Question 1

Marks and percentage points: In a test of 400 marks, A scored 10 percentage points more than B, and B scored 5 percentage points more than C. If C scored 300 marks out of 400, find A’s marks.

Accepted Answer

Introduction / Context: The phrase “scored x% more than” can be ambiguous. In many exam contexts, it is used to mean percentage points (i.e., absolute increase on the percent scale). We use that interpretation here, which produces an integer answer aligned with the options. Given Data / Assumptions: Total marks = 400. C scored 300 ⇒ C% = 300/400 * 100% = 75%. B scored 5 percentage points more than C ⇒ B% = 80%. A scored 10 percentage points more than B ⇒ A% = 90%. Concept / Approach: Convert the given marks to percentage for C, adjust B and A by percentage points (not relative multipliers), then convert A’s percentage back to marks using the total of 400. Step-by-Step Solution: C% = 75%.B% = 75% + 5% = 80%.A% = 80% + 10% = 90%.A’s marks = 90% of 400 = 0.90 * 400 = 360. Verification / Alternative check: If one (incorrectly) treats “% more” as relative increase, then B = 1.05*C and A = 1.10*B, giving 346.5, which is not among the options. The percentage-point reading resolves the ambiguity and matches a listed choice. Why Other Options Are Wrong: 310, 325, 350: Do not correspond to 90% of 400. Common Pitfalls: Confusing percentage points with relative percent increases. Read carefully: exam comparisons often mean absolute percentage points for simplicity of scoring. Final Answer: 360

Question 2

Overall exam percentage: Mathew scored Biology 42, Chemistry 51, Mathematics 58, Physics 35, and English 48. The maximum in each subject is 60. What is Mathew’s overall percentage?

Accepted Answer

Introduction / Context: Overall percentage is computed as total obtained marks divided by total maximum marks multiplied by 100%. Summing subject-wise scores first avoids repeated percentage calculations and minimizes rounding error. Given Data / Assumptions: Scores: 42, 51, 58, 35, 48. Maximum per subject = 60; five subjects total maximum = 5 * 60 = 300. Concept / Approach: Compute the aggregate and then the fraction of maximum. This one-step aggregate method is clearer than averaging individual percentages and then averaging again. Step-by-Step Solution: Total obtained = 42 + 51 + 58 + 35 + 48 = 234.Total maximum = 300.Overall percentage = (234 / 300) * 100% = 78%. Verification / Alternative check: Average per subject = 234/5 = 46.8. As a fraction of 60, 46.8/60 = 0.78 = 78%, consistent. Why Other Options Are Wrong: 76%, 82%, 68%: Do not match 234/300 exactly. Common Pitfalls: Miscalculating the total or maximum, or averaging percentages per subject incorrectly. Always compute with totals when all subjects have the same maximum. Final Answer: 78%

Question 3

Measurement error on rectangle area: A student measures the length 20% less and the breadth 10% more than actual. If the true area is 200 sq cm, what area results from this mistaken measurement?

Accepted Answer

Introduction / Context: Area of a rectangle equals length * breadth. When each dimension is scaled by a factor, area scales by the product of those factors. This question focuses on compounded percentage changes in orthogonal dimensions. Given Data / Assumptions: True area A = L * B = 200 sq cm. Measured length = 80% of L (20% less) ⇒ 0.8L. Measured breadth = 110% of B (10% more) ⇒ 1.1B. Concept / Approach: New measured area = (0.8L) * (1.1B) = 0.8 * 1.1 * (L * B) = 0.88 * A. Multiply the true area by 0.88 to get the resulting area; no need to know L and B separately. Step-by-Step Solution: Scale factors: length 0.8, breadth 1.1.Combined area factor = 0.8 * 1.1 = 0.88.Measured area = 0.88 * 200 = 176 sq cm. Verification / Alternative check: Consider L = 20, B = 10 ⇒ true area 200. Measured: 0.8*20 = 16 and 1.1*10 = 11 ⇒ area = 176, confirming. Why Other Options Are Wrong: 206, 226, 316: Do not equal 0.88 * 200. They reflect incorrect handling of percentage compounding. Common Pitfalls: Adding/subtracting percentages from the area directly (e.g., 20% − 10% = 10%) which ignores compounding across dimensions. Always multiply the dimension factors for area. Final Answer: 176 sq cm

Question 4

Composition percentage after removal: A box has 100 blue, 50 red, and 50 black balls. After removing 25% of the blue balls and 50% of the red balls, what percentage of the balls are black?

Accepted Answer

Introduction / Context: When parts of some categories are removed from a collection, the total count changes and so do category percentages. We carefully recompute the new totals and then find the proportion of black balls in the updated mixture. Given Data / Assumptions: Initial counts: Blue = 100, Red = 50, Black = 50. Remove 25% of Blue: 0.25 * 100 = 25 removed. Remove 50% of Red: 0.50 * 50 = 25 removed. Black balls are unchanged. Concept / Approach: Compute post-removal counts for each color and sum to get the new total. Then compute the percentage of black balls as (black / total) * 100%. Step-by-Step Solution: Blue after removal = 100 − 25 = 75.Red after removal = 50 − 25 = 25.Black remains = 50.New total = 75 + 25 + 50 = 150.Percentage black = 50 / 150 * 100% = 1/3 * 100% = 33 1/3%. Verification / Alternative check: Fractional view: colors in ratio 75 : 25 : 50 = 3 : 1 : 2. Black is 2 parts out of 6 ⇒ 1/3 ⇒ 33 1/3%. Why Other Options Are Wrong: 25%: Would be true only if black were 1/4 of the new total, which it is not. 40% or 50%: Overstate black’s presence relative to the recalculated total. Common Pitfalls: Using the original total of 200 when calculating percentages after removal. Always recompute the total after changes. Final Answer: 33 1/3%

Question 5

Maintaining a price ratio after changes: Prices of two articles are in the ratio 3 : 4. If the first increases by 10% and the second increases by ₹4 so that the original ratio remains 3 : 4, find the original price of the second article.

Accepted Answer

Introduction / Context: This is a ratio preservation problem with a percentage change on one item and an absolute change on the other. We set unknowns for original prices, apply changes, and force the ratio to remain 3 : 4, then solve for the scale factor. Given Data / Assumptions: Original prices: First = 3x, Second = 4x. Change: First increases by 10% ⇒ 3x becomes 3.3x. Change: Second increases by ₹4 ⇒ 4x becomes 4x + 4. Post-change ratio remains 3 : 4. Concept / Approach: Impose 3.3x : (4x + 4) = 3 : 4. Cross-multiply to get a linear equation in x and solve. The second’s original price is 4x. Step-by-Step Solution: 3.3x / (4x + 4) = 3/44 * 3.3x = 3 * (4x + 4)13.2x = 12x + 12 ⇒ 1.2x = 12 ⇒ x = 10Original second price = 4x = ₹40. Verification / Alternative check: New prices: First = 33, Second = 44. Ratio 33 : 44 simplifies to 3 : 4, matching the condition. Why Other Options Are Wrong: ₹10, ₹30, ₹35: Do not satisfy the preserved ratio when tested with the given changes. Common Pitfalls: Applying 10% to the ratio instead of the first price, or forgetting that ₹4 is an absolute change, not percentage-based. Final Answer: ₹ 40

Question 6

Comparing year-over-year changes: Sales decreased by 80% from 2008 to 2009. If sales in 2010 returned to the 2008 level, by what percentage did sales increase from 2009 to 2010?

Accepted Answer

Introduction / Context: This problem shows the asymmetry of percentage increase and decrease. A large fall followed by a recovery to the original base can imply a very large percentage increase, because the reference base (the lower intermediate value) is smaller. Given Data / Assumptions: Let 2008 sales = S. 2009 sales = S decreased by 80% ⇒ 0.20S. 2010 sales = back to S. Concept / Approach: Percentage increase from 2009 to 2010 is (New − Old)/Old * 100%. Here, (S − 0.20S)/0.20S = 0.80S/0.20S = 4 ⇒ 400% increase. Step-by-Step Solution: 2009 = 0.20S.2010 = S.Increase% = (S − 0.20S) / 0.20S * 100% = (0.80S/0.20S) * 100% = 4 * 100% = 400%. Verification / Alternative check: Numeric check: If S = 100 units, 2009 = 20 units. Jump to 100 units is a gain of 80 units; 80/20 * 100% = 400%. Why Other Options Are Wrong: 80%, 100%, 120%: Underestimate the increase because they compare to the wrong base (often mistakenly to the original S rather than the reduced 0.20S). Common Pitfalls: Assuming the recovery percentage equals the initial drop (80%). Reverse changes are computed relative to the new base, not the old one. Final Answer: 400%

Question 7

Election arithmetic with valid/invalid votes: In a constituency, 75% of voters cast votes; 2% of those were rejected as invalid. The winner got 75% of the valid votes and polled 9,261 votes. Find the total number of voters.

Accepted Answer

Introduction / Context: This is a layered percentage problem: turnout rate, rejection rate of cast ballots, and the winner’s share of valid votes. We track each stage multiplicatively from total registered voters to the winner’s votes, then invert to find the total electorate. Given Data / Assumptions: Total voters = T. Cast votes = 75% of T = 0.75T. Rejected = 2% of cast ⇒ valid votes = 98% of cast = 0.98 * 0.75T = 0.735T. Winner’s share = 75% of valid = 0.75 * 0.735T = 0.55125T. Winner’s votes = 9,261. Concept / Approach: Set 0.55125T = 9261 and solve for T. Working with decimals or their fractional equivalents both lead to the same result. Step-by-Step Solution: 0.55125T = 9261T = 9261 / 0.55125Compute: T = 16800 (since 16800 * 0.55 = 9240 and 16800 * 0.00125 = 21; sum = 9261). Verification / Alternative check: Forward check with T = 16800: Cast = 12600; Valid = 12348; Winner’s 75% = 9261. All figures align exactly. Why Other Options Are Wrong: 14500, 18900, 24000: Do not satisfy the composite proportion 0.55125T = 9261. Common Pitfalls: Applying 2% rejection to total voters instead of to cast votes, or compounding percentages in the wrong order. Always multiply stage by stage. Final Answer: 16800

Question 8

Comparing scored marks and given percentage: Aryan scored 350 marks. Vidya scored 76% in the same test and obtained 296 marks more than Aryan. What were the maximum marks of the test?

Accepted Answer

Introduction / Context: Here we relate an absolute marks difference to a percentage score. Knowing one candidate’s raw score and the other candidate’s percentage plus the stated difference allows us to back out the maximum marks of the exam. Given Data / Assumptions: Aryan’s marks = 350. Vidya’s marks = 76% of maximum = 0.76M. Vidya scored 296 more than Aryan ⇒ 0.76M = 350 + 296 = 646. Concept / Approach: Set up a single linear equation in the unknown maximum M using Vidya’s percentage relationship and the given difference with Aryan. Solve by dividing. Step-by-Step Solution: 0.76M = 646M = 646 / 0.76M = 64600 / 76 = 850 Verification / Alternative check: Check Vidya’s marks: 76% of 850 = 0.76 * 850 = 646. Aryan’s marks = 350; difference = 646 − 350 = 296, matching the problem statement. Why Other Options Are Wrong: 650, 900, 950: None of these yield Vidya’s marks as 646 at 76% or the stated difference of 296 when tested. Common Pitfalls: Adding 296 to 76% or mixing percent-of-maximum with a percent-of-Aryan’s marks. Keep the percentage tied to the maximum M, not to Aryan’s value. Final Answer: 850

Question 9

Equality of percentage expressions: Choose the expression that is numerically equal to 32% of 750.

Accepted Answer

Introduction / Context: This item checks mental percentage fluency by asking you to match 32% of 750 to an equivalent expression from a list, without direct computation for every option. Recognizing common multiples and factors is helpful. Given Data / Assumptions: Target value: 32% of 750. Four candidate equivalences involving percents of other bases. Concept / Approach: Compute the target value, then compare quickly to each option. For speed, notice that 75% of 320 is exactly 3/4 of 320 = 240, which is also 0.32 * 750. Step-by-Step Solution: 32% of 750 = 0.32 * 750 = 240.Test options:23% of 600 = 0.23 * 600 = 138 (not equal).46% of 207 ≈ 95.22 (not equal).98% of 250 = 245 (close but not equal).75% of 320 = 0.75 * 320 = 240 (equal). Verification / Alternative check: Note 750 * 0.32 = (3/10 of 750) + (2/100 of 750) = 225 + 15 = 240, reinforcing the calculation. Why Other Options Are Wrong: The other percentages compute to values different from 240 upon quick evaluation. Common Pitfalls: Rounding 98% of 250 incorrectly to 240; it is 245. Always compute precisely. Final Answer: 75% of 320

Question 10

Compare class-wise first-class rates: Class X has 30 students with 24 first-class; Class Y has 35 students with 28 first-class. Which class has a higher percentage of first-class students?

Accepted Answer

Introduction / Context: Comparing counts across groups of different sizes requires converting to percentages or proportions. Here, we compare first-class rates in two classes of unequal strength. Given Data / Assumptions: Class X: 24 first-class out of 30. Class Y: 28 first-class out of 35. Concept / Approach: Compute each percentage as (first-class / total) * 100%. Alternatively, reduce each fraction to simplest terms or compare cross-products. Step-by-Step Solution: Class X % = 24/30 * 100% = 0.8 * 100% = 80%.Class Y % = 28/35 * 100% = 0.8 * 100% = 80%.Hence, both classes have equal first-class percentages. Verification / Alternative check: Reduce ratios: 24:30 simplifies to 4:5; 28:35 simplifies to 4:5; equal proportions confirm identical percentages. Why Other Options Are Wrong: Class X more / Class Y more: Both claim a difference that does not exist; both are 80%. None of the above: There is a correct specific choice—equality. Common Pitfalls: Comparing raw counts (24 vs 28) without considering class sizes. Always normalize by the total when comparing groups. Final Answer: Both classes have equal percentage of students getting first class

Question 11

Monthly-to-annual budgeting: Veena spends 25% of her monthly income on household expenses. Her annual income is ₹4.32 lakh. What is the total amount she spends on household expenses in 8 months?

Accepted Answer

Introduction / Context: We convert annual income to monthly, apply a fixed spending percentage per month, and then scale to 8 months. This is a direct proportional calculation without compounding. Given Data / Assumptions: Annual income = ₹4.32 lakh = ₹432,000. Monthly income = ₹432,000 / 12 = ₹36,000. Household spend = 25% of monthly income. We need 8 months’ total spend. Concept / Approach: Compute monthly spend first (a constant 25% each month), then multiply by 8. No seasonal or variable spending is implied, so linear scaling is valid. Step-by-Step Solution: Monthly spend = 0.25 * 36,000 = ₹9,000.Eight-month total = 9,000 * 8 = ₹72,000. Verification / Alternative check: Direct fraction method: 8 months is 8/12 of the year; annual spend at 25% = 0.25 * 432,000 = 108,000; 8/12 of 108,000 = 72,000. Same result. Why Other Options Are Wrong: ₹71,000; ₹73,000; ₹74,000: Off by rounding or arithmetic slips; the exact figure is ₹72,000. Common Pitfalls: Mistaking ₹4.32 lakh for ₹432,00 (missing a zero) or applying 25% to the annual total and then mishandling the 8/12 factor. Final Answer: ₹ 72000

Question 12

Successive monthly increases: An employee’s salary increases every month by 4%. If his salary in August was ₹6,300, what is his approximate salary in October of the same year?

Accepted Answer

Introduction / Context: This is a successive percentage increase problem (compound growth). Two monthly increases of 4% are applied to reach October from August (August → September → October). Given Data / Assumptions: August salary = ₹6,300. Monthly increase = 4%. Two steps of increase to reach October. Concept / Approach: Use the compound growth factor (1.04) for each month. After two months, the multiplier is (1.04)^2. Then round to the nearest rupee if required by the options. Step-by-Step Solution: September salary = 6300 * 1.04 = 6552.October salary = 6552 * 1.04 = 6814.08.Approximately ₹6,814 (matching the option). Verification / Alternative check: Direct two-step factor: October = 6300 * (1.04)^2 = 6300 * 1.0816 = 6814.08. Same result. Why Other Options Are Wrong: ₹6552: That is September’s salary, not October’s. ₹6627 and ₹6967: Do not correspond to exactly two 4% increases from ₹6300. Common Pitfalls: Applying a flat 8% once (ignoring compounding) would give 6300*1.08 = 6804, close but not equal. Successive increases require multiplying by 1.04 twice. Final Answer: ₹ 6814

Question 13

Dilution with added water: An initial 5 L alcohol–water solution has 40% alcohol. If 1 L of water is added, what is the percentage strength of alcohol in the new 6 L solution?

Accepted Answer

Introduction / Context: When only water is added to an alcohol–water solution, the amount of alcohol remains constant while total volume increases. Hence the concentration (percentage strength) decreases predictably based on new total volume. Given Data / Assumptions: Initial volume = 5 L. Alcohol fraction initially = 40% = 0.40. Water added = 1 L. Alcohol volume does not change upon adding water. Concept / Approach: Compute alcohol volume first: 0.40 * 5 = 2 L. New total volume = 5 + 1 = 6 L. New strength = alcohol volume / total volume * 100%. Step-by-Step Solution: Initial alcohol = 0.40 * 5 = 2 L.New volume = 6 L; alcohol remains 2 L.New strength = (2 / 6) * 100% = 1/3 * 100% = 33 1/3%. Verification / Alternative check: As a fraction, 2 parts alcohol out of 6 total is indeed one-third, confirming the percentage 33 1/3%. Why Other Options Are Wrong: 30% and 33%: Rounding or estimation errors; exact value is 33 1/3%. 33 2/3%: Would require alcohol 2 of 5.94 approx; not this case. Common Pitfalls: Recomputing alcohol amount after adding water or using 40% of 6 L. The alcohol quantity stays 2 L; only the denominator (total volume) changes. Final Answer: 33 1/3%

Question 14

Inclusion–exclusion with pass percentages: 49% failed in English, 36% failed in Hindi, and 15% failed in both. If 450 students passed overall, how many students appeared for the exam in total?

Accepted Answer

Introduction / Context: When two failure sets overlap, inclusion–exclusion helps compute the union (failed in at least one subject). Passing percentage is then 100% minus this union. Knowing the count of passed students lets us back-solve for the total cohort. Given Data / Assumptions: Failed English = 49%. Failed Hindi = 36%. Failed both = 15%. Passed students count = 450. Concept / Approach: Inclusion–exclusion for failure: Fail(any) = 49% + 36% − 15% = 70%. Therefore, Pass = 30% of total. If 30% equals 450, total = 450 / 0.30. Step-by-Step Solution: Fail(any) = 49 + 36 − 15 = 70%.Pass = 100 − 70 = 30%.Let Total = T. 0.30T = 450 ⇒ T = 450 / 0.30 = 1500. Verification / Alternative check: 30% of 1500 is 450; remaining 1050 students failed at least one, consistent with the 70% union of failures. Why Other Options Are Wrong: 2000, 1800, 1100: 30% of these do not equal 450, hence they do not satisfy the given pass count. Common Pitfalls: Adding the failure rates without subtracting the overlap, which would incorrectly give 85% failure and 15% pass. Final Answer: 1500

Question 15

Income split and total constraint: Bina’s monthly income is 90% of Anita’s monthly income. The sum of their monthly incomes equals Mr. Sen’s monthly income. Mr. Sen’s annual income is ₹7,75,200. Find Bina’s monthly income.

Accepted Answer

Introduction / Context: We relate three incomes: two individual monthly incomes whose sum equals a third person’s monthly income. The third person’s annual income is provided, so first convert it to monthly, then allocate between Anita and Bina according to the 90% relation. Given Data / Assumptions: Mr. Sen’s annual income = ₹7,75,200 ⇒ monthly = 7,75,200 / 12 = ₹64,600. Bina = 90% of Anita ⇒ B = 0.9A. A + B = 64,600. Concept / Approach: Express B in terms of A and substitute into the sum to solve for A. Then compute B = 0.9A. This is a basic linear equation in one variable after substitution. Step-by-Step Solution: A + 0.9A = 64,600 ⇒ 1.9A = 64,600.A = 64,600 / 1.9 = 34,000.B = 0.9A = 0.9 * 34,000 = ₹30,600. Verification / Alternative check: Sum check: 34,000 + 30,600 = 64,600 equals Mr. Sen’s monthly income; annualizing 64,600 * 12 = 7,75,200 as given. Why Other Options Are Wrong: ₹34,000 and ₹36,000: These are Anita’s or unrelated values, not Bina’s monthly income. ₹30,000: Slight underestimate; 90% of 34,000 is exactly ₹30,600. Common Pitfalls: Forgetting to convert annual to monthly, or misapplying the 90% to the combined income rather than to Anita’s income alone. Final Answer: ₹ 30600

Question 16

Passing threshold and shortfall: To pass, 296 marks are required. A student secured 222 marks and is declared failed by 10% of the maximum marks. What are the maximum aggregate marks?

Accepted Answer

Introduction / Context: This problem combines a fixed pass mark with a stated shortfall expressed as a percentage of the total maximum. The key is interpreting “failed by 10% marks” as being short by 10% of maximum marks, not 10% of the pass mark. Given Data / Assumptions: Pass marks = 296. Obtained = 222. Shortfall = 296 − 222 = 74. Shortfall equals 10% of maximum M. Concept / Approach: Set 0.10M = 74 and solve for M. The pass threshold is provided only to determine the shortfall. Once M is found, no further adjustments are needed. Step-by-Step Solution: Shortfall = 296 − 222 = 74.74 = 10% of M ⇒ 0.10M = 74.M = 74 / 0.10 = 740. Verification / Alternative check: 10% of 740 = 74, which matches the stated shortfall; the student is indeed 74 below the required 296. Why Other Options Are Wrong: 830, 810, 780: None make 10% equal to 74, so they cannot satisfy the “failed by 10%” condition. Common Pitfalls: Interpreting “failed by 10%” as 10% of the pass mark (296), which would give 29.6—incorrect by the problem’s phrasing that references maximum marks. Final Answer: 740

Question 17

Weighted voting by gender split: In an office, 40% of the staff are female and 60% are male. If 40% of the females and 60% of the males voted for me, what percentage of the entire staff voted for me?

Accepted Answer

Introduction / Context: This is a weighted average problem across two groups with different sizes and different within-group support rates. We must weight each group’s support by its proportion in the total staff before summing to find the overall support percentage. Given Data / Assumptions: Female proportion = 40% of staff. Male proportion = 60% of staff. Support within females = 40%. Support within males = 60%. Concept / Approach: Overall support = (female share * female support) + (male share * male support). Convert all to decimals and multiply accordingly. Step-by-Step Solution: Female contribution = 0.40 * 0.40 = 0.16 (16%).Male contribution = 0.60 * 0.60 = 0.36 (36%).Total support = 0.16 + 0.36 = 0.52 = 52%. Verification / Alternative check: Assume a 100-person office: 40 females → 16 supporters; 60 males → 36 supporters; total = 52 supporters → 52%. Why Other Options Are Wrong: 24%: Counts only female supporters wrongly as overall. 42% or 50%: Arithmetic slips or averaging 40% and 60% without weights. Common Pitfalls: Averaging 40% and 60% to get 50% without weighting by 40% and 60% group sizes. Always apply weighted averages when groups differ in size. Final Answer: 52%

Question 18

Price drop and quantity gained: A 20% reduction in the price of wheat allows a buyer to purchase 3.5 kg more for a fixed budget of ₹770. What was the original price per kilogram (₹/kg)?

Accepted Answer

Introduction / Context: This percentage-and-proportion problem asks you to convert a stated percentage reduction in price into an equivalent increase in quantity for a fixed budget. By linking price and quantity through the identity Money = Price * Quantity, we can solve for the original price per kilogram. Given Data / Assumptions: Budget is fixed at ₹770. Price of wheat is reduced by 20% (new price = 80% of old price). Due to the reduction, the buyer gets 3.5 kg more than before. No other discounts or taxes are involved. Concept / Approach: Let the original price be p (₹/kg). Then original quantity = 770 / p. New price = 0.8 * p, so new quantity = 770 / (0.8p) = (770/p) / 0.8 = 1.25 * (770/p). The difference between new and old quantities equals 3.5 kg. Solve for p directly. Step-by-Step Solution: Original qty = 770 / pNew qty = 770 / (0.8p) = 1.25 * (770 / p)Increase in qty = (1.25 − 1) * (770 / p) = 0.25 * (770 / p)0.25 * (770 / p) = 3.5 ⇒ 770 / p = 14 ⇒ p = 770 / 14 = 55 Verification / Alternative check: At ₹55/kg, original qty = 770/55 = 14 kg. New price = 0.8*55 = ₹44/kg; new qty = 770/44 = 17.5 kg; increase = 3.5 kg, consistent. Why Other Options Are Wrong: ₹45, ₹65, ₹37, ₹50 do not satisfy the increase of exactly 3.5 kg when applying a 20% price cut on a ₹770 budget. Common Pitfalls: Confusing 20% reduction with 20% more quantity directly. The correct link arises from Money = Price * Quantity and inverse proportionality of price and quantity for fixed money. Final Answer: ₹ 55 per kg

Question 19

Price reduced by 5%: For the same money that used to buy 50 kg of rice, how many kilograms can now be purchased after the 5% price cut?

Accepted Answer

Introduction / Context: When price decreases, a fixed amount of money buys more quantity. This is a classic inverse-proportion scenario with a single percentage change in price and a constant budget. Given Data / Assumptions: Original quantity purchasable = 50 kg (at old price). New price = 95% of old price (5% reduction). Budget is unchanged. Concept / Approach: Quantity at fixed money is inversely proportional to price. Therefore, new quantity = old quantity / 0.95. Compute 50 / 0.95 to determine the updated quantity. Step-by-Step Solution: New quantity = 50 / 0.9550 / 0.95 = 5000 / 95 = 52.631578…Rounded to two decimals ≈ 52.63 kg Verification / Alternative check: If old price is P, then budget = 50P. New price is 0.95P; quantity = (50P)/(0.95P) = 50/0.95, confirming the result. Why Other Options Are Wrong: 50.40 and 51.00 understate the inverse effect; 60.50 and 42.30 are far from the correct fractional increase implied by 5%. Common Pitfalls: Adding 5% to 50 kg instead of dividing by 0.95. Quantity scales inversely with price for a fixed budget, not additively. Final Answer: 52.63 kg

Question 20

Price of rice increased by 25%: A buyer can purchase 20 kg less for ₹400 after the increase. What is the new price per kilogram?

Accepted Answer

Introduction / Context: This question links a percentage increase in price to a decrease in quantity for a fixed budget. By modeling prices before and after the change and using the observed reduction in quantity, we can find the increased price per kilogram. Given Data / Assumptions: Budget = ₹400, unchanged before and after. Price increases by 25% (new price = 1.25 * old price). Quantity falls by 20 kg due to the increase. Concept / Approach: Let old price be p. Then original quantity = 400/p. New price = 1.25p, so new quantity = 400/(1.25p). The observed reduction is 20 kg. Set up the equation and solve for p, then compute the new price 1.25p. Step-by-Step Solution: 400/p − 400/(1.25p) = 20400/p * (1 − 1/1.25) = 201 − 1/1.25 = 1 − 0.8 = 0.2(400/p) * 0.2 = 20 ⇒ 80/p = 20 ⇒ p = 4New price = 1.25 * 4 = ₹5 per kg Verification / Alternative check: Old qty = 400/4 = 100 kg. New qty = 400/5 = 80 kg. Reduction = 20 kg, matching the statement. Why Other Options Are Wrong: ₹6 and ₹10 mismatch the 20 kg drop; ₹4 is the old price, not the new one; ₹8 is not supported by the fixed-budget math. Common Pitfalls: Confusing new price with old price or adding absolute rupees instead of working proportionally with price and quantity. Final Answer: ₹ 5

Percentage Questions