Question 1

Weighted average of two classes: Two classes have average marks of 25 and 40 respectively. When the students of both classes are combined, the overall average becomes 30 marks. If the second class has 30 students, determine how many students are in …

Accepted Answer

Introduction / Context: This problem tests the weighted-average concept. When two groups with different averages are combined, the overall average is a weighted mean, where each group average is weighted by its group size. We use this to back-calculate the unknown number of students. Given Data / Assumptions: Average of class 1 = 25. Average of class 2 = 40. Overall (combined) average = 30. Class 2 has 30 students; class 1 size is unknown (say x). Concept / Approach: Weighted average formula: overall average = (sum of totals from each class) / (total students). If A1, A2 are averages and n1, n2 are sizes, then overall = (A1*n1 + A2*n2) / (n1 + n2). Solve for n1 when other values are known. Step-by-Step Solution: Let x = number of students in class 1.25*x + 40*30 is the combined total marks.Overall average 30 ⇒ (25x + 1200) / (x + 30) = 30.25x + 1200 = 30x + 900 ⇒ 300 = 5x ⇒ x = 60. Verification / Alternative check: Totals: class 1 = 25*60 = 1500; class 2 = 40*30 = 1200. Combined = 2700 over 90 students → 2700/90 = 30, as required. Why Other Options Are Wrong: 45, 70, 80, and 40 do not satisfy the weighted-average equation when substituted; only 60 produces an overall average of 30. Common Pitfalls: Mixing simple average with weighted average; forgetting to divide by total students x + 30; or incorrectly moving terms while solving the linear equation. Final Answer: 60

Question 2

Milk–water adjustment: A 60 L mixture has milk and water in the ratio 2 : 1 (milk 40 L, water 20 L). How many litres of water must be added so that the final ratio becomes 1 : 2 (milk : water)? Show the proportion setup explicitly.

Accepted Answer

Introduction / Context: Ratio-change problems in mixtures require keeping one component fixed if it is not removed or added. Here only water is added, so milk stays constant while water increases until the target ratio is reached. Given Data / Assumptions: Total mixture = 60 L with milk : water = 2 : 1. Milk = 40 L; Water = 20 L. Add y litres of water; milk remains 40 L. Target ratio milk : water = 1 : 2. Concept / Approach: Use the target ratio 40 : (20 + y) = 1 : 2. Cross-multiply to solve for y while keeping milk constant because only water is added. Step-by-Step Solution: 40 : (20 + y) = 1 : 2 ⇒ 2*40 = 20 + y.80 = 20 + y ⇒ y = 60 L. Verification / Alternative check: New water = 20 + 60 = 80 L; milk = 40 L. Ratio = 40 : 80 = 1 : 2, correct. Why Other Options Are Wrong: 20 L, 30 L, 40 L yield water totals 40, 50, 60 which produce ratios 40:40, 40:50, 40:60, none equal to 1:2; 50 L is also incorrect. Common Pitfalls: Recomputing milk incorrectly or trying to scale both parts without acknowledging that milk volume does not change when only water is added. Final Answer: 60 L

Question 3

School strength split: In a certain school, the ratio of boys to girls is 7 : 5 and the total number of students is 2400. Determine the exact number of girls by applying the part-of-whole method.

Accepted Answer

Introduction / Context: A two-part ratio and a total allow direct computation of each subgroup by converting the ratio into fractional shares of the whole. This is a staple ratio-to-quantity application. Given Data / Assumptions: Boys : Girls = 7 : 5. Total students = 2400. Concept / Approach: Sum of parts = 7 + 5 = 12. Girls represent 5/12 of the total. Multiply the total by 5/12 to get the number of girls. Step-by-Step Solution: Girls = (5/12) * 2400.2400 / 12 = 200 ⇒ Girls = 5 * 200 = 1000. Verification / Alternative check: Boys = (7/12)*2400 = 1400. Sum = 1400 + 1000 = 2400 matches the total; the ratio 1400:1000 reduces to 7:5. Why Other Options Are Wrong: 500, 700, 800, and 1200 are not equal to 5/12 of 2400 and do not preserve the 7 : 5 split with 2400 total. Common Pitfalls: Dividing 2400 by only one of the ratio numbers or using 5/7 rather than 5/12; always divide by the sum of the ratio parts first. Final Answer: 1000

Question 4

Three-sum ratio system: If (a + b) : (b + c) : (c + a) = 6 : 7 : 8 and also a + b + c = 14, find the value of c. Show how to convert the chained ratio into individual variables.

Accepted Answer

Introduction / Context: Ratios involving sums of variables can be converted into linear equations by introducing a common factor. Summing the three pairwise sums cleverly yields a direct relationship to a + b + c, unlocking individual values. Given Data / Assumptions: (a + b) : (b + c) : (c + a) = 6 : 7 : 8. a + b + c = 14. Concept / Approach: Let a + b = 6k, b + c = 7k, c + a = 8k. Then (a + b) + (b + c) + (c + a) = 2(a + b + c) = (6 + 7 + 8)k = 21k. This gives k and then c from a simple combination of the three sums. Step-by-Step Solution: 2(a + b + c) = 21k ⇒ 2*14 = 21k ⇒ 28 = 21k ⇒ k = 4/3.c = [(b + c) + (c + a) − (a + b)] / 2 = (7k + 8k − 6k)/2 = (9k)/2.With k = 4/3 ⇒ c = (9 * 4/3)/2 = (12)/2 = 6. Verification / Alternative check: Compute a + b = 6*(4/3) = 8, b + c = 28/3, c + a = 32/3; sum/2 gives 14. c = 6 satisfies all relations. Why Other Options Are Wrong: 7, 8, 10, 5 do not satisfy the three simultaneous relations when checked against a + b + c = 14. Common Pitfalls: Forgetting that the sum of the three pairwise sums equals 2(a + b + c), or misapplying the half step when isolating a single variable. Final Answer: 6

Question 5

Percent-to-ratio conversion between two numbers: One-half of a certain first number equals 65% of a second number. Determine the ratio of the first number to the second number (first : second), showing the percentage-to-decimal transition.

Accepted Answer

Introduction / Context: Converting a relation containing a fraction and a percentage into a clean ratio is a standard proportional skill. Translate percent to decimal and then isolate the ratio between the two numbers to simplest integers. Given Data / Assumptions: (1/2)*First = 65% of Second = 0.65*Second. Concept / Approach: Start from (1/2)A = 0.65B. Multiply both sides to isolate A/B. Express as a fraction and then reduce to the smallest integer ratio. Step-by-Step Solution: (1/2)A = 0.65B ⇒ A = 1.30B.Therefore A : B = 1.30 : 1 = 13 : 10. Verification / Alternative check: Let B = 10 ⇒ A = 13. Then (1/2)A = 6.5 and 65% of B = 6.5, consistent. Why Other Options Are Wrong: 10 : 13, 8 : 13, 13 : 8, 5 : 13 invert or distort the 1.30 multiplier connecting A and B. Common Pitfalls: Treating 65% as 0.65 of the first number or mixing the sides; ensure the fraction applies to the correct variable. Final Answer: 13 : 10

Question 6

Ratio shift after equal increments: Two numbers are in the ratio 3 : 4. If 3 is added to each number, the new ratio becomes 4 : 5. Find the difference between the original numbers and show your algebra clearly.

Accepted Answer

Introduction / Context: Changing ratios after adding equal amounts to each term can be solved by expressing the original numbers with a common multiplier and then imposing the new ratio. This leads to a simple linear equation in the multiplier. Given Data / Assumptions: Original numbers: 3x and 4x. Add 3 to each → (3x + 3) and (4x + 3). New ratio = 4 : 5. Concept / Approach: Set (3x + 3)/(4x + 3) = 4/5 and solve for x. Then compute the original difference 4x − 3x = x. Step-by-Step Solution: (3x + 3)/(4x + 3) = 4/5 ⇒ 5(3x + 3) = 4(4x + 3).15x + 15 = 16x + 12 ⇒ x = 3.Difference between original numbers = 4x − 3x = x = 3. Verification / Alternative check: Original numbers 9 and 12; after adding 3 → 12 and 15, ratio 12 : 15 simplifies to 4 : 5, as required. Why Other Options Are Wrong: 2, 5, 7, 9 result from arithmetic slips when solving the proportion or misreading the original ratio. Common Pitfalls: Cross-multiplying incorrectly or computing the difference after the increment rather than for the original pair. Final Answer: 3

Question 7

Relative subtraction and resulting multiple: From each of two numbers, half the smaller is subtracted. After this, the larger becomes exactly 4 times the smaller. Determine the ratio of the original numbers (larger : smaller).

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Introduction / Context: This problem encodes a before-and-after transformation. By writing the post-subtraction values explicitly and applying the given multiplicative relation, you can form a linear equation connecting the original numbers, then express the ratio. Given Data / Assumptions: Original larger = L; original smaller = S (L > S > 0). Subtract (1/2)S from each. Resulting larger = L − (1/2)S; resulting smaller = S − (1/2)S = (1/2)S. After subtraction: resulting larger = 4 * resulting smaller. Concept / Approach: Set L − (1/2)S = 4 * (1/2)S and solve for L in terms of S. Then read the ratio L : S. Step-by-Step Solution: L − (1/2)S = 2S ⇒ L = 2S + (1/2)S = (5/2)S.Therefore L : S = (5/2)S : S = 5 : 2. Verification / Alternative check: Example: let S = 2; then L = 5. After subtraction: new larger = 5 − 1 = 4; new smaller = 2 − 1 = 1; indeed 4 = 4 * 1. Why Other Options Are Wrong: 1 : 4 and 4 : 5 invert or misorder; 4 : 1 ignores the half-subtraction; 3 : 2 does not satisfy the post-subtraction multiple condition. Common Pitfalls: Subtracting half of the larger instead of half of the smaller from both numbers, or misidentifying which becomes 4 times which after the operation. Final Answer: 5 : 2

Question 8

Ratio update after adding a constant: Two numbers are in the ratio 2 : 3. After adding 9 to each number, the new ratio becomes 3 : 4. Compute the product of the original two numbers.

Accepted Answer

Introduction / Context: Like other ratio-shift problems, we parameterize the original numbers, apply the given increment, and enforce the new ratio to solve for the scaling factor. Once the originals are known, compute their product directly. Given Data / Assumptions: Original numbers = 2x and 3x. Add 9 to each ⇒ (2x + 9), (3x + 9). New ratio = 3 : 4. Concept / Approach: Set (2x + 9)/(3x + 9) = 3/4. Solve for x, then compute (2x)*(3x) as the product of the original pair. Step-by-Step Solution: (2x + 9)/(3x + 9) = 3/4 ⇒ 4(2x + 9) = 3(3x + 9).8x + 36 = 9x + 27 ⇒ x = 9.Original numbers: 18 and 27 ⇒ product = 18 * 27 = 486. Verification / Alternative check: After adding 9 → 27 and 36; ratio 27 : 36 = 3 : 4, confirming the setup. Why Other Options Are Wrong: 360, 432, 480, 512 are products derived from incorrect x or arithmetic errors when multiplying. Common Pitfalls: Forgetting to add 9 to both numbers or cross-multiplying incorrectly in the proportion. Final Answer: 486

Question 9

Income–expense with equal savings: Two persons have weekly incomes in the ratio 7 : 3 and weekly expenses in the ratio 5 : 2. Each saves ₹ 300 per week. Find the weekly income of the first person by solving the pair of linear equations.

Accepted Answer

Introduction / Context: When incomes and expenses are given in separate ratios and savings (income minus expense) are fixed, represent each with scaled variables and solve the resulting simultaneous equations. This is a classic linear system application in ratio problems. Given Data / Assumptions: Incomes: 7x and 3x. Expenses: 5y and 2y. Savings for each = ₹ 300. Concept / Approach: Write equations for each person: 7x − 5y = 300 and 3x − 2y = 300. Solve the system for x and y, then compute the first person’s income 7x. Step-by-Step Solution: 7x − 5y = 300 … (1)3x − 2y = 300 … (2)Multiply (2) by 5: 15x − 10y = 1500.Multiply (1) by 2: 14x − 10y = 600.Subtract: x = 900 ⇒ first income = 7x = ₹ 6300. Verification / Alternative check: From (2), 3*900 − 2y = 300 ⇒ y = 1200. Expenses: 5y = 6000 and 2y = 2400. Savings: 6300 − 6000 = 300 and 2700 − 2400 = 300, consistent. Why Other Options Are Wrong: ₹ 5400, ₹ 4500, ₹ 7500, ₹ 7000 do not satisfy both savings equations simultaneously. Common Pitfalls: Treating the same scaling variable for both income and expense ratios or incorrectly eliminating variables when solving the system. Final Answer: ₹ 6300

Question 10

Investment split from a total: Amit and Sudesh invest in the ratio 4 : 7. If the sum of their investments is ₹ 49,500, compute Sudesh’s investment by converting the ratio to parts of the whole.

Accepted Answer

Introduction / Context: Turning a ratio into absolute amounts requires finding the value of one part. Once the sum corresponds to the total parts, multiply by the relevant number of parts to get an individual’s share. Given Data / Assumptions: Amit : Sudesh = 4 : 7. Total = ₹ 49,500. Concept / Approach: Total parts = 4 + 7 = 11. One-part value = 49,500 / 11. Sudesh’s share = 7 parts. Step-by-Step Solution: One part = 49,500 / 11 = 4,500.Sudesh’s investment = 7 * 4,500 = ₹ 31,500. Verification / Alternative check: Amit’s share = 4 * 4,500 = ₹ 18,000. Sum 31,500 + 18,000 = 49,500, matching the total and preserving 4 : 7. Why Other Options Are Wrong: ₹ 18,00 (typo), ₹ 18,500, ₹ 31,000, or ₹ 28,000 do not correspond to 7/11 of the total. Common Pitfalls: Dividing the total by just one of the ratio numbers or forgetting to multiply by the correct number of parts afterwards. Final Answer: ₹ 31500

Question 11

Mixed coins, fixed total value: A bag contains ₹1, 50 paise, and 25 paise coins in the ratio 8 : 9 : 11. If the total money is ₹ 366, find how many 25 paise coins are in the bag. Use paise to avoid decimal confusion.

Accepted Answer

Introduction / Context: When coin counts are in a ratio but denominations differ, compute total value in a single unit (paise) and express counts as multiples of a common variable. This yields a single linear equation in that variable. Given Data / Assumptions: Denominations: ₹1 = 100p, 50p, and 25p. Count ratio = 8 : 9 : 11. Total value = ₹ 366 = 36,600 paise. Concept / Approach: Let counts be 8x, 9x, and 11x. Total value V = 100*(8x) + 50*(9x) + 25*(11x). Set V = 36,600 and solve for x to get the 25p coin count = 11x. Step-by-Step Solution: V = 800x + 450x + 275x = 1,525x.1,525x = 36,600 ⇒ x = 36,600 / 1,525 = 24.25p coins = 11x = 11 * 24 = 264. Verification / Alternative check: Value check: 100*192 + 50*216 + 25*264 = 19,200 + 10,800 + 6,600 = 36,600 paise = ₹ 366. Why Other Options Are Wrong: 364, 241, 245, 220 would produce incorrect totals and violate the 8 : 9 : 11 count ratio when verified. Common Pitfalls: Mixing number ratio with value ratio or forgetting to convert rupees to paise before forming the equation. Final Answer: 264

Question 12

Fractional ratio conversion and split: ₹ 1664 is divided between P and Q in the ratio 1/3 : 1/5. Convert this to an integer-part ratio and determine P’s share precisely.

Accepted Answer

Introduction / Context: Ratios given as fractions can be scaled to whole-number parts by clearing denominators. Then the standard part-of-whole method yields each person’s share from the total amount. Given Data / Assumptions: P : Q = (1/3) : (1/5). Total = ₹ 1664. Concept / Approach: Multiply both ratio terms by the least common multiple of denominators (15) to get whole parts: 5 : 3. Then find one-part value and scale for P’s part (5 parts). Step-by-Step Solution: (1/3) : (1/5) ⇒ multiply by 15 ⇒ 5 : 3.Total parts = 5 + 3 = 8 ⇒ one part = 1664 / 8 = 208.P’s share = 5 parts = 5 * 208 = ₹ 1040. Verification / Alternative check: Q’s share = 3 * 208 = ₹ 624. Sum = 1040 + 624 = 1664, matching the given total and preserving the 5 : 3 split. Why Other Options Are Wrong: 1085, 1015, 1090, and 980 do not correspond to 5/8 of ₹ 1664. Common Pitfalls: Adding numerators/denominators directly or interpreting 1/3 : 1/5 as 1 : 1; remember to clear denominators to get integer parts. Final Answer: 1040

Question 13

Ratio-constrained scores with a cap: Marks of A, B, and C are in the ratio 10 : 12 : 15. If the maximum possible mark is 100, identify the range in which B’s mark cannot lie.

Accepted Answer

Introduction / Context: When scores maintain a fixed ratio, all three are scaled by a single positive factor k. Imposing a maximum score cap restricts k, which in turn restricts the allowable ranges for each student’s marks. We test intervals for feasibility under this cap. Given Data / Assumptions: A : B : C = 10 : 12 : 15 ⇒ A = 10k, B = 12k, C = 15k. Maximum per paper = 100 marks. Concept / Approach: Since C = 15k ≤ 100, we must have k ≤ 100/15 = 6.666…. Therefore B = 12k ≤ 12 * 6.666… = 80. Thus B cannot exceed 80 under the common scaling factor constraint. Step-by-Step Solution: C ≤ 100 ⇒ 15k ≤ 100 ⇒ k ≤ 100/15 ≈ 6.666…B = 12k ≤ 12 * 6.666… = 80.Hence any value strictly between 80 and 90 is impossible for B. Verification / Alternative check: Feasible examples: k = 2 ⇒ B = 24 (20–30 possible); k = 4 ⇒ B = 48 (40–50 possible); k ≈ 6 ⇒ B = 72 (70–80 possible). No k produces B within 80–90 strictly. Why Other Options Are Wrong: The intervals 20–30, 40–50, 70–80 each contain at least one attainable multiple of 12k with k ≤ 6.666…, but 80–90 does not. Common Pitfalls: Allowing k to vary separately for each student or forgetting that all three marks must share the same scaling factor. Final Answer: 80 - 90

Question 14

Linear increment chain and ratio: A total of ₹ 300 is divided among P, Q, and R such that Q gets ₹ 30 more than P, and R gets ₹ 60 more than Q. Determine the ratio of their shares (P : Q : R) and verify against the total.

Accepted Answer

Introduction / Context: When sequential differences are given, define one variable and express the others relatively. The ratio emerges naturally after solving for the base amount using the total sum constraint, and then simplifying the three values to a ratio. Given Data / Assumptions: Total = ₹ 300. Q = P + 30; R = Q + 60 = P + 90. Concept / Approach: Let P = p. Then Q = p + 30 and R = p + 90. Use p + (p + 30) + (p + 90) = 300 to find p. Then compute P : Q : R and simplify to lowest terms. Step-by-Step Solution: 3p + 120 = 300 ⇒ 3p = 180 ⇒ p = 60.P, Q, R = 60, 90, 150 ⇒ ratio = 60 : 90 : 150.Divide by 30 ⇒ 2 : 3 : 5. Verification / Alternative check: Check differences: Q − P = 30 and R − Q = 60; sum is 60 + 90 + 150 = 300, consistent. Why Other Options Are Wrong: 3 : 2 : 5, 2 : 5 : 3, 5 : 3 : 2, and 4 : 5 : 6 do not meet the stated difference conditions simultaneously. Common Pitfalls: Reversing the difference directions (e.g., subtracting rather than adding) or forgetting to apply both increments before forming the ratio. Final Answer: 2 : 3 : 5

Question 15

Chained percentages into a three-term ratio: Salary of Mr. X is 80% of Mr. Y’s salary, and salary of Mr. Z is 120% of Mr. X’s salary. What is the ratio X : Y : Z? Select the correct option or “None of these” if no option matches.

Accepted Answer

Introduction / Context: Converting chained percentage relationships to a direct ratio requires expressing each person’s salary in terms of a single base (here, Y). Then scale to integers and compare with the given options. If none match, “None of these” is appropriate. Given Data / Assumptions: X = 80% of Y ⇒ X = 0.8Y. Z = 120% of X ⇒ Z = 1.2X. Concept / Approach: Substitute X into Z to express all in terms of Y, then read off X : Y : Z and simplify. Compare with the listed choices. Step-by-Step Solution: X = 0.8Y.Z = 1.2X = 1.2 * 0.8Y = 0.96Y.Therefore X : Y : Z = 0.8Y : 1Y : 0.96Y = 0.8 : 1 : 0.96.Scale by 25 to clear decimals ⇒ 20 : 25 : 24 (no further simplification). Verification / Alternative check: 20 : 25 : 24 equals 0.8 : 1 : 0.96. Comparing to options shows no exact match; thus “None of these”. Why Other Options Are Wrong: 4 : 5 : 6 corresponds to 0.8 : 1 : 1.2 (Z too large). Other options similarly fail to represent Z = 0.96Y. Common Pitfalls: Assuming Z is 120% of Y directly or mis-scaling decimals when converting to integers. Final Answer: None of these

Question 16

Two-digit number with digit swap: When the digits of a two-digit number are interchanged, the new number is smaller by 45 than the original. Also, the ratio (new : original) is 3 : 8. Find the original number.

Accepted Answer

Introduction / Context: Let the original number be 10a + b and the swapped number be 10b + a. The problem gives both a difference and a ratio condition, which combine to determine the original number uniquely using algebraic substitution. Given Data / Assumptions: New = Original − 45. New : Original = 3 : 8. Concept / Approach: Translate the ratio to New = (3/8)*Original and combine with the difference condition. Solve for Original. Then verify that the digits indeed swap correctly to satisfy both conditions. Step-by-Step Solution: Let O = original, N = new.N = O − 45 and N = (3/8)O ⇒ (3/8)O = O − 45.O − (3/8)O = 45 ⇒ (5/8)O = 45 ⇒ O = 72. Verification / Alternative check: New = 27. Difference: 72 − 27 = 45. Ratio: 27 : 72 = 3 : 8. Both conditions hold, digits 7 and 2 swap as expected. Why Other Options Are Wrong: 61, 94, and 54 do not satisfy the combined difference and ratio conditions upon swapping digits; only 72 meets both. Common Pitfalls: Using 8 : 3 instead of 3 : 8, or forgetting that the swapped number must be smaller according to the problem statement. Final Answer: 72

Question 17

Two relations on three shares: ₹ 5625 is divided among A, B, and C such that A receives half as much as B and C together, and B receives one-fourth as much as A and C together. Find (A + B).

Accepted Answer

Introduction / Context: With two linear relations among three shares and a known total, set up simultaneous equations and solve. Express one or two shares in terms of a single variable to simplify the algebra and then compute the requested sum (A + B). Given Data / Assumptions: A = (1/2)(B + C). B = (1/4)(A + C). A + B + C = 5625. Concept / Approach: From A = 0.5(B + C), write B in terms of A and C; combine with B = 0.25(A + C) to link A and C directly. Then find A, B, and hence A + B using the total. Step-by-Step Solution: A = (1/2)(B + C) ⇒ 2A = B + C ⇒ B = 2A − C.B = (1/4)(A + C) ⇒ 2A − C = (1/4)(A + C).Multiply by 4: 8A − 4C = A + C ⇒ 7A = 5C ⇒ C = (7/5)A.Then B = 2A − C = 2A − (7/5)A = (3/5)A.Total: A + B + C = A + (3/5)A + (7/5)A = 3A = 5625 ⇒ A = 1875; B = 1125.A + B = 1875 + 1125 = ₹ 3000. Verification / Alternative check: Check relations: A = 0.5(B + C) ⇒ 1875 = 0.5(1125 + 2625) = 0.5*3750 = 1875; B = 0.25(A + C) ⇒ 1125 = 0.25(1875 + 2625) = 0.25*4500 = 1125. Why Other Options Are Wrong: ₹ 5000, ₹ 15000, ₹ 9000, ₹ 2500 contradict at least one relation or the total when computed. Common Pitfalls: Treating “half as much as” incorrectly or forgetting that both equations must hold simultaneously with the same A, B, C. Final Answer: ₹ 3000

Question 18

Linear offsets in a three-way split: ₹ 710 is divided among A, B, and C such that A has ₹ 40 more than B and C has ₹ 30 more than A. Determine C’s share exactly.

Accepted Answer

Introduction / Context: When amounts differ by fixed increments, assign a base variable and build the others from it. Sum to the total to find the base, then recover each share directly. This linear approach is quick and robust. Given Data / Assumptions: A = B + 40. C = A + 30 = B + 70. A + B + C = 710. Concept / Approach: Let B = x. Then A = x + 40 and C = x + 70. Sum and solve for x. Then compute C. Step-by-Step Solution: x + (x + 40) + (x + 70) = 710 ⇒ 3x + 110 = 710.3x = 600 ⇒ x = 200.C = x + 70 = 270. Verification / Alternative check: A = 240, B = 200, C = 270. Differences: A − B = 40 and C − A = 30; sum 240 + 200 + 270 = 710. Why Other Options Are Wrong: ₹ 300, ₹ 135, ₹ 235, ₹ 250 do not fit the linear offsets with the total 710 when computed. Common Pitfalls: Misreading “more than” or forgetting to add both offsets when expressing C in terms of B. Final Answer: ₹ 270

Question 19

Fare ratio vs passenger ratio with total collection: The 1st-class to 2nd-class train fare ratio between two stations is 3 : 1. The number of passengers in 1st : 2nd class is 1 : 50. If ₹ 2650 was collected in total on a day, find the amount collect…

Accepted Answer

Introduction / Context: Revenue equals fare * number of passengers. With two classes, express each revenue using the given ratios, sum to the total, and solve for the fare unit. Then compute the second-class revenue directly from its share. Given Data / Assumptions: Fare ratio (1st : 2nd) = 3 : 1 ⇒ fares = 3k and k. Passenger ratio (1st : 2nd) = 1 : 50. Total collection = ₹ 2650. Concept / Approach: Revenue_1st = 3k * 1 = 3k; Revenue_2nd = k * 50 = 50k. Total = 3k + 50k = 53k. Solve for k and then find the second-class revenue 50k. Step-by-Step Solution: 53k = 2650 ⇒ k = 2650 / 53 = 50.Second-class collection = 50k = 50 * 50 = ₹ 2500. Verification / Alternative check: First-class collection = 3k = 150; 150 + 2500 = 2650, matching the total. Why Other Options Are Wrong: ₹ 2800, ₹ 3000, ₹ 3500, ₹ 2000 are not equal to 50/53 of the total and fail the ratio constraints. Common Pitfalls: Multiplying the fare and passenger ratios together incorrectly; remember revenue is the product for each class, then sum. Final Answer: ₹ 2500

Question 20

Class sections A and B have different student counts. If 10 students shift from section B to section A, the strength of A becomes 3 times the strength of B. However, if instead 10 students shift from section A to section B, then both sections become…

Accepted Answer

Introduction / Context: This classic ratio puzzle tests simultaneous linear equations and careful reading of shifting scenarios between two groups. We must form equations for two different transfers and solve for the original strengths of sections A and B. Given Data / Assumptions: Initially, let A = a and B = b. Scenario 1: 10 move B → A, so A becomes 3 times B: a + 10 = 3(b − 10). Scenario 2: 10 move A → B, so A and B become equal: a − 10 = b + 10. All numbers are whole students, and transfers are feasible. Concept / Approach: Translate each scenario into an equation and solve the two-variable system. Finally, express the ratio A : B in simplest terms. Step-by-Step Solution: From equality scenario: a − 10 = b + 10 ⇒ a − b = 20 ⇒ a = b + 20. From triple scenario: a + 10 = 3(b − 10) ⇒ a + 10 = 3b − 30 ⇒ a = 3b − 40. Equate a: b + 20 = 3b − 40 ⇒ 2b = 60 ⇒ b = 30. Then a = b + 20 = 50. Hence A : B = 50 : 30 = 5 : 3. Verification / Alternative check: Move 10 from B to A → A = 60, B = 20 ⇒ A is 3× B (60 vs 20). Move 10 from A to B → A = 40, B = 40 ⇒ equal. Both conditions hold. Why Other Options Are Wrong: 2 : 1, 3 : 1, 9 : 4, and 7 : 5 fail at least one scenario when tested, so they cannot be the initial ratio. Common Pitfalls: Mixing the direction of transfer or forgetting to subtract/add the 10 in the correct place. Always update both groups consistently. Final Answer: 5 : 3

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