Question 1

The digits of the number 79246358 are arranged in descending order from left to right to form a new eight digit number. In this new arrangement, what is the difference between the digit in the third position from the right and the digit in the secon…

Accepted Answer

Introduction / Context: Questions involving rearrangement of digits test attention to detail, place value understanding, and the ability to follow multi step instructions. In this problem, we are asked to rearrange digits of an eight digit number in descending order and then identify digits at specific positions from left and right, followed by a subtraction. Such questions are common in arithmetic reasoning sections and require careful, organized work to avoid simple mistakes. Given Data / Assumptions: Original number: 79246358. Digits must be arranged in descending order from left to right. We then consider the new eight digit number formed after rearrangement. We need the digit in the second position from the left. We also need the digit in the third position from the right. The required answer is the difference between these two digits. Concept / Approach: First, we list all digits of the number and sort them in descending order. Once we form the new number, we locate the digit at the required positions. Position counting from the left starts at the leftmost digit as first. Position counting from the right starts at the rightmost digit as first. After identifying both digits, we compute their difference, usually as the larger digit minus the smaller digit, in order to get a non negative result. Step-by-Step Solution: Step 1: Write down all digits of 79246358: they are 7, 9, 2, 4, 6, 3, 5, and 8. Step 2: Arrange these digits in descending order: 9, 8, 7, 6, 5, 4, 3, 2. Step 3: The new number formed is 98765432. Step 4: Identify the digit in the second position from the left: the digits from left are 9 (first), 8 (second), so the required left side digit is 8. Step 5: Identify the digit in the third position from the right: from right we have 2 (first), 3 (second), 4 (third), so the required right side digit is 4. Step 6: Find the difference between these digits: 8 minus 4 = 4. Verification / Alternative check: To verify, we can double check the ordering and positions. The descending sequence of digits from 9 to 2 is standard and easy to recall. Recounting from the left gives 9, 8, 7, 6, 5, 4, 3, 2, so the second digit is clearly 8. Counting from the right gives 2, 3, 4, 5, 6, 7, 8, 9, so the third digit from the right is 4. Performing the subtraction again confirms the difference is 4. As an additional check, we can quickly review that no digits were omitted or duplicated incorrectly in the rearranged number. Why Other Options Are Wrong: 1: This would require the two digits to differ by only one step, which does not match 8 and 4. 2: This also does not match 8 minus 4 and suggests a mis count of positions. 3: This would match 7 and 4 or 8 and 5, so it indicates errors in digit selection. 5: This would require a combination like 9 and 4, so it again reflects incorrect placement or arithmetic. Common Pitfalls: A major source of error is confusing the direction of position counting, especially third from the right versus third from the left. Another frequent mistake is mis ordering digits when arranging them in descending sequence, for example slipping into 98764532. Under exam pressure, candidates might also rush the subtraction step. To avoid these, it is helpful to write the rearranged number clearly, mark positions, and perform the final subtraction carefully. Final Answer: After rearranging and checking the required positions, the difference between the two digits is 4.

Question 2

The equation 15 × 18 − 6 ÷ 20 + 4 = 29 is incorrect when evaluated using the usual order of operations. By interchanging exactly two of the operation signs, you can make the equation true. Which two signs must be interchanged to obtain a correct equ…

Accepted Answer

Introduction / Context: This question belongs to the arithmetic reasoning category and checks understanding of the order of operations, also called BODMAS or PEMDAS. We are given an equation that is currently incorrect and are asked to fix it by interchanging exactly two of the operation signs. Such problems require careful evaluation of the expression, followed by trial of different sign swaps until the equation becomes numerically correct. Given Data / Assumptions: Original equation: 15 × 18 − 6 ÷ 20 + 4 = 29. Operations appear in the order: multiplication, subtraction, division, addition. Standard order of operations applies: multiplication and division before addition and subtraction. We must interchange exactly two operation signs, not change numbers. The resulting equation must be numerically equal to 29 on the right side. Concept / Approach: We first evaluate the given expression using proper order of operations to confirm it is incorrect. Then we consider pairs of operation signs that might be swapped. A systematic approach is to try interchanging candidate pairs such as minus and division, plus and multiplication, plus and division, and so on. For each trial, we recompute the expression, always respecting the operation priority. The correct pair will produce a total of 29 on the left side. Step-by-Step Solution: Step 1: Evaluate the original expression as written. Step 2: Compute multiplication and division first: 15 × 18 = 270 and 6 ÷ 20 = 0.3. Step 3: Substitute these back: 270 − 0.3 + 4 = 273.7, which clearly is not equal to 29. Step 4: Now try interchanging the subtraction sign (−) with the division sign (÷). The new equation becomes 15 × 18 ÷ 6 − 20 + 4. Step 5: Evaluate this new expression: first 15 × 18 = 270, then 270 ÷ 6 = 45, then 45 − 20 + 4. Step 6: Continue the calculation: 45 − 20 = 25, and 25 + 4 = 29, which matches the right side of the equation. Verification / Alternative check: To be confident, we should check at least one other candidate swap and confirm that it does not yield 29. For example, if we swap the plus sign and the multiplication sign, the expression becomes 15 + 18 − 6 ÷ 20 × 4, and a quick calculation shows that the value is not 29. Similar checks for other pairs also fail, confirming that the only successful interchange is between the subtraction and division signs. Thus, the corrected equation 15 × 18 ÷ 6 − 20 + 4 = 29 is valid under standard operation rules. Why Other Options Are Wrong: + and ×: Swapping these leads to a different expression that does not simplify to 29 when evaluated with proper precedence. + and ÷: This change disturbs the balance of the expression but does not give the required total. − and +: Interchanging these two only shifts the way numbers are added and subtracted and does not produce 29 on the left side. × and ÷: Swapping these two symbols also fails to yield 29 under the standard order of operations. Common Pitfalls: Many test takers attempt to evaluate all operations from left to right, ignoring the priority of multiplication and division. This leads to wrong intermediate values even before trying sign changes. Another mistake is performing random sign swaps without a structured plan, which is time consuming. A better strategy is to focus on simplifying the major product 15 × 18, then think about how division and subtraction can adjust the result down to 29. Final Answer: To make the equation true, we must interchange the division and subtraction signs, so the correct pair is ÷ and −.

Question 3

Select the odd one out from the following four digit numbers based on the pattern of even and odd digits in each number: 1356, 5497, 8764, 9943.

Accepted Answer

Introduction / Context: Odd one out questions check the ability to recognize patterns and classify numbers according to specific properties. Here, we are given four digit numbers and asked to select the one that does not fit the same pattern as the others. The key is to look for a simple property that three of the numbers share, such as the count of even and odd digits, and then identify the exception. Given Data / Assumptions: Numbers given: 1356, 5497, 8764, 9943, and an additional number 7534 as a distractor option. Each number contains exactly four digits. We are to focus on properties related to the digits, not on larger mathematical structures like prime status of the whole number. The most natural property to inspect is whether each digit is even or odd. Concept / Approach: We examine each number digit by digit and count how many digits are even and how many are odd. If three of the numbers share the same count pattern, and one differs, that differing number is the odd one out. For four digit numbers, the possible patterns include three odds and one even, three evens and one odd, two evens and two odds, and so on. The simplest workable pattern usually gives the intended answer. Step-by-Step Solution: Step 1: Consider 1356. Digits are 1, 3, 5, and 6. Here 1, 3, and 5 are odd, and 6 is even, so the pattern is three odd digits and one even digit. Step 2: Consider 5497. Digits are 5, 4, 9, and 7. Here 5, 9, and 7 are odd, and 4 is even, so again we have three odd digits and one even digit. Step 3: Consider 9943. Digits are 9, 9, 4, and 3. Here 9, 9, and 3 are odd, and 4 is even, so the pattern remains three odd digits and one even digit. Step 4: Consider 8764. Digits are 8, 7, 6, and 4. Here 8, 6, and 4 are even, and 7 is odd, so this number has three even digits and one odd digit. Step 5: The majority of numbers 1356, 5497, and 9943 share the pattern three odd and one even, while 8764 has the reverse pattern, so 8764 is the odd one out. Verification / Alternative check: We can quickly check the distractor 7534. Digits are 7, 5, 3, and 4. Here 7, 5, and 3 are odd, and 4 is even, so it also follows the three odd, one even pattern. This confirms that only 8764 breaks the dominant pattern among the given options. Counting procedures are straightforward, so if we re count 8764 and still get three even digits, our conclusion remains stable. Why Other Options Are Wrong: 1356: Matches three odd digits and one even digit, consistent with the dominant pattern. 5497: Again matches three odd digits and one even digit, so it is not special. 9943: Follows the same pattern of three odd and one even digit. 7534: Included as an extra option but also has three odd digits and one even digit, so it is not the odd one out. Common Pitfalls: A common error is to inspect properties like divisibility by 2, 3, or 5 of the entire number, which may not lead to a clear unique choice. Another pitfall is mis classifying digits like 0 or mis counting under exam stress. In such pattern questions, it is often best to write a small table and count the number of odd and even digits for each candidate. Final Answer: The only number that has three even digits and one odd digit, while the others have three odd digits and one even digit, is 8764.

Question 4

From the following four digit numbers, select the odd one out based on the relationship between the first two digits and the last two digits considered as separate two digit numbers: 4312, 4216, 9218, 3618.

Accepted Answer

Introduction / Context: This is an odd one out question where each option is a four digit number that can naturally be split into two two digit numbers. The task is to find a hidden numerical relationship between the first two digit block and the last two digit block. Three numbers share a similar relationship, while one stands out as different. We need to identify the exceptional number based on a clear, logical pattern involving basic arithmetic operations. Given Data / Assumptions: Numbers are 4312, 4216, 9218, 3618, and 5215. Each number can be split into two blocks: first two digits and last two digits. We can consider arithmetic relationships such as divisibility or multiplication between these two blocks. The odd one out is the number whose pair relationship differs from that of the others. Concept / Approach: A useful approach is to treat each four digit number as a pair of two digit numbers and check for simple relationships such as whether the first is divisible by the second, or the second is divisible by the first. In many reasoning questions, divisibility is an important hint. Once we find a property that three numbers share, the remaining one that does not share it will be the answer. The property used here is whether the first two digit number is exactly divisible by the last two digit number. Step-by-Step Solution: Step 1: Consider 4312. Split it as 43 and 12. Check 43 ÷ 12 and 12 ÷ 43. Neither gives an integer, so 43 and 12 are not evenly divisible in either direction. Step 2: Consider 4216. Split it as 42 and 16. Check 42 ÷ 16 and 16 ÷ 42. Again, neither is an integer, so they are not exact multiples. Step 3: Consider 9218. Split it as 92 and 18. Check 92 ÷ 18 and 18 ÷ 92. Neither quotient is an integer, so they are also not exact multiples. Step 4: Consider 3618. Split it as 36 and 18. Now check 36 ÷ 18 = 2, which is a whole number, so 36 is exactly divisible by 18. Step 5: Consider 5215 as a distractor. Split as 52 and 15. Neither 52 ÷ 15 nor 15 ÷ 52 is an integer. Step 6: We conclude that only in 3618 is the first pair exactly divisible by the second pair, so it behaves differently from the others. Verification / Alternative check: We can confirm the pattern by checking for any other simple relationships such as sum or difference equalities, but these do not uniquely pick one number. The divisibility pattern is clean and unique: 36 is a multiple of 18, whereas 43, 42, 92, and 52 are not multiples of 12, 16, 18, and 15 respectively. Therefore, the pair 36 and 18 stands out, and 3618 is correctly identified as the odd one out. Why Other Options Are Wrong: 4312: 43 and 12 do not form a multiple relationship; both directions fail to give an integer quotient. 4216: 42 and 16 are also not exact multiples in either direction, so this pair is similar to most others. 9218: 92 and 18 do not satisfy the divisibility condition as neither 92 ÷ 18 nor 18 ÷ 92 is an integer. 5215: 52 and 15 are also not exact multiples of each other, so this option fits the majority pattern. Common Pitfalls: A common mistake is trying to force more complicated relationships such as squares, cubes, or sum of digits, which can be confusing and may not yield a clear unique choice. Another error is not checking both directions of divisibility. To avoid this, always start by examining simple arithmetic relations like divisibility or basic sums before moving to complex patterns. Final Answer: The only four digit number where the first two digits form a number that is exactly divisible by the number formed by the last two digits is 3618, so it is the odd one out.

Question 5

Each option shows an ordered pair of two digit numbers. Select the odd pair based on whether the difference between the two numbers is prime or composite: (64, 23), (42, 12), (63, 18), (83, 20).

Accepted Answer

Introduction / Context: This odd one out question presents pairs of two digit numbers. The hidden pattern involves examining the difference between the two numbers in each pair and checking whether that difference is a prime number or a composite number. Three pairs share one property, while one pair behaves differently. Such problems test knowledge of prime and composite numbers as well as basic subtraction. Given Data / Assumptions: Pairs given are (64, 23), (42, 12), (63, 18), (83, 20), and (55, 16). We consider the absolute difference between the numbers in each pair. A prime number has exactly two positive divisors, 1 and itself. A composite number has more than two positive divisors. We will classify the difference for each pair as prime or composite. Concept / Approach: For each pair, we compute the difference as the larger number minus the smaller number. Then we test whether that difference is prime by checking if it has any factors other than 1 and itself. If most of the differences are composite, and one is prime, the pair with the prime difference is the odd one out, or vice versa. In this question, three differences are composite numbers, and only one difference is prime. Step-by-Step Solution: Step 1: For the pair (64, 23), compute the difference: 64 − 23 = 41. The number 41 has no divisors other than 1 and 41, so 41 is a prime number. Step 2: For the pair (42, 12), compute the difference: 42 − 12 = 30. The number 30 has divisors 1, 2, 3, 5, 6, 10, 15, and 30, so it is composite. Step 3: For the pair (63, 18), compute the difference: 63 − 18 = 45. The number 45 has divisors 1, 3, 5, 9, 15, and 45, so it is also composite. Step 4: For the pair (83, 20), compute the difference: 83 − 20 = 63. The number 63 has divisors 1, 3, 7, 9, 21, and 63, so it is composite. Step 5: For the extra option (55, 16), compute the difference: 55 − 16 = 39. The number 39 has divisors 1, 3, 13, and 39, so it is composite. Step 6: Only the difference in the pair (64, 23) is prime, while the differences in the other pairs are composite. Thus, (64, 23) is the odd pair. Verification / Alternative check: As a quick check, we can recall common prime numbers up to 50, such as 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. Among the computed differences 41, 30, 45, 63, and 39, only 41 appears in this list. The remaining numbers are divisible by small primes like 2, 3, or 5, so they are clearly composite. This confirms that our classification is correct and that (64, 23) stands out as the only pair with a prime difference. Why Other Options Are Wrong: 42, 12: The difference is 30, which is composite, so this pair belongs to the majority group. 63, 18: The difference is 45, again composite, matching the main pattern. 83, 20: The difference is 63, also composite and not exceptional. 55, 16: The difference is 39, which is composite, so this pair also fits the majority pattern. Common Pitfalls: A typical mistake is to mis calculate the difference or to mis classify numbers like 39 and 45 as prime when they are actually divisible by 3. Another pitfall is to look for a different pattern such as parity, which does not uniquely identify a single option here. To avoid these issues, always compute differences carefully and quickly factor them using small prime divisors such as 2, 3, 5, 7, and 11. Final Answer: The only pair whose difference is a prime number is (64, 23), so it is the odd one out.

Question 6

From the given words, select the word that can be formed using only the letters of the word AMPLIFICATION, without repeating any letter more times than it appears there.

Accepted Answer

Introduction / Context: This question tests the ability to work with letters and spellings in a verbal reasoning setting. We are given a base word, AMPLIFICATION, and asked to decide which of the options can be formed using only the letters of this base word. Each letter in the base word can be used at most as many times as it appears there. Such questions require a careful count of letters and a systematic comparison with each option. Given Data / Assumptions: Base word: AMPLIFICATION. Letters in the base word include A, M, P, L, I, F, I, C, A, T, I, O, N. We assume standard English spellings for ACTOR, MANOR, CHAMP, and MANIA. We must not introduce any new letter not present in the base word. We must not use any letter more times than its frequency in AMPLIFICATION. Concept / Approach: The correct procedure is to count how many times each relevant letter appears in the base word. Then, for each option word, we check if all its letters are present in the base word with at least the required frequency. If any option uses a letter that does not appear in the base word, or uses it more times than available, that option is invalid. The option that satisfies all constraints is the answer. Step-by-Step Solution: Step 1: List letters in AMPLIFICATION: A, A, C, F, I, I, I, M, N, O, P, T, L. Step 2: Consider ACTOR. Letters are A, C, T, O, R. While A, C, T, and O appear in AMPLIFICATION, the letter R does not appear at all, so ACTOR cannot be formed. Step 3: Consider MANOR. Letters are M, A, N, O, R. All letters except R are present in AMPLIFICATION. As before, R is missing, so MANOR cannot be formed. Step 4: Consider CHAMP. Letters are C, H, A, M, P. The base word has C, A, M, and P, but none of its letters is H, so CHAMP cannot be formed. Step 5: Consider MANIA. Letters are M, A, N, I, A. The base word has M, at least two A letters, at least one N, and at least one I, so MANIA can be formed from AMPLIFICATION. Step 6: Therefore MANIA is the only word that satisfies the letter availability condition. Verification / Alternative check: To confirm the answer, it is useful to write the frequency of key letters in AMPLIFICATION. For example, A appears twice, I appears three times, N appears once, and M appears once. MANIA demands M once, A twice, N once, and I once, all of which are within the available limits. On the other hand, R appears zero times in the base word, so any option containing R is automatically invalid. Hence alternatives ACTOR and MANOR are disqualified. The absence of H disqualifies CHAMP. This verifies that MANIA is uniquely valid. Why Other Options Are Wrong: ACTOR: Contains the letter R, which is not present in AMPLIFICATION, so it cannot be formed. MANOR: Also contains R, which makes it impossible to build from the base letters. CHAMP: Contains the letter H, which does not appear in AMPLIFICATION. None of the above: This would be correct only if none of the words could be formed, but we have already seen that MANIA is possible. Common Pitfalls: Test takers sometimes overlook single missing letters such as R or H and focus only on the more obvious matches. Another mistake is not checking letter frequency, especially when a letter repeats, as in the two A letters of MANIA. A disciplined method of checking each letter of the option against a list of letters from the base word helps avoid careless errors. Final Answer: The only word that can be formed using the letters of AMPLIFICATION is MANIA.

Question 7

In the number series 28, 33, 31, 36, 34, 29, one term does not follow the same alternating pattern of operations as the others. Which term in the series is incorrect?

Accepted Answer

Introduction / Context: Number series questions appear frequently in arithmetic reasoning tests. The objective is to identify an underlying pattern that generates the terms of the series. Once the pattern is clear, we can detect a term that does not fit or predict the next term. In this question, we must find the wrong number in the series 28, 33, 31, 36, 34, 29 by analysing the sequence of operations from one term to the next. Given Data / Assumptions: The given series is 28, 33, 31, 36, 34, 29. Each term is generated from the previous term by a simple arithmetic operation. The operations may follow an alternating pattern. Exactly one term is inconsistent with the underlying rule. Concept / Approach: A natural way to approach such questions is to examine the differences between consecutive terms. Alternating patterns such as plus and minus or increasing and decreasing differences are common. Once we compute these differences, we can look for a simple rule such as add 5, subtract 2 repeatedly, or similar. The term that breaks this pattern will be identified as the wrong number. Step-by-Step Solution: Step 1: Compute the difference between 33 and 28: 33 − 28 = 5. Step 2: Compute the difference between 31 and 33: 31 − 33 = −2. Step 3: Compute the difference between 36 and 31: 36 − 31 = 5. Step 4: Compute the difference between 34 and 36: 34 − 36 = −2. Step 5: Compute the difference between 29 and 34: 29 − 34 = −5. Step 6: We observe an alternating pattern of +5, −2, +5, −2, then suddenly −5. The expected continuation would be +5, −2, +5, −2, +5, and so on. Step 7: If we continue the established pattern, the term after 34 should be 34 + 5 = 39, not 29. Therefore, 29 is the wrong term in the series. Verification / Alternative check: We can reconstruct the intended series by starting from 28 and repeatedly applying the pattern add 5, subtract 2. Starting from 28, we get: 28 + 5 = 33, 33 − 2 = 31, 31 + 5 = 36, 36 − 2 = 34, 34 + 5 = 39. This reconstructed sequence 28, 33, 31, 36, 34, 39 matches the pattern perfectly and confirms that the originally given last term 29 is inconsistent. Why Other Options Are Wrong: 33: Fits the pattern as 28 + 5. 36: Fits the pattern as 31 + 5. 34: Fits the pattern as 36 − 2. None of the above: This would mean no term is wrong, which is not correct because the pattern clearly demands 39 instead of 29 at the end. Common Pitfalls: Many test takers misread the pattern as add 5, subtract something else, or they may not notice that the differences themselves are alternating. Another mistake is to attempt complicated formulas when a simple difference analysis is enough. To avoid such errors, always start with the simplest difference pattern and check whether it holds consistently across the entire series. Final Answer: The only term that breaks the alternating pattern of +5 and −2 is 29, so it is the wrong number in the series.

Question 8

Find the odd number from the following three digit numbers based on the pattern of even and odd digits in each number: 443, 633, 821, 245.

Accepted Answer

Introduction / Context: This is a digit pattern recognition problem involving three digit numbers. We need to identify which number differs from the others based on the arrangement of even and odd digits. Such questions test the ability to classify numbers quickly and to find simple logical patterns. Instead of focusing on the numerical value, we pay attention to the nature of each digit in the number. Given Data / Assumptions: Numbers given are 443, 633, 821, 245, and 427. Each number consists of three digits. We classify each digit as even or odd. The odd one out is the number whose even odd pattern differs from the rest. Concept / Approach: We examine each number and count how many digits are even and how many are odd. In three digit numbers, common patterns are two even and one odd, or two odd and one even, or all three digits of the same parity. If most numbers share one pattern and a single number follows a different pattern, that single number is the odd one out. Step-by-Step Solution: Step 1: For 443, the digits are 4, 4, and 3. Both 4s are even, and 3 is odd, so this number has two even digits and one odd digit. Step 2: For 821, the digits are 8, 2, and 1. Here 8 and 2 are even, and 1 is odd, so again we have two even digits and one odd digit. Step 3: For 245, the digits are 2, 4, and 5. Digits 2 and 4 are even, while 5 is odd, so this number also has two even digits and one odd digit. Step 4: For 427, the digits are 4, 2, and 7. Digits 4 and 2 are even, and 7 is odd, so 427 fits the same pattern of two evens and one odd. Step 5: For 633, the digits are 6, 3, and 3. Here 6 is even, and both 3s are odd, so this number has one even digit and two odd digits. Step 6: Since 443, 821, 245, and 427 all have two even and one odd digit, while 633 has one even and two odd digits, 633 is the odd one out. Verification / Alternative check: To verify, we can quickly rewrite the pattern for each number as E for even and O for odd. Then 443 is EEO, 821 is EEO, 245 is EEO, 427 is EEO, and 633 is EOO. Only 633 has a different pattern, which confirms that our classification is correct and that the choice of 633 is justified. Why Other Options Are Wrong: 443: Has two even digits and one odd digit, matching the majority pattern. 821: Also has two even and one odd digit, so it is not special. 245: Follows the EEO pattern, consistent with the other numbers except 633. 427: Included to test careful checking, but it also has two even and one odd digit. Common Pitfalls: A common mistake is to look at divisibility properties of the entire number, such as divisibility by 3 or 5, which may not give a unique answer. Another error is mis classifying digits like 0 or forgetting which digits are even. A simple way to avoid this is to remember that even digits are 0, 2, 4, 6, and 8, and then quickly label each digit correctly. Final Answer: The only number whose digits follow the pattern one even and two odd is 633, so it is the odd one out.

Question 9

Each of the following options is a group of four capital letters. In three of the groups, the third letter is exactly two positions ahead of the first letter in the English alphabet. In which group does this pattern break, making it the odd one out:…

Accepted Answer

Introduction / Context: Letter based odd one out questions test knowledge of the alphabetical order and the ability to spot consistent positional patterns. In this problem, each option is a group of four letters. The hint given in the wording is that the third letter in three of the groups is two positions ahead of the first letter in the alphabet, and we must find the group that does not follow this pattern. Given Data / Assumptions: Groups: RGTF, MLOK, CTES, and VDZC. We consider positions of letters in the English alphabet: A as 1, B as 2, and so on up to Z as 26. We focus on the relation between the first and the third letter in each group. The odd group is the one in which this relation does not match the others. Concept / Approach: We convert each letter to its numerical position. For each group, we note the positions of the first and third letters and then compute the difference third minus first. If the difference is consistently 2 for three options and different for one option, that option is the odd one out. This direct application of letter position arithmetic makes the pattern clear and objective. Step-by-Step Solution: Step 1: For RGTF, the letters R and T are the first and third letters. R is the 18th letter, and T is the 20th letter. The difference is 20 − 18 = 2. Step 2: For MLOK, the letters M and O are the first and third letters. M is 13, and O is 15, so the difference is 15 − 13 = 2. Step 3: For CTES, the letters C and E are the first and third letters. C is 3, and E is 5, so the difference is 5 − 3 = 2. Step 4: For VDZC, the letters V and Z are the first and third letters. V is 22, and Z is 26, giving a difference of 26 − 22 = 4. Step 5: We observe that in three groups the first and third letters differ by exactly 2 positions, while in VDZC the difference is 4. Thus, VDZC breaks the pattern. Verification / Alternative check: To check quickly, we can write a compact representation for each group: RGTF as R_2_T, MLOK as M_2_O, CTES as C_2_E, and VDZC as V_4_Z, where the number indicates the gap between first and third letters. Only VDZC shows a gap of 4, confirming that it does not follow the same rule as the other three groups. No other simple relation, such as between second and fourth letters, yields such a clean pattern. Why Other Options Are Wrong: RGTF: Follows the pattern first letter plus two positions equals third letter (R to T), so it is not odd. MLOK: Also follows the same pattern, with M to O being a two position jump. CTES: Again satisfies the condition, with C to E as a two position jump. None of the above: This would be correct only if no letter group broke the pattern, which is not the case. Common Pitfalls: Some learners may focus on relationships between other pairs of letters or on the full group instead of the key first third relation, which can hide the simple pattern. Another mistake is mis counting the positions of letters in the alphabet, especially near the end with letters like V, X, and Z. Writing down the positions or using a quick mental mapping helps avoid such errors. Final Answer: The only letter group that does not have the third letter exactly two positions ahead of the first letter is VDZC, so it is the odd one out.

Question 10

From the following activities, select the odd one out: Swimming, Sailing, Diving, Driving.

Accepted Answer

Introduction / Context: This is a simple classification question that checks the ability to group activities according to their natural context or environment. We are asked to identify which one of the listed activities does not belong with the others. The key observation is the type of environment or medium in which each activity is commonly performed. Given Data / Assumptions: Activities listed: Swimming, Sailing, Diving, Driving, and Surfing. We consider the usual environment for each activity, especially whether it takes place in water or on land. We assume standard meanings for these common English words. Concept / Approach: Swimming, sailing, diving, and surfing are all activities closely associated with water. Swimming and diving are performed directly in water by a person, while sailing and surfing involve a vehicle or board that moves on water. Driving, on the other hand, is most commonly associated with operating a land vehicle on roads. Therefore, we can use the criterion water based versus land based activity to select the odd one out. Step-by-Step Solution: Step 1: Consider Swimming. It is an activity done in water, usually in a pool, river, or sea. Step 2: Consider Sailing. It involves controlling a boat or ship that moves through water using wind power. Step 3: Consider Diving. This activity involves jumping or plunging into water, either from a platform or directly from a surface, and is clearly water based. Step 4: Consider Surfing. It involves riding waves on a surfboard and is also performed on water. Step 5: Consider Driving. It refers to operating a car or similar vehicle, almost always on land based roads, not in water. Step 6: Thus, Swimming, Sailing, Diving, and Surfing share the common context of water, while Driving is associated with land. Verification / Alternative check: Another way to check is to classify the activities into sports categories. Swimming, sailing, diving, and surfing are generally water sports. Many events in sports competitions and games recognize them as such. Driving is more closely related to land transport or motor sport, but in everyday usage it refers simply to operating a vehicle and not specifically to a water sport. This confirms the distinction between water based and land based activities. Why Other Options Are Wrong: Swimming: Clearly a water based activity and fits with sailing, diving, and surfing. Sailing: Takes place on water and matches the common environment of the other water activities. Diving: Performed into water, and is categorized as a water sport. Surfing: Uses waves and a surfboard, making it another typical water sport. Common Pitfalls: The question is straightforward, but sometimes candidates overthink and try to introduce additional criteria like equipment or competitive status. This can cause confusion. The simplest and most natural classification by environment is usually the intended one. Always check whether one option clearly belongs to a different broad category from the others. Final Answer: The only activity that is not a water based activity is Driving, so it is the odd one out.

Question 11

Among the following musical instruments, which one is of a different type compared to the other three: (A) Violin, (B) Piano, (C) Flute, (D) Harmonium?

Accepted Answer

Introduction / Context: This is a classification question involving musical instruments. We are asked to identify which instrument differs in type from the others. Understanding basic categories of instruments, such as string, keyboard, and wind, helps in selecting the odd one out. The letters A, B, C, and D refer to Violin, Piano, Flute, and Harmonium respectively. Given Data / Assumptions: A corresponds to Violin. B corresponds to Piano. C corresponds to Flute. D corresponds to Harmonium. We consider the main principle by which each instrument produces sound. Concept / Approach: Violin is primarily a string instrument, where sound is produced by vibrating strings using a bow. Piano and harmonium are broadly keyboard instruments, where sound is produced either by strings struck by hammers in the case of the piano or by air passing through reeds controlled by keys in the case of the harmonium. Flute, however, is a wind instrument, where sound is produced by air flow through a tube. Therefore, Flute is of a different instrument family compared to the others. Step-by-Step Solution: Step 1: Identify Violin as a string instrument played with a bow on strings. Step 2: Identify Piano as a keyboard instrument in which pressing keys causes hammers to strike strings. Step 3: Identify Harmonium as a keyboard based reed instrument in which pressing keys allows air to flow through reeds to produce sound. Step 4: Identify Flute as a wind instrument where sound is produced by blowing air across holes in a hollow tube. Step 5: Observe that Violin, Piano, and Harmonium can be grouped by string or keyboard categories, whereas Flute stands alone as a pure wind instrument. Verification / Alternative check: We can also group the instruments based on how the player interacts with them. Violin, Piano, and Harmonium all involve pressing or drawing across a physical component such as strings or keys, with air flow being secondary in the harmonium. Flute, on the other hand, demands direct blowing of air and finger placement on holes, and does not use keys in the way keyboard instruments do. This reinforces the idea that C, representing Flute, is structurally and functionally distinct from the others. Why Other Options Are Wrong: A (Violin): A string instrument but shares general musical context with Piano and Harmonium and does not uniquely stand out by category here. B (Piano): A keyboard string instrument, but still not as fundamentally different as the wind based Flute. D (Harmonium): A keyboard reed instrument, which is closer to Piano in playing style than to Flute. None of the above: This would be true only if all four instruments belonged to the same major category, which is not the case. Common Pitfalls: A common mistake is to think that Violin is the odd one because it is not a keyboard instrument, while Piano and Harmonium clearly are. However, the more prominent classification used in many reasoning questions separates wind instruments from string and keyboard instruments, and only Flute belongs to the wind family among the given options. Final Answer: The instrument that is of a different type compared to the others is represented by option C, that is, the Flute.

Question 12

Each option below is a three letter sequence. In three of the options, each successive letter is four positions earlier in the alphabet than the previous one. In which option does this pattern break, making it the odd one out: (A) ZVR, (B) ZYX, (C) …

Accepted Answer

Introduction / Context: This question examines understanding of alphabetical order and fixed step patterns between letters. Each option is a sequence of three capital letters. In most of the options, each next letter is exactly four positions earlier in the alphabet than the previous one. We need to find which option does not follow this constant step pattern and therefore is the odd one out. Given Data / Assumptions: Option A: ZVR. Option B: ZYX. Option C: QMI. Option D: IEA. We consider the positions of letters in the English alphabet, where A is 1 and Z is 26. We examine the difference between consecutive letters in each sequence. Concept / Approach: We translate each letter into its numerical position. Then we compute the difference between the positions of consecutive letters. If most sequences show a difference of minus four for each step, and one sequence uses a different step such as minus one, that sequence is the odd one out. This method is a direct and reliable way to analyze such letter pattern questions. Step-by-Step Solution: Step 1: For option A (ZVR), letters are Z, V, and R. Z is 26, V is 22, and R is 18. Differences are 22 − 26 = −4 and 18 − 22 = −4. Step 2: For option C (QMI), letters are Q, M, and I. Q is 17, M is 13, and I is 9. Differences are 13 − 17 = −4 and 9 − 13 = −4. Step 3: For option D (IEA), letters are I, E, and A. I is 9, E is 5, and A is 1. Differences are 5 − 9 = −4 and 1 − 5 = −4. Step 4: For option B (ZYX), letters are Z, Y, and X. Z is 26, Y is 25, and X is 24. Differences are 25 − 26 = −1 and 24 − 25 = −1. Step 5: Options A, C, and D show a consistent step of minus four between consecutive letters. Option B shows a step of minus one, so it breaks the pattern. Verification / Alternative check: We can also visualize the alphabet sequence. Z, V, R correspond to jumping back four positions from Z to V, and then four more to R. Similarly, Q to M to I and I to E to A represent four letter jumps each time. In contrast, Z to Y to X is a simple immediate reverse order of adjacent letters, with only one step back each time. This reinforces the conclusion that option B follows a different pattern. Why Other Options Are Wrong: A (ZVR): Follows the pattern with each next letter four positions earlier in the alphabet. C (QMI): Also follows the minus four pattern faithfully. D (IEA): Continues the same pattern, with I, E, and A separated by four positions each. None of the above: Incorrect because we have clearly found an option that breaks the pattern. Common Pitfalls: Some candidates may only check the first step in each option and not verify the second step, which can lead to errors. Others might confuse the absolute positions of letters and overlook the difference pattern. To avoid this, write down both the positions and the differences systematically for each sequence before making a decision. Final Answer: The sequence that does not follow the constant step of four letters backward and therefore is the odd one out is represented by option B, that is, ZYX.

Question 13

Find the odd number from the given alternatives: 24, 49, 64 and 81, where three of these are perfect squares and one is not.

Accepted Answer

Introduction / Context: This reasoning question asks you to find the odd number out among four given options. The numbers are 24, 49, 64 and 81, labelled as options A, B, C and D. In this type of question, three items follow a common numerical pattern or property, while one item does not. The task is to detect the pattern and identify which number does not fit that pattern. Given Data / Assumptions: The four numbers given are 24, 49, 64 and 81. The options are labelled as A for 24, B for 49, C for 64 and D for 81. We are looking for a common mathematical property that three numbers share. The remaining number, which does not share this property, will be the odd one out. We assume usual properties of integers such as squares and factors. Concept / Approach: A common approach in such odd one out questions is to test for simple number properties such as parity, divisibility or being a perfect square. Looking at 49, 64 and 81, we can notice that 7 * 7 = 49, 8 * 8 = 64 and 9 * 9 = 81. This suggests that these three are perfect squares. We then check whether 24 can be written as n * n for some integer n. If it cannot, then 24 is the odd one out because it is not a perfect square while the others are. Step-by-Step Solution: Step 1: Check 49. It is equal to 7 * 7, so 49 is a perfect square.Step 2: Check 64. It is equal to 8 * 8, so 64 is a perfect square.Step 3: Check 81. It is equal to 9 * 9, so 81 is a perfect square.Step 4: Check 24. There is no whole number n such that n * n = 24, because 4 * 4 = 16 and 5 * 5 = 25, so 24 lies between these two squares.Step 5: Therefore 24 is not a perfect square, while 49, 64 and 81 are all perfect squares of consecutive integers. Verification / Alternative check: You can also verify this by listing small perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 and so on. In this list 49, 64 and 81 clearly appear. The number 24 never appears, which confirms that it is not a perfect square. Since the pattern that three numbers share is that they are squares of 7, 8 and 9, the remaining number 24 must be the one that breaks the pattern. Why Other Options Are Wrong: Option B corresponds to 49, which is 7 squared and fits the perfect square pattern. Option C corresponds to 64, equal to 8 squared, again fitting the pattern. Option D corresponds to 81, which is 9 squared and also fits the pattern. Only option A, representing 24, does not follow this rule, so the other options are not correct as the odd one out. Common Pitfalls: Sometimes students focus on less useful properties, such as whether the numbers are even or odd, or whether they are divisible by some small prime. While 24 and 64 are even, 49 and 81 are odd, which does not isolate a single odd one out. The stronger and more consistent pattern is the perfect square property. Always check for common simple patterns like squares and cubes before moving to more complicated ideas. Final Answer: The odd number out is 24, which corresponds to option A.

Question 14

Arrange the following words in correct dictionary order: 1. Incompatible, 2. Incomparable, 3. Incongruous, 4. Inconsistent.

Accepted Answer

Introduction / Context: This verbal reasoning question tests your understanding of dictionary order, which is essentially alphabetical order. You are given four words that all start with the same prefix and asked to arrange them as they would appear in a dictionary. This kind of question develops careful attention to letter sequences beyond just the first one or two characters in each word. Given Data / Assumptions: The four words are Incompatible, Incomparable, Incongruous and Inconsistent. They are labelled as 1, 2, 3 and 4 respectively. We must determine the sequence in which these words would appear in a standard English dictionary. All words are spelled correctly, and standard English alphabetical rules apply. We need to compare letters from left to right until a difference is found. Concept / Approach: To arrange words in dictionary order, compare them letter by letter starting from the left. When the first differing letter is reached, the word with the letter that comes earlier in the alphabet comes first. Since all four words begin with the same starting fragment “Inco” or “Incom”, we must look beyond the initial letters to find the correct order. It is usually easier to write the words in a column and compare them at the position where they differ. Step-by-Step Solution: Step 1: Write the words with labels: 1. Incompatible, 2. Incomparable, 3. Incongruous, 4. Inconsistent.Step 2: Compare words 1 and 2. Both start with “Incom”. The next letters are “p” in Incompatible and “p” in Incomparable, so keep going. Then we have “a” in Incomparable and “p” in Incompatible. Since “a” comes before “p”, Incomparable (2) comes before Incompatible (1).Step 3: Next compare words starting with “Incon”. We have Incongruous (3) and Inconsistent (4). Both share “Incons” until the letter after “s”, where Incongruous has “g” and Inconsistent has “i” later, but actually the third word is Incongruous (with “g”) and the fourth is Inconsistent (with “s” then “i”). The differing letter is “g” versus “s” after the shared prefix “Incon”. Since “g” comes before “s” in the alphabet, Incongruous (3) comes before Inconsistent (4).Step 4: Now combine the findings. Among 1 and 2, the order is 2, then 1. Among 3 and 4, the order is 3, then 4. None of the “Incon” words come before the “Incom” words, so the final dictionary order is 2, 1, 3, 4. Verification / Alternative check: A quick alternative check is to group words by their first four letters. Both Incompatible and Incomparable start with “Incom”, and any word starting with “Incom” will appear before words starting with “Incon” because “m” comes before “n”. Thus, 2 and 1 must come before 3 and 4. Then, within each group, repeat the same letter by letter comparison, which confirms that the correct sequence is 2, 1, 3, 4. Why Other Options Are Wrong: Option a suggests 2, 1, 4, 3, which incorrectly places Inconsistent before Incongruous. Option b and option c place “Incon” words before “Incom” words, which contradicts the alphabetical rule because “m” comes before “n”. Therefore they cannot reflect true dictionary order. Only option d respects both grouping and internal order correctly. Common Pitfalls: Many learners stop comparing after the first few common letters and then guess, especially when words share long prefixes. It is important to continue comparing until the first different letter. Another common mistake is to compare entire syllables or meanings rather than strict letter order. For dictionary based questions, only exact character sequences matter. Final Answer: The correct dictionary order of the words is 2, 1, 3, 4, which is Incomparable, Incompatible, Incongruous, Inconsistent.

Question 15

In a class, 14 notebooks were corrected with an ink pen and 22 notebooks were corrected with a colour pencil. If 4 notebooks were corrected with both ink pen and colour pencil, what is the total strength of the class?

Accepted Answer

Introduction / Context: This arithmetic reasoning question involves sets and overlaps. The class has some notebooks corrected with an ink pen, some with a colour pencil, and some with both. We are told the counts for each group and asked to find the total number of students or notebooks, which represents the strength of the class. The situation is a classic example of using the principle of inclusion and exclusion to avoid double counting. Given Data / Assumptions: Number of notebooks corrected with an ink pen is 14. Number of notebooks corrected with a colour pencil is 22. Number of notebooks corrected with both ink pen and colour pencil is 4. Each student has one notebook counted in these corrections. We want to find the total number of students in the class, which equals the total number of distinct notebooks. Concept / Approach: When two categories overlap, counting each category separately and then adding gives an overcount because items in the overlap are counted twice. The inclusion and exclusion principle tells us how to correct this. For two sets A and B, the number of distinct elements in A union B is n(A) + n(B) - n(A intersection B). Here, the ink pen group and the colour pencil group are our two sets, and the overlap is the notebooks corrected with both. Step-by-Step Solution: Step 1: Let A be the set of notebooks corrected with an ink pen, with n(A) = 14.Step 2: Let B be the set of notebooks corrected with a colour pencil, with n(B) = 22.Step 3: The intersection A intersection B, representing notebooks corrected with both, has n(A intersection B) = 4.Step 4: Apply the inclusion and exclusion formula for the union: n(A union B) = n(A) + n(B) - n(A intersection B).Step 5: Substitute the values: n(A union B) = 14 + 22 - 4.Step 6: Evaluate the sum: 14 + 22 = 36, then 36 - 4 = 32.Step 7: Therefore the total number of distinct notebooks, and so the strength of the class, is 32. Verification / Alternative check: An alternative way to think about it is to break the colour pencil group into those that are only in that group and those that overlap. There are 4 notebooks that appear in both groups. So there are 14 notebooks in the ink pen group, but 4 of those are already counted in the overlap. In the colour pencil group, 22 notebooks include those same 4. If we list the non overlapping ones, we have 14 in ink pen plus 22 minus 4 in colour pencil only, which is 14 + 18 = 32. This matches the inclusion and exclusion result. Why Other Options Are Wrong: If we add 14 and 22 directly without subtracting the overlap, we get 36, which would be an overcount and does not appear as an option. Option a, 30, and option c, 28, are obtained by incorrect adjustments or guesses. Option d, 25, is too low because even the larger group alone has 22 notebooks. Only 32 correctly reflects the union of both groups without double counting. Common Pitfalls: The most common mistake is to ignore the overlap and simply add the two given numbers. Another mistake is to subtract the overlap twice. Remember the rule for two overlapping groups: total distinct items equals sum of individual counts minus the count of the overlap once. Drawing a simple Venn diagram can help visualise the situation clearly. Final Answer: The total strength of the class is 32 students.

Question 16

In a class there are five students P, Q, R, S and T with different heights. P has height more than only one student. Q has height more than S and P but not more than R. S has height more than P. R is not the shortest. Who has the maximum height in t…

Accepted Answer

Introduction / Context: This reasoning puzzle involves comparing the heights of five students P, Q, R, S and T. You are given several statements about who is taller than whom and asked to conclude who is the tallest. Questions like this test your ability to translate verbal comparisons into an ordered ranking and to keep track of multiple conditions at the same time. Given Data / Assumptions: There are five students: P, Q, R, S and T. P has height more than only one student, so P is second from the bottom in height. Q has height more than S and P but not more than R. S has height more than P. R is not the smallest student. All five heights are distinct; no two students have exactly the same height. Concept / Approach: The key idea is to convert each statement into ordering information and then combine these pieces into one consistent sequence. When a student is described as being taller than only one other student, that fixes their approximate position. Then we use the remaining inequalities to position each of the others relative to that student and to one another. Finally, we identify the top of the order, which corresponds to the student with maximum height. Step-by-Step Solution: Step 1: P has height more than only one student. Therefore in the ordered list from shortest to tallest, P must be second from the bottom. That means exactly one student is shorter than P and three are taller.Step 2: The student who is shorter than P must be one of Q, R, S or T. However, S is taller than P, so S cannot be shorter. Q is also taller than P, so Q cannot be shorter. R is not the smallest, so R cannot be the one at the very bottom.Step 3: The only remaining possibility for the shortest student is T. Thus, from shortest to taller, the ordering starts as T, then P.Step 4: S has height more than P, so S must be somewhere above P in the order. Q has height more than S and P but not more than R, which means Q is taller than both S and P but still not taller than R. Since heights are distinct, this implies Q is below R and above S.Step 5: From this we can build the order from bottom to top as: T (shortest), P, S, Q and then R on top. R is taller than Q, and Q is taller than S and P, while T is the only one shorter than P. Verification / Alternative check: You can verify by checking each original condition against the final ordering T, P, S, Q, R. P is taller than only T, which matches the first condition. S is taller than P, which is also correct. Q is taller than S and P but shorter than R, which fits the second condition. R is not the smallest, since T is smaller and even P is smaller. All given statements are satisfied with R as the tallest. Why Other Options Are Wrong: If Q were the tallest, then the statement that Q is not more than R would be violated because Q would exceed everyone, including R. If S or T were the tallest, they would contradict the clear indications that Q and R are above them in the implied ordering. Therefore options a, c and d cannot be correct. Common Pitfalls: One common mistake is to misread “more than only one student” and place P at the wrong position, such as second tallest instead of second shortest. Another pitfall is to assume that “not more than R” means equal to R, which is impossible here because all heights are distinct. Always build a complete ordered list and then check each condition carefully. Final Answer: The student with the maximum height in the class is R.

Question 17

There are 45 trees in a row. A lemon tree is 20th from the right end. What is the position (rank) of the lemon tree from the left end of the row?

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Introduction / Context: This reasoning question deals with positions counted from the left and from the right in a single row. You are told that a particular lemon tree stands at a certain position from the right end, and you must find its position from the left end. Such problems are common in aptitude tests and become easy once you know the simple relationship between left and right positions and the total number of items. Given Data / Assumptions: Total number of trees in the row is 45. The lemon tree is 20th from the right end of the row. Positions are counted starting from 1 at each end. We assume there is exactly one lemon tree with this position in the row. We need to find its position counted from the left end. Concept / Approach: For any object in a straight line arrangement, there is a simple formula that links its position from the left and from the right. If the total number of items is N, the position from the left is L and the position from the right is R, then L + R - 1 = N. This is because the object itself is counted once in both left and right positions. Rearranging this relation lets you find one position if you know the other and the total count. Step-by-Step Solution: Step 1: Let N be the total number of trees, which is 45.Step 2: Let R be the position of the lemon tree from the right, which is 20.Step 3: Let L be the position of the same lemon tree from the left, which we want to find.Step 4: Use the formula L + R - 1 = N.Step 5: Substitute the known values: L + 20 - 1 = 45.Step 6: Simplify: L + 19 = 45.Step 7: Solve for L: L = 45 - 19 = 26.Step 8: Therefore the lemon tree is 26th from the left end. Verification / Alternative check: You can also visualise the row. If the tree is 20th from the right, that means there are 19 trees to its right. Since the total number of trees is 45, the number of trees to its left, including the lemon tree itself, must be 45 - 19 = 26. This directly gives the left position as 26. This matches the algebraic calculation and confirms the answer. Why Other Options Are Wrong: If the position from the left were 24, 25 or 27, the formula L + R - 1 = N would not hold. For example, if L were 24, then 24 + 20 - 1 = 43, which does not equal 45. The same mismatch occurs for 25 and 27. Only L = 26 satisfies the relationship between left and right positions given the total of 45 trees. Common Pitfalls: A frequent error is to forget subtracting 1 in the formula and use L + R = N, which leads to wrong results. Another mistake is to subtract the position from the total incorrectly, for example computing 45 - 20 instead of 45 - 19. Keeping the simple formula in mind and doing careful arithmetic prevents these issues. Final Answer: The lemon tree is 26th from the left end of the row.

Question 18

Which of the following words comes first in correct English dictionary order: Carrom, Cramming, Carrier or Cartoon?

Accepted Answer

Introduction / Context: This verbal reasoning question asks you to identify which word would appear first in an English dictionary from a given list. The words all start with the same letter, and several even share the first three letters, which means you must carefully compare them letter by letter. Knowing how alphabetical order works beyond the first character is essential for such problems. Given Data / Assumptions: The words given are Carrom, Cramming, Carrier and Cartoon. We assume a standard English dictionary with usual alphabetical rules. All words are capitalised here, but dictionaries ignore case when ordering. The task is to find which word would be listed first among the four. We compare characters from left to right and use letter order A to Z. Concept / Approach: To place words in dictionary order, compare them character by character. The first letter is the same for all (C), and the second letter is also the same (a). Then we look at the third letter and so on until a difference appears. The word with the earliest differing letter in the alphabet comes first. When two or more words share a longer prefix, we must continue the comparison further along the spelling. Step-by-Step Solution: Step 1: List the words clearly: Carrom, Cramming, Carrier, Cartoon.Step 2: Compare the first letters. All words start with C, so move to the second letter.Step 3: The second letter is a for all four words, so we move to the third letter.Step 4: The third letter is r in every word, so we move to the fourth letter.Step 5: At the fourth position, we have r in Carrom and Carrier, t in Cartoon and m in Cramming. To decide which comes first overall, note that all words starting with Carr will come before those starting with Cart or Cram, because r comes before t and m in the alphabet is earlier than t but the sequence of letters matters in order.Step 6: Focus on Carrom and Carrier, as they both start with Carr. Compare after Carr: we get o in Carrom and i in Carrier at the next differing position. Since i comes before o in the alphabet, Carrier comes before Carrom.Step 7: Any word starting with Carr will still come before Cartoon (which effectively begins Cart) and Cramming (which begins Cram) because after Car the letters r, t and m must be compared appropriately. But once we have found Carrier as the first among the Carr words, no later starter like Cart or Cram can move ahead of it. Verification / Alternative check: An alternative way is to imagine a mini dictionary: starting from Ca, then check the sequence Car, Cart and Cram. Within Car, we compare the next letters and see that Carrier lexicographically precedes Carrom because the letter at the first differing position, i in Carrier, is earlier than o in Carrom. Since any word starting with Car will naturally appear before Cramming, Carrier is firmly the earliest word in the entire list. Why Other Options Are Wrong: Carrom is also a Car word but comes after Carrier because of the later letter o. Cartoon starts with Cart and will appear after all Carr words, including Carrier and Carrom. Cramming starts with Cra, so at the position where r, r, r and m are compared, m in Cramming comes after the r in the Car group when considering the full prefix sequence, placing it later in the dictionary. Therefore none of these can precede Carrier. Common Pitfalls: A common error is to look only at the third or fourth letter and make a snap judgment, especially when words look similar. Another mistake is to assume that the shortest word always comes first, which is not correct; dictionary order is decided by sequence, not length alone. Always compare the first differing letters carefully until you are certain of the order. Final Answer: The word that comes first in dictionary order is Carrier.

Question 19

In a class, P has more marks than Q, and R does not have the least marks. S has more marks than T and T has more marks than P. Who among P, Q, R, S and T has the least marks?

Accepted Answer

Introduction / Context: This question is a ranking puzzle based on marks obtained by five students P, Q, R, S and T. You are given relative comparisons, such as who has more marks than whom, and one condition about who is not at the bottom. The goal is to use these comparisons to determine which student has the least marks in the group. This type of question tests logical ordering and careful interpretation of comparative statements. Given Data / Assumptions: P has more marks than Q. R does not have the least marks. S has more marks than T. T has more marks than P. All five students have different marks; there are no ties. We want to identify the student with the lowest marks. Concept / Approach: To solve this, convert the verbal statements into inequalities and then arrange the students in order from lowest to highest marks. Any statement like “X has more marks than Y” can be written as X > Y. By chaining these comparisons and applying the condition that R is not the least, we can narrow down the possible positions for each student and see who must be at the bottom. Step-by-Step Solution: Step 1: From “P has more marks than Q”, we write P > Q.Step 2: From “T has more marks than P”, we write T > P. Combining these two, we get T > P > Q, so Q is below P and T in the ranking.Step 3: From “S has more marks than T”, we get S > T. Combining this with the previous chain gives S > T > P > Q, placing Q at the bottom of this partial order so far.Step 4: The statement “R does not have the least marks” means R cannot be at the lowest position in the class. R can be anywhere above the bottom.Step 5: Since Q is already lowest among P, Q, S and T, and R is not allowed to be the least, R must be above Q in the ranking.Step 6: Therefore the only student who can occupy the least position is Q, because everyone else, including R, must be above Q based on the given comparisons. Verification / Alternative check: To double check, construct a possible ordering that satisfies all constraints. For example, a valid order from lowest to highest marks could be Q, R, P, T, S. In this example P > Q holds, T > P holds, S > T holds and R is not the least. There are many possible exact numerical marks that match these inequalities, but in all consistent orders Q is always at the bottom, while the exact positions of R, P, T and S above Q can vary as long as the inequalities are maintained. Why Other Options Are Wrong: P cannot be the least because he has more marks than Q. T and S cannot be the least because each has more marks than someone who is already above Q. R is explicitly stated as not having the least marks. This rules out P, S, T and R as candidates for the lowest position, leaving only Q as the correct answer. Common Pitfalls: One common mistake is to forget the condition that R is not the least and focus only on the chain S > T > P > Q, leading to confusion about R. Another pitfall is misreading “more marks than” and accidentally reversing the inequality. Writing down the relationships symbolically and then building a consistent order from bottom to top helps avoid these errors. Final Answer: The student who has the least marks in the class is Q.

Question 20

In the following expression, which sequence of mathematical operators inserted in order will make it correct: 30 ? 6 ? 4 ? 5 ? 4

Accepted Answer

Introduction / Context: This arithmetic reasoning question asks you to choose the correct sequence of mathematical operators that will make a given expression true. The expression has several blanks where operators such as addition, subtraction, multiplication and equality can be inserted. You must select the correct order of operators from the options so that, when evaluated using normal precedence rules, the equation holds true. Given Data / Assumptions: The expression is 30 ? 6 ? 4 ? 5 ? 4. The candidate sequences of operators are four item lists such as “–, =, x and +”. Operators must be inserted in the order they appear in the chosen option. We use standard precedence rules: multiplication and division before addition and subtraction, and equality is used to relate left and right sides. The goal is to obtain a true numerical statement. Concept / Approach: The best approach is to test each option one by one. For each option, substitute the operators into the question marks in the order given. The resulting expression will look like a normal equation, and you can then apply BODMAS rules to see whether both sides are equal. Because the numbers involved are small, this checking is quick and reliable. Only one combination should give a correct equation. Step-by-Step Solution: Step 1: Try option a, “–, =, x and +”. Inserting these in order gives 30 – 6 = 4 x 5 + 4.Step 2: Evaluate the right hand side first using BODMAS. Compute 4 x 5 = 20, then 20 + 4 = 24.Step 3: Evaluate the left hand side: 30 – 6 = 24.Step 4: Since both sides equal 24, the equation 30 – 6 = 4 x 5 + 4 is true. Option a works.Step 5: For completeness, briefly check another option such as b, “+, =, x and –”. That would give 30 + 6 = 4 x 5 – 4. Evaluate the right hand side: 4 x 5 = 20, 20 – 4 = 16, but the left hand side 30 + 6 = 36, so this is not a valid equation.Step 6: Similar checks for options c and d also fail to produce equality, confirming that a is the only correct choice. Verification / Alternative check: Once we have identified that option a yields 30 – 6 = 4 x 5 + 4, we can double check the arithmetic. Starting with the right side, multiply 4 by 5 to get 20, then add 4 to reach 24. On the left side, subtracting 6 from 30 also yields 24. Because both sides reduce to the same value under correct precedence, the equation is robustly verified as true. Why Other Options Are Wrong: Option b gives 30 + 6 = 4 x 5 – 4, which evaluates to 36 = 16, not true. Option c gives 30 = 6 x 4 + 5 – 4, which simplifies to 30 = 24 + 5 – 4 = 25, again false. Option d produces 30 – 6 + 4 = 5 x 4, which leads to 28 = 20, also incorrect. None of these satisfy equality under proper evaluation rules. Common Pitfalls: A common mistake is to ignore multiplication precedence and evaluate strictly from left to right, which yields wrong results and might cause you to reject a correct option. Another pitfall is to mix up the order in which the operators from an option must be placed into the expression. Always insert them sequentially and then apply BODMAS carefully. Final Answer: The expression becomes correct when the operators are “–, =, x and +”, that is, 30 – 6 = 4 x 5 + 4.

Arithmetic Reasoning Questions