Question 1

In the letter pairs VT, FD, PN and JM, find the odd pair based on a constant backward step in the alphabet.

Accepted Answer

Introduction / Context: Letter pair questions are a core part of verbal reasoning. They usually test whether the learner can translate letters to their numerical positions in the alphabet and then spot a consistent pattern, such as a fixed forward or backward step. In this problem we have four pairs of letters, and our task is to identify the pair that does not follow the same positional rule as the others. Given Data / Assumptions: The letter pairs given are VT, FD, PN and JM. We use the standard English alphabet where A = 1, B = 2, ..., Z = 26. We compare the first and second letters in each pair by looking at their numerical positions. We focus on the size and direction of the step from the first letter to the second letter. Concept / Approach: The key idea is to examine how far we move in the alphabet when we go from the first letter of a pair to the second. For each pair, we convert both letters to their numerical positions and compute the difference. If most pairs move backward by a fixed number of positions, while one pair either moves forward or has a different gap, that pair is the odd one out. In this question, three pairs move backward by 2 positions, and one pair does not follow that rule. Step-by-Step Solution: Step 1: For VT, V is the 22nd letter and T is the 20th letter. The difference is 22 minus 20 which equals 2, so T is 2 steps backward from V. Step 2: For FD, F is the 6th letter and D is the 4th letter. The difference is 6 minus 4 which equals 2, so D is 2 steps backward from F. Step 3: For PN, P is the 16th letter and N is the 14th letter. The difference is 16 minus 14 which equals 2, so N is 2 steps backward from P. Step 4: For JM, J is the 10th letter and M is the 13th letter. Here the movement is from 10 to 13, which is forward by 3 positions, not backward by 2. Step 5: Therefore, VT, FD and PN all show a backward step of 2 in the alphabet, while JM shows a forward step of 3, so JM is the odd pair. Verification / Alternative check: We can reconvert quickly: V = 22, T = 20, F = 6, D = 4, P = 16, N = 14, J = 10 and M = 13, and recompute the differences to confirm our earlier calculations. The repeated backward step of 2 for three pairs and a clear forward step for JM shows a strong and consistent pattern. No other simple rule, such as total sum of positions, gives such a clean separation between three options and one option. Why Other Options Are Wrong: VT is not the odd one out because it moves backward by 2 positions from V to T. FD is not the odd one out because it also moves backward by 2 positions from F to D. PN is not the odd one out because it again moves backward by 2 positions from P to N. JM is the only pair that moves forward by 3 positions instead of backward by 2, which makes it different from the other three. Common Pitfalls: Some learners check only the absolute size of the gap and ignore the direction, which can lead them to think that a gap of 3 is similar to 2 in difficulty, but the direction is the actual key here. Others may try to see if the letters form familiar abbreviations, which is not the intention in abstract letter pair questions. To succeed in such problems, it is important to develop the habit of always checking both the size and the direction of alphabetic movement. Final Answer: The letter pair that does not follow the common backward step of 2 and is therefore the odd one out is JM.

Question 2

Among the numbers 13981, 93172, 47542 and 67325, identify the odd one out based on the sum of digits.

Accepted Answer

Introduction / Context: Odd one out questions with large numbers often hide a pattern in the digits rather than in the overall magnitude of the number. A very common and exam friendly idea is to look at the sum of the digits, sometimes called the digit sum, and see whether it satisfies a particular property such as being equal in several options or being a multiple of a fixed value. In this problem, we have four five digit numbers and need to decide which one is different when we look at the sum of its digits. Given Data / Assumptions: The numbers given are 13981, 93172, 47542 and 67325. We compute the sum of the digits of each number separately. We compare these sums to see whether three of them share a common value and one does not. All calculations use simple addition of individual digits. Concept / Approach: The approach is to compute the digit sum of each number and then compare them. If three numbers have the same digit sum and one number has a different digit sum, then the number with the different sum is the odd one out. This method is straightforward to implement and generates small results that are easy to compare mentally, which is why exam setters frequently use it as the hidden logic behind such questions. Step-by-Step Solution: Step 1: For 13981, the digits are 1, 3, 9, 8 and 1. The sum of digits is 1 + 3 + 9 + 8 + 1 = 22. Step 2: For 93172, the digits are 9, 3, 1, 7 and 2. The sum of digits is 9 + 3 + 1 + 7 + 2 = 22. Step 3: For 47542, the digits are 4, 7, 5, 4 and 2. The sum of digits is 4 + 7 + 5 + 4 + 2 = 22. Step 4: For 67325, the digits are 6, 7, 3, 2 and 5. The sum of digits is 6 + 7 + 3 + 2 + 5 = 23. Step 5: We observe that 13981, 93172 and 47542 all have a digit sum of 22, while 67325 has a digit sum of 23. Step 6: Therefore 67325 is the number that does not match the common digit sum and is the odd one out. Verification / Alternative check: We can quickly recompute each sum to avoid any addition mistakes, especially with five digit numbers. Seeing the same value 22 for three different numbers strongly suggests that the examiner intentionally chose these values to form a group. Since only 67325 breaks this otherwise perfect match, and there is no other simple pattern that separates three numbers from one, the digit sum property is clearly the intended logic. Why Other Options Are Wrong: 13981 is not the odd one out because its digit sum is 22, which matches the sums of 93172 and 47542. 93172 is not the odd one out because it also has a digit sum of 22 and belongs to the same group. 47542 is not the odd one out because its digit sum again equals 22, continuing the pattern. Only 67325 has a digit sum of 23, which is different from 22 and sets it apart from the others. Common Pitfalls: A common mistake is to try complex divisibility tests or factorization on these large numbers, which is time consuming and usually not necessary in exam conditions. Another pitfall is to miscalculate the digit sum of one of the numbers under time pressure and then draw a wrong conclusion. To avoid such issues, it is wise to double check simple additions and to always consider digit sums as a first idea when dealing with odd one out questions involving multi digit numbers. Final Answer: The number whose digit sum is different from the others and is therefore the odd one out is 67325.

Question 3

Among the numbers 5712, 6212, 3811 and 7831, identify the odd one out based on the count of odd and even digits.

Accepted Answer

Introduction / Context: In some odd one out questions, the hidden pattern is not about divisibility or digit sums but about the parity pattern of the digits, that is, whether the digits are odd or even. Such questions are popular in reasoning sections because they can be solved quickly once you look at the correct feature. In this problem, we have four four digit numbers that at first seem unrelated, but one of them has a different combination of odd and even digits compared to the others. Given Data / Assumptions: The numbers given are 5712, 6212, 3811 and 7831. Each number has four digits. We classify each digit as odd or even. We compare the count of odd digits and even digits in each number. Concept / Approach: The concept is to look at the parity pattern across all four digits in each number. In three of the numbers, the pattern is three odd digits and one even digit. In one of the numbers, the pattern is reversed with three even digits and one odd digit. The number with the different pattern becomes the odd one out. This method does not require heavy calculations and relies only on recognizing whether each digit is odd or even, which is ideal for fast reasoning in competitive exams. Step-by-Step Solution: Step 1: Consider 5712. The digits are 5, 7, 1 and 2. Here 5 is odd, 7 is odd, 1 is odd and 2 is even. That means this number has three odd digits and one even digit. Step 2: Consider 3811. The digits are 3, 8, 1 and 1. Here 3 is odd, 8 is even and the two 1s are odd, so again we have three odd digits and one even digit. Step 3: Consider 7831. The digits are 7, 8, 3 and 1. Here 7, 3 and 1 are odd, while 8 is even, so this number also has three odd digits and one even digit. Step 4: Consider 6212. The digits are 6, 2, 1 and 2. In this case, 6 is even, 2 is even, the next 1 is odd and the last 2 is even. So there are three even digits and one odd digit. Step 5: Therefore, 5712, 3811 and 7831 share a common pattern of three odd digits and one even digit, while 6212 has the opposite pattern of three even digits and one odd digit. So 6212 is the odd one out. Verification / Alternative check: We can verify by simply recounting for each number: 5712 has odd, odd, odd, even; 3811 has odd, even, odd, odd; 7831 has odd, even, odd, odd; and 6212 has even, even, odd, even. This confirms that only 6212 does not fit the three odd one even pattern. No simple divisibility or digit sum rule separates the numbers as neatly as this parity pattern, which makes it the intended logic for exam use. Why Other Options Are Wrong: 5712 is not the odd one out because it has three odd digits and one even digit, just like 3811 and 7831. 3811 is not the odd one out because its parity pattern matches that of 5712 and 7831. 7831 is not the odd one out because it also has three odd digits and one even digit. 6212 is different from all three because it has three even digits and one odd digit, which reverses the common pattern. Common Pitfalls: Many learners first try divisibility by 2, 3, 5 or 11 for such number sets and may waste time without finding a clear distinction. Another common mistake is to look only at the first or last digit and ignore the total pattern across all digits. To avoid these issues, it is useful to quickly scan for patterns in parity across all digits whenever you see multi digit odd one out questions. Final Answer: The number whose combination of odd and even digits is different and is therefore the odd one out is 6212.

Question 4

In this reasoning question, select the odd word from the following group of shelter related terms: Igloo, Den, Stables and Rock.

Accepted Answer

Introduction / Context: Odd one out questions test how well you can see the underlying category or property that three items share and identify the fourth that does not fit. In this verbal reasoning question, all the given words are related to places or structures that can be associated with living beings, but one of them is not used as a proper shelter or dwelling. Understanding the meaning and typical usage of each word is the key to answering correctly in competitive exams and aptitude tests. Given Data / Assumptions: The options are: Igloo, Den, Stables and Rock. We assume the question is asking for the word which does not represent a regular shelter or dwelling place. Common English usage and general knowledge are to be used, not any rare or technical meanings. Concept / Approach: The standard approach in such questions is to group the items by category and see which one does not match the dominant group. Here, we look for words that are clearly recognized as man made or naturally used shelters for humans or animals. Then we compare the remaining word and decide which one does not function as a shelter in usual language. Semantics and everyday meaning matter more than literal possibility. Step-by-Step Solution: Step 1: Consider the word “Igloo”. An igloo is a dome shaped house made of blocks of ice, traditionally used by people in polar regions. It is clearly a shelter. Step 2: Consider the word “Den”. A den is the natural home or lair of wild animals such as lions, foxes or bears. It is also a kind of shelter. Step 3: Consider the word “Stables”. Stables are buildings where horses and sometimes other animals are housed and cared for. Again, this is a proper animal shelter. Step 4: Consider the word “Rock”. A rock is simply a large stone or solid mineral mass. It is an object in nature, not normally used as a recognized dwelling or shelter name. Step 5: Therefore three words are names of shelters, while rock is not. So rock is the odd one out. Verification / Alternative check: We can also ask: if each of these words were used in the sentence “The animal returned to its ____”, which ones sound like normal shelter names. “Returned to its den” and “returned to its stables” make sense, and even “people returned to their igloo” makes sense. However “returned to its rock” does not sound like the name of a proper shelter, which confirms that rock is different from the others. Why Other Options Are Wrong: Igloo is wrong as the odd one out because it is a clearly recognized house or shelter type for humans in cold regions. Den is wrong as the odd one out because it is a natural shelter for many wild animals. Stables is wrong as the odd one out because it is a building used specifically to shelter horses and other animals. Common Pitfalls: Some learners may overthink the presence of humans or animals and try to treat igloo as special because it is for human beings while den and stables are for animals. However the question is about the common idea of shelter, not about the species that uses it. Others may be distracted by singular and plural forms, but that is not logically relevant. Always focus on the broad category and function of the word rather than minor grammatical details. Final Answer: The word that does not belong with the others because it is not a recognized shelter is Rock.

Question 5

In this question on verbal reasoning, choose the group of letters that is least like the others: CUD, SIP, SET and NTR.

Accepted Answer

Introduction / Context: This odd one out question asks you to compare small groups of letters that also form short everyday English words. Three of the options are meaningful words, while one option is simply a random letter sequence with no meaning. Identifying the meaningful word groups and contrasting them with the meaningless group is a common pattern in verbal reasoning exams and helps to test vocabulary awareness. Given Data / Assumptions: The given options are: CUD, SIP, SET and NTR. We assume standard English vocabulary usage. The task is to find the option that does not belong with the others on a logical basis. Concept / Approach: One familiar approach is to check whether each group of letters forms a valid, commonly used English word. If three items are meaningful words and one is not, the meaningless one will be the odd man out. In verbal odd man out questions, meaning and usability in normal language are usually more important than deep numerical patterns or letter position arithmetic. Step-by-Step Solution: Step 1: Check “CUD”. Cud is a real English word, referring to partly digested food that some animals chew again. Step 2: Check “SIP”. Sip is also a real word, meaning to drink something in small quantities. Step 3: Check “SET”. Set is a very common English word with many meanings, such as a group of things, or to place something. Step 4: Check “NTR”. This does not form any standard common English word. It looks like a random combination of letters, unlike the other three which are meaningful words. Step 5: Since CUD, SIP and SET are meaningful words, and NTR is not, NTR is the odd one out. Verification / Alternative check: Another quick check is to mentally place each option in a simple sentence. For example, “I took a small sip of tea” and “The teacher gave us a set of questions” are normal sentences. “The cow chewed its cud” is also meaningful. Trying to use “NTR” in a sentence reveals that it is not a word at all. This confirms that NTR does not share the common property of being a proper English word. Why Other Options Are Wrong: CUD is not the odd one out because it is a valid English noun referring to food chewed by animals. SIP is not the odd one out because it is a common verb or noun related to drinking. SET is not the odd one out because it is a widely used word with many accepted meanings. Common Pitfalls: A common mistake is to start looking for complicated numerical patterns using letter positions in the alphabet without first checking the simple meaning based property. Verbal reasoning questions often rely on everyday vocabulary, so ignoring meaning can waste time. Another pitfall is assuming every group must hide some arithmetic pattern in the letters, which is not necessary here. Final Answer: The group of letters that does not form a meaningful word and therefore is the odd one out is NTR.

Question 6

In this aptitude question on odd numbers, select the number that does not follow the same property as the others: 132, 145, 187 and 144.

Accepted Answer

Introduction / Context: Numerical odd one out questions check whether you can quickly classify numbers based on mathematical properties such as being a perfect square, perfect cube, prime number or composite number. In this problem, you are given four numbers and asked to find the one which does not share the common characteristic that the remaining three have. Such questions are frequent in bank exams, SSC exams and other competitive tests. Given Data / Assumptions: The four alternatives are: 132, 145, 187 and 144. We are asked to select the odd number from the given alternatives. We assume standard school level number theory concepts like squares and cubes. Concept / Approach: A natural approach is to check if any of the numbers is a perfect square, perfect cube or prime. Often, only one number will have a special property while the others are ordinary composite numbers. Here, 144 is immediately familiar because it appears in square tables. We will test whether 144 is a perfect square and whether the other numbers are squares as well. Step-by-Step Solution: Step 1: Check 144. We know that 12 * 12 = 144, so 144 is a perfect square. Step 2: Check 132. The square of 11 is 121 and the square of 12 is 144. Since 132 lies between 121 and 144, it cannot be a perfect square. Step 3: Check 145. Again, 144 is 12^2 and 169 is 13^2. 145 is between these two squares, so it is not a perfect square. Step 4: Check 187. The squares near it are 169 (13^2) and 196 (14^2). Since 187 is between them, it is not a perfect square. Step 5: Therefore, only 144 is a perfect square, while 132, 145 and 187 are not. That makes 144 the odd one out. Verification / Alternative check: Another way to verify is to factorise the numbers. 144 factors into 2 * 2 * 2 * 2 * 3 * 3, which can be grouped into equal pairs of prime factors, confirming it is a perfect square. The other numbers do not factorise into completely paired primes, so they are not squares. This reinforces our conclusion that 144 has a special property which the others lack. Why Other Options Are Wrong: 132 is wrong as the odd one out because it is not a perfect square and behaves similarly to 145 and 187 in this respect. 145 is wrong as the odd one out because, like 132 and 187, it lies between two consecutive squares and is not itself a square. 187 is wrong as the odd one out because it is another non square composite number without the special square property. Common Pitfalls: A common mistake is to focus only on parity, that is, whether a number is odd or even. In this set, 132 and 144 are even while 145 and 187 are odd, which does not give a clear majority pattern. Another pitfall is looking for complicated divisibility relations when a simple and well known property like being a perfect square is sufficient. Always start by testing basic patterns such as squares, cubes and primes before attempting advanced ideas. Final Answer: The only perfect square among the given numbers, and therefore the odd one out, is 144.

Question 7

In this family relationship reasoning question, select the one term that does not belong to the group: Niece, Nephew, Aunt and Sister.

Accepted Answer

Introduction / Context: Odd man out questions based on family relationships test your understanding of how different relatives are classified, often by gender or generation. In this problem, four relationship terms are given. Three of them share a common property and one is different. Such questions are common in verbal reasoning and general intelligence sections of competitive exams because they check both vocabulary and logical grouping. Given Data / Assumptions: The four options are: Niece, Nephew, Aunt and Sister. All four are close family relation terms used in everyday language. We use standard meanings: niece and nephew are children of a sibling, aunt is a sister of a parent, and sister is a female sibling. Concept / Approach: One clean way to classify these words is by gender. Reasoning questions often form a group where three items refer to one gender while the fourth refers to the other gender. Here we check whether each term refers to a male or female relative. Once that is clear, the term with different gender from the majority will be the odd one out. Step-by-Step Solution: Step 1: A niece is the daughter of one's brother or sister, so niece is always a female relative. Step 2: An aunt is the sister of one's father or mother, or the wife of one's uncle, so the term aunt also clearly refers to a female relative. Step 3: A sister is a female sibling, so this word again refers only to a female person. Step 4: A nephew is the son of one's brother or sister, which makes the nephew a male relative. Step 5: Thus, niece, aunt and sister are female relatives, while nephew is a male relative. This makes nephew the odd one out. Verification / Alternative check: We can double check by thinking of sample sentences. We say “She is my niece”, “She is my aunt” and “She is my sister”, and in each case the pronoun she fits naturally, which confirms they are female. For nephew, we say “He is my nephew”, and the pronoun he is correct. Therefore, nephew differs in gender from the other three terms and must be the odd item out. Why Other Options Are Wrong: Niece is wrong as the odd one out because it matches the female gender property shared with aunt and sister. Aunt is wrong as the odd one out because it is also a female relative, consistent with niece and sister. Sister is wrong as the odd one out because it is again clearly a female relation, not unique among the group. Common Pitfalls: Some students may get distracted by generational differences, noting that a niece and nephew belong to a younger generation, aunt belongs to an older generation, and sister belongs to the same generation. However, this does not create a clear majority property to isolate one term. Focusing on gender produces a simple and unique pattern, which is usually what exam setters intend in such questions. Final Answer: The only male relation among the four options, and therefore the odd one out, is Nephew.

Question 8

In this odd one out question based on letters, select the pair that does not follow the same pattern as the others: AE, IO, UA and GH.

Accepted Answer

Introduction / Context: Reasoning questions involving small groups of letters often rely on basic properties such as vowels and consonants. In this problem, four two letter combinations are given, and you are asked to find the one that does not belong to the same family as the others. This is a classic example of an odd one out question that tests your understanding of the English alphabet and classification of letters. Given Data / Assumptions: The options are: AE, IO, UA and GH. We treat A, E, I, O, U as vowels and the remaining letters as consonants. The task is to pick the pair whose composition of vowels and consonants differs from the rest. Concept / Approach: A straightforward technique is to check whether each pair consists of vowels, consonants, or a mix of both. Patterns like both vowels, both consonants or vowel consonant combinations are frequently used to build such questions. Once we identify the majority type among the options, the pair that does not share that structure is declared the odd one out. Step-by-Step Solution: Step 1: Examine AE. A is a vowel and E is also a vowel. So AE is a pair of vowels. Step 2: Examine IO. I is a vowel and O is a vowel, so IO is also a vowel pair. Step 3: Examine UA. U is a vowel and A is a vowel, so UA again forms a pair made only of vowels. Step 4: Examine GH. G is a consonant and H is a consonant, so GH is a consonant pair, not a vowel pair. Step 5: Since AE, IO and UA each contain two vowels, while GH contains two consonants, GH is the odd one out. Verification / Alternative check: We can confirm by simply writing V for vowel and C for consonant. The pairs become: AE = VV, IO = VV, UA = VV and GH = CC. Three pairs are VV and one pair is CC. The single CC pair is clearly different. There is no need for any deeper arithmetic or positional logic in this particular question, because the vowel consonant pattern already gives a clean and unique classification. Why Other Options Are Wrong: AE is wrong as the odd one out because, like IO and UA, it is formed by two vowels. IO is wrong as the odd one out because it also has the same property of being a vowel pair. UA is wrong as the odd one out because it is yet another pair made up entirely of vowels. Common Pitfalls: Students sometimes overcomplicate such questions by checking letter positions or trying to find numerical patterns in the alphabet indexes. While that is sometimes useful, here the simplest method is to just look at whether the letters are vowels or consonants. Another pitfall is assuming that every question must involve a complex rule, which can lead to wasted time during an exam. Final Answer: The only pair that consists entirely of consonants and therefore does not match the vowel pairs is GH.

Question 9

In this numerical reasoning question on prime numbers, select the odd number from the following alternatives: 17, 23, 29 and 33.

Accepted Answer

Introduction / Context: Odd one out questions involving numbers frequently use the concept of prime and composite numbers. In this problem, four numbers are given and you are asked to identify which one does not share the same property as the others. Recognising prime numbers quickly is an important skill for aptitude tests and forms the core idea tested here. Given Data / Assumptions: The numbers given are 17, 23, 29 and 33. We use the standard definition of a prime number: a number greater than 1 that has exactly two distinct positive divisors, namely 1 and itself. We need to select the number that is not prime if the others are prime, or vice versa. Concept / Approach: The typical strategy is to test each number for divisibility by small primes such as 2, 3, 5 and 7. If a number has no divisors other than 1 and itself, it is prime. If it has more divisors, it is composite. When three numbers turn out to be primes and one is composite, the composite number becomes the odd one out in the group. Step-by-Step Solution: Step 1: Check 17. It is not divisible by 2, 3 or 5. The only divisors are 1 and 17, so 17 is prime. Step 2: Check 23. It is not divisible by 2, 3 or 5. The only divisors are 1 and 23, so 23 is also prime. Step 3: Check 29. It is not divisible by 2, 3 or 5. The only divisors are 1 and 29, so 29 is prime as well. Step 4: Check 33. It is divisible by 3, because 3 * 11 = 33. Therefore 33 has more than two divisors and is a composite number, not a prime number. Step 5: Hence 17, 23 and 29 are primes, while 33 is composite. So 33 is the odd one out. Verification / Alternative check: An alternative check is to use the digit sum method for divisibility by 3. For 33, the sum of digits is 3 + 3 = 6, which is divisible by 3, so 33 is divisible by 3. For 17, 23 and 29, the digit sums are 8, 5 and 11 respectively, none of which are divisible by 3. This confirms that 33 alone is composite and clearly different in nature from the other three numbers. Why Other Options Are Wrong: 17 is wrong as the odd one out because it is a prime number, sharing the same property as 23 and 29. 23 is wrong as the odd one out because it is also prime and behaves like 17 and 29. 29 is wrong as the odd one out because it is prime and not uniquely different in this list. Common Pitfalls: One common mistake is to rely only on familiarity and assume that all two digit numbers ending with 3 or 7 are automatically prime. While many are, each number should still be checked for divisibility. Another pitfall is ignoring simple divisibility rules, such as the rule for 3, which makes identifying 33 as composite very quick. Carefully applying basic rules avoids such errors in timed exams. Final Answer: The only composite number among the given options, and therefore the odd one out, is 33.

Question 10

In this number puzzle on digit patterns, select the four digit number that does not follow the same rule as the others: 7214, 3618, 3519 and 9436.

Accepted Answer

Introduction / Context: Odd one out questions with four digit numbers often hide a pattern inside the digits rather than in the whole number. The pattern can involve sums, differences or products of certain digits. This question gives four numbers and expects you to identify which number follows a different rule. Detecting these internal digit relationships is a useful skill in quantitative aptitude and logical reasoning tests. Given Data / Assumptions: The four digit numbers are 7214, 3618, 3519 and 9436. We are looking for a common digit based pattern followed by three numbers but not by one. Only basic operations on digits, such as addition, are expected. Concept / Approach: A common technique is to split each four digit number into two pairs of digits: the first two and the last two. Then we compare sums of these pairs. Often, three numbers will not satisfy a special relation and one number will. Sometimes the odd item is the one that satisfies the special property instead of the ones that do not. We must observe which interpretation creates a clear majority pattern. Step-by-Step Solution: Step 1: For 7214, add the first two digits: 7 + 2 = 9. Add the last two digits: 1 + 4 = 5. These sums are not equal. Step 2: For 3519, add the first two digits: 3 + 5 = 8. Add the last two digits: 1 + 9 = 10. These sums are not equal. Step 3: For 9436, add the first two digits: 9 + 4 = 13. Add the last two digits: 3 + 6 = 9. Again, the sums are not equal. Step 4: For 3618, add the first two digits: 3 + 6 = 9. Add the last two digits: 1 + 8 = 9. Here, the sums are equal. Step 5: Thus, 7214, 3519 and 9436 do not satisfy the property “sum of first two digits equals sum of last two digits”, while 3618 does satisfy it. Therefore, 3618 is the odd one out, since it behaves differently from the majority. Verification / Alternative check: We can quickly verify our conclusion by writing each number in the form AB CD and checking AB sum versus CD sum: 72 14 gives 9 and 5, 35 19 gives 8 and 10, 94 36 gives 13 and 9, and 36 18 gives 9 and 9. Only for 3618 are the two sums equal. Exam setters sometimes choose the unique item as the odd one out, even though it satisfies a more regular pattern. Here that is clearly the case. Why Other Options Are Wrong: 7214 is wrong as the odd one out because its first pair and last pair of digits add to different totals, just like most of the other numbers. 3519 is wrong as the odd one out because it does not satisfy the equal sum condition and behaves similarly to 7214 and 9436. 9436 is wrong as the odd one out because, like 7214 and 3519, it also fails the equal sum test. Common Pitfalls: Many students initially test divisibility or prime properties of the whole numbers and see no clear pattern, which can be confusing. Others may try to overcomplicate the problem by using multiplication or difference operations on the digits. In these puzzles, the simplest operations such as adding digit pairs usually lead to the intended pattern. Remember that sometimes the number which follows a neat internal rule is the odd one out, which is the opposite of what students first expect. Final Answer: The only number for which the sum of the first two digits equals the sum of the last two digits, and hence the odd one out, is 3618.

Question 11

In this astronomy based odd one out question, choose the object that does not belong with the others: Sirius, Proxima Centauri, Deimos and Alpha Centauri.

Accepted Answer

Introduction / Context: General knowledge odd one out questions often draw on basic astronomy. This problem lists four celestial objects and asks you to select the one that does not belong to the same category as the others. Recognising which objects are stars and which are moons or planets is a key part of school level space science and is frequently tested in competitive exams. Given Data / Assumptions: The given names are Sirius, Proximacentauri, Deimos and Alpha centauri. Sirius is known as a bright star in the night sky. Proxima Centauri and Alpha Centauri are also stars, though written with minor spacing variations here. Deimos is one of the natural satellites of Mars. Concept / Approach: The central idea is to classify each object as either a star or a moon. Three of them are stars belonging to star systems, while one is a small natural satellite. Once we identify which is which, the lone moon among the group will be the odd one out. Category based classification is the standard method for such general knowledge reasoning questions. Step-by-Step Solution: Step 1: Sirius is a well known bright star in the constellation Canis Major. It is definitely a star. Step 2: Proxima Centauri is the closest known star to the Sun, part of the Alpha Centauri star system. It is a star as well. Step 3: Alpha Centauri is another star system, often referred to as one of the nearest star neighbours of our Sun. Hence it is also classified under stars. Step 4: Deimos, in contrast, is not a star. It is one of the two natural moons of the planet Mars, the other being Phobos. Step 5: Since Sirius, Proxima Centauri and Alpha Centauri are stars, and Deimos is a moon, Deimos does not fit into the majority category and is the odd one out. Verification / Alternative check: To verify, think about how each name is used in basic astronomy descriptions. Sirius is cited as a prominent star, Proxima Centauri as the nearest star to the Sun, and Alpha Centauri as a nearby star system. Deimos, on the other hand, is described together with Phobos as the two small moons of Mars. No source treats Deimos as a star. This supports the classification difference and confirms our selection. Why Other Options Are Wrong: Sirius is wrong as the odd one out because it is a star, like Proxima Centauri and Alpha Centauri. Proximacentauri is wrong as the odd one out because it is again a star, not a moon, and matches the dominant category. Alpha centauri is wrong as the odd one out because it names a star system and not a satellite, fitting with Sirius and Proxima Centauri. Common Pitfalls: Students unfamiliar with astronomy may misidentify Deimos or confuse it with a star name. Another pitfall is ignoring the slight spelling inconsistency in Proximacentauri and Alpha centauri and getting distracted by formatting rather than focusing on meaning. In reasoning and general knowledge questions, minor spelling or spacing issues are not the focus; the category of the object is what matters for the correct answer. Final Answer: The only natural satellite in the list, and therefore the odd one out, is Deimos.

Question 12

In this letter sequence reasoning question, choose the group that does not follow the same reverse alphabetical order as the others: PON, SRQ, XYZ and VUT.

Accepted Answer

Introduction / Context: Many odd one out questions are based on alphabetical patterns. Letter triplets can be arranged in ascending or descending order, and the question asks you to spot which combination does not follow the main pattern. Here, four three letter sequences are given, and your task is to identify the one that is not arranged in the same way as the others. This is a common concept in logical and verbal reasoning tests. Given Data / Assumptions: The options are: PON, SRQ, XYZ and VUT. We use the standard English alphabet order from A to Z. We must check whether the letters move forward or backward in alphabetical order within each triplet. Concept / Approach: The key idea is to examine the order of letters in each group. A pattern may reflect a sequence of three consecutive letters in reverse (for example CBA), or in normal forward order (ABC). If three of the options show a descending sequence and one shows an ascending sequence, the ascending one will be the odd one out. So we convert each letter to its position in the alphabet to identify the pattern. Step-by-Step Solution: Step 1: For PON, P, O and N are consecutive letters in reverse order: P (16), O (15), N (14). The positions decrease by 1 each time. Step 2: For SRQ, S, R and Q are also consecutive in reverse order: S (19), R (18), Q (17). Again the positions decrease by 1. Step 3: For VUT, V, U and T are consecutive letters going backwards: V (22), U (21), T (20), showing the same descending pattern. Step 4: For XYZ, X, Y and Z form a sequence in normal forward order: X (24), Y (25), Z (26), where positions increase by 1. Step 5: Therefore three groups PON, SRQ and VUT show three consecutive letters in reverse order, while XYZ shows three consecutive letters in forward order. XYZ is thus the odd one out. Verification / Alternative check: You can verify quickly by simply rewriting each group forward: PON corresponds to NOP reversed, SRQ to QRS reversed and VUT to TUV reversed, all of which are familiar consecutive triplets. XYZ, however, already appears in direct alphabetical order and not as a reversal. This simple inspection confirms that XYZ does not follow the reverse order pattern. Why Other Options Are Wrong: PON is wrong as the odd one out because it is clearly a reverse sequence of three consecutive letters. SRQ is wrong as the odd one out because it follows the same reverse consecutive pattern. VUT is wrong as the odd one out because, like PON and SRQ, it also represents three letters arranged in descending order. Common Pitfalls: Some learners may focus on where the letters lie in the alphabet (beginning, middle or end) instead of the direction of the sequence. Others may be confused by the fact that XYZ is a very familiar cluster and overlook that it is actually in ascending order while all others are descending. Always convert letter groups into positions or visualize the alphabet line to avoid such mistakes. Final Answer: The only group of letters arranged in forward alphabetical order, and therefore the odd one out, is XYZ.

Question 13

In this digit sum based odd one out question, choose the four digit number that does not share the same digit sum property as the others: 1919, 5656, 6761 and 7760.

Accepted Answer

Introduction / Context: Number puzzles that ask you to find the odd one out often hide patterns in the sum of digits or relationships between digit groups. This problem presents four four digit numbers and expects you to identify the one whose digit sum behaves differently from the others. Learning to quickly compute and compare digit sums is useful for many aptitude questions, including divisibility tests and pattern recognition. Given Data / Assumptions: The four numbers are 1919, 5656, 6761 and 7760. We will use simple addition of digits to compute the digit sum of each number. The goal is to spot the number that breaks the common pattern observed in the other three. Concept / Approach: A neat way to approach this is to find the sum of digits of each number and compare them. If three numbers share the same total digit sum and one number has a different total, then the number with the different digit sum can be considered the odd one out. This is a standard style in reasoning questions where entire numbers are not special, but their digit patterns are. Step-by-Step Solution: Step 1: For 1919, add the digits: 1 + 9 + 1 + 9 = 20. So the digit sum of 1919 is 20. Step 2: For 6761, add the digits: 6 + 7 + 6 + 1 = 20. The digit sum of 6761 is also 20. Step 3: For 7760, add the digits: 7 + 7 + 6 + 0 = 20. Again, the digit sum of 7760 is 20. Step 4: For 5656, add the digits: 5 + 6 + 5 + 6 = 22. The digit sum of 5656 is 22, which is different from 20. Step 5: Therefore three numbers (1919, 6761 and 7760) share the same digit sum of 20, while 5656 has digit sum 22. So 5656 is the odd one out. Verification / Alternative check: A quick verification is to observe that 1919, 6761 and 7760 all have digit sums that are multiples of 20, while 5656 gives a higher sum. Because the pattern is very clean (exact equality of digit sums for three numbers), and 5656 is the only number that breaks it, no further complex checks are needed. This confirms that 5656 is uniquely different in terms of digit sum. Why Other Options Are Wrong: 1919 is wrong as the odd one out because its digit sum 20 matches the pattern shared with 6761 and 7760. 6761 is wrong as the odd one out because it also has digit sum 20, consistent with 1919 and 7760. 7760 is wrong as the odd one out because its digit sum 20 again follows the same rule. Common Pitfalls: Students may start by testing divisibility or looking for some complex arithmetic relation between the whole numbers and feel stuck. Another mistake is to check only the first or last two digits and ignore the full four digit structure. In many such questions, the digit sum is the intended pattern because it is easy to compute and provides a clear distinction, as it does here. Final Answer: The only number whose digit sum is different from the others, and therefore the odd one out, is 5656.

Question 14

In this perfect cube based odd one out question, select the number that is not a perfect cube among 2890, 3375, 1728 and 1331.

Accepted Answer

Introduction / Context: Many numerical odd one out questions rely on basic powers such as squares and cubes. In this problem you are given four numbers, three of which are perfect cubes of integers, while one is not. Recognising standard cubes is an important skill for quantitative aptitude because it speeds up both mental calculations and pattern recognition in puzzles. Given Data / Assumptions: The given numbers are 2890, 3375, 1728 and 1331. We use the definition of a perfect cube: a number of the form n^3 for some integer n. The task is to identify the number that is not a perfect cube. Concept / Approach: The approach here is to recall standard cube values of integers around 10, 11, 12, 13, 14 and 15 and match them with the given numbers. If a number exactly matches one of these cube values, it is a perfect cube. The number that does not correspond to any integer cube will be the odd one out. This is a direct application of memorised cube tables. Step-by-Step Solution: Step 1: Recall that 11^3 = 1331. Therefore, 1331 is a perfect cube. Step 2: Recall that 12^3 = 1728. So 1728 is also a perfect cube. Step 3: Recall that 15^3 = 3375. Thus 3375 is again a perfect cube. Step 4: Now consider 2890. Check nearby cube values: 13^3 = 2197, 14^3 = 2744 and 15^3 = 3375. The number 2890 does not match any integer cube among these and lies between 2744 and 3375. Step 5: Hence 2890 is not a perfect cube, while 3375, 1728 and 1331 are perfect cubes. Therefore 2890 is the odd one out. Verification / Alternative check: To verify, you could factor 2890. It equals 2 * 5 * 17 * 17. For a number to be a perfect cube, each prime factor must appear with an exponent that is a multiple of 3. Here, the factor 17 appears with exponent 2, which is not a multiple of 3, so 2890 cannot be a perfect cube. The other three values are standard cubes and appear frequently in cube tables, which confirms our conclusion. Why Other Options Are Wrong: 3375 is wrong as the odd one out because it is 15^3, a perfect cube. 1728 is wrong as the odd one out because it is 12^3, another perfect cube. 1331 is wrong as the odd one out because it is 11^3, again a perfect cube. Common Pitfalls: Some test takers may not remember cube values and therefore start performing time consuming repeated multiplication. Others may confuse 2890 with a near cube because it lies between 14^3 and 15^3. The safest method is to recall or quickly derive nearby cubes 2197, 2744 and 3375 and notice that 2890 does not equal any of them. Practice with cube tables helps avoid these mistakes in exams. Final Answer: The only number that is not a perfect cube and therefore the odd one out is 2890.

Question 15

In this Indian culture based odd one out question, identify the option that is not a classical or folk dance form among Kuchipudi, Kathak, Bhangra and Pongal.

Accepted Answer

Introduction / Context: General awareness questions about Indian culture often ask you to pick the odd one out from dance forms, festivals and other cultural elements. In this question, three of the options are dance styles, while one is an important festival. Identifying which option belongs to a different cultural category is a common task in reasoning and GK sections of exams. Given Data / Assumptions: The options are Kuchipudi, Kathak, Bhangra and Pongal. Kuchipudi is a classical dance form from India. Kathak is also a classical Indian dance form. Bhangra is a lively folk dance associated mainly with Punjab. Pongal is a harvest festival celebrated predominantly in Tamil Nadu. Concept / Approach: The key idea is to classify each term as either a dance form or a festival. When three items fall into one category and one item belongs to a clearly different category, that single different item is the odd one out. Here, you simply need basic cultural knowledge about India, not any numerical or letter based pattern. Step-by-Step Solution: Step 1: Kuchipudi is a well known classical dance that originated in Andhra Pradesh. It clearly belongs to the category of dance forms. Step 2: Kathak is another classical Indian dance, historically associated with North India and known for its spins and storytelling gestures. It is again a dance form. Step 3: Bhangra is a popular folk dance of Punjab, traditionally performed during harvest festivals and celebrations. It is not a festival itself, but a dance. Step 4: Pongal, however, is primarily the name of a harvest festival celebrated in Tamil Nadu. It is a cultural and religious celebration, not a style of dance. Step 5: Therefore, Kuchipudi, Kathak and Bhangra are all dance forms, whereas Pongal is a festival. So Pongal is the odd one out. Verification / Alternative check: Another way to verify is to ask whether the word fits into the sentence “He or she is performing ____ on stage.” It makes sense to say “performing Kuchipudi”, “performing Kathak” or “performing Bhangra”, because all of them are dance styles. Saying “performing Pongal” sounds incorrect, as Pongal is celebrated and not performed as a dance form. This language based test confirms that Pongal belongs to a different cultural category. Why Other Options Are Wrong: Kuchipudi is wrong as the odd one out because it is clearly an Indian classical dance form. Kathak is wrong as the odd one out because it is also a classical dance style from North India. Bhangra is wrong as the odd one out because it is a very popular folk dance and not a festival. Common Pitfalls: A possible confusion is thinking that because Pongal is associated with celebrations, it might somehow also be a dance. Another pitfall is not remembering that Bhangra is a dance and assuming it is only a music style or festival. To avoid such errors, always recall whether people say they dance that item or celebrate that item, which usually clarifies whether it is a festival or a dance form. Final Answer: The only option that is a festival rather than a dance form, and therefore the odd one out, is Pongal.

Question 16

In this letter pair reasoning question, select the pair of letters that does not follow the same positional difference pattern as the others: PE, MV, GP and DM.

Accepted Answer

Introduction / Context: Odd one out questions using letter pairs often hide a fixed difference in alphabetical positions between the first and second letter. In this problem, you are given four such pairs and asked to identify which pair does not share the same difference. These questions check your familiarity with alphabetical positions and your ability to spot simple numeric patterns within letters. Given Data / Assumptions: The letter pairs are PE, MV, GP and DM. We assign positions to letters in the English alphabet: A = 1, B = 2, C = 3, and so on. We find the difference between the position of the second letter and the first letter in each pair. Concept / Approach: A standard approach is to convert each letter into its position and calculate the difference. If three pairs show the same difference (for example plus 9), and one pair shows a different value, then that different pair is the odd one out. This method transforms a verbal pattern into a simple arithmetic pattern on numbers. Step-by-Step Solution: Step 1: For PE, P is the 16th letter and E is the 5th letter. The difference is 16 - 5 = 11 in absolute value. Step 2: For MV, M is the 13th letter and V is the 22nd letter. The difference is 22 - 13 = 9. Step 3: For GP, G is the 7th letter and P is the 16th letter. The difference is 16 - 7 = 9. Step 4: For DM, D is the 4th letter and M is the 13th letter. The difference is 13 - 4 = 9. Step 5: Thus three pairs (MV, GP and DM) have a difference of 9 between first and second letter, while PE has a difference of 11. Hence PE is the odd one out. Verification / Alternative check: We can cross verify by noting that moving 9 steps forward from M gives V, from G gives P, and from D gives M. However, moving 9 steps from P would not give E, and moving 11 steps back from P gives E. The consistent pattern for three pairs is a fixed jump of 9 alphabet positions, which PE breaks. This confirms that PE is the unique non matching pair. Why Other Options Are Wrong: MV is wrong as the odd one out because the letters are separated by 9 positions, matching the main pattern. GP is wrong as the odd one out because it also shows a difference of 9 and fits the same rule. DM is wrong as the odd one out because it continues the same relationship with a difference of 9. Common Pitfalls: Students sometimes look for forward or backward order alone and ignore the exact numeric difference. Others may calculate position differences incorrectly by miscounting letters. A good way to avoid errors is to write down positions or at least check them quickly in your mind using known reference points such as A = 1, M = 13 and Z = 26. Once the differences are computed carefully, the pattern becomes obvious. Final Answer: The only pair that does not have a difference of 9 between the letters, and therefore the odd one out, is PE.

Question 17

In this perfect cube pattern question, select the number which is not a perfect cube among 512, 216, 343 and 719.

Accepted Answer

Introduction / Context: This odd one out problem again focuses on perfect cubes, a very common concept in quantitative aptitude. Recognising cube numbers quickly helps in mental math and pattern detection. You are given four numbers, three of which are cubes of integers, and you need to identify the non cube number that does not fit the pattern. Given Data / Assumptions: The numbers are 512, 216, 343 and 719. We use the definition that a perfect cube is n^3 for some integer n. We recall or compute standard cube values for small integers. Concept / Approach: The common technique is to match each number with known cubes: 6^3, 7^3, 8^3, 9^3 and so on. If the number exactly equals one of these cubes, it is a perfect cube. Otherwise, it is not. Three of the given values will fit into cube tables, and one will not, which becomes the odd one out. Step-by-Step Solution: Step 1: Check 216. Since 6 * 6 * 6 = 216, we have 6^3 = 216, so 216 is a perfect cube. Step 2: Check 343. Since 7 * 7 * 7 = 343, we have 7^3 = 343, so 343 is also a perfect cube. Step 3: Check 512. Since 8 * 8 * 8 = 512, we have 8^3 = 512, so 512 is another perfect cube. Step 4: Check 719. Nearby cubes are 8^3 = 512 and 9^3 = 729. The number 719 lies between 512 and 729 and does not match any exact cube value. Step 5: Therefore 512, 216 and 343 are perfect cubes, while 719 is not. Hence 719 is the odd one out. Verification / Alternative check: We can verify further by trying to factor 719. 719 is not divisible by 2, 3 or 5. Testing 7 gives remainder 5, and 11, 13 or 17 do not divide it evenly. Its prime factorisation cannot be arranged into triples of the same prime, so it cannot be written as n^3. This prime based check aligns with the observation that it lies between 8^3 and 9^3 without being equal to any cube. Why Other Options Are Wrong: 512 is wrong as the odd one out because it is 8^3 and clearly a perfect cube. 216 is wrong as the odd one out because it is 6^3, a cube number. 343 is wrong as the odd one out because it is 7^3 and also a perfect cube. Common Pitfalls: Sometimes candidates do not remember cube values and guess based on size alone. Another error is to treat 719 as special simply because it is not an obvious product of small numbers, without confirming the cube status of the others. Learning standard cubes at least up to 15^3 makes questions like this very straightforward and avoids such confusion. Final Answer: The only number that is not a perfect cube, and therefore the odd one out, is 719.

Question 18

In this four digit number pattern question, find the number that does not follow the same digit sum rule as the others: 2543, 2192, 9362 and 3713.

Accepted Answer

Introduction / Context: This problem is another example of an odd one out question that hides a pattern within the digits of each four digit number. Often, exam setters choose a property like the sum of digits being equal or following a particular value. Here, three numbers share the same digit sum while one number does not, and you must identify that different one. Given Data / Assumptions: The options are 2543, 2192, 9362 and 3713. We will compute the sum of digits of each number. The number with a digit sum that differs from the common value will be the odd one out. Concept / Approach: Digit sum patterns are very common in reasoning questions. The method is simply to add the digits of each four digit number and compare the resulting sums. If three numbers give the same total and one gives a different total, the latter is the odd man out. This approach does not require advanced maths, only careful addition. Step-by-Step Solution: Step 1: For 2543, add the digits: 2 + 5 + 4 + 3 = 14. Step 2: For 2192, add the digits: 2 + 1 + 9 + 2 = 14. Step 3: For 3713, add the digits: 3 + 7 + 1 + 3 = 14. Step 4: For 9362, add the digits: 9 + 3 + 6 + 2 = 20. Step 5: We see that 2543, 2192 and 3713 all have digit sum 14, while 9362 has digit sum 20. Therefore 9362 is the odd one out. Verification / Alternative check: To verify, we note that if there were any arithmetic or divisibility pattern based on this common digit sum, these three numbers might share it. However, the question only requires us to identify the differing item. The clear split between 14 and 20 shows one number standing apart from the others. No additional complex checks are necessary to confirm that 9362 is the unique element. Why Other Options Are Wrong: 2543 is wrong as the odd one out because it has the same digit sum 14 as two of the other numbers. 2192 is wrong as the odd one out because it also sums to 14 and follows the main pattern. 3713 is wrong as the odd one out because its digit sum equals 14, matching 2543 and 2192. Common Pitfalls: Some candidates look only at the first or last two digits or try to identify a pattern based on multiplication of digits. Others do not compute all the digit sums carefully and may mis add one of them, leading to a wrong conclusion. Since this question is built purely on digit sums, careful and accurate addition is essential for solving it correctly. Final Answer: The only number whose digit sum is different from the others, and hence the odd one out, is 9362.

Question 19

In this vocabulary based odd one out question on clarity of appearance, select the word that does not describe a state of reduced visibility: Hazy, Cloudy, Translucent and Transparent.

Accepted Answer

Introduction / Context: This odd one out question focuses on English vocabulary describing how clearly we can see through a material or atmosphere. Three of the words describe conditions where visibility is blocked or reduced, and one word describes a state of complete clarity. Identifying which term belongs to the opposite group is a common way of testing word meaning in verbal reasoning. Given Data / Assumptions: The options are Hazy, Cloudy, Translucent and Transparent. We interpret these words in the context of visibility through air or materials. The aim is to separate words that refer to reduced clarity from the word that refers to full clarity. Concept / Approach: The approach is to recall the definitions of each word. Hazy and cloudy generally refer to conditions where objects cannot be seen clearly. Translucent refers to materials that allow some light through but not clear images. Transparent refers to materials that allow objects to be seen clearly. Thus, most words here describe partial or poor visibility, while one describes complete clarity. Step-by-Step Solution: Step 1: Hazy describes a situation in which the air or image is blurred, misty or not clearly visible. Visibility is reduced. Step 2: Cloudy similarly describes something that is covered with clouds or has many particles that block light, making sight unclear. Step 3: Translucent describes a material through which light can pass, but you cannot see sharp details clearly. Examples include frosted glass, where visibility is partial but not perfect. Step 4: Transparent is different. A transparent material allows light and images to pass through so clearly that you can see objects on the other side distinctly, like clear glass or pure water. Step 5: Therefore, Hazy, Cloudy and Translucent describe reduced or partial clarity, while Transparent describes complete clarity. So Transparent is the odd one out. Verification / Alternative check: You can verify by asking how easily you can see objects through a medium described by each word. Through a hazy or cloudy medium, visibility is poor. Through a translucent sheet, shapes and colours may be visible but details are blurred. Through a transparent window, you can see everything sharply. Transparent clearly stands apart as describing full clarity, which confirms our reasoning. Why Other Options Are Wrong: Hazy is wrong as the odd one out because it indicates reduced clarity and fits with the partially obstructed group. Cloudy is wrong as the odd one out because it also suggests blocked or disturbed visibility. Translucent is wrong as the odd one out because it likewise refers to a state where some light passes through but objects are not seen clearly. Common Pitfalls: Learners may confuse translucent with transparent and assume both mean fully clear. However, translucent allows only partial visibility, which is closer to hazy and cloudy. Another pitfall is focusing only on everyday weather usage (cloudy day, hazy day) and ignoring the deeper meaning related to light passing through materials. Always compare how each word affects clarity to identify the odd one out correctly. Final Answer: The only word that describes complete clarity instead of reduced or partial visibility, and therefore the odd one out, is Transparent.

Question 20

In this letter group reasoning question, choose the group that does not follow the same vowel consonant starting pattern as the others: IDD, AGG, UTT and REE.

Accepted Answer

Introduction / Context: Simple odd one out questions with letter groups often involve checking whether each group starts with a vowel or a consonant, or whether the letters follow some repeated pattern. In this problem, you are given four three letter groups, all of which end with double letters. However, the first letters differ in type. Recognising the consistent starting pattern for most options helps you find the odd one out. Given Data / Assumptions: The options are IDD, AGG, UTT and REE. We classify vowels as A, E, I, O, U and the remaining letters as consonants. We are looking for the group that does not share the same starting letter type as the others. Concept / Approach: Notice that all groups have the structure consonant or vowel followed by a double consonant or double vowel. The repeated letters at the end are common, so the main difference arises from the first letter. The standard approach is to check whether that first letter is a vowel or a consonant. If three groups start with vowels and one starts with a consonant, the consonant starting group becomes the odd one out. Step-by-Step Solution: Step 1: For IDD, the first letter I is a vowel. D and D are consonants, but the starting letter is vowel. Step 2: For AGG, the first letter A is a vowel. G and G are consonants, with a vowel at the beginning. Step 3: For UTT, the first letter U is a vowel. T and T are consonants, again giving a vowel at the start. Step 4: For REE, the first letter R is a consonant. E and E are vowels. Here, unlike the others, the group starts with a consonant. Step 5: Therefore IDD, AGG and UTT start with vowels, while REE starts with a consonant. So REE is the odd one out. Verification / Alternative check: We can label each group with V for vowel and C for consonant at the start. That gives IDD = VCC, AGG = VCC, UTT = VCC and REE = CVV. Three groups share a VCC pattern and one group has CVV. The different starting pattern of REE confirms that it is uniquely different from the other three options. Why Other Options Are Wrong: IDD is wrong as the odd one out because it begins with the vowel I, matching the starting pattern of AGG and UTT. AGG is wrong as the odd one out because it begins with the vowel A, like IDD and UTT. UTT is wrong as the odd one out because it begins with the vowel U and fits the dominant pattern. Common Pitfalls: Some students may be distracted by the double letters at the end and think that this repetition is the key pattern, but since all four groups have repetition, it does not differentiate them. Another mistake is not confirming vowel or consonant status carefully and misclassifying letters. A quick mental check using A, E, I, O, U as vowels avoids such confusion. Final Answer: The only letter group that starts with a consonant rather than a vowel, and therefore the odd one out, is REE.

Odd Man Out and Series Questions