Free Topic-Wise General Studies MCQs
Build a strong foundation with Ranking, Ordering, and Comparisons MCQs. High-yield practice covering relative positioning and analytical reasoning to boost your readiness for the UPSC CSAT examination.
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Explanation: Use the rank conversion formula: Rank from bottom = Total students โ Rank from top + 1. Substitute the given values and compute directly.
Explanation: First determine B's original position from the left using: Position from left = Total โ Position from right + 1. After interchange, B occupies A's original position. Then convert B's new left position to a right position using the same formula rearranged: Position from right = Total โ New position from left + 1.
Explanation: Apply the overlap formula for dual ranking: Total number of students = Rank from left + Rank from right โ 1. This accounts for the student being counted in both directions.
Explanation: Convert both positions to the same directional reference (either both from left or both from right). Then apply: Students between = |Position of A โ Position of B| โ 1.
Explanation: First compute the total weight of the original five boxes using Total = Average ร Count. Then compute the new total weight with six boxes. The sixth box's weight equals the difference between the two totals. Compare this value with the known maximum of the original set.
Explanation: Build the complete age chain by transitivity: start with the direct comparisons and link them sequentially. C > A and A > B gives C > A > B. D > C extends this to D > C > A > B. E < B places E at the end. The complete order is D > C > A > B > E.
Explanation: Construct the height hierarchy by chaining the direct comparisons using transitivity. R > P and P > Q yields R > P > Q. S > R extends to S > R > P > Q. T > S gives T > S > R > P > Q. U < Q places U below Q. The full order is T > S > R > P > Q > U.
Explanation: First compute the total of A, B, and C using Total = Average ร 3. Subtract C's score to obtain A + B. Since A > B and both are integers, deduce the maximum possible value of B. Then apply the chain C > A > B > D > E to find the maximum integer value D can take while remaining strictly less than B and strictly greater than E.
Explanation: Determine the number of students above A in each subject. Then apply the principle of inclusion-exclusion for the minimum overlap: Minimum in both = (Above in Math) + (Above in Science) โ (Total other students). This ensures the union does not exceed the available pool.
Explanation: First determine the day of the week for the 15th by counting forward from the given reference day. Then calculate the number of days between the 15th and the 31st. Find the remainder when divided by 7 and advance that many days from the 15th's day.
Explanation: Compute each departure time sequentially. Add or subtract the given intervals from the reference times. Convert all times to a common format (minutes since midnight) for comparison, then identify the maximum value.
Explanation: Calculate X's time to complete the full race using Time = Distance รท Speed. Y covers 100 metres less distance and finishes 20 seconds later than X. Apply Speed = Distance รท Time to Y's run, using Y's distance and Y's total time.
Explanation: Represent the present ages as 3k and 5k using the given ratio. After 8 years, both ages increase by 8. Set up the new ratio equation, cross-multiply to eliminate fractions, solve for k, then multiply by 3 to obtain A's present age.
Explanation: Apply the inclusion-exclusion principle for percentages: Percentage passing at least one subject = Physics% + Chemistry% โ Both%. Equate this percentage to the given absolute number, then solve for the total by dividing the absolute number by the resulting percentage (converted to decimal).
Explanation: Choose a convenient base value for C's income (such as 100). Compute B's income by increasing the base by 25%. Then compute A's income by increasing B's result by 20%. Finally, compare A's income to C's base using the percentage increase formula: ((A โ C) / C) ร 100.
Explanation: Compare each fraction to 1 by computing the difference 1 โ fraction. The fraction with the smallest difference from 1 is the largest. Convert all differences to a common denominator for direct comparison.
Explanation: Align all numbers to the same number of decimal places by appending zeros where needed. Then compare digit by digit from left to right. After sorting, identify the element at the middle position of the odd-length list.
Explanation: Calculate the cutoff rank for Grade A by finding 25% of the total number of students. Compare Ravi's rank to this cutoff. If his rank is numerically less than or equal to the cutoff, he falls in the top group.
Explanation: Compute the qualifying cutoff rank by taking 20% of the total students. The required improvement equals the difference between the student's current rank and the cutoff rank.
Explanation: Let the five numbers in ascending order be a, b, c, d, e. Use the given sums to express a + b and d + e. The total sum equals the average multiplied by 5. Substitute the known pair sums into the total equation and solve for c, which is the median of an odd-count ordered set.
Explanation: When students to the left of a given position leave, the position shifts left by exactly the number of students who departed. Subtract the number of departing students from the original rank.
Explanation: First determine B's original rank from the right using the formula: Right rank = Total โ Left rank + 1. When new students are added to the right end, the left rank of existing students remains unchanged but the total increases. Recompute the right rank with the new total, or simply add the number of new students to the original right rank.
Explanation: For seven numbers in ascending order, the median is the 4th number. Compute the sum of the first three using their average, the sum of the last three using their average, and add the median. Divide the total sum by 7 to obtain the overall average.
Explanation: When A completes 100 m, B completes 90 m. When B completes 100 m, C completes 90 m. Scale C's distance proportionally to B's 90 m: C covers 90 ร (90/100) = 81 m. Therefore, when A finishes 100 m, C has run 81 m. The gap is 100 โ 81.
Explanation: Chain the inequalities by transitivity. A > B and C > A gives C > A > B. D > C extends to D > C > A > B. E > D gives E > D > C > A > B. The leftmost element in the complete chain is the highest scorer.
Explanation: When students below a given rank leave, the rank from the top remains unchanged because the number of students above is unaffected. Compute the new total by subtracting the departing students from the original total. Then apply the conversion formula: Rank from bottom = New total โ Rank from top + 1.
Explanation: Factor the product of the two smallest numbers into pairs of positive integers, keeping only the pair where both are less than the median. Sum all five numbers and subtract the known values to obtain the sum of the two largest. Use the product and sum of the two largest to form a quadratic equation whose roots are those two numbers. Select the larger root.
Explanation: Let x be the percentage that passed in both. Express the percentage that passed only History as 70 โ x and only Geography as 60 โ x. The four groups (only H, only G, both, neither) must sum to 100%. Set up the equation and solve for x.
Explanation: Establish the height order by transitivity: T > S > R > P > Q > U. Compute the total height of all six from the overall average. Subtract the sums of P+R and S+T (derived from their averages) to obtain Q+U. Use the chain constraints Q < P and U < Q to bound Q. Since Q+U is fixed and U < Q, express U in terms of Q and find the maximum integer U that satisfies all strict inequalities.
Explanation: First find the combined work rate per day: 1/15 + 1/10 = 1/6. In 2 days working together, they complete 2 ร (1/6) = 1/3 of the work. The remaining work is 2/3. A's individual rate is 1/15 per day. Divide the remaining work by A's rate: (2/3) รท (1/15) = (2/3) ร 15 = 10 days.
Explanation: Divide the class into three equal groups by rank to determine the grade boundaries. Convert Rahul's bottom rank to a top rank using the conversion formula. Compare both students' top ranks to determine their relative standing and grade assignment.
Explanation: After interchange, A occupies B's original position. Therefore B's original left rank equals A's new left rank. After interchange, B occupies A's original position. Convert B's new right rank to a left rank and set it equal to A's original left rank. Solve for the total.
Explanation: Convert all positions to the same directional reference. Compute the gap between each adjacent pair using the formula: Students between = |Position difference| โ 1. Sum the individual gaps to obtain the combined count.
Explanation: Compute the total weight before and after using Total = Average ร Count. The difference gives the combined weight of the two people who left. Subtract the known heaviest person's weight to find the second person's weight. To minimize their rank from the heaviest, maximize the number of people heavier than them among the remaining six. Use the remaining total and count to verify how many can exceed this weight while keeping all weights positive.
Explanation: Establish the age order by transitivity: E > A > B > C > D. From the future sum condition, derive the present sum of A and B. Since A > B and both are integers, determine the maximum possible value of B. Then apply the constraint C < B along with C > D to find the maximum integer value C can take.
Explanation: Count the exact number of days from the reference date to the target date, accounting for the full month lengths of March (31) and April (30). Divide the total day count by 7 and advance the remainder from the starting day.
Explanation: Compute each train's travel time and derive their speeds as fractions of the total distance per hour. Determine how far the first train has traveled by the time the second train starts. The remaining distance is covered by both trains approaching each other; divide this remaining distance by the sum of their speeds to find the meeting time after the second train's departure.
Explanation: Two runners meet at the starting point only when the elapsed time is an exact multiple of both lap times. Compute the least common multiple of the two lap times by prime factorizing each, taking the highest power of each prime, and multiplying.
Explanation: Compare fractions by cross-multiplying pairs. For two fractions a/b and c/d, compare ad and bc. If ad < bc, then a/b < c/d. Systematically compare each fraction against the others to identify the minimum.
Explanation: Let the four numbers in descending order be a, b, c, d. Compute the sum of the first pair from their average and the sum of the last pair from their average. The total sum of all four numbers is simply the sum of these two pairs. Divide by 4 to obtain the overall average.
Explanation: When students join at the left end, every existing student's left rank increases by that number. When students join at the right end, the total increases but existing students' left ranks are unaffected. Compute the new left rank and new total, then apply the conversion formula. Alternatively, add only the right-end arrivals to the original right rank.
Explanation: Represent the four ages in arithmetic progression as a, a+d, a+2d, a+3d. Set the youngest equal to the given value. Express the future ages of the eldest and youngest, compute their average, and equate it to the present age of the second eldest. Solve for the common difference d, then compute the eldest sibling's present age.
Explanation: Express the three numbers as 2k, 3k, and 5k using the given ratio. Square each term and sum them. Set the result equal to the given total and solve for kยฒ, then take the positive root. Multiply by 5 to obtain the largest number.
Explanation: Compute the rank boundaries for each category by taking the respective percentages of the total. Determine the student's new rank after the improvement. Compare the new rank to the category boundaries to identify the new classification.
Explanation: Compute the total weight of all seven people. Compute the total after the heaviest leaves by multiplying the new average by the new count. The heaviest person's weight is the difference between the original total and this new total.
Explanation: When both positions are given from the same directional reference, subtract the smaller rank from the larger rank and subtract one additional unit to exclude the endpoints themselves.
Explanation: Calculate the cutoff rank for the merit certificate by finding 10% of the total number of students. The required improvement is the difference between the student's current rank and this cutoff rank.
Explanation: Compute each traveller's speed using Speed = Distance รท Time. Since they move in the same direction, the relative speed is the difference of their individual speeds. Divide the required lead distance by this relative speed to obtain the time.
Explanation: Compute the cutoff ranks for each category by taking one-fourth and one-third of the total. Determine the student's category in each subject separately by comparing their rank to these cutoffs. The better rank corresponds to the better category, which becomes the final classification.
Explanation: Convert all values to a common format, either all decimals or all fractions with a common denominator. For decimal conversion, divide the numerator by the denominator for each fraction. Then compare the decimal values digit by digit.
Explanation: After interchange, A occupies B's original position, so B's original left rank equals A's new left rank. Apply the rank conversion formula to B's original left rank to obtain B's original right rank: Original right = Total โ Original left + 1.
Explanation: Let the four numbers be a, b, c, d in ascending order. Compute the total sum from the average. Subtract the sum of the first two to obtain c + d. Use the given product of the last two to form a quadratic equation. Solve for c and d, selecting the smaller root as the third number.
Explanation: The 85th percentile indicates that 85% of the students scored below him. Calculate this number: 85% of 200 = 170 students. Since 170 students are below him, his rank from the bottom is 171st. Convert this to a top rank using the standard formula: Total students โ Rank from bottom + 1 (200 โ 171 + 1 = 30).
Explanation: Compute each runner's speed in metres per minute. For same-direction motion on a circular track, the faster runner gains one full lap when the distance gained equals the track length. Divide the track length by the relative speed to find the time required, then divide by the faster runner's lap time.
Explanation: When students leave from the left end, subtract that number from the original left rank. When students leave from the right end, subtract that number from the original right rank. Alternatively, compute the new total and new left rank, then apply the conversion formula.
Explanation: For seven values in ascending order, the median is the 4th value. Compute the total height of all seven from the overall average. Compute the total of the tallest three and the shortest three from their respective averages. Subtract both from the grand total to isolate the median.
Explanation: Track the position of the target student through each interchange step by step. After the first swap, note the intermediate positions. After the second swap, determine the final left position. Convert this final left position to a right position using the total count.
Explanation: Express the three numbers as 4k, 5k, and 6k. Write the given condition as an equation: (Largest + Smallest) = Middle + 50. Substitute the expressions in terms of k and solve. Then compute the sum of all three terms.
Explanation: Identify the average of the remaining 36 students. Add the given margin to obtain the 13th ranked student's marks. Subtract this value from the average of the top 12 to find the required difference.
Explanation: Establish the mark order: E > D > C > A > B. Compute the totals of the two groups from their averages. From the second group, derive the constraint D + E = 240 โ C. Since D > C and E > D with distinct integers, D + E โฅ 2C + 3, which bounds C from above. From the first group, A + B = 210 โ C. To maximize B, minimize A and C while maintaining A > B and C > A. Test the boundary values to find the maximum integer B that satisfies all strict inequalities and the total mark constraint.