Question 1

Among four books, Book 1 is twice as heavy as Book 2. The weight of Book 3 is half of the weight of Book 2. Book 4 is 60 grams more than Book 2 but 60 grams less than Book 1. Which book is the heaviest?

Accepted Answer

Introduction / Context:This arithmetic reasoning question involves comparing weights of four books based on given relationships rather than direct numerical values. You are told how each book weight relates to the others and asked to determine which book is heaviest. The problem tests your ability to convert verbal descriptions into algebraic relationships and then reason about relative sizes. Given Data / Assumptions:Book 1 is twice as heavy as Book 2.The weight of Book 3 is half of the weight of Book 2.Book 4 is 60 grams more than Book 2.Book 4 is also 60 grams less than Book 1.All books have positive weights.We need to identify which book has the greatest weight. Concept / Approach:A convenient way to handle such problems is to assign a variable to one book and express all other weights in terms of that variable. Here, Book 2 is a natural choice. Using the given relationships, we can derive expressions for the weights of the other books. Even if we do not compute exact numbers, the relative sizes become clear. However, in this case, we can also find the exact weight of Book 2 by using the two conditions on Book 4 and solving a simple equation. Step-by-Step Solution:Step 1: Let the weight of Book 2 be x grams.Step 2: Book 1 is twice as heavy as Book 2, so its weight is 2x grams.Step 3: The weight of Book 3 is half of Book 2, so its weight is x / 2 grams.Step 4: Book 4 is 60 grams more than Book 2, so one expression for Book 4 weight is x + 60 grams.Step 5: Book 4 is also 60 grams less than Book 1. Since Book 1 weighs 2x, another expression for Book 4 weight is 2x - 60 grams.Step 6: Set the two expressions for Book 4 equal to each other: x + 60 = 2x - 60.Step 7: Solve this equation: bring x to one side and constants to the other. Subtract x from both sides to get 60 = x - 60. Add 60 to both sides to obtain x = 120.Step 8: Now substitute back. Book 2 weighs 120 grams, Book 1 weighs 2 * 120 = 240 grams, Book 3 weighs 120 / 2 = 60 grams and Book 4 weighs 120 + 60 = 180 grams.Step 9: Comparing 240, 120, 60 and 180, Book 1 has the largest weight. Verification / Alternative check:Even without solving for x, we can see that Book 1 is twice Book 2, whereas Book 4 is only 60 grams above Book 2 but still 60 grams below Book 1. This shows Book 1 must be heavier than both Book 2 and Book 4. Book 3 is half of Book 2, so it is clearly lighter than both Book 2 and Book 1. This reasoning also makes it clear that Book 1 is the heaviest book, consistent with the numeric calculation. Why Other Options Are Wrong:Book 2 cannot be the heaviest because Book 1 is twice its weight and Book 4 is even heavier than Book 2. Book 3 is explicitly half of Book 2, making it the lightest. Book 4, while heavier than Book 2, is still 60 grams lighter than Book 1. Therefore it cannot be the heaviest either. Only Book 1 has no heavier competitor. Common Pitfalls:Some learners might misinterpret the phrases “more than” and “less than” and reverse an inequality, for example thinking Book 4 is heavier than Book 1 instead of lighter by 60 grams. Others may try to guess without writing equations, leading to confusion. Writing all relationships in terms of a single variable keeps the reasoning clear and avoids mistakes. Final Answer:The heaviest book is Book 1.

Question 2

Sumitra remembers that her mother birthday is after 13th February but before 16th February, while her brother remembers that the birthday is after 14th February but before 17th February. On which exact date will their mother birthday be celebrated?

Accepted Answer

Introduction / Context:This logical reasoning question asks you to determine an exact date based on two overlapping pieces of partial information. Two siblings remember their mother birthday only within certain ranges of dates. The goal is to combine these ranges to find the single date that satisfies both memories. It is a simple but important example of using intersections of intervals to find a precise result. Given Data / Assumptions:Sumitra remembers that the birthday is after 13th February but before 16th February.Her brother remembers that the birthday is after 14th February but before 17th February.Dates are in February of the same year.The birthday is on exactly one specific day, not spread over several days.The wording “after” and “before” means the endpoints themselves are excluded. Concept / Approach:Each child is giving a date range. Sumitra narrows the birthday down to a small set of possible dates, and her brother narrows it down to another set. The correct date must be one that lies inside both ranges simultaneously. Mathematically, it is the intersection of the two sets of possible dates. So we list the dates allowed by each memory and then find the common date. Step-by-Step Solution:Step 1: Interpret Sumitra statement. The birthday is after 13th February but before 16th February. That means it can be on 14th or 15th February, because 14 and 15 are between 13 and 16, not including the endpoints.Step 2: So according to Sumitra, the possible dates are {14th, 15th} February.Step 3: Now interpret the brother statement. The birthday is after 14th February but before 17th February. That means it can be on 15th or 16th February.Step 4: So according to the brother, the possible dates are {15th, 16th} February.Step 5: The actual birthday must belong to both sets of possible dates. Therefore we take the intersection of the sets {14th, 15th} and {15th, 16th}.Step 6: The only date common to both sets is 15th February.Step 7: Hence the birthday will be celebrated on 15th February. Verification / Alternative check:We can also visualise the dates on a number line. Mark 13, 14, 15, 16 and 17 as points. Sumitra valid region is between 13 and 16, excluding both, which covers 14 and 15 only. The brother valid region is between 14 and 17, excluding both, which covers 15 and 16 only. The overlap of these two highlighted segments is the single point at 15. This confirms that all given information points to the same date. Why Other Options Are Wrong:13th February and 14th February are not possible because Sumitra said the birthday is after 13th and the brother said it is after 14th, so neither 13 nor 14 fits the second statement. 17th February is ruled out because both children said “before” 16th or 17th, excluding those end dates. Only 15th February lies inside both specified ranges. Common Pitfalls:A typical mistake is to include the endpoints when the phrases clearly say “after” and “before”. Another error is to take the union of the ranges instead of the intersection, which would enlarge rather than narrow the set of possible dates. Always pay close attention to whether endpoints are included or excluded and whether you need common elements or just any element from either range. Final Answer:Sumitra mother birthday will be celebrated on 15th February.

Question 3

There are five energy drinks Red, Moto, Energy, Lion and Bull with different sugar contents. Moto has sugar content more than all other drinks. Energy has sugar content only more than Lion. Bull does not have sugar content more than Red. Which drink…

Accepted Answer

Introduction / Context:This reasoning question compares the sugar content of five energy drinks named Red, Moto, Energy, Lion and Bull. You are not given exact sugar values but a set of comparative statements. The aim is to determine which drink is second in the ranking when the drinks are ordered from highest to lowest sugar content. Such problems test your ability to build and interpret an ordered list from relational information. Given Data / Assumptions:Drinks are Red (R), Moto (M), Energy (E), Lion (L) and Bull (B).Moto has sugar content more than all other drinks.Energy has sugar content only more than Lion, meaning E is above L but below R, B and M.Bull does not have sugar content more than Red, which implies B is less than Red.All sugar contents are distinct; no two drinks have the exact same sugar level.We are asked to identify the drink with the second highest sugar content. Concept / Approach:First, convert the sentences into inequalities. Moto being above all others makes it the top of the list. Energy being only above Lion places both Energy and Lion near the bottom. The condition about Bull and Red, combined with the information about Energy, allows us to fix the full order. Once the complete ranking is known, the second position from the top gives the drink with the second highest sugar content. Step-by-Step Solution:Step 1: From “Moto having sugar content more than all other drinks”, we conclude M is the highest. So M is at the top of the order.Step 2: From “Energy having the sugar content only more than the Lion”, we see that E > L, and E is below every other drink except Lion. Thus R and B must both have more sugar than E.Step 3: This places Lion at the very bottom, because the only drink below Energy is Lion. So the partial order from lowest upwards is L, then E, and above E lie B, R and finally M at the top.Step 4: From “Bull is not having sugar content more than the Red”, we know B is not above R. Since all values are distinct and we already know both B and R are above E, the only possibility consistent with this statement is that R > B.Step 5: Combine all pieces. From lowest to highest, the full order is L, E, B, R, M.Step 6: From highest to lowest, the order is M, R, B, E, L. Therefore the drink with the second highest sugar content is Red. Verification / Alternative check:You can assign hypothetical values consistent with the inequalities to verify. For example, let L have sugar level 1, E have 2, B have 3, R have 4 and M have 5. These values satisfy E > L, R > E, B > E, B all others. In this scenario, Red is clearly second highest after Moto, which supports the logical ranking found earlier. Why Other Options Are Wrong:Moto is explicitly the highest, so it cannot be second highest. Energy is only above Lion and is below Red and Bull, so it is near the bottom. Lion is at the very bottom as everyone else has more sugar than Lion. Bull has less sugar than Red, so it cannot be above Red in the list and thus cannot be second highest. Only Red fits the position directly below Moto. Common Pitfalls:A common mistake is to misread “only more than Lion” for Energy and think that Energy is second highest. In reality, this phrase means Energy is just one level above Lion and below everyone else. Another pitfall is misinterpreting “Bull is not having sugar content more than the Red” as equality rather than a strict lower ranking. Remember that all sugar levels are different, so “not more than” must mean “less than” in this context. Final Answer:The drink with the second highest sugar content is Red.

Question 4

The weights of four boxes are 30, 70, 60 and 90 kilograms. Which of the following cannot be the total weight, in kilograms, of any combination of these boxes if each box can be used at most once in a combination?

Accepted Answer

Introduction / Context:This question is a subset sum puzzle involving four box weights. You are given their individual weights and several candidate totals, and you must decide which total cannot be formed by adding the weights of some selection of these boxes. Each box can appear at most once in any combination. This type of problem checks your ability to reason systematically about all possible sums of a small set of numbers. Given Data / Assumptions:The four box weights are 30 kilograms, 70 kilograms, 60 kilograms and 90 kilograms.Each box can be either included or not included in a combination, but cannot be used more than once.Candidate total weights are 250, 200, 190 and 220 kilograms.We need to determine which one of these totals cannot be obtained from any subset of the four weights.All additions are simple integer sums. Concept / Approach:Because there are only four boxes, we can list all possible subset sums. There are 2^4 = 16 subsets including the empty one, but we only care about positive sums. By calculating the sums of all one box, two box, three box and four box combinations, we obtain a complete list of achievable totals. Any candidate total not in this list is impossible. We can also reason quickly about the largest sum and values close to it. Step-by-Step Solution:Step 1: List single box sums: 30, 70, 60 and 90 kilograms.Step 2: List all two box sums: 30 + 70 = 100, 30 + 60 = 90, 30 + 90 = 120, 70 + 60 = 130, 70 + 90 = 160 and 60 + 90 = 150.Step 3: List all three box sums: 30 + 70 + 60 = 160, 30 + 70 + 90 = 190, 30 + 60 + 90 = 180 and 70 + 60 + 90 = 220.Step 4: List the four box sum: 30 + 70 + 60 + 90 = 250.Step 5: Collect distinct achievable totals: 30, 60, 70, 90, 100, 120, 130, 150, 160, 180, 190, 220 and 250.Step 6: Compare the candidate options. We see that 250 is achievable by taking all four boxes, 190 is achievable using 30 + 70 + 90 and 220 is achievable using 70 + 60 + 90.Step 7: The total 200 kilograms does not appear in the list of subset sums, so it cannot be formed by any allowed combination. Verification / Alternative check:As a faster check, note that the maximum possible total is 250 kilograms using all four boxes. To make 200, we would need to leave out exactly 50 kilograms worth of boxes in some combination. But none of the individual weights or pair sums equal 50 (the available individual weights are 30, 70, 60, 90 and their pair sums are 100, 90, 120, 130, 160 and 150). Since we cannot remove exactly 50 from 250, 200 cannot be reached. This reasoning agrees with the full subset enumeration. Why Other Options Are Wrong:250 kilograms is the sum of all four boxes: 30 + 70 + 60 + 90. 190 kilograms is the sum of 30, 70 and 90. 220 kilograms is the sum of 70, 60 and 90. Since each of these totals can be formed by valid combinations under the given rules, they are not correct answers to the question. Common Pitfalls:Learners sometimes forget certain two box or three box combinations and conclude too quickly that a total is impossible. Others may incorrectly think that because 200 is near 190 and 220, it must be reachable. Systematically listing all sums or using the idea of subtracting from the maximum total helps avoid these mistakes. Final Answer:The total weight that cannot be obtained from any combination of the four boxes is 200 kilograms.

Question 5

Which of the following interchanges of signs and numbers will make the equation 8 x 6 + 4 = 52 correct?

Accepted Answer

Introduction / Context:This question involves an incorrect arithmetic equation and asks you to correct it by interchanging certain operation signs and numbers. The original equation is 8 x 6 + 4 = 52. Each option specifies a pair of signs and a pair of numbers to exchange positions. After performing the specified swaps, the equation should become true under standard operator precedence. This problem tests careful manipulation and evaluation of algebraic expressions. Given Data / Assumptions:The incorrect starting equation is 8 x 6 + 4 = 52.We may interchange one pair of operation signs and one pair of numbers as specified in the chosen option.The set of operations involved is multiplication and addition.We evaluate expressions using normal precedence, where multiplication is done before addition.Our goal is to find the option that produces a valid equality. Concept / Approach:For each option, we apply the indicated interchanges step by step. First, swap the signs in all positions where they occur, then swap the indicated numbers. After performing both swaps, we evaluate the new left hand side and compare it with the right hand side. Only those changes that make the equality true are acceptable. Because the expression is short, we can test all options reliably. Step-by-Step Solution:Step 1: Consider option c, which suggests interchanging the signs x and +, and the numbers 8 and 4.Step 2: Start from 8 x 6 + 4 = 52. First swap x and +. This gives 8 + 6 x 4 = 52.Step 3: Next, interchange the numbers 8 and 4. In the expression 8 + 6 x 4 = 52, the 8 on the left becomes 4, and the 4 in the product becomes 8, giving 4 + 6 x 8 = 52.Step 4: Evaluate the left hand side using BODMAS. Compute 6 x 8 = 48 first, then add 4 to 48 to get 52.Step 5: Thus the new equation 4 + 6 x 8 = 52 is true, since both sides equal 52.Step 6: For comparison, check option a briefly. Swapping x and + and then 8 and 52 leads to 52 + 6 x 4 = 8, which evaluates to 52 + 24 = 76 on the left, not equal to 8 on the right, so it is incorrect.Step 7: Options b and d similarly fail when tested, leaving option c as the only correct choice. Verification / Alternative check:We can verify the final corrected equation independently. Starting from 4 + 6 x 8, perform the multiplication first: 6 x 8 = 48. Then add 4 to get 52. This exactly matches the right hand side of 52, confirming that the equation is valid after the described interchanges. No other sequence of swaps from the options yields such a clean equality. Why Other Options Are Wrong:Option a gives 52 + 6 x 4 = 8, which simplifies to 76 = 8, clearly false. Option b, after swapping + and x and then 6 and 4, results in 8 + 4 x 6 = 52, which simplifies to 8 + 24 = 32, not 52. Option d gives 6 + 8 x 4 = 52, which simplifies to 6 + 32 = 38, also incorrect. Only option c correctly transforms the equation into a true statement. Common Pitfalls:One frequent mistake is to perform the swaps in the wrong order or to forget that both the sign pair and number pair must be swapped in all their occurrences. Another pitfall is evaluating the resulting expression from left to right without respecting multiplication precedence. Carefully applying each change and then using BODMAS avoids these issues. Final Answer:The equation becomes correct when we interchange x and +, and 8 and 4, corresponding to option c.

Question 6

Arnav birthday is on Tuesday 14th March in a particular year. In the same year, on which day of the week will Pranay birthday fall if Pranay was born on 13th September?

Accepted Answer

Introduction / Context:This calendar based reasoning question provides the day of the week for one date and asks you to determine the day for another date in the same year. Arnav birthday is on Tuesday 14th March, and you must find the weekday for 13th September of the same year, which is Pranay birthday. The problem tests your ability to count days across months and use modular arithmetic to track day of week shifts. Given Data / Assumptions:Arnav birthday is Tuesday 14th March.Pranay birthday is 13th September of the same year.The year is not specified, but we only need the relative day, not the actual calendar year.Month lengths are taken as usual: March has 31 days, April 30, May 31, June 30, July 31, August 31 and September 30.We assume a standard Gregorian calendar and Arnav date as reference. Concept / Approach:The day of the week advances by one for each day that passes, wrapping around every seven days. To find the day for 13th September, compute the number of days between 14th March and 13th September, then take that number modulo 7. A remainder of 0 means the same day, 1 means one day later, and so on. Because February is before March, the leap year issue does not affect the count between these two dates. Step-by-Step Solution:Step 1: Count the days remaining in March after 14th March. March has 31 days, so there are 31 - 14 = 17 days from 15th March to 31st March.Step 2: Add the days in the full months between March and September. April has 30 days, May 31, June 30, July 31 and August 31.Step 3: For September, we count from 1st September up to 13th September, which is 13 days.Step 4: Sum all these days: 17 (rest of March) + 30 (April) + 31 (May) + 30 (June) + 31 (July) + 31 (August) + 13 (September).Step 5: Compute the total: 17 + 30 = 47, 47 + 31 = 78, 78 + 30 = 108, 108 + 31 = 139, 139 + 31 = 170, 170 + 13 = 183 days.Step 6: Find 183 modulo 7. Since 7 * 26 = 182, we get a remainder of 1.Step 7: A remainder of 1 means that 13th September is one day after the day on 14th March. Since 14th March is Tuesday, one day later is Wednesday. Verification / Alternative check:Another way to think about it is to note that 28 days is exactly 4 weeks, so contributes zero net shift in day of week. We can try to adjust the 183 days to the nearest multiple of 28 or 7. Subtracting 182, which is exactly 26 weeks, leaves 1 day. This confirms that we have advanced one weekday from the reference date. Because the reference day is Tuesday, the target day must be Wednesday. Both methods align and strengthen confidence in the result. Why Other Options Are Wrong:Monday would represent a shift of minus one day or six days forward, which does not match a remainder of 1. Tuesday would require the remainder to be 0 or a multiple of 7, meaning the total day difference would have to be divisible by 7, which it is not. Thursday would represent a shift of two days forward, which would come from a remainder of 2. Only Wednesday corresponds correctly to a remainder of 1 day after Tuesday. Common Pitfalls:Common mistakes include miscounting the number of days in a month, forgetting to exclude the starting day appropriately or making arithmetic slips in the total sum. Some students also confuse whether to include both endpoints or only one. Here we count the days that pass between 14th March and 13th September, and the modulo 7 calculation handles the wrap around. Careful month by month counting prevents errors. Final Answer:Pranay birthday on 13th September of the same year will fall on Wednesday.

Question 7

The weight of C is twice the weight of M. The weight of M is one ninth of the weight of R. The weight of R is five by two times the weight of K. The weight of K is one third of the weight of G. Who is the second lightest among C, M, K, G and R?

Accepted Answer

Introduction / Context:This problem compares the weights of five people or objects, named C, M, K, G and R, using a chain of proportional relationships. You are not given explicit weights in kilograms, but instead how each weight relates to another. The task is to find who is the second lightest, which requires building an ordering from lightest to heaviest based on these ratios. It is a good exercise in translating fractional relationships into a consistent ranking. Given Data / Assumptions:C is twice as heavy as M.M is one ninth of the weight of R.R is five by two times the weight of K.K is one third of the weight of G.All weights are positive and distinct.We want to identify the second lightest among C, M, K, G and R. Concept / Approach:A clear strategy is to express all weights in terms of a single base variable, such as M. Once each person weight is written as a multiple of M, it is straightforward to compare them numerically. Because multipliers are all positive, the order of the coefficients directly reflects the order of the actual weights. We can then sort these coefficients from smallest to largest to find the lightest and second lightest individuals. Step-by-Step Solution:Step 1: Let the weight of M be M units. We will express all other weights in terms of M.Step 2: From “C is twice the weight of M”, we get C = 2M.Step 3: From “M is one ninth of the weight of R”, we have M = R / 9, so R = 9M.Step 4: From “R is five by two times the weight of K”, we have R = (5 / 2) * K. Substituting R = 9M gives 9M = (5 / 2) * K.Step 5: Solve this for K: K = (2 / 5) * 9M = (18 / 5) * M, which is 3.6M.Step 6: From “K is one third of the weight of G”, we have K = G / 3, so G = 3K.Step 7: Substitute K = (18 / 5) * M into G = 3K to get G = 3 * (18 / 5) * M = (54 / 5) * M, which is 10.8M.Step 8: Now list all weights in terms of M: M = 1M, C = 2M, K = 3.6M, R = 9M, G = 10.8M.Step 9: Order these coefficients from smallest to largest: 1, 2, 3.6, 9 and 10.8, corresponding to M, C, K, R and G respectively.Step 10: The lightest is M and the second lightest is C. Verification / Alternative check:You can assign a concrete value to M, for example M = 10 kilograms, and compute all weights explicitly. Then C = 20, K = 36, R = 90 and G = 108 kilograms. Clearly the order from lightest to heaviest is M (10), C (20), K (36), R (90), G (108). The second lightest remains C. Any positive choice for M will preserve the ordering because all relations are linear and proportional. Why Other Options Are Wrong:M is the lightest, not the second lightest, since all other weights are multiples greater than 1 times M. K is heavier than both M and C because 3.6M is greater than 2M. R and G are much heavier, being 9M and 10.8M, so they are nowhere near the bottom end of the ranking. Therefore options b, c and d cannot represent the second lightest individual. Common Pitfalls:Some learners confuse phrases like “one ninth of” and “five by two times” and invert the fractions incorrectly. Another common error is trying to reason purely in words without expressing the relationships algebraically, which can lead to inconsistent or mistaken comparisons. Writing each expression in terms of one base weight and then comparing numerical coefficients is the safest approach. Final Answer:The second lightest among the five is C.

Question 8

Four boxes have weights 90 kg, 40 kg, 80 kg and 50 kg. Each box can be used at most once in a combination. Which of the following values cannot represent the total weight, in kilograms, of any possible combination of these four boxes?

Accepted Answer

Introduction / Context: This quantitative aptitude question tests understanding of combinations and simple arithmetic. The aim is to decide which given total weight is not achievable when selecting some or all of the boxes exactly once. Such questions are common in banking and SSC exams because they check both logical completeness and careful calculation. Given Data / Assumptions: There are four boxes with weights 90 kg, 40 kg, 80 kg and 50 kg. Any box can be used at most once in a combination. We can form totals using one, two, three or all four boxes. We need to identify which of the given totals cannot be obtained from any valid combination. Concept / Approach: The key idea is to list or logically derive all possible sums of these weights. Since there are only four boxes, the number of different non-empty subsets is limited. We systematically consider totals of one box, two boxes, three boxes and all four boxes. After that, we compare the list of achievable totals with the options and see which option does not appear in the list. Step-by-Step Solution: Step 1: Single-box totals are 90, 40, 80 and 50 kilograms. Step 2: Two-box totals are 90 + 40 = 130, 90 + 80 = 170, 90 + 50 = 140, 40 + 80 = 120, 40 + 50 = 90 and 80 + 50 = 130 kilograms. Step 3: Three-box totals are 90 + 40 + 80 = 210, 90 + 40 + 50 = 180, 90 + 80 + 50 = 220 and 40 + 80 + 50 = 170 kilograms. Step 4: Four-box total is 90 + 40 + 80 + 50 = 260 kilograms. Step 5: Collecting all distinct totals, we obtain 40, 50, 80, 90, 120, 130, 140, 170, 180, 210, 220 and 260 kilograms. Step 6: Compare these values with the options 200, 260, 180 and 170. Verification / Alternative check: We can also reason directly for 200 kg. The largest total less than or equal to 200 is 90 + 80 + 40 = 210, which is already more than 200. Any three-box total either equals 210, 180 or 170, none of which is 200. Testing all two-box combinations like 90 + 50, 90 + 40, 80 + 50 and 80 + 40 also shows that none equals 200. Therefore, 200 kg cannot be obtained by any allowed combination. Why Other Options Are Wrong: Option 260 kg is achievable by using all four boxes together: 90 + 40 + 80 + 50 = 260. Option 180 kg is achievable via the three-box combination 90 + 40 + 50 = 180. Option 170 kg is achievable via either 90 + 80 = 170 or 40 + 80 + 50 = 170. Common Pitfalls: Candidates often forget to include three-box combinations or accidentally reuse a box twice, which is not allowed. Another common mistake is to miss a valid combination or perform an incorrect addition, leading to a wrong conclusion about which total is impossible. Writing the combinations carefully and checking sums avoids these errors. Final Answer: The total weight that cannot be formed is 200 kilograms.

Question 9

In a certain year, Nasir has a birthday on Thursday, 18 May. In the same calendar year, Rehan was born on 19 August. On which day of the week does Rehan's birthday fall in that year?

Accepted Answer

Introduction / Context: This reasoning question checks understanding of calendars and day counting within a single year. We are given the day of the week for one specific date and must find the day for another date later in the same year. It is a classic exam problem that tests careful counting of days and working with remainders modulo seven. Given Data / Assumptions: Nasir has a birthday on Thursday, 18 May of a particular year. Rehan has a birthday on 19 August of the same year. We assume a normal Gregorian calendar with May having 31 days, June 30 days, July 31 days and August 31 days. We need only the difference in days between the two dates within the same year, so leap year status does not affect the answer. Concept / Approach: The day of the week advances by one for each extra day. Therefore, if the second date is n days after the first date, we simply add n (modulo 7) to the known starting day. To apply this, we compute the exact number of days between 18 May and 19 August and then use the remainder when divided by seven to shift from Thursday to the required day. Step-by-Step Solution: Step 1: Count days remaining in May from 18 May to 31 May, which is 31 - 18 = 13 days. Step 2: June has 30 days, so add 30 days. Step 3: July has 31 days, so add another 31 days. Step 4: From 1 August to 19 August inclusive of 19 August and exclusive of 18 May gives 19 days. Step 5: Total difference in days = 13 + 30 + 31 + 19 = 93 days. Step 6: Compute 93 modulo 7. Since 7 * 13 = 91, the remainder is 93 - 91 = 2. Step 7: Starting from Thursday and moving forward 2 days gives: Friday (1 day), Saturday (2 days). Verification / Alternative check: We can verify by grouping days into complete weeks. Ninety three days consist of 13 full weeks (91 days) plus 2 extra days. Full weeks do not change the day of the week, so only the extra 2 days matter, again confirming that the weekday moves from Thursday to Saturday. Why Other Options Are Wrong: Wednesday would correspond to moving backward rather than forward, which contradicts the fact that August is after May. Friday would be correct only if the difference were 1 day modulo seven, which is not true. Thursday would require the difference to be a multiple of seven days with no remainder, which is also not the case. Common Pitfalls: A common mistake is to count both starting and ending dates or to miscount days in a month. Another error is forgetting to use modulo seven and instead using the raw day difference. Writing each month's contribution separately and then using the remainder eliminates such errors in calendar questions. Final Answer: Rehan's birthday in that year falls on Saturday.

Question 10

The weights of four boxes are 30 kg, 40 kg, 50 kg and 100 kg. Each box can be used at most once in a combination. Which of the following values cannot be the total weight, in kilograms, of any possible combination of these boxes?

Accepted Answer

Introduction / Context: This is another combination and arithmetic reasoning question. We are given four fixed weights and asked to decide which of the given totals cannot be produced using any subset of these weights when each box is used at most once. Exam questions of this type train students to systematically consider all feasible combinations rather than guess. Given Data / Assumptions: Box weights are 30 kg, 40 kg, 50 kg and 100 kg. Any box can appear in a combination either once or not at all. We may use one, two, three or all four boxes in making a total. We must find which option cannot be expressed as a valid sum. Concept / Approach: The best method is to compute or logically derive all distinct possible sums from the four given numbers. Because there are only four boxes, the total number of non-empty subsets is small. Once we list all possible totals, we simply compare them with the options and identify the one value that does not occur in our list. Step-by-Step Solution: Step 1: Single-box totals are 30, 40, 50 and 100 kilograms. Step 2: Two-box totals are 30 + 40 = 70, 30 + 50 = 80, 30 + 100 = 130, 40 + 50 = 90, 40 + 100 = 140 and 50 + 100 = 150 kilograms. Step 3: Three-box totals are 30 + 40 + 50 = 120, 30 + 40 + 100 = 170, 30 + 50 + 100 = 180 and 40 + 50 + 100 = 190 kilograms. Step 4: The four-box total is 30 + 40 + 50 + 100 = 220 kilograms. Step 5: Collecting all distinct totals we get 30, 40, 50, 70, 80, 90, 100, 120, 130, 140, 150, 170, 180, 190 and 220. Step 6: Compare with the options 190, 180, 160 and 140. Note that 160 kilograms does not appear in the achievable totals. Verification / Alternative check: To be sure about 160 kg, try constructing it directly. Two-box combinations such as 100 + 50, 100 + 40, 100 + 30, 50 + 40 and 50 + 30 give totals of 150, 140, 130, 90 and 80 kilograms. Three-box combinations such as 100 + 40 + 30 or 100 + 50 + 30 give 170 and 180 kilograms. No combination yields exactly 160 kilograms, confirming that it is impossible. Why Other Options Are Wrong: Option 190 kg is obtainable as 40 + 50 + 100 = 190. Option 180 kg is obtainable as 30 + 50 + 100 = 180. Option 140 kg is obtainable as 40 + 100 = 140. Common Pitfalls: Students often forget one or more combinations or assume that if two nearby totals are possible then all intermediate values are also possible, which is not true when only specific weights are available. Careful listing or structured reasoning is essential for correctness in such discrete combination questions. Final Answer: The total weight that cannot be formed is 160 kilograms.

Question 11

Among four people A, B, C and E, C has more money than E, and E has more money than B. C is the second richest person after A. Who has the least amount of money among A, B, C and E?

Accepted Answer

Introduction / Context: This is a simple logical reasoning question about ranking people by the amount of money they have. Such questions appear frequently in aptitude tests, where candidates must convert verbal comparisons into an ordered list and then answer queries about positions like richest, poorest, or second richest. Given Data / Assumptions: There are four people: A, B, C and E. C has more money than E. E has more money than B. C is the second richest person after A. No other people are involved besides A, B, C and E. Concept / Approach: The statements describe relative order, not exact amounts. We translate each comparison into inequality relations and then merge them into a single ranking from richest to poorest. When we know that one person is the second richest, we can place that person and then position the remaining individuals above or below based on the other comparisons. Finally, we look for the person at the bottom of the order, which corresponds to the least amount of money. Step-by-Step Solution: Step 1: From “C has more money than E”, write C > E. Step 2: From “E has more money than B”, write E > B. Step 3: Combining these two relations gives C > E > B, meaning C is above E and E is above B. Step 4: From “C is the second richest person after A”, we infer that A is the richest, and C is placed immediately after A in the order. Step 5: Now we can write the complete order from richest to poorest as A > C > E > B. Step 6: The least amount of money corresponds to the person at the end of this chain, which is B. Verification / Alternative check: We can imagine approximate amounts that satisfy all comparisons, for example A = 100, C = 90, E = 60 and B = 40. These values satisfy C > E, E > B and C as second richest after A. With any such consistent assignment, B always appears last, so B is inevitably the poorest. Why Other Options Are Wrong: Person C cannot be the poorest because C is explicitly richer than both E and B. Person A is stated to be richer than C, so A is the richest, not the poorest. Person E has more money than B, so E also cannot be the least. Common Pitfalls: Candidates sometimes misread “second richest” and place that person incorrectly, or they forget to combine all given inequalities. Drawing a simple arrow diagram or chain from richest to poorest helps avoid confusion. Final Answer: The person with the least money is B.

Question 12

In a certain town, the average temperature for the first six days of a month was 41 °C. The sum of the temperatures recorded during the first five days alone was 201 °C. What was the temperature, in degrees Celsius, on the sixth day of that month?

Accepted Answer

Introduction / Context: This question tests basic understanding of averages and sums. Average temperature questions are common in competitive exams because they require candidates to relate total sums, number of observations and missing values. Here, we know the average over six days and the sum of the first five days, and we must find the missing sixth value. Given Data / Assumptions: Average temperature for the first six days of the month = 41 °C. Sum of temperatures for the first five days of the month = 201 °C. Temperatures are measured in degrees Celsius. We assume that the “average” is the arithmetic mean. Concept / Approach: The average of a set of numbers is defined as total sum divided by the number of observations. If we know the average and the number of days, we can compute the total sum. Once we know the total for six days and the total for the first five days, the sixth day's temperature is simply the difference between these two sums. Step-by-Step Solution: Step 1: Use the formula average = total sum / number of observations. Step 2: Total sum for six days = average * number of days = 41 * 6 = 246 °C. Step 3: Sum of temperatures for the first five days = 201 °C (given). Step 4: Temperature on the sixth day = total for six days - total for first five days. Step 5: Sixth day temperature = 246 - 201 = 45 °C. Verification / Alternative check: We can check our result by imagining actual numbers. Assume the first five days have temperatures that add exactly to 201 °C and the sixth day is 45 °C. Then the total for six days is 201 + 45 = 246 °C. Dividing 246 by 6 gives 41 °C, which matches the given average, confirming the correctness of the calculation. Why Other Options Are Wrong: Option 40 °C would give a total sum of 201 + 40 = 241 °C, and the average would become 241 / 6, which is not 41 °C. Option 46 °C would give a total sum of 201 + 46 = 247 °C, again not consistent with the given average. Option 50 °C would make the total 201 + 50 = 251 °C, which also fails to give an average of 41 °C. Common Pitfalls: A common error is to average the options directly or to divide the sum of given information incorrectly. Always start from the basic formula total = average * number of observations and work step by step to avoid arithmetic mistakes. Final Answer: The temperature on the sixth day of the month was 45 °C.

Question 13

Five boys A, B, C, D and E are standing in a line according to their heights. A is taller than E but shorter than D. B is shorter than E, and C is the tallest among all five boys. Who occupies the middle position when they are arranged from tallest …

Accepted Answer

Introduction / Context: This question belongs to the category of ranking and ordering based on height. We are given pairwise comparisons among five boys and must construct a full height order. Once the order is known, identifying the middle position becomes straightforward. Such problems strengthen logical deduction skills. Given Data / Assumptions: Five boys: A, B, C, D and E. A is taller than E but shorter than D, so D > A > E. B is shorter than E, so E > B. C is the tallest of all the boys. We want the boy in the central position when the boys are ordered from tallest to shortest. Concept / Approach: We interpret each comparison as an inequality and combine them to build a single chain from tallest to shortest. Knowing that C is tallest immediately fixes the top position. Then we insert D, A, E and B into the chain in a way that satisfies all inequalities. The boy who ends up with exactly two boys taller and two boys shorter is in the middle. Step-by-Step Solution: Step 1: From “C is the tallest”, place C at the top of the height order. Step 2: From “A is taller than E but shorter than D”, we write D > A > E. Step 3: From “B is shorter than E”, we know E > B. Step 4: Combining everything, we have C at the top, D below C, then A, then E, then B, giving C > D > A > E > B. Step 5: When arranged from tallest to shortest, the sequence is C, D, A, E, B. The middle position (third position) is occupied by A. Verification / Alternative check: We can quickly check each relation in the final order. A is indeed shorter than D and taller than E. B is at the bottom and therefore shorter than E. C is at the top and therefore taller than all others. Since all conditions are satisfied, the order is consistent. Why Other Options Are Wrong: C is the tallest and therefore at the first position, not in the middle. D is second from the top because he is shorter than C but taller than A. E is fourth in the order, since both C and D and A are taller than E, and E is taller than only B. Common Pitfalls: Candidates may misread “shorter than” and reverse inequalities or forget to include B's relation. Drawing the order as a vertical list from tallest to shortest or vice versa greatly reduces confusion. Final Answer: The boy standing in the middle position is A.

Question 14

Five friends Amrita, Nikita, Nishu, Pooja and Riya are standing in a line according to their heights. Nishu is taller than Riya but shorter than Pooja. Amrita is the shortest. Riya is shorter than Nishu but taller than Nikita. Who is the second tall…

Accepted Answer

Introduction / Context: This ranking question requires building a height order from multiple comparative statements. The friends are compared pairwise, and students must combine the information to derive a unique sequence from tallest to shortest. The goal is to identify the second tallest person, which is a very common type of reasoning task in competitive exams. Given Data / Assumptions: Friends: Amrita, Nikita, Nishu, Pooja and Riya. Nishu is taller than Riya but shorter than Pooja. Amrita is the shortest among all five. Riya is shorter than Nishu but taller than Nikita. All comparisons are consistent and refer to the same group of five people. Concept / Approach: We convert the verbal statements into inequalities describing heights: “taller than” and “shorter than”. By chaining these inequalities, we place each person in the correct relative position. Once we get the complete order, it is easy to identify the second tallest friend. Step-by-Step Solution: Step 1: From “Nishu is taller than Riya but shorter than Pooja”, we write Pooja > Nishu > Riya. Step 2: From “Riya is shorter than Nishu but taller than Nikita”, we write Nishu > Riya > Nikita. Step 3: Combining these, we obtain Pooja > Nishu > Riya > Nikita as a partial order. Step 4: Amrita is given as the shortest, so Amrita must come below Nikita in the final order. Step 5: Therefore, the full order from tallest to shortest is Pooja, Nishu, Riya, Nikita and Amrita. Step 6: The second tallest person is the one just below Pooja in this order, that is, Nishu. Verification / Alternative check: Check each original condition against the final order. Nishu is indeed shorter than Pooja and taller than Riya. Riya is taller than Nikita but shorter than Nishu. Amrita is last, so she is the shortest. All conditions match the derived order, confirming its correctness. Why Other Options Are Wrong: Pooja is the tallest, not the second tallest. Riya appears in the middle of the order, so she is neither tallest nor second tallest. Amrita is explicitly said to be the shortest, so she cannot be second tallest. Common Pitfalls: Some students may forget that “shortest” fixes Amrita at the bottom, or they might mistakenly swap Riya and Nikita. Sketching a simple vertical line and placing names step by step in the correct order helps to avoid such confusion in ranking questions. Final Answer: The second tallest friend in the group is Nishu.

Question 15

In a jungle racing event involving five animals — deer, dog, elephant, lion and tiger — the dog ran faster than the elephant but slower than the tiger. The deer was the fastest of all. The lion ran faster than the tiger. Which animal finished second…

Accepted Answer

Introduction / Context: This is a classic ordering and comparison question framed in a story about animal racing. We are given relative speeds and must determine the finishing order. The aim is to identify the animal that comes second in the race. Such problems train candidates to convert verbal comparisons into a clear ranked sequence. Given Data / Assumptions: Animals in the race: deer, dog, elephant, lion and tiger. The dog ran faster than the elephant but slower than the tiger. The deer was the fastest among all five animals. The lion ran faster than the tiger. No two animals tied; all rankings are strict. Concept / Approach: Speed order translates directly into finishing positions: the fastest animal finishes first, the second fastest finishes second, and so on. From the given comparisons, we build an inequality chain of speeds. After constructing the complete order, we read off which animal is second in that list. Step-by-Step Solution: Step 1: From “the deer was the fastest”, we place deer at the top of the speed order. Step 2: The dog ran faster than the elephant but slower than the tiger, so we have tiger > dog > elephant. Step 3: The lion ran faster than the tiger, so lion > tiger. Step 4: Combining all, the speed order from fastest to slowest becomes deer > lion > tiger > dog > elephant. Step 5: Therefore, the second place in the race corresponds to the second fastest animal, which is the lion. Verification / Alternative check: We can quickly verify that all given relations are satisfied by this order. The deer is indeed first. The dog is faster than the elephant but slower than the tiger. The lion is placed above the tiger, so it is faster than the tiger. Consequently, this order is consistent and unique. Why Other Options Are Wrong: The deer is explicitly the fastest and thus finished first, not second. The dog appears in the middle of the order, slower than both lion and tiger, so it cannot be second. The elephant is the slowest animal according to the chain and must finish last, not second. Common Pitfalls: Candidates may misread “faster than” versus “slower than” and accidentally reverse the order of two animals. It is helpful to write inequalities in one direction, for example always from fastest to slowest, to prevent this common mistake. Final Answer: The animal that finished second in the race is the lion.

Question 16

Five friends Anjum, Ankit, Amit, Priya and Ram are sitting on a bench in a row facing north. Ankit is sitting immediately to the right of Anjum. Amit is sitting to the left of Priya and immediately to the right of Ram. Ram is sitting somewhere to th…

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Introduction / Context: This is a seating arrangement question in which five friends sit on a bench facing north. The problem provides relative positions using phrases like “immediately to the right” and “to the left of”. We must use these clues together to reconstruct the unique arrangement and identify who sits at the extreme right end. Given Data / Assumptions: Persons: Anjum, Ankit, Amit, Priya and Ram. They sit in a single row, all facing north. Ankit sits immediately to the right of Anjum. Amit sits to the left of Priya and immediately to the right of Ram. Ram sits somewhere to the right of Ankit. Positions are fixed, and no two people share the same seat. Concept / Approach: When everyone faces north, the right side is to the east and appears as the right side on the page. We treat the row as ordered from left to right. “Immediately to the right” means adjacent and one position to the right. We construct the seating positions step by step, ensuring all conditions hold simultaneously, and then determine who occupies the last (rightmost) position. Step-by-Step Solution: Step 1: Let the bench positions from left to right be numbered 1 to 5. Step 2: From “Ankit is sitting immediately to the right of Anjum”, we get a block (Anjum, Ankit) in consecutive positions, with Ankit to the right. Step 3: From “Amit is sitting to the left of Priya and immediately to the right of Ram”, we form another block (Ram, Amit, Priya) where Amit is in the middle and Priya is to the right of Amit. Step 4: Ram is to the right of Ankit, so the block (Ram, Amit, Priya) must appear to the right of Ankit. Combining both blocks yields the sequence Anjum, Ankit, Ram, Amit, Priya. Step 5: Therefore, from left to right, the arrangement is Anjum (1), Ankit (2), Ram (3), Amit (4) and Priya (5). Step 6: The person at the extreme right end (position 5) is Priya. Verification / Alternative check: Verify each statement: Ankit is immediately to the right of Anjum (positions 1 and 2). Ram is to the right of Ankit (position 3). Amit is immediately to the right of Ram (positions 3 and 4) and to the left of Priya (positions 4 and 5). All conditions are satisfied, so the arrangement is consistent. Why Other Options Are Wrong: Amit sits in the second position from the right, not at the extreme right end. Ankit is between Anjum and Ram, so he cannot be at either extreme end. Anjum sits at the extreme left, not at the extreme right. Common Pitfalls: Candidates may forget that all are facing north and accidentally reverse left and right. Drawing labeled boxes from left to right and marking each condition makes it much easier to avoid such orientation errors. Final Answer: The person sitting at the extreme right end of the bench is Priya.

Question 17

The weights of four boxes are 20 kg, 90 kg, 40 kg and 60 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?

Accepted Answer

Introduction / Context: This problem is another example of finding impossible totals from a given set of weights. We must determine which option cannot be formed by adding one or more of the four given box weights, with each box used at most once. Systematic enumeration or careful reasoning is key to avoiding missed combinations. Given Data / Assumptions: Box weights: 20 kg, 90 kg, 40 kg and 60 kg. Any box may be chosen once or not at all in a combination. We may use one, two, three or all four boxes. We want the option that cannot match any valid sum. Concept / Approach: Because there are only four boxes, listing all possible sums is manageable. We compute totals for single boxes, then for each pair, each triple, and finally for all four boxes together. Once we have the full set of achievable totals, we compare them with the provided options and see which value is missing. Step-by-Step Solution: Step 1: Single-box totals are 20, 40, 60 and 90 kilograms. Step 2: Two-box totals: 20 + 40 = 60, 20 + 60 = 80, 20 + 90 = 110, 40 + 60 = 100, 40 + 90 = 130 and 60 + 90 = 150. Step 3: Three-box totals: 20 + 40 + 60 = 120, 20 + 40 + 90 = 150, 20 + 60 + 90 = 170 and 40 + 60 + 90 = 190. Step 4: Four-box total: 20 + 40 + 60 + 90 = 210 kilograms. Step 5: Collecting all distinct totals gives 20, 40, 60, 80, 90, 100, 110, 120, 130, 150, 170, 190 and 210 kilograms. Step 6: Compare with options 210, 170, 190 and 200. The only value not present in the list is 200. Verification / Alternative check: To double-check, try to construct 200 kg directly. The largest total less than or equal to 200 from three boxes is 40 + 60 + 90 = 190. Including the remaining box makes the total 210, which is too large. Testing all two-box combinations such as 90 + 60, 90 + 40 and 60 + 40 yields 150, 130 and 100 kilograms respectively, none of which is 200. Therefore 200 kg is genuinely impossible. Why Other Options Are Wrong: Option 210 kg is the sum of all four boxes: 20 + 40 + 60 + 90. Option 170 kg is obtained by 20 + 60 + 90. Option 190 kg is obtained by 40 + 60 + 90. Common Pitfalls: One frequent error is to skip a combination or reproduce the same total twice and mistakenly believe all values are covered. Writing the combinations in a systematic order, such as sorting by number of boxes used, helps ensure that no possible sum is overlooked. Final Answer: The total weight that cannot be formed is 200 kilograms.

Question 18

The weights of four boxes are 30 kg, 20 kg, 50 kg and 90 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?

Accepted Answer

Introduction / Context: This question is another variant of the subset-sum style problems using small integer weights. We must determine which of the given totals cannot be made by adding together some or all of the four given box weights, with each box used at most once. Such problems test careful enumeration and number sense. Given Data / Assumptions: Box weights are 30 kg, 20 kg, 50 kg and 90 kg. Each box can appear at most once in a combination. We may select one, two, three or all four boxes. We have to find which option cannot be expressed as a valid sum. Concept / Approach: Since the number of boxes is small, we can compute all possible sums explicitly. Listing all single-box, two-box, three-box and four-box totals guarantees that we do not miss any combination. Finally, we compare this list to the options to see which total is missing and therefore impossible. Step-by-Step Solution: Step 1: Single-box totals are 20, 30, 50 and 90 kilograms. Step 2: Two-box totals: 20 + 30 = 50, 20 + 50 = 70, 20 + 90 = 110, 30 + 50 = 80, 30 + 90 = 120 and 50 + 90 = 140 kilograms. Step 3: Three-box totals: 20 + 30 + 50 = 100, 20 + 30 + 90 = 140, 20 + 50 + 90 = 160 and 30 + 50 + 90 = 170 kilograms. Step 4: Four-box total is 20 + 30 + 50 + 90 = 190 kilograms. Step 5: Collecting all distinct sums gives 20, 30, 50, 70, 80, 90, 100, 110, 120, 140, 160, 170 and 190 kilograms. Step 6: The options provided are 200, 190, 140 and 160. The only value not in the list is 200, so it cannot be formed. Verification / Alternative check: We can also reason directly about 200 kg. The largest possible sum using all four boxes is 190 kg, which is already less than 200 kg. Therefore, there is no way to reach 200 kg without exceeding the available weights. Because the total of all four boxes is below 200, no subset of these boxes can possibly reach or exceed 200 kg, confirming that 200 kg is impossible. Why Other Options Are Wrong: Option 190 kg is equal to the sum of all four boxes: 20 + 30 + 50 + 90. Option 140 kg can be formed either as 50 + 90 or as 20 + 30 + 90. Option 160 kg is formed as 20 + 50 + 90. Common Pitfalls: Some candidates may incorrectly assume that if all weights are less than 100 then any large total is impossible, overlooking that sums of multiple boxes can exceed 100. Others may forget to test the total of all boxes. Carefully checking both logic and arithmetic helps avoid these mistakes. Final Answer: The total weight that cannot be formed is 200 kilograms.

Question 19

The weights of four boxes are 30 kg, 20 kg, 60 kg and 70 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?

Accepted Answer

Introduction / Context: This question continues the theme of identifying impossible totals from a small set of weights. Candidates must be comfortable enumerating possible sums and recognizing which target value cannot be formed when each box can be used at most once. It is a typical numerical reasoning task. Given Data / Assumptions: Box weights: 30 kg, 20 kg, 60 kg and 70 kg. Any box may appear at most once in a combination. We can use one, two, three or all four boxes. We must find which option cannot match any valid sum. Concept / Approach: We again use systematic listing: compute all sums of one box, then of each pair, then of each triple, and finally of all four boxes together. After collecting all distinct totals, we compare them with the answer choices. The number that does not appear in the list is impossible. Step-by-Step Solution: Step 1: Single-box totals are 20, 30, 60 and 70 kilograms. Step 2: Two-box totals: 20 + 30 = 50, 20 + 60 = 80, 20 + 70 = 90, 30 + 60 = 90, 30 + 70 = 100 and 60 + 70 = 130 kilograms. Step 3: Three-box totals: 20 + 30 + 60 = 110, 20 + 30 + 70 = 120, 20 + 60 + 70 = 150 and 30 + 60 + 70 = 160 kilograms. Step 4: Four-box total: 20 + 30 + 60 + 70 = 180 kilograms. Step 5: Collecting distinct totals gives 20, 30, 50, 60, 70, 80, 90, 100, 110, 120, 130, 150, 160 and 180 kilograms. Step 6: The options are 180, 170, 120 and 150. Only 170 kilograms does not appear in the achievable totals. Verification / Alternative check: To confirm that 170 kg is impossible, try constructing it. The largest three-box total is 30 + 60 + 70 = 160 kg. Adding the remaining 20 kg gives 180 kg, which exceeds 170. No two-box combination equals 170 either, because 70 + 60 = 130 and 70 + 30 = 100. Hence, 170 kg cannot be formed using these weights once each. Why Other Options Are Wrong: Option 180 kg is equal to the sum of all four boxes. Option 120 kg is obtained as 20 + 30 + 70. Option 150 kg can be formed as 20 + 60 + 70. Common Pitfalls: Some test takers may assume that any number between the smallest and largest total is possible, which is not true when only specific weights are available. Others may forget three-box combinations. A neat table of all one, two, three and four box sums avoids such mistakes. Final Answer: The total weight that cannot be formed is 170 kilograms.

Question 20

In a company hierarchy, Kathir is senior to Ganesh, and Ganesh is senior to Apparu. Apparu is junior to Raju, while Raju is junior to Ganesh. Who is the most senior person among Kathir, Ganesh, Apparu and Raju?

Accepted Answer

Introduction / Context: This ranking question deals with seniority in an organizational hierarchy. Statements describe who is senior or junior compared to whom. The goal is to determine the most senior person among the four given individuals. Such problems test a student's ability to combine multiple relational statements into a consistent order. Given Data / Assumptions: Kathir is senior to Ganesh. Ganesh is senior to Apparu. Apparu is junior to Raju, which means Raju is senior to Apparu. Raju is junior to Ganesh, so Ganesh is senior to Raju. We consider only these four people: Kathir, Ganesh, Apparu and Raju. Concept / Approach: We translate each statement into a simple inequality representing order of seniority. Seniority can be seen as a ranking where higher positions mean greater authority. By combining all inequalities into a single chain, we can clearly see who is at the top of the hierarchy and who is at the bottom. Step-by-Step Solution: Step 1: From “Kathir is senior to Ganesh”, write Kathir > Ganesh. Step 2: From “Ganesh is senior to Apparu”, write Ganesh > Apparu. Step 3: From “Apparu is junior to Raju”, deduce Raju > Apparu. Step 4: From “Raju is junior to Ganesh”, deduce Ganesh > Raju. Step 5: Now combine these: Kathir > Ganesh, Ganesh > Raju and Raju > Apparu, which gives a complete order Kathir > Ganesh > Raju > Apparu. Step 6: The most senior person is the one at the start of this chain, which is Kathir. Verification / Alternative check: We can imagine assigning ranks: Kathir as rank 1, Ganesh as rank 2, Raju as rank 3 and Apparu as rank 4. All statements are satisfied: Kathir is above Ganesh, Ganesh above Apparu, Raju above Apparu and Ganesh above Raju. There is no contradiction, so the order is consistent. Why Other Options Are Wrong: Ganesh is senior to Raju and Apparu but junior to Kathir, so he is not the most senior. Raju is senior only to Apparu and is specifically said to be junior to Ganesh and therefore also junior to Kathir. Apparu is at the bottom of the chain, junior to both Ganesh and Raju, so he cannot be the most senior. Common Pitfalls: A frequent error is to misinterpret “junior to” and reverse the direction of the relation. Remember that “X is junior to Y” means Y holds a higher rank. Carefully translating each phrase into an inequality helps prevent such mistakes. Final Answer: The most senior person in the group is Kathir.

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