Question 1

Seven boys A, B, C, D, E, F and G are standing in a straight line. The following information is given: (i) G stands somewhere between A and E. (ii) F and A have exactly one boy standing between them. (iii) E and C have exactly two boys standing betw…

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Introduction / Context: This is a multi-condition linear arrangement puzzle involving seven boys. The problem provides several constraints about relative positions, such as “between”, “immediately to the right of” and “with a fixed number of boys between”. We must integrate all these conditions to find the unique valid arrangement and then identify who stands second from the left. Given Data / Assumptions: Boys: A, B, C, D, E, F and G. They stand in a straight line facing the same direction. (i) G stands between A and E (either A – G – E or E – G – A). (ii) Exactly one boy stands between F and A. (iii) Exactly two boys stand between E and C. (iv) D is immediately to the right of F. (v) Exactly three boys stand between C and B. Concept / Approach: We treat the line as seven positions from left to right. Each condition restricts possible positions. We often start with constraints involving exact numbers of people between, because they sharply limit positions. By trial, elimination and cross-checking, we arrive at a single arrangement that satisfies all statements. Step-by-Step Solution: Step 1: Label the seven positions from left to right as 1, 2, 3, 4, 5, 6 and 7. Step 2: Condition (v) says there are exactly three boys between C and B, so their positions differ by 4 (for example, 1 and 5, 2 and 6 or 3 and 7 in either order). Step 3: Condition (iii) says there are exactly two boys between E and C, so their positions differ by 3. Step 4: Using these two conditions together significantly restricts which positions C, B and E can take. A consistent solution emerges when C, A, etc., are placed so that all differences match. Step 5: Condition (ii) says F and A have exactly one boy between them, so their positions differ by 2. Condition (iv) states that D is immediately to the right of F, so D sits one position to the right of F. Step 6: Finally, condition (i) requires that G is between A and E in the line. Step 7: When all constraints are applied systematically, the only arrangement that fits every condition is: B, E, G, A, C, F, D from left to right. Verification / Alternative check: Check each condition against B, E, G, A, C, F, D. G is between E and A, so condition (i) holds. There is exactly one boy (G) between F and A because A is in position 4 and F in position 6 with C in between, giving a gap of two, satisfying condition (ii). There are exactly two boys between E (position 2) and C (position 5), satisfying condition (iii). D is immediately right of F (positions 6 and 7), satisfying (iv). Finally, there are exactly three boys between C (position 5) and B (position 1), satisfying condition (v). All clues are satisfied. Why Other Options Are Wrong: In the final arrangement, position 2 from the left is occupied by E, not by C, G or A. Options A, C and G refer to boys in positions 4, 5 and 3 respectively, so none of them is second from the left. Common Pitfalls: Students often overlook at least one condition when trying arrangements, leading to partially correct but invalid orders. Another common issue is miscounting “number of boys between”, which counts only boys strictly in between, not the endpoints themselves. Writing position numbers and checking each condition explicitly helps maintain accuracy. Final Answer: The boy standing second from the left is E.

Question 2

A man is hired to plant trees for a continuous period of 10 hours. He can plant 10 trees in every hour of work, but after each hour of planting he takes a rest for 30 minutes before resuming. Assuming this pattern continues throughout the 10-hour pe…

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Introduction / Context: This time and work question involves periodic rest intervals. The man plants trees at a constant rate, but he does not work continuously, because he takes a break after every hour of planting. The challenge is to correctly interpret the 10-hour period and count how many full working hours fit into it, rather than simply multiplying rate by 10. Given Data / Assumptions: Total engagement period for the job = 10 hours. During each full working hour, the man plants 10 trees. After every 1 hour of planting, he takes a rest of 30 minutes before starting the next working hour. The 10-hour period includes both working time and rest time. Concept / Approach: Each complete cycle consists of 1 hour of work followed by 0.5 hour of rest, for a total of 1.5 hours. During each cycle he plants 10 trees. We must find how many full working hours, and thus how many cycles, fit into 10 hours. In the final part of the period, if there is enough time for an additional working hour but not for another rest, he is still able to work that extra hour, as rest is taken after completing an hour of planting. Step-by-Step Solution: Step 1: One work–rest cycle length is 1 hour work + 0.5 hour rest = 1.5 hours. Step 2: In each such cycle he plants 10 trees. Step 3: Compute how many full cycles fit into 10 hours: 10 / 1.5 ≈ 6 full cycles (since 6 * 1.5 = 9 hours). Step 4: After 6 complete cycles, he has used 9 hours and planted 6 * 10 = 60 trees. Step 5: There is still 1 hour remaining (10 - 9 = 1 hour). He can use this remaining hour for planting, because the rest comes only after completing the hour of work. Step 6: In this final hour he plants another 10 trees. Total trees planted = 60 + 10 = 70. Verification / Alternative check: We can create a rough timetable. He works from hour 0 to 1, rests from 1 to 1.5, works again from 1.5 to 2.5, rests from 2.5 to 3, and so on. After six such cycles he reaches time 9 hours. From 9 to 10 hours he can perform one more full working hour. This matches the cycle-based calculation and confirms that he completes 7 working hours in the 10-hour period. Why Other Options Are Wrong: Option 100 assumes he works all 10 hours without rest, ignoring the breaks. Option 50 would correspond to only five working hours, which undercounts the available time. Option 45 corresponds to 4.5 hours of work and does not align with the given schedule pattern of full working hours followed by breaks. Common Pitfalls: A very common mistake is to multiply 10 hours by 10 trees per hour and ignore rest, or to assume every 1.5 hours is a fixed cycle without considering that the last working hour does not need a break within the given time window. Carefully modeling the timeline helps avoid these errors. Final Answer: The man plants a total of 70 trees in 10 hours.

Question 3

The weights of four boxes are 40 kg, 70 kg, 80 kg and 50 kg. Each box can be used at most once in any combination. Which of the following values cannot be the total weight, in kilograms, of any possible combination of these boxes?

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Introduction / Context: This final problem in the set continues the pattern of subset-sum reasoning with four given box weights. Again, we must determine which listed total weight cannot be obtained by selecting some or all of the boxes, with each used at most once. The question tests careful enumeration and the ability to rule out impossible totals. Given Data / Assumptions: Box weights: 40 kg, 70 kg, 80 kg and 50 kg. Each box can be used at most once. Possible combinations include one, two, three or all four boxes. We must decide which given total weight is not achievable. Concept / Approach: As with previous questions of this type, the straightforward method is to list all possible sums of the four weights. Because there are only four boxes, the total number of non-empty subsets is limited and manageable. Once we have the full list, we compare it against the answer options and pick the value that does not appear. Step-by-Step Solution: Step 1: Single-box totals are 40, 50, 70 and 80 kilograms. Step 2: Two-box totals: 40 + 50 = 90, 40 + 70 = 110, 40 + 80 = 120, 50 + 70 = 120, 50 + 80 = 130 and 70 + 80 = 150 kilograms. Step 3: Three-box totals: 40 + 50 + 70 = 160, 40 + 50 + 80 = 170, 40 + 70 + 80 = 190 and 50 + 70 + 80 = 200 kilograms. Step 4: Four-box total: 40 + 50 + 70 + 80 = 240 kilograms. Step 5: Collect all distinct totals: 40, 50, 70, 80, 90, 110, 120, 130, 150, 160, 170, 190, 200 and 240 kilograms. Step 6: Compare these with options 240, 160, 200 and 220. Only 220 kilograms is missing from the list, so it cannot be formed. Verification / Alternative check: To verify, try constructing 220 kg directly. Three-box totals are 160, 170, 190 and 200 kg, none of which equals 220. Adding the remaining box to any three-box combination always yields either 240 kg (the four-box total) or a value greater than 220, which is not possible here. Therefore, 220 kg cannot be obtained by any allowed combination. Why Other Options Are Wrong: Option 240 kg is the sum of all four box weights: 40 + 50 + 70 + 80. Option 160 kg is achieved as 40 + 50 + 70. Option 200 kg is achieved as 50 + 70 + 80. Common Pitfalls: Examinees may overlook one of the triple combinations or may incorrectly assume that if a number lies between the smallest and largest possible totals it must be achievable. This is not true when selections are limited to specific discrete weights. Careful calculation of each combination prevents such errors. Final Answer: The total weight that cannot be formed is 220 kilograms.

Question 4

The weights of four boxes are 70 kg, 50 kg, 30 kg and 90 kg. Which of the following cannot be obtained as the total weight, in kilograms, of any combination of these boxes if each box is used at most once?

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Introduction / Context: This arithmetic reasoning question tests your understanding of how to form different possible totals by combining a small set of distinct weights. You are given four box weights and asked to decide which one of the given totals cannot be formed when each box can be used at most once. Such questions are very common in reasoning and bank or SSC type competitive exams, and they quietly check both systematic counting and basic number sense. Given Data / Assumptions: Available box weights: 70 kg, 50 kg, 30 kg, 90 kg. Each box can be used at most one time in a combination. We must check the totals: 220 kg, 240 kg, 210 kg, 170 kg. We assume there are no hidden conditions such as partial use of a box. Every used box contributes its full given weight. Concept / Approach: The key idea is to list or logically derive all distinct possible sums that can be formed by adding these four weights using each box at most once. For four numbers, there are combinations of one box, two boxes, three boxes, and all four boxes. Once we know these sums, we simply check which option is missing from this list. Working systematically avoids missing any combination or double counting a total. Step-by-Step Solution: Step 1: Write the weights: 70, 50, 30, 90.Step 2: Sums using one box: 30, 50, 70, 90.Step 3: Sums using two boxes: 30 + 50 = 80, 30 + 70 = 100, 30 + 90 = 120, 50 + 70 = 120, 50 + 90 = 140, 70 + 90 = 160.Step 4: Sums using three boxes: 30 + 50 + 70 = 150, 30 + 50 + 90 = 170, 30 + 70 + 90 = 190, 50 + 70 + 90 = 210.Step 5: Sum using all four boxes: 30 + 50 + 70 + 90 = 240.Step 6: Collect all distinct sums: 30, 50, 70, 80, 90, 100, 120, 140, 150, 160, 170, 190, 210, 240.Step 7: Compare the options: 220, 240, 210, 170 with this list of possible totals. Verification / Alternative check: A quick alternative check is to see whether 220 can be written as a sum of any subset of the four numbers. Try the largest combinations: 90 + 70 + 50 = 210, which is already less than 220 and adding 30 would overshoot to 240. Next, 90 + 70 + 30 = 190, and adding 50 gives 240. Other pairs: 90 + 50 = 140, 90 + 70 = 160, 90 + 30 = 120, 70 + 50 = 120, 70 + 30 = 100, 50 + 30 = 80. None of these give 220, confirming that 220 is not reachable. Why Other Options Are Wrong: 240 kg is possible by using all four boxes: 30 + 50 + 70 + 90 = 240.210 kg is possible by using 50 kg, 70 kg and 90 kg: 50 + 70 + 90 = 210.170 kg is possible by using 30 kg, 50 kg and 90 kg: 30 + 50 + 90 = 170.The option that says None of these totals is incorrect because some of the given totals are indeed achievable. Common Pitfalls: Many learners try random additions and may miss a combination, leading them to wrongly believe that a valid sum is impossible or an impossible sum is valid. Another mistake is to assume that the question asks for the largest or smallest total, rather than the one that cannot be formed. Some students also misread the condition about using each box at most once and may accidentally reuse a weight, which completely changes the set of possible totals. Final Answer: The only total that cannot be formed from any combination of the given four box weights is 220 kilograms.

Question 5

The weights of four boxes are 20 kg, 40 kg, 80 kg and 90 kg. Which of the following cannot be obtained as the total weight, in kilograms, of any combination of these boxes if each box is used at most once?

Accepted Answer

Introduction / Context: This problem again focuses on reasoning with combinations of fixed weights. You are given four distinct box weights and asked to determine which of the listed totals cannot be produced if each weight is used at most one time. Such questions are popular in aptitude tests because they examine systematic thinking rather than simply mechanical calculation. Given Data / Assumptions: Box weights: 20 kg, 40 kg, 80 kg, 90 kg. Each box can be used at most once in any combination. Candidate totals: 220 kg, 230 kg, 150 kg, 210 kg. No partial weights are allowed; a box is either fully used or not used at all. Concept / Approach: We should list the distinct sums that can be formed by using one, two, three, or all four boxes. For four numbers, the total number of non empty subsets is limited, so it is feasible to enumerate them or at least reason through them carefully. Once we know the set of possible totals, we only need to check which option is missing from that set. This avoids guesswork and ensures a logically correct conclusion. Step-by-Step Solution: Step 1: Note the weights: 20, 40, 80, 90.Step 2: Sums using one box: 20, 40, 80, 90.Step 3: Sums using two boxes: 20 + 40 = 60, 20 + 80 = 100, 20 + 90 = 110, 40 + 80 = 120, 40 + 90 = 130, 80 + 90 = 170.Step 4: Sums using three boxes: 20 + 40 + 80 = 140, 20 + 40 + 90 = 150, 20 + 80 + 90 = 190, 40 + 80 + 90 = 210.Step 5: Sum using all four boxes: 20 + 40 + 80 + 90 = 230.Step 6: Collect all distinct sums: 20, 40, 60, 80, 90, 100, 110, 120, 130, 140, 150, 170, 190, 210, 230.Step 7: Compare each option with this set of sums. Verification / Alternative check: Check each option directly. For 150, we already saw 20 + 40 + 90 = 150, so 150 is possible. For 210, we have 40 + 80 + 90 = 210, so that is possible. For 230, using all four boxes 20 + 40 + 80 + 90 gives 230, so 230 is also possible. For 220, try to write it as a sum of these weights. The largest three weights give 40 + 80 + 90 = 210, and adding 20 gives 230, which is too large. No other combination reaches exactly 220, so it is not obtainable. Why Other Options Are Wrong: 150 kg is achievable by the combination 20 + 40 + 90.210 kg is achievable by the combination 40 + 80 + 90.230 kg is achievable by using all four boxes: 20 + 40 + 80 + 90.Therefore, these totals cannot be the answer to the question about impossibility. Common Pitfalls: Learners sometimes overlook a valid combination such as 20 + 40 + 90 or 40 + 80 + 90 and may wrongly conclude that those totals are impossible. Another common error is assuming that if a number is close to the sum of all four weights then it must be achievable, without actually checking each subset. Approaching the problem randomly instead of systematically is the main reason for mistakes in this type of question. Final Answer: The only total that cannot be obtained from any combination of the four given box weights is 220 kilograms.

Question 6

At a college party, five girls are sitting in a row. P is to the left of M and to the right of O, and R is sitting to the right of N but to the left of O. Who is sitting exactly in the middle position of the row?

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Introduction / Context: This is a classic linear arrangement puzzle from verbal reasoning. The question tests your ability to interpret positional relationships, visualise a row of people and deduce who occupies a specific position, here the middle seat. Such problems frequently appear in competitive exams to assess logical thinking and careful reading of conditions. Given Data / Assumptions: There are exactly five girls sitting in a single row. The girls are P, M, O, R and N. P is to the left of M. P is to the right of O, so the order is O, then P, then M from left to right. R is to the right of N. R is to the left of O. We assume all five girls occupy distinct positions with no empty seats in between. Concept / Approach: In linear arrangement questions, it is helpful to convert each sentence into partial order information. Once we know that O is left of P and P is left of M, we can group them as O P M from left to right. The information about R and N relative to O lets us fit them into the sequence. When all conditions are respected, the complete order clearly reveals who sits in the middle position, which is the third seat in a row of five. Step-by-Step Solution: Step 1: From P is to the left of M and to the right of O, we obtain the sequence O P M from left to right.Step 2: From R is to the right of N but to the left of O, we obtain N R O in that order from left to right.Step 3: Combine both conditions. If N R O must come in that order and O P M must also appear in order, then the full arrangement is N R O P M from left to right.Step 4: In a row of five seats labelled 1 to 5 from left to right, the middle seat is seat number 3.Step 5: In the sequence N R O P M, the third position is occupied by O. Verification / Alternative check: You can test this ordering by reading the conditions back. P is to the right of O and to the left of M in N R O P M, which is correct. R is to the right of N and to the left of O, which is also correct. There is no alternative arrangement with five distinct seats that satisfies both sets of constraints simultaneously. Therefore, N R O P M is the unique valid sequence up to reversal, and the middle position is firmly fixed as O. Why Other Options Are Wrong: R is in the second position from the left, not the middle.P is in the fourth position from the left, so it is to the right of the middle seat.M is in the fifth position from the left, and cannot be in the middle.N is in the first position from the left and is clearly not in the middle. Common Pitfalls: Students often misread left and right or forget that left and right are always judged with respect to a fixed direction for the row, not from the perspective of each person. Another frequent error is to guess the answer without fully combining the two separate sets of conditions, which can lead to choosing P or R incorrectly. Drawing a simple diagram with five boxes and filling them according to the constraints removes most of these mistakes. Final Answer: The girl sitting in the exact middle of the row is O.

Question 7

Ram leaves his house at 20 minutes to seven in the morning and reaches the house of Kunal in 25 minutes. They finish their breakfast in another 15 minutes and then leave together for their office, which takes another 35 minutes to reach. At what tim…

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Introduction / Context: This question is a straightforward time calculation problem. It checks whether you can correctly interpret phrases like 20 minutes to seven, add travel and activity durations, and identify the exact clock time when a particular event occurs. These types of problems appear frequently in arithmetic reasoning sections of competitive exams. Given Data / Assumptions: Ram leaves his house at 20 minutes to seven in the morning. 20 minutes to seven means 6:40 A.M. He takes 25 minutes to reach the house of Kunal. They spend 15 minutes on breakfast at Kunal's house. After breakfast they leave for their office and that final travel takes 35 minutes, but we are only asked about the time when they leave the house of Kunal. Concept / Approach: The core approach is to convert all time expressions into standard clock times and then add or subtract durations step by step. We carefully track the timeline: departure from Ram's house, arrival at Kunal's house, completion of breakfast, and final departure for the office. Since the question asks specifically for the departure time from the house of Kunal, we focus on the moment just after breakfast finishes. Step-by-Step Solution: Step 1: 20 minutes to seven means 7:00 A.M minus 20 minutes, which is 6:40 A.M.Step 2: Ram takes 25 minutes to reach Kunal, so arrival time is 6:40 A.M + 25 minutes = 7:05 A.M.Step 3: They spend 15 minutes having breakfast at Kunal's house.Step 4: Breakfast ends at 7:05 A.M + 15 minutes = 7:20 A.M.Step 5: Immediately after breakfast, they leave for their office. Therefore, the departure time from Kunal's house is 7:20 A.M. Verification / Alternative check: To verify, construct a full timeline including the last travel leg. From 7:20 A.M they take 35 minutes to reach the office, so they arrive at 7:55 A.M. This value is not directly needed for the answer, but it confirms the arithmetic is consistent. There are no hour crossings between 6:40 and 7:55 that would cause confusion, so the calculations remain simple and stable. Why Other Options Are Wrong: 7:40 A.M would be 20 minutes after they leave Kunal's house, which does not match the 15 minute breakfast duration.7:45 A.M would imply a 40 minute breakfast, which contradicts the given 15 minutes.8:15 A.M is much too late and could be confused with the possible time of office arrival if calculations were done incorrectly.7:30 A.M assumes either a different departure time or incorrect addition of minutes. Common Pitfalls: Many students misinterpret 20 minutes to seven as 7:20 A.M instead of 6:40 A.M, which reverses the meaning of the phrase. Another error is mixing up which time the question is asking for, such as accidentally giving the office arrival time instead of the departure time from Kunal's house. Writing down the timeline in order and labelling each event is an effective way to avoid confusion. Final Answer: They leave the house of Kunal to go to their office at 7:20 A.M.

Question 8

If A is taller than B, B is shorter than C and D is shorter than A, then which of the following statements is definitely true about their heights?

Accepted Answer

Introduction / Context: This is a verbal reasoning problem based on comparison of heights. The question checks your ability to interpret inequality statements like A is taller than B and combine them logically to identify which conclusion must always hold, regardless of the exact heights. Such problems are extremely common in reasoning sections of competitive exams because they test pure logical consistency rather than calculation. Given Data / Assumptions: A is taller than B. B is shorter than C, which also means C is taller than B. D is shorter than A. No information is given about the direct comparison of C with A or D, and no equal heights are explicitly mentioned. Concept / Approach: The approach is to rewrite each statement as a clear inequality and then see what relationships are guaranteed. We can write A taller than B as A greater than B. B shorter than C becomes B less than C. D shorter than A becomes D less than A. The task is to see which of the options is implied by these three inequalities in every possible arrangement that respects them. If an option could be false in some valid arrangement, it is not definitely true. Step-by-Step Solution: Step 1: Translate statements: A taller than B means A > B. B shorter than C means B < C. D shorter than A means D < A.Step 2: From A > B we know A is taller than B in every valid arrangement.Step 3: From D < A we know A is taller than D in every valid arrangement.Step 4: Therefore, A is taller than both B and D, so the statement A is taller than both B and D must always be true.Step 5: Examine A and C. Since we only know B < C and A > B, it is possible that C is taller than A or A is taller than C. Both scenarios can satisfy all the given inequalities, so no fixed relation between A and C is guaranteed.Step 6: There is no information linking D directly to C or B besides the relation through A, so conclusions involving D compared directly with B or C may not always hold. Verification / Alternative check: You can test with specific numerical heights. Take B = 150 cm, A = 160 cm, C = 170 cm, D = 155 cm. Here A is taller than B and D, and B is shorter than C. Now choose another valid arrangement: B = 150 cm, A = 160 cm, C = 155 cm, D = 140 cm. In both cases A is still taller than B and D, but in the first case C is taller than A, and in the second case A is taller than C. This confirms that A greater than B and D is fixed, while relations involving C can change. Why Other Options Are Wrong: D is taller than C is not guaranteed; we can pick C as very tall and D as quite short.A and C are of equal height is never stated and cannot be concluded.D is shorter than B is not forced, because D could be between A and B in height while still being shorter than A.C shorter than A is not guaranteed, because C can be chosen taller than A while still satisfying the conditions. Common Pitfalls: A typical mistake is to assume transitivity inappropriately, for example concluding that since B is between A and C in some imagined order, A and C must have a specific relation. Another error is to think that missing comparisons, such as between C and A, always follow a particular pattern, even when they are not constrained by the conditions. To avoid these issues, work with actual example values that satisfy the conditions and see which options always remain true. Final Answer: The statement that is definitely and always true is A is taller than both B and D.

Question 9

In a row of people, the position of Raju is 13th from the front side and 6th from the back side. How many people are standing in that row altogether?

Accepted Answer

Introduction / Context: This is a standard rank and position problem in arithmetic reasoning. You are given the position of one person from the front and from the back in a single row, and you must determine the total number of people in that row. Questions like this are common in exams and are based on a simple but very important formula. Given Data / Assumptions: Raju is 13th from the front. Raju is 6th from the back. All people stand in one straight row. Each person is counted only once, with no gaps or missing positions. Concept / Approach: When the position of the same person from the front and from the back is known, the total number of people in the row can be found using a simple relationship. That relationship is: total number of people = position from the front + position from the back - 1. The subtraction of 1 is necessary because the counted person is included in both the front and back counts. Step-by-Step Solution: Step 1: Note that Raju is 13th from the front, so there are 12 people in front of him.Step 2: Raju is 6th from the back, so there are 5 people behind him.Step 3: The total number of people is equal to number in front of Raju + Raju himself + number behind Raju.Step 4: That is 12 + 1 + 5 = 18.Step 5: Using the standard formula, total = 13 + 6 - 1 = 18, which matches the previous calculation. Verification / Alternative check: Imagine a row of 18 people numbered from 1 to 18 from the front. The 13th person from the front is at position 13. From the back, the positions are counted as 18, 17, 16, and so on. The person at position 13 from the front is at position 18 - 13 + 1 = 6 from the back, which exactly matches the given condition. This confirms that the total must be 18. Why Other Options Are Wrong: 21, 20 and 19 all violate the formula total = front position + back position - 1, which gives 18.For 21, the person that is 13th from the front would be 21 - 13 + 1 = 9th from the back, not 6th.For 20, the person that is 13th from the front would be 20 - 13 + 1 = 8th from the back.For 19, the person would be 7th from the back.22 similarly gives a mismatch between front and back positions. Common Pitfalls: Students often forget to subtract 1 in the formula and mistakenly add the two positions directly, giving 19 instead of 18 in this case. Another mistake is to confuse the number of people in front or behind with the rank itself. Carefully visualising that the same person is counted in both directions prevents these errors. Writing the small formula and plugging in the values is the fastest and safest method. Final Answer: The total number of people standing in that row is 18.

Question 10

A is taller than B. C is taller than A. D is taller than E but shorter than B. Who is the tallest among A, B, C, D and E?

Accepted Answer

Introduction / Context: This height comparison problem is a direct application of inequality reasoning. You are given several relative height statements and are asked to identify the tallest person among five individuals. It teaches you how to combine several pieces of comparative information into a single consistent ordering. Given Data / Assumptions: A is taller than B. C is taller than A. D is taller than E but shorter than B. We have five people in total: A, B, C, D and E. No two people are stated to have equal heights. Concept / Approach: We translate the verbal comparison statements into inequalities and then build a height order. When we see phrases like C is taller than A and A is taller than B, we can chain them to deduce C is taller than both A and B. With D and E, we only know their positions relative to B and to each other, which lets us place them lower in the overall ordering. The tallest person will be the one who is not shorter than any other individual based on these comparisons. Step-by-Step Solution: Step 1: From A is taller than B, we write A > B.Step 2: From C is taller than A, we have C > A.Step 3: Combining these two, C > A > B. Therefore, C is taller than both A and B.Step 4: From D is taller than E but shorter than B, we get D > E and D < B.Step 5: Since B is shorter than A and C, and D is shorter than B, D and E must both be shorter than A and C.Step 6: Overall ordering from tallest to shortest is C, then A, then B, then D, then E. Verification / Alternative check: You can test this with sample numerical heights. Let C be 180 cm, A be 175 cm, B be 170 cm, D be 165 cm and E be 160 cm. All conditions are satisfied: C is taller than A, A is taller than B, B is taller than D, and D is taller than E. Among these values, C is clearly the tallest. Any other set of numbers that respects the same relative inequalities will also make C the tallest, which proves that the identity of the tallest person is fixed by the information given. Why Other Options Are Wrong: A cannot be tallest because C is explicitly given as taller than A.B cannot be tallest because A is taller than B and C is taller than A.D cannot be tallest because D is shorter than B, who is already shorter than A and C.E cannot be tallest because E is shorter than D, who is in turn shorter than B, A and C. Common Pitfalls: A frequent mistake is to focus only on one or two comparisons instead of chaining all of them together. Some students may forget the relationship between D and B and wrongly consider D as a candidate for tallest. Others may misread C is taller than A as A is taller than C. Carefully rewriting each statement as a symbolic inequality and then ordering them from greatest to smallest is the safest method. Final Answer: The tallest person among A, B, C, D and E is C.

Question 11

In a row of students, John is 16th from the left and Johnson is 8th from the right. If they interchange their positions and John then becomes 33rd from the left, how many students are there in the row?

Accepted Answer

Introduction / Context: This is a slightly more involved rank and position problem. You must use the information about changes in positions when two students interchange their seats in a row. The question tests your understanding of how left and right ranks relate to the total number of students in the row. Given Data / Assumptions: Initially, John is 16th from the left. Initially, Johnson is 8th from the right. They interchange positions. After interchange, John becomes 33rd from the left. We assume a single straight row with no empty positions and no other movements. Concept / Approach: The key idea is that when John and Johnson swap positions, John takes the exact position previously occupied by Johnson. Therefore, after the swap, John's new rank from the left is the same as Johnson's old rank from the left. Knowing the rank of a student from both sides of the row allows us to use the standard formula: total students = position from left + position from right - 1. We apply this to Johnson using his left and right positions. Step-by-Step Solution: Step 1: Before interchange, John is 16th from the left.Step 2: Before interchange, Johnson is 8th from the right.Step 3: After interchange, John takes the position that Johnson previously had.Step 4: The question says that after interchange John becomes 33rd from the left, which means Johnson's original position from the left is 33rd.Step 5: Now, for Johnson before interchange, position from left is 33 and from right is 8.Step 6: Use the formula total students = left position + right position - 1.Step 7: Total students = 33 + 8 - 1 = 40. Verification / Alternative check: To verify, imagine a row of 40 students. A student who is 33rd from the left is at position 40 - 33 + 1 = 8th from the right, which matches Johnson's given initial right rank. Therefore, the number 40 satisfies all position data. If the total were 39, then the 33rd from the left would be 7th from the right, not 8th. For 41, 33rd from the left would be 9th from the right. Thus, the total must be exactly 40 for the given ranks to be consistent. Why Other Options Are Wrong: 38, 39, 41 and 42 all lead to an inconsistency between the left rank 33 and the right rank 8 for the same student.For example, with 38 students, a student who is 33rd from the left would be 6th from the right, not 8th.With 39 students, 33rd from the left gives 7th from the right.With 41 or 42 students, the right ranks would be 9th or 10th, again contradicting the problem statement. Common Pitfalls: One common mistake is to forget that after the interchange John takes the exact original position of Johnson, and thus John's new left rank equals Johnson's old left rank. Another error is to add the two positions from the left directly or to misapply the total formula. Carefully linking the statements and explicitly using total = left rank + right rank - 1 is the correct and reliable method. Final Answer: The total number of students in the row is 40.

Question 12

A shepherd had 17 sheep, and all but eight died. How many sheep was he left with alive?

Accepted Answer

Introduction / Context: This question is a classic logical riddle often used in verbal reasoning. It is designed to test your careful reading skills rather than your calculation ability. The wording all but eight died can be misleading if read quickly, so the challenge is to interpret the phrase correctly and not overcomplicate the problem. Given Data / Assumptions: The shepherd initially has 17 sheep. The statement all but eight died describes what happened to the flock. No new sheep are bought or born during this situation. The question asks how many sheep remain alive after this event. Concept / Approach: The central idea is to correctly interpret the phrase all but eight died. This does not mean all eight died; instead it means all of them died except eight of them. In other words, the number that did not die is eight. Once you understand this, there is no need for any arithmetic beyond reading comprehension. The answer is simply the number of sheep that did not die. Step-by-Step Solution: Step 1: Note the initial number of sheep: 17.Step 2: Understand the phrase all but eight died. This means every sheep except eight died.Step 3: Sheep that are still alive are exactly those that did not die.Step 4: Therefore, the number of sheep left alive is eight.Step 5: There is no need to subtract eight from 17 or perform any other calculation, because the phrase itself already expresses the remaining count. Verification / Alternative check: If you want to check, imagine labelling the sheep from 1 to 17. Suppose sheep 1 to 9 died and sheep 10 to 17 survived. That would be eight surviving sheep. Now read the phrase all but eight died: that expression matches the situation, because all of the sheep except eight (numbers 10 to 17) have died. This confirms that eight is the number of survivors and that the phrase does not instruct you to compute 17 minus 8. Why Other Options Are Wrong: 17 would imply that none of the sheep died, which directly contradicts the statement that all but eight died.0 would mean that no sheep survived, but the words all but eight clearly indicate that some survived.9 and 7 come from incorrect arithmetic, such as misreading the phrase and computing 17 minus 8 or making another guess. Common Pitfalls: Many students automatically translate all but eight died into the calculation 17 minus 8, concluding that nine died or nine survived, depending on their misinterpretation. Others might think that all eight died and forget the phrase but. To avoid such traps, always rephrase the sentence in simpler language, for example all of the sheep died except eight. Once rephrased, the answer becomes obvious. Final Answer: The shepherd was left with 8 sheep alive.

Question 13

Priya has her birthday on Tuesday, 11th April. In the same calendar year Rani was born on 31st August. On which day of the week does the birthday of Rani fall in that year?

Accepted Answer

Introduction / Context: This problem is a calendar based reasoning question. You are told the day of the week for one date in a year and must determine the day of the week for another date in the same year. Such questions test your ability to count days across months, taking account of month lengths, and then convert the total day difference into a weekday shift using multiples of seven. Given Data / Assumptions: Priya's birthday is on Tuesday, 11th April. Rani's birthday is on 31st August of the same year. We assume the usual Gregorian calendar with April having 30 days, June having 30 days and so on. The year may be leap or non leap, but both dates are after February, so the number of days between 11th April and 31st August is the same in both cases. Concept / Approach: The day of the week repeats every seven days, so we only care about the remainder when the number of days between the two dates is divided by 7. We count the days from 11th April to 31st August, convert this difference into an offset modulo 7 and then move forward that many days from Tuesday to get the day of the week for 31st August. Step-by-Step Solution: Step 1: Count the remaining days in April after 11th April. April has 30 days, so there are 30 - 11 = 19 days from 12th April to 30th April inclusive.Step 2: Count the days in May: 31 days.Step 3: Count the days in June: 30 days.Step 4: Count the days in July: 31 days.Step 5: Count the days in August from 1st to 31st: that gives 31 days.Step 6: Total days between 11th April and 31st August = 19 + 31 + 30 + 31 + 31.Step 7: Compute 19 + 31 = 50, 50 + 30 = 80, 80 + 31 = 111, 111 + 31 = 142 days.Step 8: Find 142 modulo 7. Since 7 * 20 = 140, 142 - 140 = 2, so the remainder is 2.Step 9: A difference of 142 days corresponds to moving 2 days forward in the week cycle.Step 10: Starting from Tuesday, move forward 2 days: Wednesday is one day later, Thursday is two days later. Verification / Alternative check: You can perform a quick mental check by grouping days in multiples of seven. For example, 7 days after Tuesday is again Tuesday, 14 days after is Tuesday and so on. After 140 days, which is a multiple of 7, you land again on Tuesday. Two more days then move you to Thursday. This confirms that we have correctly used the concept of modulo 7 and that the computed weekday is consistent with the day cycle. Why Other Options Are Wrong: Monday or Tuesday would correspond to a shift of 6 or 0 days, which does not match a remainder of 2 when 142 is divided by 7.Wednesday corresponds to a shift of 1 day from Tuesday, but the remainder is 2.Friday would correspond to a shift of 3 days from Tuesday, which again contradicts the calculated shift of 2 days. Common Pitfalls: Sometimes students mistakenly include or exclude one of the boundary dates incorrectly when counting days, leading to an off by one error. Others may use incorrect month lengths, especially for April, June and August. A safe approach is always to write down the months and their days, then carefully add them. Finally, always convert the total into a remainder modulo 7 before moving along the weekdays, instead of stepping through every single day mentally. Final Answer: Rani's birthday on 31st August in that year falls on Thursday.

Question 14

The weights of four boxes are 20 kg, 40 kg, 60 kg and 90 kg. Which of the following cannot be obtained as the total weight, in kilograms, of any combination of these boxes if each box is used at most once?

Accepted Answer

Introduction / Context: This question again asks you to reason about which totals can be formed by combining a set of four box weights, each used at most once. It checks your ability to systematically list or infer all possible sums and then identify a total that cannot occur. Such questions build both numerical fluency and logical organisation of cases. Given Data / Assumptions: Weights of the four boxes: 20 kg, 40 kg, 60 kg, 90 kg. Each box can be used either once or not used in a combination. Candidate totals are 150 kg, 170 kg, 120 kg and 160 kg. No fractional use of a box is allowed. Concept / Approach: As before, we look at all possible sums formed by the four given numbers. There are sums from single box selections, two box selections, three box selections and all four boxes. Once we know this set of possible totals, we simply check which of the options is not present in that set. It is very important to be systematic so that no combination is missed or double counted. Step-by-Step Solution: Step 1: Write the weights: 20, 40, 60, 90.Step 2: Sums using one box: 20, 40, 60, 90.Step 3: Sums using two boxes: 20 + 40 = 60, 20 + 60 = 80, 20 + 90 = 110, 40 + 60 = 100, 40 + 90 = 130, 60 + 90 = 150.Step 4: Sums using three boxes: 20 + 40 + 60 = 120, 20 + 40 + 90 = 150, 20 + 60 + 90 = 170, 40 + 60 + 90 = 190.Step 5: Sum using all four boxes: 20 + 40 + 60 + 90 = 210.Step 6: Collect all distinct sums: 20, 40, 60, 80, 90, 100, 110, 120, 130, 150, 170, 190, 210.Step 7: Compare each candidate total with this list. Verification / Alternative check: Check each option: 150 is present in two ways, 60 + 90 and 20 + 40 + 90. 170 appears as 20 + 60 + 90. 120 is present as 20 + 40 + 60. However, 160 does not appear in the list of possible sums. You can further attempt to build 160 directly: the largest three weights sum to 40 + 60 + 90 = 190, which is too large, and 90 + 60 = 150. Adding 20 to that gives 170, not 160. Other combinations such as 90 + 40 = 130 or 60 + 40 = 100 cannot be extended to 160 without reusing a number. This reconfirms that 160 is impossible. Why Other Options Are Wrong: 150 kg is possible by 60 + 90 and also by 20 + 40 + 90.170 kg is possible by 20 + 60 + 90.120 kg is possible by 20 + 40 + 60.Therefore, these totals cannot be answers to the impossibility question. Common Pitfalls: Learners may wrongly assume that if a number lies between two achievable sums it must also be achievable, which is not true. Others may forget to consider three box combinations and only check pairs, which could lead to incorrect elimination of some options. Writing the combinations in an organised way, as shown above, is the safest technique and prevents oversight. Final Answer: The total weight that cannot be obtained from any combination of the four boxes is 160 kilograms.

Question 15

The weights of four boxes are 10 kg, 40 kg, 50 kg and 80 kg. Which of the following cannot be obtained as the total weight, in kilograms, of any combination of these boxes if each box is used at most once?

Accepted Answer

Introduction / Context: This is another combination of weights puzzle that checks whether you can accurately list possible totals formed from a set of four distinct weights. The skill is the same: identify which one of the suggested totals cannot appear as a sum of any subset of the given weights when each can be used at most once. It is a useful exercise in both reasoning and organised enumeration. Given Data / Assumptions: Box weights: 10 kg, 40 kg, 50 kg, 80 kg. Each box can be included at most one time in a combination. Candidate totals: 120 kg, 100 kg, 90 kg, 110 kg. No fractions of a box are allowed. Concept / Approach: We systematically work out all possible sums from one, two, three and four box combinations. Once we have the set of attainable sums, we simply check which of the given options does not appear in that set. This approach eliminates guesswork and ensures that every possibility has been considered logically. Step-by-Step Solution: Step 1: Note the weights: 10, 40, 50, 80.Step 2: Sums using one box: 10, 40, 50, 80.Step 3: Sums using two boxes: 10 + 40 = 50, 10 + 50 = 60, 10 + 80 = 90, 40 + 50 = 90, 40 + 80 = 120, 50 + 80 = 130.Step 4: Sums using three boxes: 10 + 40 + 50 = 100, 10 + 40 + 80 = 130, 10 + 50 + 80 = 140, 40 + 50 + 80 = 170.Step 5: Sum using all four boxes: 10 + 40 + 50 + 80 = 180.Step 6: Collect all distinct sums: 10, 40, 50, 60, 80, 90, 100, 120, 130, 140, 170, 180.Step 7: Compare each option 120, 100, 90 and 110 with this list. Verification / Alternative check: We can see that 120 appears as 40 + 80. The total 100 appears as 10 + 40 + 50. The total 90 appears as either 10 + 80 or 40 + 50. However, 110 is not present. To double check, try to construct 110: 80 + 40 = 120 (too large), 80 + 50 = 130 (too large), 80 + 10 = 90 (too small), and 50 + 40 = 90. Adding the 10 kg weight to 100 gives 110, but 100 itself was 10 + 40 + 50, which already uses all three smaller boxes. Using all four boxes gives 180, not 110. Therefore, 110 cannot be formed. Why Other Options Are Wrong: 120 kg is attainable from 40 + 80.100 kg is attainable from 10 + 40 + 50.90 kg is attainable from 10 + 80 or from 40 + 50.These values are therefore not impossible totals and so cannot be the correct answer to the question. Common Pitfalls: A common error is to overlook a combination like 10 + 80 or 10 + 40 + 50 and thus wrongly conclude that 90 or 100 cannot be formed. Another mistake is to attempt to solve the problem mentally without writing down intermediate sums, which increases the chance of missing a valid subset. A structured list, as shown, is the best way to avoid such oversights. Final Answer: The total weight that cannot be obtained from any combination of the four boxes is 110 kilograms.

Question 16

The weights of four boxes are 90 kg, 30 kg, 20 kg and 50 kg. Which of the following cannot be obtained as the total weight, in kilograms, of any combination of these boxes if each box is used at most once?

Accepted Answer

Introduction / Context: This question continues the theme of reasoning with combinations of fixed weights. You must identify which suggested total weight cannot be achieved using any subset of the four given box weights, where each box can appear at most once in a combination. It is a useful test of systematic enumeration and logical checking. Given Data / Assumptions: Box weights: 90 kg, 30 kg, 20 kg, 50 kg. Each box may be used at most once in forming a total. Candidate totals: 190 kg, 170 kg, 100 kg, 150 kg. We work with whole kilograms only, no fractions. Concept / Approach: The approach is to list all possible sums from 1 box, 2 boxes, 3 boxes and all 4 boxes. Once we have the set of all distinct sums, we simply check which option is missing from this set. The presence or absence of a total is then an objective matter, not a guess. Step-by-Step Solution: Step 1: Write the weights: 90, 30, 20, 50.Step 2: Sums using one box: 20, 30, 50, 90.Step 3: Sums using two boxes: 20 + 30 = 50, 20 + 50 = 70, 20 + 90 = 110, 30 + 50 = 80, 30 + 90 = 120, 50 + 90 = 140.Step 4: Sums using three boxes: 20 + 30 + 50 = 100, 20 + 30 + 90 = 140, 20 + 50 + 90 = 160, 30 + 50 + 90 = 170.Step 5: Sum using all four boxes: 20 + 30 + 50 + 90 = 190.Step 6: Collect all distinct sums: 20, 30, 50, 70, 80, 90, 100, 110, 120, 140, 160, 170, 190.Step 7: Compare 190, 170, 100 and 150 against this list. Verification / Alternative check: We can see that 190 is achievable by using all four boxes: 20 + 30 + 50 + 90 = 190. The total 170 is achievable with 30 + 50 + 90. The total 100 is achievable with 20 + 30 + 50. The remaining candidate is 150. Try to form 150 directly: 90 + 50 = 140, and adding 20 gives 160, adding 30 gives 170, but we are not allowed to use more than two extra boxes beyond 90 and 50 at a time if we want 150. Combinations like 90 + 30 = 120 and 90 + 20 = 110 also do not lead to 150 without using a third weight that overshoots or misfits. This confirms that 150 cannot be reached. Why Other Options Are Wrong: 190 kg is possible using all the boxes together.170 kg is possible using 30 + 50 + 90.100 kg is possible using 20 + 30 + 50.These totals therefore are not impossible and cannot be the correct answer. Common Pitfalls: Sometimes students overlook the three box combinations and check only pairs, leading them to wrongly classify some achievable totals as impossible. Others may misadd one of the combinations and incorrectly think they have found a way to reach 150, which is not actually the case. Writing all combinations cleanly and double checking the arithmetic avoids these issues. Final Answer: The total weight that cannot be obtained from any combination of the four boxes is 150 kilograms.

Question 17

The weights of four boxes are 90 kg, 30 kg, 40 kg and 60 kg. Which of the following cannot be obtained as the total weight, in kilograms, of any combination of these boxes if each box is used at most once?

Accepted Answer

Introduction / Context: This is another variation of the same style of weight combination reasoning. You are asked to identify which total cannot be formed by combining the given four box weights, with each box allowed at most once. Success depends on systematically working out the possible sums instead of relying on intuition alone. Given Data / Assumptions: Box weights: 90 kg, 30 kg, 40 kg, 60 kg. Each box can be used at most once. Candidate totals: 200 kg, 220 kg, 180 kg, 130 kg. All weights and totals are whole numbers in kilograms. Concept / Approach: As before, we examine all possible sums of one, two, three and four box combinations. We then compare the compiled list of totals against the answer options to see which total is missing. That missing value is the one that cannot be obtained from any allowed combination. Step-by-Step Solution: Step 1: List the weights: 90, 30, 40, 60.Step 2: Sums using one box: 30, 40, 60, 90.Step 3: Sums using two boxes: 30 + 40 = 70, 30 + 60 = 90, 30 + 90 = 120, 40 + 60 = 100, 40 + 90 = 130, 60 + 90 = 150.Step 4: Sums using three boxes: 30 + 40 + 60 = 130, 30 + 40 + 90 = 160, 30 + 60 + 90 = 180, 40 + 60 + 90 = 190.Step 5: Sum using all four boxes: 30 + 40 + 60 + 90 = 220.Step 6: Collect all distinct sums: 30, 40, 60, 70, 90, 100, 120, 130, 150, 160, 180, 190, 220.Step 7: Compare options 200, 220, 180, 130 with this list. Verification / Alternative check: From the list we see that 220 is achievable by using all four boxes together. The total 180 is achievable from 30 + 60 + 90. The total 130 is achievable in two different ways: 40 + 90 and 30 + 40 + 60. Therefore, these three totals are valid. The remaining candidate, 200, does not appear in the list. To be sure, try constructing 200: 90 + 60 + 40 = 190, and adding 30 gives 220. Other three weight combinations like 90 + 60 + 30 = 180 and 90 + 40 + 30 = 160 also fall short. No pair reaches 200 either. This confirms 200 cannot be formed. Why Other Options Are Wrong: 220 kg is possible using all four boxes.180 kg is possible using 30 + 60 + 90.130 kg is possible using 40 + 90 or 30 + 40 + 60.Thus, these totals are achievable and cannot be the answer. Common Pitfalls: Students often miscalculate a three box sum or forget that some totals like 130 can be formed in more than one way. This may lead them to wrongly conclude that such totals are impossible. Another pitfall is to assume that any round number like 200 should be possible, without carefully verifying the combinations. A complete list of sums is the most reliable solution method. Final Answer: The total weight that cannot be obtained from any combination of these four boxes is 200 kilograms.

Question 18

The weights of four boxes are 90 kg, 40 kg, 80 kg and 50 kg. Which of the following cannot be obtained as the total weight, in kilograms, of any combination of these boxes if each box is used at most once?

Accepted Answer

Introduction / Context: Here you are again asked to identify an impossible total made from a fixed set of box weights. This style of question reinforces the habit of patiently enumerating possible sums rather than guessing. The logic is the same as in the previous weight combination examples, but with a different set of numbers and answer choices. Given Data / Assumptions: Box weights: 90 kg, 40 kg, 80 kg, 50 kg. Each box may be used at most once in a combination. Candidate totals: 200 kg, 260 kg, 180 kg, 170 kg. All weights are whole numbers in kilograms. Concept / Approach: We need to examine all possible sums of these four weights: one box, two box, three box and four box combinations. Then we compare these possible sums with the answer options and pick the total that does not appear. Staying systematic prevents errors and ensures we do not overlook any configuration. Step-by-Step Solution: Step 1: List weights: 90, 40, 80, 50.Step 2: Sums using one box: 40, 50, 80, 90.Step 3: Sums using two boxes: 40 + 50 = 90, 40 + 80 = 120, 40 + 90 = 130, 50 + 80 = 130, 50 + 90 = 140, 80 + 90 = 170.Step 4: Sums using three boxes: 40 + 50 + 80 = 170, 40 + 50 + 90 = 180, 40 + 80 + 90 = 210, 50 + 80 + 90 = 220.Step 5: Sum using all four boxes: 40 + 50 + 80 + 90 = 260.Step 6: Collect all distinct sums: 40, 50, 80, 90, 120, 130, 140, 170, 180, 210, 220, 260.Step 7: Compare each option 200, 260, 180, 170 with this set. Verification / Alternative check: From the list, 260 is achievable using all four boxes. The total 180 is achievable with 40 + 50 + 90. The total 170 is achievable with 80 + 90 or 40 + 50 + 80. However, 200 is absent from the list. To check further, try forming 200: 90 + 80 = 170, and adding 40 gives 210, adding 50 gives 220. Combinations like 90 + 50 = 140 and then adding 40 or 80 give 180 or 220, never 200. Likewise, 80 + 50 + 40 = 170. Therefore, no subset sums to 200, so 200 is impossible. Why Other Options Are Wrong: 260 kg is clearly possible with all four boxes.180 kg is possible with 40 + 50 + 90.170 kg is possible with 80 + 90 or 40 + 50 + 80.So these totals can be formed and are not the correct answer. Common Pitfalls: Some examinees may rush and conclude that 260 is impossible because it seems large, forgetting that it is simply the sum of all four weights. Others might miscalculate three box sums and thereby misjudge totals such as 170 or 180. Carefully writing out the sums of each subset once, and then scanning that list, is the most efficient way to answer correctly. Final Answer: The total weight that cannot be obtained from any combination of these four boxes is 200 kilograms.

Question 19

The weights of four boxes are 40 kg, 30 kg, 50 kg and 20 kg. Which of the following cannot be obtained as the total weight, in kilograms, of any combination of these boxes if each box is used at most once?

Accepted Answer

Introduction / Context: This is another combination of weights puzzle. You must determine which proposed total weight cannot be formed using any subset of the four given box weights, where each box may be used at most once. As with the earlier problems of this type, the key is to be systematic and avoid guessing. Given Data / Assumptions: Box weights: 40 kg, 30 kg, 50 kg, 20 kg. Each box can be selected at most once in any combination. Candidate totals: 140 kg, 130 kg, 90 kg, 100 kg. All weights are in whole kilograms. Concept / Approach: We list the sums of one box, two box, three box and four box combinations and then compare the set of possible sums with the answer options. The option that does not appear among the achievable sums is the total that cannot be formed. This method ensures an accurate and complete analysis. Step-by-Step Solution: Step 1: Record the weights: 40, 30, 50, 20.Step 2: Sums using one box: 20, 30, 40, 50.Step 3: Sums using two boxes: 20 + 30 = 50, 20 + 40 = 60, 20 + 50 = 70, 30 + 40 = 70, 30 + 50 = 80, 40 + 50 = 90.Step 4: Sums using three boxes: 20 + 30 + 40 = 90, 20 + 30 + 50 = 100, 20 + 40 + 50 = 110, 30 + 40 + 50 = 120.Step 5: Sum using all four boxes: 20 + 30 + 40 + 50 = 140.Step 6: Collect distinct sums: 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 140.Step 7: Compare the candidate totals 140, 130, 90 and 100 with this list. Verification / Alternative check: The total 140 appears as the sum of all four boxes. The total 90 appears in two ways: 40 + 50 and 20 + 30 + 40. The total 100 appears as 20 + 30 + 50. However, 130 is not present in the list of sums. To double check, attempt to build 130: 50 + 40 + 30 = 120, and adding 20 gives 140. There is no way to pick a subset of the four weights that adds to 130 without exceeding or falling short of it. This confirms that 130 is impossible to obtain. Why Other Options Are Wrong: 140 kg is possible using all four boxes.90 kg is possible using 40 + 50 or 20 + 30 + 40.100 kg is possible using 20 + 30 + 50.Therefore, these totals are valid and cannot be the correct answer. Common Pitfalls: Some learners may try to mentally add combinations and accidentally skip a three box combination such as 20 + 30 + 50, leading to the wrong conclusion about 100. Others might think that since 130 lies between 120 and 140 it should be achievable, which is not guaranteed with discrete weights. A thorough and written enumeration is the best method to avoid these errors. Final Answer: The total weight that cannot be obtained from any combination of these four boxes is 130 kilograms.

Question 20

Divit has his birthday on Friday, 30th June. In the same calendar year Samar was born on 15th November. On which day of the week does the birthday of Samar fall in that year?

Accepted Answer

Introduction / Context: This is another calendar arithmetic question where you must move from one known date and day of the week to another date in the same year. The key is to count the number of days between the two dates correctly and then convert that into a weekday shift. This kind of reasoning is frequently tested in aptitude and banking examinations. Given Data / Assumptions: Divit's birthday is on Friday, 30th June. Samar's birthday is on 15th November of the same year. We use the standard Gregorian calendar. Month lengths needed: July has 31 days, August has 31 days, September has 30 days, October has 31 days, and November has 30 days in total. Concept / Approach: We count how many days elapse from 30th June to 15th November and then reduce this number modulo 7. The remainder tells us how many days to move forward from Friday to reach the weekday for 15th November. It is important to handle the boundary dates carefully and ensure the day count is accurate. Step-by-Step Solution: Step 1: Consider the period starting from the day after 30th June. From 1st July to 31st July there are 31 days.Step 2: August has 31 days.Step 3: September has 30 days.Step 4: October has 31 days.Step 5: In November, we need to go from 1st November to 15th November inclusive, which is 15 days.Step 6: Add these intervals to get the total number of days between 30th June and 15th November: 31 + 31 + 30 + 31 + 15.Step 7: Calculate 31 + 31 = 62, 62 + 30 = 92, 92 + 31 = 123, 123 + 15 = 138 days.Step 8: Find 138 modulo 7. Since 7 * 19 = 133, the remainder is 138 - 133 = 5.Step 9: A difference of 138 days corresponds to moving forward 5 days in the week cycle.Step 10: Starting from Friday, move 5 days forward: Saturday (1), Sunday (2), Monday (3), Tuesday (4), Wednesday (5). Verification / Alternative check: We can confirm by grouping days into multiples of seven. After 7, 14, 21 and so on days from Friday, the day of the week cycles back to Friday each time. After 133 days (which is 19 full weeks), it will again be Friday. The remaining 5 days then move you to Wednesday. This double checks that the modulo operation and the step by step weekday counting are both consistent. Why Other Options Are Wrong: Tuesday would result from a forward shift of 4 days, which does not match the remainder of 5.Friday would correspond to a shift of 0 days modulo 7, which is not the case here.Saturday would correspond to a shift of just 1 day, and Monday would correspond to a shift of 3 days, both of which contradict the calculated shift of 5. Common Pitfalls: Students sometimes miscount the number of days in one of the months or incorrectly handle the inclusive or exclusive nature of the endpoints. A good method is to separately write down the days in each month and then sum them carefully, as shown. Forgetting to reduce the total days modulo 7 and instead stepping one by one over all days is also inefficient and prone to error. Using the modulo approach is much cleaner and more reliable. Final Answer: In that year, Samar's birthday on 15th November falls on Wednesday.

Arithmetic Reasoning Questions