s 24 marks equal 0.40 * N. Solve for N directly. Step-by-Step Solution: 24 = 0.40 * N ⇒ N = 24 / 0.40 = 60. Verification / Alternative check: 40% of 60 is 24, which matches the given score, confirming consistency. Why Other Options Are Wrong: 30, 40, 54, 70 would correspond to different percentage relationships; none make 24 equal exactly 40% of total. Common Pitfalls: Confusing ”40% marks obtained” with ”40% questions attempted.” Here, full marks on each attempted question means a direct proportion. Final Answer: 60

Percentage Questions