Question 1

Vishal is 24th from each end of a row of boys; that is, he is 24th from the left and also 24th from the right. How many boys are there in the row in total?

Accepted Answer

Introduction / Context: This is a simple ranking question involving a boy who has the same position from both ends of a row. Such problems are very common in aptitude tests and are designed to test your understanding of how positions from the left and right relate to the total number of people in the row. Given Data / Assumptions: - Vishal is 24th from the left end of the row. - Vishal is also 24th from the right end of the row. - The row is fully occupied; every position from 1 to N is taken by exactly one boy. Concept / Approach: If a person is at position L from the left and position R from the right in a row of N people, then the total number of people is given by N = L + R - 1. This comes from counting the person once and counting the number of people to the left and right of him. In this question, L and R are equal (both 24), which makes the calculation very straightforward. Step-by-Step Solution: Step 1: Let N be the total number of boys in the row. Step 2: Vishal is 24th from the left, so L = 24. Step 3: Vishal is also 24th from the right, so R = 24. Step 4: Use the formula N = L + R - 1. Step 5: Substitute the values: N = 24 + 24 - 1. Step 6: Calculate N = 48 - 1 = 47. Step 7: Therefore, there are 47 boys in the row. Verification / Alternative check: To verify, imagine a row of 47 boys. If Vishal is 24th from the left, then there are 23 boys to his left. Similarly, if he is 24th from the right, there must be 23 boys to his right. Counting everyone gives 23 (left) + 1 (Vishal) + 23 (right) = 47, which matches the total N. This confirms that the formula N = L + R - 1 is correctly applied and that 47 is the correct total. Why Other Options Are Wrong: 48: For N = 48, someone who is 24th from one end would be 25th from the other, not 24th from both ends. 49: For N = 49, a person who is 24th from one end would be 26th from the other, so the positions cannot match. 46: For N = 46, a person 24th from one end would be 23rd from the other, again inconsistent with the question. Common Pitfalls: A common mistake is to simply add the two ranks and forget to subtract 1, providing L + R instead of L + R - 1. Another error is to confuse the notion of “from either end” and assume two different people rather than the same person with mirrored positions. Remembering the derivation of the formula helps avoid such mistakes in similar problems. Final Answer: The total number of boys in the row is 47.

Question 2

In the following series of numbers, how many times does the pattern 1, 7, 3 or 3, 7, 1 occur, with 7 in the middle and 1 and 3 on either side: 2 9 7 3 1 7 3 7 7 1 3 3 1 7 3 8 5 7 1 3 7 7 1 7 3 9 0 6?

Accepted Answer

Introduction / Context: Pattern recognition in number sequences is a common type of reasoning question. Here, you are asked to identify how many times a specific three number pattern appears with a fixed middle number. The middle number is always 7, and the numbers on either side must be 1 and 3 in any order (either 1, 7, 3 or 3, 7, 1). The goal is to scan the entire sequence and count all such occurrences accurately. Given Data / Assumptions: - The series is: 2 9 7 3 1 7 3 7 7 1 3 3 1 7 3 8 5 7 1 3 7 7 1 7 3 9 0 6. - We seek triples of consecutive numbers where the middle number is 7. - The two neighbours must be 1 and 3, in any order. - Overlapping patterns are allowed as long as each triple satisfies the condition. Concept / Approach: The approach is to slide a window of three consecutive numbers across the sequence and inspect the middle element. Whenever the middle element is 7, check if the left and right elements are exactly 1 and 3 in any order. If they are, we count one valid occurrence. Proceeding this way from the start to the end ensures that no potential triple is missed and that the count is accurate. Step-by-Step Solution: Step 1: Rewrite the sequence with indices to avoid confusion: Index: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Value: 2 9 7 3 1 7 3 7 7 1 3 3 1 7 3 8 5 7 1 3 7 7 1 7 3 9 0 6 Step 2: Consider triples (i - 1, i, i + 1) where the middle index i runs from 2 to 27. Step 3: For each triple, check if the middle value is 7. Step 4: When the middle is 7, check whether the other two numbers are 1 and 3 in any order. Step 5: At indices (4, 5, 6): 3, 1, 7 is not valid because 7 is not in the middle. Step 6: Look at indices (5, 6, 7): 1, 7, 3. Here the middle is 7, and neighbours are 1 and 3. Count 1. Step 7: At indices (13, 14, 15): 1, 7, 3. Again the middle is 7 with neighbours 1 and 3. Count 2. Step 8: At indices (21, 22, 23): 7, 7, 1 has middle 7 but neighbours are 7 and 1, not 1 and 3, so do not count. Step 9: At indices (22, 23, 24): 7, 1, 7 has neighbours 7 and 7, so not valid. Step 10: At indices (23, 24, 25): 1, 7, 3 again has middle 7 with neighbours 1 and 3. Count 3. Step 11: No other triple satisfies the pattern, so the total count is 3. Verification / Alternative check: To quickly verify, list all triples where the middle value is 7: (1, 7, 3) at indices 5–7, (1, 7, 3) at indices 13–15, and (1, 7, 3) at indices 23–25 are the only ones with neighbours 1 and 3. Triples like (3, 7, 7), (7, 7, 1), or (7, 1, 7) fail the neighbour condition. This re scan confirms that there are exactly three valid patterns. Why Other Options Are Wrong: 4: Overcounts by including at least one triple where the neighbours are not 1 and 3. 5: Greatly overestimates and assumes many more matching triples than actually appear. More than 5: Clearly impossible once the sequence is examined carefully. Common Pitfalls: A typical error is to count any occurrence of 1, 3, and 7 in any order without insisting that 7 must be in the middle. Another is to double count overlapping triples or to misread the sequence and skip a term. Working with indices and checking each triple systematically avoids these mistakes. Final Answer: The pattern with 7 in the middle and 1 and 3 on either side occurs 3 times in the series.

Question 3

If the series 7, 3, 9, 7, 0, 3, 8, 4, 6, 2, 1, 0, 5, 11, 13 is written in reverse order, which number will be fourth to the right of the seventh number from the left in the reversed series?

Accepted Answer

Introduction / Context: This question tests your ability to handle positions in a sequence after reversing its order. Reversing a series flips the positions of all elements, and then you must apply positional instructions like “seventh from the left” and “fourth to the right.” Careful indexing is essential to avoid confusion once the series has been reversed. Given Data / Assumptions: - Original series: 7, 3, 9, 7, 0, 3, 8, 4, 6, 2, 1, 0, 5, 11, 13. - The series is to be written in reverse order. - In the reversed series, we must find the seventh number from the left. - Then we must locate the number that is fourth to the right of this seventh number. Concept / Approach: First, construct the reversed series by writing the numbers from last to first. Once we have the reversed sequence, we treat it as a fresh series and label positions from the left starting at 1. Identify the seventh element from the left, then move four positions to the right (increasing the index by four) to locate the required element. No additional pattern recognition is needed; only careful positional counting is required. Step-by-Step Solution: Step 1: Write the original series: 7, 3, 9, 7, 0, 3, 8, 4, 6, 2, 1, 0, 5, 11, 13. Step 2: Reverse it by writing from last to first: 13, 11, 5, 0, 1, 2, 6, 4, 8, 3, 0, 7, 9, 3, 7. Step 3: Label positions in the reversed series from the left: 1: 13, 2: 11, 3: 5, 4: 0, 5: 1, 6: 2, 7: 6, 8: 4, 9: 8, 10: 3, 11: 0, 12: 7, 13: 9, 14: 3, 15: 7. Step 4: Identify the seventh number from the left. From the list, position 7 holds the number 6. Step 5: Now move four positions to the right of position 7. That means we go to position 7 + 4 = 11. Step 6: At position 11 in the reversed series, the number is 0. Step 7: Therefore, the number that is fourth to the right of the seventh from the left is 0. Verification / Alternative check: To verify, you can re write the reversed series in a table and mark the seventh element and then count four steps to the right. Starting at position 7 (6), the positions 8, 9, 10, and 11 contain 4, 8, 3, and 0 respectively. The last in this list is indeed at position 11 and is 0. This reconfirmation ensures that there was no miscount in the initial pass. Why Other Options Are Wrong: 5: This number appears at position 3 in the reversed sequence, not at the specified relative position. 11: This is at position 2 from the left, again not fourth to the right of the seventh element. 9: This appears at position 13, which is two places to the right of 11 and not the desired position. Common Pitfalls: A common error is to try to apply the “seventh from the left” and “fourth to the right” instructions to the original sequence without actually reversing it. Another mistake is mis counting positions after reversal, especially when switching back and forth between the two orders in your head. Writing down the reversed sequence and indexing it explicitly is the safest strategy. Final Answer: In the reversed series, the number that is fourth to the right of the seventh number from the left is 0.

Question 4

In the arrangement 1 8 5 9 4 7 1 2 5 8 3 6 5 9 2 7 6 4 5 2 9 2 6 4 1 2 3 5 1 4 2 8 3, which digit is fifth to the right of the twelfth digit counted from the right end?

Accepted Answer

Introduction / Context: This problem involves positional reasoning within a long string of digits. You are asked to combine two different ways of counting: first from the right end of the arrangement, then from the left, after identifying a particular reference position. Such questions appear frequently in competitive exams and require a clear method for converting positions counted from the right into standard left to right indices. Given Data / Assumptions: - The arrangement is: 1 8 5 9 4 7 1 2 5 8 3 6 5 9 2 7 6 4 5 2 9 2 6 4 1 2 3 5 1 4 2 8 3. - We first locate the twelfth digit from the right end. - Then we find the digit that is fifth to the right of that digit (toward the right end). - Positions are counted within one continuous row of digits. Concept / Approach: Let N be the total number of digits. The k-th digit from the right is the same as the (N - k + 1)-th digit from the left. After we compute this equivalent left side position, we can easily move “fifth to the right” by adding 5 to the index. This method converts all instructions into one left to right indexing system, which is simpler to manage accurately. Step-by-Step Solution: Step 1: Count the total number of digits in the arrangement. Step 2: There are 33 digits in total. Step 3: The twelfth digit from the right is at position N - 12 + 1 from the left, which is 33 - 12 + 1 = 22 from the left. Step 4: Now list the digits with left to right indices around this position: 21: 9, 22: 2, 23: 6, 24: 4, 25: 1, 26: 2, 27: 3, 28: 5, 29: 1, 30: 4, 31: 2, 32: 8, 33: 3. Step 5: At position 22 from the left, the digit is 2. This is the twelfth digit from the right. Step 6: We now need the digit that is fifth to the right of this position. Step 7: Moving five places to the right means adding 5 to the index: 22 + 5 = 27. Step 8: At position 27 in the above list, the digit is 3. Step 9: Therefore, the digit that is fifth to the right of the twelfth digit from the right is 3. Verification / Alternative check: We can also verify by counting directly from the right. The twelfth from the right is at index 12 in a right to left list, then moving five places toward the right end corresponds exactly to five additional positions toward the right edge of the original sequence, which is the same as moving to index 17 from the right. Converting index 17 from the right back to a left index also lands on position 27, confirming that the digit there is indeed 3. Why Other Options Are Wrong: 1: This digit appears nearby but not at the required position after both steps of counting. 2: This is the digit at the reference position (twelfth from the right), not the one five steps further to the right. 7: Although present in the arrangement, it is not located at the specific index calculated. Common Pitfalls: A common mistake is to misapply the formula and use N - k instead of N - k + 1 when converting from right side indexing to left side indexing. Another error is to confuse “fifth to the right of” with “fifth from the right,” which are different instructions. Keeping all counts on a single left to right index line after conversion minimizes such confusion. Final Answer: The digit that is fifth to the right of the twelfth digit from the right end is 3.

Question 5

From the number 31549786, the second, fifth and eighth digits are taken to form a three-digit number. If it is possible to form a number that is the perfect square of a two-digit even number, what will be the second digit of that two-digit even numb…

Accepted Answer

Introduction / Context: This question combines digit selection with properties of perfect squares. You are asked to extract certain digits from an eight digit number, rearrange them to form a three digit number, and then determine whether that three digit number can be a perfect square of a two digit even number. Once that two digit even number is identified, you must report its second digit. This tests both numerical reasoning and knowledge of common perfect squares. Given Data / Assumptions: - Original number: 31549786. - We take the 2nd, 5th, and 8th digits of this number. - These digits may be rearranged to form a three digit number. - That three digit number must be the perfect square of a two digit even number. - We must find the second digit of that two digit even number, if such a square can be formed. Concept / Approach: First, identify the required digits from 31549786 by position. Next, list all distinct three digit numbers that can be formed using those digits. Then compare this set with the list of three digit perfect squares of two digit even numbers. Once a match is found, the corresponding two digit even number is determined, and we can simply read off its second digit. This is a finite search problem rather than an algebraic one. Step-by-Step Solution: Step 1: Label the digits of 31549786 by position: Positions: 1 2 3 4 5 6 7 8 Digits: 3 1 5 4 9 7 8 6 Step 2: The second digit is 1, the fifth digit is 9, and the eighth digit is 6. Step 3: Thus the available digits are 1, 9, and 6. Step 4: Possible three digit numbers formed using 1, 9, and 6 are: 196, 169, 916, 961, 619, and 691. Step 5: List the perfect squares of two digit even numbers that are three digit numbers. Even numbers from 10 to 30 give squares like 10^2 = 100, 12^2 = 144, 14^2 = 196, 16^2 = 256, 18^2 = 324, 20^2 = 400, 22^2 = 484, 24^2 = 576, 26^2 = 676, 28^2 = 784, 30^2 = 900. Step 6: Compare our candidate numbers with this list of squares. Step 7: The number 196 appears both in our candidate list and in the perfect square list. Step 8: 196 is the square of 14, that is, 14^2 = 196. Step 9: The two digit even number is 14. Its first digit is 1 and its second digit is 4. Step 10: Therefore, the required second digit is 4. Verification / Alternative check: We can confirm no other candidate fits. Numbers like 169 are 13^2, but 13 is not even. Likewise, 961 is 31^2 and 916, 619, and 691 are not squares of any two digit integer. Among the even squares list, only 196 can be made using digits 1, 9, and 6. This guarantees that 14 is the unique valid two digit even number and that 4 is the unique second digit answer. Why Other Options Are Wrong: 1: This is the first digit of 14, not the second digit. 6: No square of an even two digit number formed from these digits would give a second digit of 6. None of these: This would only be correct if no suitable square existed, but 196 = 14^2 clearly satisfies all conditions. Common Pitfalls: One common mistake is to forget the requirement that the two digit number must be even, leading candidates to choose 13 because 169 is a familiar square. Another is to not systematically check all permutations of the digits 1, 9, and 6, causing some learners to overlook 196. Building the full list of permutations and then comparing against known squares is the surest way to avoid missing the correct combination. Final Answer: The second digit of the two digit even number whose square is formed from these digits is 4.

Question 6

In the following list of number pairs, three pairs follow the same numerical rule between the first and the second number. Which number pair is the odd one out that does not follow the common rule?

Accepted Answer

Introduction / Context: This question tests pattern recognition in pairs of numbers. In many reasoning examinations, you are given several ordered pairs where three follow a hidden rule and one does not. Your task is to carefully identify the rule that connects the first and second element in most pairs, and then pick the odd pair that breaks this rule. This kind of question strengthens analytical thinking and the ability to work with squares and simple operations like addition. Given Data / Assumptions: The number pairs given are: 4 : 18, 7 : 51, 9 : 83, 6 : 34, and 8 : 66. Exactly one pair is odd and does not follow the common pattern. We assume the same rule should connect the first and second number in all normal pairs. We look for a simple arithmetic rule such as using squares and small additions. Concept / Approach: A common approach in such pair questions is to check whether the second number can be derived from the first by applying a standard operation like squaring, doubling, or adding a fixed constant. Since the second number is significantly larger than the first, squaring the first number is a good first guess. After squaring, we check if adding or subtracting a small number such as 1, 2 or 3 converts the squared value into the second number. If the same rule works for most pairs, the one pair that fails this rule will be the odd pair. Step-by-Step Solution: Check the pair 4 : 18. Square the first number: 4^2 = 16. Add 2 to it: 16 + 2 = 18, which matches the second number. Check the pair 7 : 51. Square the first number: 7^2 = 49. Add 2 to it: 49 + 2 = 51, which again matches the second number. Check the pair 9 : 83. Square the first number: 9^2 = 81. Add 2 to it: 81 + 2 = 83, which matches perfectly. Check the pair 8 : 66. Square the first number: 8^2 = 64. Add 2 to it: 64 + 2 = 66, which also fits the same rule. Now check the pair 6 : 34. Square the first number: 6^2 = 36. Add 2 to it: 36 + 2 = 38, but the given second number is 34, so this pair does not follow the rule. Thus the common rule is: second number = (first number)^2 + 2, and only one pair fails to satisfy it. Verification / Alternative check: We can verify the pattern once more clearly. For 4, 7, 8 and 9, we have second numbers 18, 51, 66 and 83 respectively. When we compute 4^2 + 2, 7^2 + 2, 8^2 + 2 and 9^2 + 2, we get exactly 18, 51, 66 and 83. This shows that the rule holds consistently for four pairs. There is no other simple rule that fits all five pairs simultaneously. Therefore, the pair where the second number deviates from this formula must be the odd one. This additional check confirms that our reasoning is robust. Why Other Options Are Wrong: The pair 4 : 18 is correct because 4^2 + 2 = 16 + 2 = 18. The pair 7 : 51 is correct since 7^2 + 2 = 49 + 2 = 51. The pair 9 : 83 is correct because 9^2 + 2 = 81 + 2 = 83. The added option 8 : 66 also fits, as 8^2 + 2 = 64 + 2 = 66. These four pairs all follow the same pattern. Only 6 : 34 does not satisfy this formula, so it is the odd one. Common Pitfalls: A common mistake is to look only at simple differences such as second minus first and try to see a pattern in those raw differences, which does not work well here. Another pitfall is to consider unrelated operations like multiplying by 3 or adding random constants without systematically testing squares. Some candidates also stop after checking only one or two pairs and misidentify the pattern. The correct method is to check all pairs, test an operation like squaring, and see if one consistent rule works for most pairs. Final Answer: The only pair that does not satisfy the rule second number = (first number)^2 + 2 is 6 : 34, so this is the odd number pair.

Question 7

Five books A, B, C, D and E have different thicknesses. A is thicker than C, B is thicker than D, E is thicker than C but not as thick as B, and D is thinner than C. Based on this information, which book is the thinnest among all five?

Accepted Answer

Introduction / Context: This question is a classic example of ordering and comparison reasoning. You are given several statements that compare the thickness of different books. From these relational clues, you need to deduce the relative order of all books and then identify the one that is thinnest. Such questions are common in verbal reasoning and help in building skills for handling inequalities and logical ordering problems. Given Data / Assumptions: A is thicker than C. B is thicker than D. E is thicker than C but not as thick as B. D is not as thick as C, so D is thinner than C. All five books A, B, C, D and E have different thicknesses. Concept / Approach: The approach is to convert all statements into inequalities and then arrange the books in increasing order of thickness. When we say one book is thicker than another, we treat it as a greater than relationship. We combine all given inequalities carefully, making sure there are no contradictions. Once we have a consistent sequence from thinnest to thickest, we can directly read off the thinnest book from the leftmost position in that order. Step-by-Step Solution: From “A is thicker than C”, we write A > C. From “B is thicker than D”, we write B > D. From “E is thicker than C but not as thick as B”, we know C < E < B. From “D is not as thick as C”, we know D < C. Now combine these: since D < C and C < E, we have D < C < E. We also know that A is thicker than C, so A > C. We do not yet know if A is thicker than E, but A is definitely thicker than C and D. We already have E < B, so the partial chain becomes D < C < E < B, and A is somewhere to the right of C in this chain. Whatever the exact ordering between A, E and B, it is clear that D is the only one that is thinner than C, and hence thinner than all others. Verification / Alternative check: To verify, try placing books in increasing order based on the information. We know D is thinner than C, and C is thinner than E. E is thinner than B. A is thicker than C. A could be between E and B or between C and E, but D remains the only book that is directly stated to be thinner than C, and no other book is stated to be thinner than D. Therefore, D occupies the first position in any valid ordering, making D the thinnest book regardless of the exact positions of A, E and B among the thicker books. Why Other Options Are Wrong: A cannot be the thinnest because A is thicker than C, and C is not the thinnest. B cannot be the thinnest because B is thicker than D and also thicker than E. C cannot be the thinnest because D is explicitly thinner than C. E cannot be the thinnest because E is thicker than C and D is thinner than C. Only D satisfies all conditions for being the thinnest. Common Pitfalls: One common mistake is to confuse “not as thick as” with “thicker than”, which reverses the inequality. Another error is to try to place all books in a strict complete order prematurely, instead of first identifying clear extremes like the thinnest or thickest. Candidates may also overlook that D is directly compared to C and is stated to be thinner, which is the key clue. Always list all inequalities clearly before drawing conclusions. Final Answer: The book that is thinnest among A, B, C, D and E is D.

Question 8

In the number 76534218, each digit is replaced by the next higher digit (for example 1 becomes 2, 2 becomes 3 and so on) and then all the new digits are arranged in ascending order from left to right. Which digit will be fifth from the left end in t…

Accepted Answer

Introduction / Context: This reasoning question checks your ability to handle digit manipulation and ordering. You first perform a transformation on each digit, then sort the resulting digits. Such problems are common in aptitude tests because they combine two skills: understanding a simple mapping on digits and then correctly arranging numbers in ascending order. Careful step-by-step work avoids errors. Given Data / Assumptions: The original number is 76534218. Each digit is replaced by the next higher digit: 1 becomes 2, 2 becomes 3, 3 becomes 4, and so on, up to 9. After replacement, all new digits are arranged in ascending order from left to right. We need the digit that appears at the fifth position from the left in the final sorted sequence. Concept / Approach: The approach has two phases. First, apply the digit transformation consistently: every original digit d is mapped to d + 1. Second, collect all transformed digits and sort them from smallest to largest. Since there are eight digits, the fifth from the left is the fifth smallest transformed digit. Working systematically ensures that none of the digits are missed or miscopied and that sorting is done correctly. Step-by-Step Solution: Write the original digits of 76534218 as a list: 7, 6, 5, 3, 4, 2, 1, 8. Apply the rule “replace each digit by the next higher digit” to each one: 7 becomes 8, 6 becomes 7, 5 becomes 6, 3 becomes 4, 4 becomes 5, 2 becomes 3, 1 becomes 2, and 8 becomes 9. The transformed list of digits is therefore: 8, 7, 6, 4, 5, 3, 2, 9. Now arrange these digits in ascending order: 2, 3, 4, 5, 6, 7, 8, 9. Count from the left. The first digit is 2, second is 3, third is 4, fourth is 5, and the fifth is 6. Thus, the digit at the fifth position from the left in the final arrangement is 6. Verification / Alternative check: An alternative quick check is to ensure that all digits from 2 to 9 appear exactly once in the transformed list, because the original digits 1 to 8 all shift up by 1. This means the sorted sequence must be 2, 3, 4, 5, 6, 7, 8, 9. Counting to the fifth position again gives 6. Since this method does not depend on any particular digit being misread, it confirms that the earlier transformation and sorting are correct. Why Other Options Are Wrong: Option 4 would be the third digit, not the fifth. Option 5 is the fourth digit in the sorted list. Options 7 and 8 are the sixth and seventh digits respectively. All these are correct digits from the sequence but at different positions. Only the digit 6 occupies the fifth place from the left in the final arrangement. Common Pitfalls: A frequent error is to misapply the rule and, for example, treat 9 as wrapping around to 0, which is not stated. Another mistake is failing to transform all digits before sorting, or accidentally sorting the original digits instead of the transformed ones. Some candidates also miscount positions, especially when in a hurry. Always write the full sorted list and then count carefully from the left to avoid such mistakes. Final Answer: The digit that will be fifth from the left end in the final ascending arrangement is 6.

Question 9

Consider the eight digit number 93217648. Using only the second, fourth and sixth digits of this number, and using each of these three digits exactly once, it is possible to form a three digit number that is the square of a two digit odd number. Whi…

Accepted Answer

Introduction / Context: This question combines digit selection with knowledge of perfect squares. You must first identify the specific digits from the given large number, then use them to construct possible three digit numbers. One of these numbers is a perfect square of a two digit odd integer. Finally, you need to identify which two digit odd number produces that square. This type of reasoning problem strengthens your ability to handle constraints and number properties simultaneously. Given Data / Assumptions: The given number is 93217648. We are allowed to use only the second, fourth and sixth digits of this number. We must use each of these three digits exactly once to form a three digit number. The resulting three digit number is the square of a two digit odd number. The candidate two digit odd numbers in the options are 17, 19, 15, 21 and 23. Concept / Approach: The approach is to first extract the required digits by position, then list all permutations of those digits to see which three digit numbers can be formed. Separately, we find the squares of the given two digit odd numbers and check whether any of these squares match a number that can be formed from the chosen digits. Using properties of squares and systematic checking of permutations ensures that we do not miss the correct answer. Step-by-Step Solution: Write the digits of 93217648 with positions: 1: 9, 2: 3, 3: 2, 4: 1, 5: 7, 6: 6, 7: 4, 8: 8. The second digit is 3, the fourth digit is 1, and the sixth digit is 6. So the three digits available are 3, 1 and 6. List all three digit numbers that can be formed using 3, 1 and 6 exactly once: 136, 163, 316, 361, 613 and 631. Now compute the squares of the given two digit odd numbers: 17^2 = 289, 19^2 = 361, 15^2 = 225, 21^2 = 441 and 23^2 = 529. Compare these squares with the list of three digit permutations. Only 361 appears in the list and is the square of 19. Therefore the required two digit odd number whose square can be formed from the digits 3, 1 and 6 is 19. Verification / Alternative check: To verify, note that 361 uses digits 3, 6 and 1 exactly once each, which is consistent with the constraint. Also, none of the other squares 289, 225, 441 or 529 can be formed from the digits 3, 1 and 6 because they contain digits 2, 4, 5 or 9 that are not allowed. Another quick check is to observe that 361 is a well known perfect square of 19, which further confirms our result. Hence there is no alternative match and 19 is uniquely correct. Why Other Options Are Wrong: Option 17 is wrong because 17^2 = 289, which uses digits 2, 8 and 9 that are not in the allowed set {1, 3, 6}. Option 15 is wrong because 15^2 = 225, which repeats the digit 2 and does not use the digit 3 at all. Option 21 is wrong because 21^2 = 441, which involves the digit 4 twice and again ignores the digits 3 and 6. Option 23 is wrong because 23^2 = 529, containing digits 5, 2 and 9, which are not the required digits. Common Pitfalls: A common mistake is to attempt to square random two digit odd numbers other than those supplied in the options, which wastes time. Another pitfall is to misread the positions of digits in the original number or to forget that each of the three digits must be used exactly once. Some candidates also forget to check all permutations of the digits and may wrongly discard 361. Working methodically through positions, permutations and squares is the safest approach. Final Answer: The two digit odd number whose square can be formed using the second, fourth and sixth digits of 93217648 is 19.

Question 10

The weights of four boxes are 20 kilograms, 30 kilograms, 50 kilograms and 70 kilograms. A combination can use any box at most once. Which of the following cannot be the total weight, in kilograms, of some combination of these boxes?

Accepted Answer

Introduction / Context: This is a subset sum type reasoning question involving simple arithmetic. You are given the weights of individual boxes and asked which total weight cannot be formed by choosing some or all of them, without repetition. This kind of question appears often in competitive exams to test whether you can systematically consider all combinations rather than guessing. Given Data / Assumptions: There are four boxes with weights 20 kg, 30 kg, 50 kg and 70 kg. Each box can be used at most once in any combination. Possible totals are found by adding any subset of these weights. The candidate totals to test are 170 kg, 160 kg, 150 kg, 120 kg and 140 kg. Concept / Approach: The straightforward approach is to list all possible sums of these four weights. There are 2^4 = 16 possible subsets, including the empty set. We do not need the empty set, so we focus on non empty subsets. After computing all unique sums, we compare them with the totals given in the options. Any option not present in our list of subset sums cannot be formed and is therefore the correct answer. Step-by-Step Solution: List the individual weights: 20, 30, 50 and 70. Compute sums using one box: 20, 30, 50, 70. Compute sums using two boxes: 20 + 30 = 50, 20 + 50 = 70, 20 + 70 = 90, 30 + 50 = 80, 30 + 70 = 100, 50 + 70 = 120. Compute sums using three boxes: 20 + 30 + 50 = 100, 20 + 30 + 70 = 120, 20 + 50 + 70 = 140, 30 + 50 + 70 = 150. Compute the sum using all four boxes: 20 + 30 + 50 + 70 = 170. Collect the distinct totals: 20, 30, 50, 70, 80, 90, 100, 120, 140, 150, 170. Now compare with options 170, 160, 150, 120 and 140. All except 160 appear in the list of possible totals. Verification / Alternative check: We can verify by specifically trying to express 160 as a sum of the given weights. Check three box combinations: 70 + 50 + 30 = 150, 70 + 50 + 20 = 140, 70 + 30 + 20 = 120, 50 + 30 + 20 = 100. None equals 160. Adding all four gives 170. Using two boxes gives at most 70 + 50 = 120, which is also not 160. Since there is no way to reach 160 using any subset, the total 160 kg is impossible, confirming our result. Why Other Options Are Wrong: 170 kg is possible by using all four boxes together. 150 kg is possible by combining 30 kg, 50 kg and 70 kg. 120 kg can be obtained either as 50 kg + 70 kg or as 20 kg + 30 kg + 70 kg. 140 kg is possible by combining 20 kg, 50 kg and 70 kg. Since these totals can be formed by valid combinations, they are not the correct answer. Common Pitfalls: A common mistake is to miss some combinations or to double count others, leading to an incorrect list of totals. Another pitfall is to rely on mental arithmetic alone for all combinations, which can cause errors under time pressure. Some candidates also assume that the largest impossible total must be the answer, which is not always true. Systematically listing and checking subset sums is the safest strategy. Final Answer: The total weight that cannot be obtained from any combination of the given boxes is 160 kilograms.

Question 11

Nitin correctly remembers that the birthday of Nidhi is after Tuesday but before Friday. Dheeraj correctly remembers that the birthday of Nidhi is after Wednesday but before Saturday. On which day of the week does the birthday of Nidhi definitely fa…

Accepted Answer

Introduction / Context: This is a day of the week reasoning question where two people give partial but correct information about an event. You need to combine their statements to find the exact day. Such questions are standard in verbal reasoning sections and test your ability to work with ranges and intersections on the days of the week. Given Data / Assumptions: Nitin remembers that Nidhi has her birthday after Tuesday but before Friday. Dheeraj remembers that Nidhi has her birthday after Wednesday but before Saturday. We assume both Nitin and Dheeraj are correct in their statements. The days of the week in order are Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday. Concept / Approach: The approach is to interpret each person’s memory as a range of possible days and then find the common day or intersection of these ranges. When someone says “after a given day but before another day”, it excludes the boundary days and includes only the days strictly between them. Once the individual sets of possible days are written, their overlap reveals the single day that satisfies both people simultaneously. Step-by-Step Solution: Translate Nitin’s statement “after Tuesday but before Friday”. The days strictly between Tuesday and Friday are Wednesday and Thursday. Thus, according to Nitin, the birthday could be on either Wednesday or Thursday. Translate Dheeraj’s statement “after Wednesday but before Saturday”. The days strictly between Wednesday and Saturday are Thursday and Friday. Thus, according to Dheeraj, the birthday could be on either Thursday or Friday. Now find the intersection of the two sets: Nitin allows {Wednesday, Thursday}; Dheeraj allows {Thursday, Friday}. The only day common to both sets is Thursday. Therefore, Nidhi’s birthday must definitely fall on Thursday. Verification / Alternative check: To verify, check each day from Monday to Saturday. Wednesday satisfies Nitin’s condition (after Tuesday and before Friday) but fails Dheeraj’s condition because Wednesday is not after Wednesday. Friday satisfies Dheeraj’s condition (after Wednesday and before Saturday) but fails Nitin’s condition because Friday is not before Friday. Thursday is after Tuesday and before Friday, and it is also after Wednesday and before Saturday. No other day meets both constraints. Hence Thursday is uniquely correct. Why Other Options Are Wrong: Monday and Tuesday are too early and do not satisfy the “after Tuesday” or “after Wednesday” parts. Wednesday satisfies Nitin but not Dheeraj. Friday satisfies Dheeraj but not Nitin. Saturday is outside both “before Friday” and “before Saturday” conditions. Only Thursday fits both people’s memories at the same time. Common Pitfalls: A common error is to treat “after Tuesday” as “Tuesday or later” or “before Friday” as “on or before Friday”, which mistakenly includes boundary days. Another pitfall is to pick any day that appears in either person’s range without checking the intersection. Always remember that both statements are correct together, so the final day must satisfy both conditions simultaneously. Final Answer: The birthday of Nidhi definitely falls on Thursday.

Question 12

Observe the following number series: 9 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 7 6 5 4 3 2 1. What is the next number that should come after this sequence?

Accepted Answer

Introduction / Context: This question checks your ability to identify patterns in a number series. The series is not a simple arithmetic progression; instead, it is built from repeated decreasing blocks. Recognizing how these blocks start and end is the key to predicting the next term. Such series questions are frequently asked in competitive exams to test numerical reasoning and pattern recognition skills. Given Data / Assumptions: The given sequence is: 9 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 7 6 5 4 3 2 1. We need the very next number that should follow the last term in this sequence. No extra hidden operations such as multiplication or addition are indicated; it appears to be a pure ordering pattern. Concept / Approach: The approach is to look for repeated blocks. Notice that the series seems to start at a particular number and then decrease step by step down to 1, then start again at a slightly smaller number and repeat the countdown. By grouping these terms, we can see how the starting number changes from one block to the next and then predict the next starting number after the last complete block. The first term in the next block will be the answer. Step-by-Step Solution: Group the numbers from the beginning. The first block is: 9 8 7 6 5 4 3 2 1. The second block is: 8 7 6 5 4 3 2 1. The third block is: 7 6 5 4 3 2 1. Observe that each block is a decreasing sequence that ends at 1. The starting number of each block is decreasing by 1: first 9, then 8, then 7. Therefore the next block should start at 6 and then go down as 6 5 4 3 2 1. The question only asks for the next single number after the given sequence, which is the first number of the next block, namely 6. Verification / Alternative check: As a check, write the start of the hypothetical continuation: after the last 1 in the question, the next block would be 6 5 4 3 2 1, followed by another block 5 4 3 2 1, and so on. This extended pattern would remain consistent with the observed rule that each block starts one less than the previous block and decreases down to 1. Since the continuation is coherent, identifying 6 as the first term of the fourth block is correct. Why Other Options Are Wrong: Option 9 is wrong because that would restart the pattern at the original highest value instead of following the decreasing starting numbers. Option 8 would be the first term of the second block, but we are already past that. Option 7 is the first term of the third block, which has also already appeared. Option 5 would be the second term of the next block, not the first. Only 6 correctly starts the next block after the given sequence. Common Pitfalls: Some candidates may focus on local differences and notice only that many numbers are decreasing by 1, missing the block structure entirely. Others might try to treat the entire series as a single long pattern without grouping, leading to confusion. It is important to look for repeated segments and how they change from one repetition to the next. Once you see that each block is a countdown starting from 9, then 8, then 7, predicting that the next block starts at 6 becomes straightforward. Final Answer: The number that should come next in the series is 6.

Question 13

In a row of trucks, Ashok is at the 17th position from the left end. Bharat is 19 places to the right of Ashok and Bharat is exactly at the centre of the row. How many trucks are there to the right of Ashok's truck?

Accepted Answer

Introduction / Context: This position and ranking question involves people or objects in a row. You are given the position of one person from one end and the relative position of another person, who is at the exact centre of the row. From this information, you must compute the total number of trucks and then find how many lie to the right of a specific truck. Such problems are standard in verbal reasoning and train you to manipulate positions from both ends consistently. Given Data / Assumptions: Ashok is 17th from the left end of the row of trucks. Bharat is 19 places to the right of Ashok. Bharat is at the exact centre of the row. All positions are counted from the left or right end in a straight line with no gaps. Concept / Approach: The first step is to determine the position of Bharat from the left end by adding the relative distance to Ashok’s known position. Since Bharat is at the centre, his position from the left is equal to his position from the right, and in an odd length row this central position is (total number of trucks + 1) / 2. Once the total number of trucks is obtained, the number of trucks to the right of Ashok can be found by subtracting Ashok’s position from the total and adjusting for inclusive counting. Step-by-Step Solution: Ashok is 17th from the left, so his position from the left is 17. Bharat is 19 places to the right of Ashok. Therefore, Bharat’s position from the left is 17 + 19 = 36. Bharat is at the exact centre of the row. In a row with an odd number N of trucks, the central position is (N + 1) / 2. Set (N + 1) / 2 = 36. Then N + 1 = 72, so N = 71. There are 71 trucks in total. To find how many trucks are to the right of Ashok, subtract his position from the total and exclude his own truck: number to the right = N - 17 = 71 - 17 = 54. Thus, there are 54 trucks to the right of Ashok’s truck. Verification / Alternative check: As a check, find Bharat’s position from the right. Since there are 71 trucks and Bharat is at position 36 from the left, his position from the right is 71 - 36 + 1 = 36. This confirms that he is indeed at the centre. Next, verify the count to the right of Ashok: positions from the left go from 1 to 71, so positions to the right of 17 are 18 through 71 inclusive, which gives 71 - 17 = 54 trucks. The calculations are consistent. Why Other Options Are Wrong: 53 and 52 would correspond to different total counts or miscalculations in subtracting positions. Option 19 is only the distance between Ashok and Bharat, not the total to the right of Ashok. Option 51 would also come from incorrect arithmetic. Only 54 matches the correct derivation based on the total number of trucks and Ashok’s position. Common Pitfalls: A frequent mistake is to forget the relation for the central position (N + 1) / 2 and instead use N / 2, which is incorrect for odd N. Another pitfall is miscounting the number of trucks to the right by including Ashok’s truck or subtracting an extra 1. Some candidates also confuse “places to the right” with direct positional numbering. Writing out the relations clearly and carefully handling inclusive counts prevents such errors. Final Answer: There are 54 trucks to the right of Ashok’s truck.

Question 14

A, B, C, D and E are standing in a row facing forward. D is the immediate neighbour of both A and E. B stands to the right of E and C is standing at the extreme right end of the row. Who is fourth to the left of C?

Accepted Answer

Introduction / Context: This question is about arranging people in a straight row based on relative position clues. You are given conditions about who stands next to whom and which positions are at the extreme ends. From these conditions, you must deduce the exact order of all individuals and then identify who is a certain number of places to the left of a specific person. Such arrangement problems are common in verbal reasoning tests. Given Data / Assumptions: Five people A, B, C, D and E stand in a single row. D is the immediate neighbour of both A and E, so the three must stand consecutively as A D E or E D A. B stands to the right of E. C is at the extreme right end of the row. We need to identify the person who is fourth to the left of C. Concept / Approach: The key idea is to treat the row as positions 1 to 5 from left to right. Because C is at the extreme right, C must occupy position 5. The triple A, D, E must appear in consecutive positions, and B must be to the right of E. By trying feasible placements of the A D E or E D A block within positions 1 to 4, while respecting the condition on B, we can deduce a unique valid arrangement. Then we count four places to the left of C to find the required person. Step-by-Step Solution: Label positions from left to right as 1, 2, 3, 4 and 5. Since C is at the extreme right, place C at position 5. D must be an immediate neighbour of both A and E, which means the three must be arranged as A D E or E D A in consecutive positions. Because C already occupies position 5, the three consecutive positions for A, D and E must be within positions 1, 2, 3 or 2, 3, 4. Try placing A D E at positions 1, 2, 3. Then E is at position 3, and B must stand to the right of E, so B must be at position 4. This gives the arrangement 1:A, 2:D, 3:E, 4:B, 5:C, which satisfies all conditions. Check the alternative pattern E D A at positions 1, 2, 3. Then E would be at position 1 and B would need to be to the right of E, but this still leaves room. However, D must be between E and A, and such arrangement combined with C at 5 cannot satisfy all conditions consistently. The earlier arrangement works perfectly and is unique. In the valid arrangement A D E B C, counting four places to the left of C at position 5 gives positions 4, 3, 2 and 1; the person at position 1 is A. Verification / Alternative check: Verify each condition in the arrangement A D E B C. D is a neighbour of A (positions 1 and 2) and also of E (positions 2 and 3). B is to the right of E (positions 4 and 3). C is at the extreme right (position 5). All clues are satisfied, and any attempt to shift the block or reverse it leads to violation of some condition. Therefore, the arrangement is correct. Counting four positions left from C at 5 clearly leads to A at position 1. Why Other Options Are Wrong: B, E, C and D do not occupy the first position in the valid arrangement, so none of them can be four places to the left of C. B is at position 4, E at 3, D at 2 and C at 5. Only A is at position 1, which is exactly four places to the left of C when counting 5 to 4, 3, 2 and 1. Common Pitfalls: Candidates often misinterpret “immediate neighbour of both” and may place D in a corner position, failing to maintain adjacency with both A and E. Another common error is to forget that C must be at the extreme right and therefore fix C incorrectly. Some also count the positions incorrectly when moving left from C. Drawing a simple row diagram and filling it step by step is the safest way to avoid such mistakes. Final Answer: In the final arrangement, the person who is fourth to the left of C is A.

Question 15

In a class, seven students P, Q, R, S, T, N and O are standing in a row from left to right. Q is standing to the left of R but to the right of P. O is standing to the right of N but to the left of P. S is standing to the right of R but to the left o…

Accepted Answer

Introduction / Context: This question involves arranging several students in a single row based on relative position clues. Each clue gives a partial ordering between some of the students, and you must combine all of them to obtain a single consistent order from left to right. Once the order is known, identifying the middle position is straightforward. Problems of this type strengthen logical deduction and ordering skills. Given Data / Assumptions: Seven students P, Q, R, S, T, N and O stand in one row from left to right. Q is to the left of R and to the right of P, so P < Q < R in positional order. O is to the right of N and to the left of P, so N < O < P. S is to the right of R and to the left of T, so R < S < T. All seven names must appear exactly once in the row. Concept / Approach: The approach is to convert each statement into an inequality involving positions, then combine these inequalities into a single chain. Each relation tells us how two or three students are ordered relative to each other. If no contradictions arise, we obtain a unique linear order. Since there are seven students, the middle position is the 4th position from the left. The student at this position is the answer. Step-by-Step Solution: From “Q is to the left of R but to the right of P”, we have P < Q < R. From “O is to the right of N and to the left of P”, we get N < O < P. From “S is to the right of R and to the left of T”, we get R < S < T. Combine the chains. From N < O < P and P < Q < R, we get N < O < P < Q < R. Further combining R < S < T, the full chain becomes N < O < P < Q < R < S < T. Thus, in the row from left to right, the order of students is N, O, P, Q, R, S, T. Since there are seven students, the middle position is the 4th from the left. The student at position 4 in this order is Q. Verification / Alternative check: Check each clue against the order N O P Q R S T. Q is right of P and left of R, so P Q R appears correctly. O is right of N and left of P, so N O P is correct. S is right of R and left of T, so R S T is correctly placed. All conditions are satisfied with no conflicts. Counting positions, we have position 1: N, 2: O, 3: P, 4: Q, 5: R, 6: S, 7: T. Therefore, Q is unquestionably at the central, 4th position. Why Other Options Are Wrong: R stands at position 5, not in the middle. P stands at position 3, left of the middle. S stands at position 6, and O at position 2. None of these are in the 4th position. Only Q occupies the exact middle position in the row. Common Pitfalls: A common mistake is to misread the direction of inequalities, for example placing Q on the left of P instead of the right. Another pitfall is to try to fix positions too early without merging all chains, which can result in contradictions later. Some candidates also confuse “middle” with an average of positions without counting carefully. Always form one complete order, then explicitly count positions from left to right to identify the central student. Final Answer: The student standing in the middle position of the row is Q.

Question 16

In a row of 74 girls, Shweta is 27th from the left end. Palak is 7th to the right of Shweta. What is the position of Palak from the right end of the row?

Accepted Answer

Introduction / Context: This is a typical position and rank problem involving a row of students. You know the position of one girl from the left, and the relative position of another girl to her. The task is to find the second girl’s position from the right. This type of question tests the ability to convert between positions from different ends of a row and to use simple arithmetic correctly. Given Data / Assumptions: There are 74 girls standing in a straight row. Shweta is 27th from the left end of the row. Palak is 7th to the right of Shweta. We must find Palak’s position from the right end. Concept / Approach: The first step is to compute Palak’s position from the left by adding the number of positions to the right of Shweta. Once Palak’s position from the left is known, we use the standard relation between left and right positions in a row of known length. If a person is at position k from the left in a row of N people, then their position from the right is N - k + 1. Applying this formula gives the required answer. Step-by-Step Solution: Shweta is at position 27 from the left. Palak is 7 positions to the right of Shweta, so Palak’s position from the left is 27 + 7 = 34. The total number of girls in the row is 74. Use the relation: position from the right = total number - position from the left + 1. Substitute the values: position of Palak from the right = 74 - 34 + 1. Compute: 74 - 34 = 40, and 40 + 1 = 41. Thus, Palak is 41st from the right end of the row. Verification / Alternative check: As a quick check, note that a girl at position 34 from the left and position 41 from the right should satisfy 34 + 41 - 1 = 74, which is the total number of girls. Compute 34 + 41 - 1 = 74, which matches the given total exactly. This reconfirms that our conversion between left and right positions is correct and that the arithmetic was performed accurately. Why Other Options Are Wrong: Options 40, 42, 43 and 44 would produce inconsistent totals when using the relation left position + right position - 1 = total. For example, 34 + 40 - 1 = 73, not 74. Similarly, 34 + 42 - 1 = 75, and so on. Only 41 satisfies the equation and matches the logic of the problem. Common Pitfalls: A very common mistake is to forget to add 1 in the formula, using N - k instead of N - k + 1, which leads to an off by one error. Another pitfall is misreading “7th to the right of Shweta”, where some candidates incorrectly add or subtract 6 instead of 7. Carefully applying the formula and rechecking the final sum against the total number helps to avoid such errors. Final Answer: Palak’s position from the right end of the row is 41.

Question 17

In a row of boys, Tarak is 18th from one end and also 18th from the other end of the row. How many boys are there in the row in total?

Accepted Answer

Introduction / Context: This reasoning question involves understanding how positions from opposite ends of a row relate to the total number of people in the row. If a person has known positions from both the left and the right, you can use a simple formula to calculate the total number. This type of problem is frequently asked in verbal reasoning exams to check conceptual clarity rather than heavy calculation. Given Data / Assumptions: Tarak is 18th from one end of the row. Tarak is also 18th from the other end of the row. We need to find the total number of boys in the row. Positions are counted starting from 1 at each end. Concept / Approach: There is a standard formula that links a person’s position from the left and right ends to the total number of people in a row. If a person is at position L from the left and at position R from the right, then the total number of people N is given by N = L + R - 1. Here, both L and R are 18. Substituting these values will directly give the total number of boys in the row. Step-by-Step Solution: Let L be Tarak’s position from the left and R be his position from the right. Given that Tarak is 18th from one end and also 18th from the other, we have L = 18 and R = 18. Use the relation N = L + R - 1 to find the total number of boys N. Substitute L = 18 and R = 18: N = 18 + 18 - 1. Compute: 18 + 18 = 36, and 36 - 1 = 35. Therefore, there are 35 boys in the row. Verification / Alternative check: Imagine labeling the boys from left to right as positions 1 to 35. If Tarak is 18th from the left, he stands at position 18. Counting from the right, positions go from 35 down to 1, so position 18 from the right corresponds to 35 - 18 + 1 = 18 from the left, which matches. This confirms that a total of 35 boys is consistent with Tarak being 18th from both ends. Why Other Options Are Wrong: For 36 boys, a boy at position 18 from the left would be at position 36 - 18 + 1 = 19 from the right, not 18. Similarly, a total of 19 or 37 boys would place the central positions differently, and Tarak would not be 18th from both ends. Only a total of 35 boys allows a single boy to occupy symmetric positions 18 from each end. Common Pitfalls: A common error is to add the positions without subtracting 1, incorrectly using N = L + R, which would yield 36 instead of 35. Another pitfall is confusing the idea of the central position in an odd length row. Remember that when a person has the same position from both ends, the total number must be an odd number, and using the formula N = L + R - 1 is the safe method. Final Answer: The total number of boys in the row is 35.

Question 18

In a row of people, Manu is 7th from the bottom end of the row. Shrey is 10 ranks above Manu. If Shrey is 8th from the top end, then how many people are there altogether in this row?

Accepted Answer

Introduction / Context: This question involves ranks from the top and bottom in a single row. You are given Manu’s position from the bottom, the vertical distance between Manu and Shrey, and Shrey’s position from the top. Using this information, you must determine the total number of people in the row. Such problems test your understanding of how positions from opposite ends relate to each other and to the total count. Given Data / Assumptions: Manu is 7th from the bottom end of the row. Shrey is 10 ranks above Manu. Shrey is 8th from the top end of the row. We must find the total number of people in the row. Concept / Approach: The key idea is to express both Manu’s and Shrey’s positions from the top and from the bottom in terms of the total number of people. If the total number of people is N, then a person who is k-th from the bottom is at position N - k + 1 from the top. We can use this to find Manu’s position from the top, then apply the information that Shrey is 10 ranks above Manu, and finally equate this to Shrey’s given position from the top to solve for N. Step-by-Step Solution: Let the total number of people in the row be N. Manu is 7th from the bottom. So Manu’s position from the top is N - 7 + 1 = N - 6. Shrey is 10 ranks above Manu, meaning Shrey is 10 positions closer to the top than Manu. Therefore, Shrey’s position from the top is (N - 6) - 10 = N - 16. We are also told that Shrey is 8th from the top, so N - 16 = 8. Solve for N: N = 8 + 16 = 24. Thus, there are 24 people in the row. Verification / Alternative check: Check the consistency of positions with N = 24. Manu is 7th from the bottom, so his position from the top is 24 - 7 + 1 = 18. Shrey is 10 ranks above Manu, so his position from the top is 18 - 10 = 8, which matches the given information. This confirms that the total number of people must be 24 and that our interpretation of “10 ranks above” is correct. Why Other Options Are Wrong: If N were 23, Manu’s position from the top would be 23 - 7 + 1 = 17, and Shrey would then be 17 - 10 = 7 from the top, not 8. Similar contradictions occur for N = 25, 26 or 27. None of these values lead to Shrey being both 10 ranks above Manu and 8th from the top. Only N = 24 satisfies all the conditions simultaneously. Common Pitfalls: A common mistake is to misunderstand “10 ranks above” and incorrectly add instead of subtracting, or to miscompute the conversion from bottom to top. Another pitfall is forgetting the +1 term when converting positions between ends, leading to off by one errors. Writing all relations algebraically and solving systematically helps avoid these mistakes. Final Answer: The total number of people in the row is 24.

Question 19

If the day after tomorrow is Saturday, then what day of the week was three days before yesterday?

Accepted Answer

Introduction / Context: This is a calendar based reasoning question. You are given information about the day after tomorrow and asked to find a day relative to yesterday. You must mentally move forwards and backwards along the days of the week in the correct order. These questions check your ability to handle relative time expressions such as “day after tomorrow” and “three days before yesterday”. Given Data / Assumptions: The day after tomorrow is Saturday. We need to find which day was three days before yesterday. The standard order of days is Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday, repeating in a cycle. There is no change of calendar or any special holiday; the days follow normally. Concept / Approach: The idea is to first determine what today is by using the information about the day after tomorrow. Once you know today, you can move to yesterday and then go three days backwards. It is important to move step by step rather than trying to jump directly from the given statement to the required day. Working with a small mental time line of days helps reduce confusion. Step-by-Step Solution: Let us denote today by T. The day after tomorrow is two days ahead of today, that is T + 2. We are told that T + 2 is Saturday. Therefore T is two days before Saturday. Counting backwards, the day before Saturday is Friday, and the day before that is Thursday. Hence today is Thursday. Yesterday is one day before today, so yesterday was Wednesday. Three days before yesterday means we move three days back from Wednesday. One day before Wednesday is Tuesday, two days before is Monday, and three days before is Sunday. Thus, three days before yesterday, the day was Sunday. Verification / Alternative check: We can verify by building a small table. If today is Thursday, then tomorrow is Friday and the day after tomorrow is Saturday, which matches the given statement. Yesterday would then be Wednesday. Going backwards from Wednesday: one day before is Tuesday, two days before is Monday and three days before is Sunday. Every step fits the normal weekly order, so the answer is consistent and correct. Why Other Options Are Wrong: Monday would be only two days before Wednesday if we count Wednesday to Tuesday, Tuesday to Monday. Tuesday is one day before Wednesday, not three. Thursday and Friday are actually at or after today, while we need a day earlier than yesterday. Only Sunday is exactly three days before Wednesday when today is Thursday. Common Pitfalls: Many candidates confuse “day after tomorrow” with “tomorrow” or forget to count carefully when moving multiple days forwards or backwards. Another common mistake is to miscount steps, for example stopping after two steps when three are required. Keeping a simple mental or written list of days and marking each movement prevents these errors. Final Answer: Three days before yesterday, the day of the week was Sunday.

Question 20

Six friends A, B, C, D, E and F joined an institute in the year 1993, each on a different day. Only E joined after both C and D. C joined immediately before A and immediately after B, forming the consecutive sequence B, C, A. F is not the first or t…

Accepted Answer

Introduction / Context: This question is a logical arrangement puzzle based on the order in which six friends joined an institute. You are given several constraints about relative joining times, including who joined consecutively, who joined after whom and which positions cannot be occupied by certain friends. From these clues you must deduce the complete order and then identify the first person who joined. Such puzzles test careful reading and multi step logical reasoning. Given Data / Assumptions: Six friends A, B, C, D, E and F joined in 1993, each on a different day. Only E joined after both C and D, implying E is the only person later than both C and D. C joined immediately before A and immediately after B, so the order B, C, A appears consecutively. F is not the first or the last to join. D is not at any corner position, so D cannot be first or last. We must find who joined first among the six friends. Concept / Approach: The key idea is to treat the joining order as six positions from 1 (first) to 6 (last). The triple B, C, A must occupy three consecutive positions. Since only E joins after both C and D, E must be the last to join because anyone after D would also be after C, and only E has this property. With E fixed at position 6, we must place B, C and A in a valid block and then position D and F without violating the corner restrictions. Systematically trying possible placements for the B, C, A block leads to a unique consistent arrangement. Step-by-Step Solution: Label the joining days as positions 1 to 6, from first to last. Since only E joined after both C and D, E must be the only person after D and after C. This is possible only if E is at position 6, the last position. C joins immediately before A and immediately after B, so the three appear as B, C, A in consecutive positions. The triple B, C, A cannot end at position 6 because that is already taken by E. Thus the possible positions for the triple are (1, 2, 3), (2, 3, 4) or (3, 4, 5). Consider the triple at positions 3, 4, 5. Then A would be at position 5 and C at 4. For E to be after C and D, E is at 6, but D must then be before C. D would also need to avoid corners, which may be possible. However, we also know F is not first or last, and with D disallowed at 1 and 6, arranging everyone consistently becomes difficult and must be checked. A more promising placement is B, C, A at positions 1, 2, 3: position 1: B, 2: C, 3: A. With E fixed at position 6, the remaining positions 4 and 5 must be taken by D and F. D cannot be in a corner, so D must be at position 5, leaving F at position 4. We now check “only E joined after both C and D”. Here, C is at 2 and D is at 5. The only person after both C (2) and D (5) is E at position 6. No one else sits after D. Thus this arrangement satisfies all conditions. Therefore the full order is B, C, A, F, D, E, and the first person to join is B. Verification / Alternative check: Verify each clue in the order B (1), C (2), A (3), F (4), D (5), E (6). C is immediately after B and immediately before A, so the block B, C, A is correct. D is not at a corner, since D is at position 5. F is at position 4, not first or last. E is at position 6 and is the only person after both C (2) and D (5). Any attempt to shift the B, C, A block to positions 2, 3, 4 or 3, 4, 5 either violates the “only E after C and D” condition or forces D or F into corner positions. Hence the found order is unique and consistent. Why Other Options Are Wrong: E cannot be first because E must join after both C and D. D cannot be first because D is not allowed to occupy a corner position. A cannot be first because it must be immediately after C. F cannot be first because F is explicitly stated not to be first or last. Only B can occupy the first position without breaking any condition. Common Pitfalls: A common error is to misread “only E joined after C and D” and place E in a non last position, which then allows other friends to be after D, violating the condition. Another pitfall is forgetting that the block B, C, A must be consecutive and in that exact order. Some candidates also overlook that both D and F have restrictions on corner positions. Drawing a six slot diagram and filling it systematically helps prevent such mistakes. Final Answer: The first person to join the institute was B.

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