Question 1

Word-formation count: Using each letter only once, how many meaningful English words can be formed from the letters I, F, E, L?

Accepted Answer

Introduction / Context: From the letter set I, F, E, L (one each), determine how many everyday English words can be formed without repeating letters. Such problems test flexible word recognition and systematic checking of permutations. Given Data / Assumptions: Letters: {I, F, E, L}. Use each letter exactly once per word. Count standard dictionary words only (no names, abbreviations). Concept / Approach: Two common words are immediately visible: LIFE and FILE. Each uses all four letters once, and both are frequent in everyday English. Other permutations such as “lief” also exist; however, “lief” is archaic/poetic and typically excluded in aptitude-test conventions favoring contemporary usage. Therefore, the accepted count is two. Step-by-Step Solution: 1) Try common arrangements: LIFE ✓, FILE ✓.2) Evaluate variants: LIEF (archaic) generally excluded from “meaningful” in test context.3) Conclude that two standard words satisfy the constraint. Verification / Alternative check: Most exam keys treat “meaningful words” as everyday vocabulary entries. Under that norm, LIFE and FILE are counted; archaic or poetic forms like LIEF are not. Why Other Options Are Wrong: “None/One” miss at least one valid word; “Three” overstates by including archaic/less-accepted forms. Common Pitfalls: Over-including rare or archaic forms; always align with contemporary, general-use dictionaries for such tests. Final Answer: Two

Question 2

Word-formation count: Using each letter only once, how many meaningful English words can be formed from the letters I, T, R, M?

Accepted Answer

Introduction / Context: From the letter set I, T, R, M, determine the number of common English words that can be formed using each letter once. This probes quick anagram spotting within a small letter bag. Given Data / Assumptions: Letters: {I, T, R, M}. Each letter used once per word. Standard English words (no proper nouns or abbreviations). Concept / Approach: The clearest common anagram is TRIM (verb/adjective). Alternatives like “mirt,” “ritm,” or “rimt” are not English words; “trim” alone satisfies both the letter-use constraint and dictionary validity. Hence, the count is one. Step-by-Step Solution: 1) Test common placements: TRIM ✓.2) Check other permutations: none yield standard words.3) Conclude only one valid word exists. Verification / Alternative check: Consulting typical exam word lists and general dictionaries confirms “trim” as the sole frequent, valid anagram. Why Other Options Are Wrong: “None” undercounts; “Two/Three” would require accepting non-words. Common Pitfalls: Overlooking “trim” due to focusing on vowel-initial starts; always scan consonant clusters too. Final Answer: One

Question 3

Alphabet-pair series: AZ, GI, MN, __, YB — identify the missing pair by discovering the rule that governs the first and second letters across the series.

Accepted Answer

Introduction / Context: In alphabet-pair series, each term consists of two letters. The rule can act on (i) the first letters across terms, (ii) the second letters across terms, or (iii) both, sometimes using symmetric or arithmetic patterns on letter positions (A=1, B=2, …, Z=26). The given series is: AZ, GI, MN, __, YB. We must determine the missing fourth pair. Given Data / Assumptions: Pairs: A Z; G I; M N; ? ?; Y B. Alphabet positions: A=1, …, Z=26. We expect a consistent progression separately for first and second letters. Concept / Approach: Analyze the first letters: A(1), G(7), M(13), ?, Y(25). The jumps are +6, +6, +6, +6. So the fourth first-letter must be 13+6 = 19, which is S. Now analyze the second letters: Z(26), I(9), N(14), ?, B(2). Notice that in the 1st, 3rd, and 5th terms, the letters are end-symmetric pairs that sum to 27 with their partners (A+Z = 1+26 = 27; M+N = 13+14 = 27; Y+B = 25+2 = 27). This symmetry strongly suggests the missing 4th pair should also sum to 27: since the first letter is S(19), the partner must be 27-19 = 8, which is H. Step-by-Step Solution: 1) First letters: A→G→M→?→Y are +6 each, so ? = S.2) Symmetry check on pairs that sum to 27: (A,Z), (M,N), (Y,B) → maintain pattern for the 4th pair.3) Compute second letter: 27 − S(19) = 8 → H.4) The missing pair is SH. Verification / Alternative check: Double-check that SH fits the observed structures: First-letter arithmetic (+6 ladder) holds; 4th pair also fits the recurring 27-sum symmetry observed among non-adjacent terms, keeping the sequence coherent. Why Other Options Are Wrong: RX, KF, TS break either the +6 rule for the first letters or fail the 27-sum symmetry constraint for the pair. Only SH satisfies both. Common Pitfalls: Focusing solely on one position or assuming both letters move by the same fixed step. Here, one track is a uniform +6, while the other is governed by a symmetric-sum rule. Final Answer: SH

Question 4

Triplet pattern decoding: J2Z, K4X, I7V, __, H16R, M22P — find the missing triplet by uncovering the independent rules for (1) the first letter, (2) the number, and (3) the last letter.

Accepted Answer

Introduction / Context: Each term in the series is a triplet: (letter)(number)(letter). Such sequences commonly apply separate but synchronized rules to each component stream: the first-letter stream, the numeric stream, and the last-letter stream. The given series is: J2Z, K4X, I7V, __, H16R, M22P. We must determine the fourth triplet. Given Data / Assumptions: First letters observed: J, K, I, ?, H, M. Numbers observed: 2, 4, 7, ?, 16, 22. Last letters observed: Z, X, V, ?, R, P. Alphabet positions: A=1 to Z=26. Concept / Approach: Last-letter stream: Z(26) → X(24) → V(22) → ? → R(18) → P(16): this is a steady −2 step each time, so the missing last letter is T(20). Numeric stream: 2 → 4 (+2) → 7 (+3) → ? (+4) → 16 (+5) → 22 (+6). This is an increasing-difference pattern (+2, +3, +4, +5, +6). Therefore the missing number is 11 (7 + 4). First-letter stream: J(10) → K(11) → I(9) → ? → H(8) → M(13). The deltas follow an alternating-signed, growing-magnitude pattern: +1, −2, +3, −4, +5. Applying +3 after I(9) yields L(12) for the fourth term; then −4 gives H(8), then +5 gives M(13), confirming the pattern. Step-by-Step Solution: 1) Deduce last letter T via −2 steps from V (22) to T (20).2) Deduce number 11 via +4 from 7.3) Deduce first letter L via +3 from I.4) Assemble the triplet: L11T. Verification / Alternative check: Project forward and backward to ensure consistency: after L11T, the next given is H16R, which matches −4 on first letters (L→H), +5 on numbers (11→16), and −2 on last letters (T→R). Then H16R to M22P continues +5, +6, −2 as expected. Why Other Options Are Wrong: I11T has the wrong first letter (breaks the +3 rule); L11S and L12T break either the −2 last-letter rule (S ≠ T) or the numeric +4 step (12 ≠ 11). Common Pitfalls: Trying to force one global arithmetic rule across mixed components. Here, each of the three sub-sequences uses its own simple, consistent pattern. Final Answer: L11T

Question 5

Digit–letter code mapping: If 1 8 5 1 4 encodes the word AHEAD (1→A, 8→H, 5→E, 1→A, 4→D), then what word is represented by 3 1 3 8 5?

Accepted Answer

Introduction / Context: Coding–decoding questions map digits to letters (or vice versa). You must infer the mapping from a given example and apply it to decode another sequence. Here, “AHEAD” is encoded as 1 8 5 1 4, which directly reveals several digit→letter assignments. Given Data / Assumptions: From AHEAD → 1 8 5 1 4 we infer: 1→A, 8→H, 5→E, 4→D. We need to decode 3 1 3 8 5. Digit 3 must map consistently to the same letter in both positions it appears. Concept / Approach: Apply known mappings first: 1→A, 8→H, 5→E. Thus 3 1 3 8 5 becomes ? A ? H E. Among the options, the only reasonable, consistent completion is “CACHE,” which corresponds to C A C H E. Therefore, 3 must map to C (3→C). This yields a coherent, letter-perfect decoding that matches one of the choices exactly. Step-by-Step Solution: 1) Write 3 1 3 8 5 as _ A _ H E using known mappings.2) Compare to options; “CACHE” fits the pattern.3) Deduce 3→C, completing the mapping. Verification / Alternative check: No other option matches the partial mapping (_ A _ H E) with consistent identical letters in slots 1 and 3. Hence “CACHE” is uniquely determined. Why Other Options Are Wrong: They either violate the repeated-letter constraint (positions 1 and 3 must be the same letter) or clash with established 1→A, 8→H, 5→E. Common Pitfalls: Assuming a positional/shift cipher; here it is a direct digit→letter substitution. Final Answer: CACHE

Question 6

Complete the sequence: 1, 11, 21, 1211, 111221, ____ (identify the next term in the standard “look-and-say” sequence).

Accepted Answer

Introduction / Context: This is the classic “look-and-say” (run-length encoding) sequence. Each term describes the previous term by counting contiguous runs of identical digits, then saying the digit. You must generate the next term from 111221. Given Data / Assumptions: Sequence so far: 1, 11, 21, 1211, 111221. Rule: read off the prior term as counts followed by digit. Concept / Approach: Break 111221 into contiguous runs: “111” (three 1s), “22” (two 2s), “1” (one 1). Convert each run to count+digit and concatenate: “31” for three 1s, “22” for two 2s, “11” for one 1, giving 312211. Step-by-Step Solution: 1) Identify runs: 111 / 22 / 1.2) Encode: 111 → 31; 22 → 22; 1 → 11.3) Concatenate: 31 22 11 → 312211. Verification / Alternative check: Check back one step: 1211 → read as “one 1, one 2, two 1s,” i.e., 111221, confirming the mechanism you just applied forward. Why Other Options Are Wrong: They miscount runs or reorder digits, which violates the deterministic rule. Common Pitfalls: Counting total occurrences rather than contiguous runs. The rule always applies to contiguous blocks. Final Answer: 312211

Question 7

From the word “CONTRIBUTION”, take the 2nd, 6th, 9th, and 12th letters. If at least one meaningful English word can be formed using those letters (each used once), what will be the last letter of such a word? If more than one such word exists, answe…

Accepted Answer

Introduction / Context: Many vocabulary puzzles ask you to extract letters by position and then determine whether a common word can be formed. If multiple words are possible, you answer “M”; if none, “X”. Otherwise, you report the last letter of the viable word. Given Data / Assumptions: Word: CONTRIBUTION. Positions: C(1) O(2) N(3) T(4) R(5) I(6) B(7) U(8) T(9) I(10) O(11) N(12). Selected letters: 2nd = O, 6th = I, 9th = T, 12th = N → set {O, I, T, N}. Use each letter exactly once to form a word. Concept / Approach: Scan for common words from the multiset {I, N, T, O}. The most immediate is “INTO,” a very common preposition. It uses all four letters exactly once and is undeniably standard English. Check for alternatives (e.g., “ton i,” “tino”)—they are either non-words or names; no second standard word exists, so we do not answer “M”. Step-by-Step Solution: 1) Extract letters: O, I, T, N.2) Form a common word: INTO.3) Identify its last letter: O. Verification / Alternative check: Try other permutations—“tino,” “tion,” “oint”—none are standalone common English words; “INTO” is unique and unambiguous here. Why Other Options Are Wrong: “T” is not the last letter of any valid full-word permutation here; “M” requires more than one distinct word; “X” would apply only if no word could be formed. Common Pitfalls: Confusing the productive suffix “-tion” for a standalone word; it is not accepted as an independent word in this context. Final Answer: O

Question 8

Alphabet test — formable word from letters of: RECOMMENDATION

Accepted Answer

Introduction / Context: In this alphabet-test item, you must find the option that can be formed by using each letter from the source word at most once. The test checks attention to letter frequency (how many times a letter appears) and whether any required letter is missing altogether. Given Data / Assumptions: Source word: RECOMMENDATION. Letters available (with counts): r(1), e(2), c(1), o(2), m(2), n(2), d(1), a(1), t(1), i(1). Rule: build a candidate only if every required letter exists in the pool with sufficient multiplicity. Concept / Approach: Construct a quick frequency map of the source word and then check each option by subtracting required letters. If any letter is missing or needed more times than available, that option is invalid. Step-by-Step Solution: Option COMMUNICATE → requires U (absent) and two C's (only one C available) → not formable. Option REMINDER → needs two R's (only one R available) → not formable. Option MEDICO → needs M, E, D, I, C, O. All exist within counts (M×2≥1, E×2≥1, D×1, I×1, C×1, O×2≥1) → formable ✔. Option MEDIATES → needs S (absent) → not formable. Option ACCOMMODATE → needs two C's (only one) and U (absent) → not formable. Verification / Alternative check: Recount letters after selection to ensure no count goes negative. MEDICO consumes at most one of each needed letter, all within limits. Why Other Options Are Wrong: They either require letters that do not occur in the source (e.g., U or S) or require a higher multiplicity (e.g., two R's or double C) than available. Common Pitfalls: Overlooking letter multiplicity (especially R, C) and assuming any semantically related word is acceptable. Meaning is irrelevant; feasibility is purely by letters and counts. Final Answer: MEDICO

Question 9

Alphabet test — formable word from letters of: MEASUREMENT

Accepted Answer

Introduction / Context: This problem asks which option can be composed from the letters of MEASUREMENT without exceeding letter counts. Such questions reward systematic frequency checking over guesswork. Given Data / Assumptions: Source letters and counts: M(2), E(3), A(1), S(1), U(1), R(1), N(1), T(1). Rule: each letter used at most as many times as it appears in the source. Concept / Approach: Create a quick inventory table and test each option against counts. Step-by-Step Solution: ASSURE → needs S(2) but S=1 only → not formable. MANTLE → needs L (absent) → not formable. MASTER → needs M(1), A(1), S(1), T(1), E(1), R(1). All available within counts → formable ✔. SUMMIT → needs I (absent) → not formable. SUGAR → needs G (absent) → not formable. Verification / Alternative check: Re-tally after selecting MASTER: remaining counts never go negative. Why Other Options Are Wrong: They require a missing letter (L, I, G) or exceed available multiplicity (two S's), making construction impossible. Common Pitfalls: Noticing obvious substrings but ignoring a single missing character (e.g., an L in MANTLE). Final Answer: MASTER

Question 10

Alphabet test — formable word from letters of: CORRESPONDING

Accepted Answer

Introduction / Context: Choose the option that can be built from the letters of CORRESPONDING. Check availability and multiplicity for each character. Given Data / Assumptions: Key letters in CORRESPONDING: C, O(2), R(2), E(1), S(1), P(1), N(2), D(1), I(1), G(1). Concept / Approach: Filter out options needing any missing letter or an extra copy beyond counts. Step-by-Step Solution: DISCERN → letters D, I, S, C, E, R, N are all present within single-count needs → formable ✔. RESPONSE → needs two E's and two S's; only E=1 and S=1 → not formable. REPENT → needs T (absent) → not formable. CORRECT → needs T (absent) and two C's (only one C) → not formable. CROSSBONE → needs B (absent) → not formable. Verification / Alternative check: Re-check DISCERN against counts; all requirements satisfied without overuse. Why Other Options Are Wrong: They require missing letters (T, B) or exceed available multiplicity (extra E or S). Common Pitfalls: Assuming any semantically related word is acceptable; the test is purely letter-feasibility. Final Answer: DISCERN

Question 11

Alphabet test — formable word from letters of: REPUTATION

Accepted Answer

Introduction / Context: Find which option can be constructed from REPUTATION using each letter no more than available. Given Data / Assumptions: REPUTATION letters: R(1), E(1), P(1), U(1), T(2), A(1), I(1), O(1), N(1). Concept / Approach: Check each candidate against the pool. Step-by-Step Solution: TUTOR → needs T(2), U(1), O(1), R(1). All available → formable ✔. PONDER → needs D (absent) → not formable. REQUIRE → needs Q (absent) and E(2) (only E=1) → not formable. RETIRE → needs E(2) (only E=1) → not formable. PRIOR → needs R(2) (only R=1) → not formable. Verification / Alternative check: Tally confirms letter counts satisfied only for TUTOR. Why Other Options Are Wrong: They require absent letters (D, Q) or exceed multiplicity (extra E or R). Common Pitfalls: Ignoring the double-T requirement; REPUTATION conveniently supplies two T's. Final Answer: TUTOR

Question 12

Alphabet test — formable word from letters of: Workshop

Accepted Answer

Introduction / Context: From the letters in “Workshop,” identify the option that can be constructed without reusing any letter beyond its count. Given Data / Assumptions: WORKSHOP letters: W(2), O(2), R(1), K(1), S(1), H(1), P(1). Concept / Approach: Test each candidate strictly against availability. Step-by-Step Solution: Show → needs H (present), S (present), O (present), W (present). However, “Show” rearranges letters but is valid only if case-insensitive. Still, more than one valid option would break uniqueness; check others. Crow → needs C (absent) → invalid. Work → needs W, O, R, K (all present) → formable ✔. Pork → needs P, O, R, K (all present) → also formable, causing ambiguity unless case or set constraints are specified. Sword → needs D (absent) → invalid. Recovery / Correction: Because both “Work” and “Pork” are formable, the current option set yields multiple correct answers. The item is defective as authored; to preserve bank quality, it must be revised or removed. Final Answer: Item invalid as written

Question 13

In the word CHAIRS, how many pairs of letters are there such that the number of letters between them in the word is the same as the number of letters between them in the English alphabet sequence?

Accepted Answer

Introduction / Context: This is an alphabet test question in which you need to count special letter pairs in a given word. The word is CHAIRS. You must identify those pairs of letters where the number of letters between the two letters in the word is equal to the number of letters between them in the normal alphabetical order. This type of problem checks attention to detail and ability to track positions in both the word and the alphabet sequence. Given Data / Assumptions: Word: C H A I R S. We use the standard English alphabet: A, B, C, D, and so on up to Z. A pair is formed by taking two letters in the word in the same order as they appear. We must count letters between that pair in the word and letters between them in the alphabet. If the counts are equal, that pair satisfies the condition. Concept / Approach: To solve such questions, we can follow a systematic method. First, write the positions of each letter in the word. Then, for each pair, count how many letters lie between them in the word. Next, note the alphabetical positions of the same letters and count how many letters lie between them in the alphabet. If both counts are equal, we count that pair. Careful counting avoids missing pairs or double counting. We only consider the direction of letters as they appear in the word, not reversing them. Step-by-Step Solution: Step 1: Write the word with positions: C(1), H(2), A(3), I(4), R(5), S(6).Step 2: Consider the pair C and A. In the word, there is one letter between them (H). In the alphabet, C is the third letter and A is the first; between A and C there is exactly one letter, B. So word gap and alphabet gap are both one. This pair satisfies the condition.Step 3: Consider the pair R and S. In the word, R is at position 5 and S at position 6, so there are zero letters between them. In the alphabet, R and S are consecutive letters, so there are also zero letters between them. This pair also satisfies the condition.Step 4: Check other pairs quickly. For example, C and H have zero letters between in the word but four letters between them in the alphabet. Similar mismatches occur for pairs such as H and I or A and R. None of these satisfy the requirement.Step 5: Conclude that only two pairs, C–A and R–S, meet the condition. Verification / Alternative check: If desired, list all possible pairs systematically and compare gaps. Doing so confirms that the only matching pairs are C with A and R with S. Therefore, the correct count is two. Double checking prevents the common error of either overlooking R–S or miscounting the gaps for other pairs such as H–I, which are consecutive in the alphabet but are separated by one letter in the word. Why Other Options Are Wrong: Option b (3) assumes one extra pair that does not actually meet the condition. Option c (1) means only one of the correct pairs is noticed, which is incomplete. Option d (4) suggests severe overcounting of pairs. Option e (0) would mean that no pair satisfies the condition, which contradicts our careful counting. Common Pitfalls: Common mistakes include counting letters from the wrong end of the alphabet, confusing the order of letters in the word, or forgetting that we count letters between the two letters rather than including them. Another pitfall is assuming that any consecutive letters in the word must satisfy the condition without checking the alphabet gap. A systematic and careful comparison avoids these errors. Final Answer: There are 2 such pairs of letters in the word CHAIRS.

Question 14

In the word CASTRAPHONE, how many pairs of letters are there such that the number of letters between them in the word is the same as the number of letters between them in the English alphabet sequence?

Accepted Answer

Introduction / Context: This alphabet test question extends the same idea used in the previous problem. The word is CASTRAPHONE, and we are asked to find how many pairs of letters have the same number of letters between them in the word as in the alphabet. Because the word is longer, there are more possible pairs, and a systematic approach is essential to avoid missing any or counting extra pairs. The problem tests visual tracking, numerical counting, and familiarity with the alphabetical order. Given Data / Assumptions: Word: C A S T R A P H O N E. We consider the standard English alphabet sequence. A pair consists of two letters in the word in their given order. For each pair, the number of letters between them in the word must equal the number of letters between them in the alphabet. We count each qualifying pair once. Concept / Approach: Because there are many possible pairs in a long word, it is best to write down the positions of letters in the word and note their alphabetical positions. For each pair, we calculate two numbers: the gap in the word and the gap in the alphabet. The gap in the word is the difference in positions minus one, and the gap in the alphabet is the difference in alphabetical positions minus one. If these two gaps match, that pair satisfies the condition. Using a structured scan ensures we do not overlook any valid pair. Step-by-Step Solution: Step 1: Write the letters of CASTRAPHONE with serial positions: C(1), A(2), S(3), T(4), R(5), A(6), P(7), H(8), O(9), N(10), E(11).Step 2: Note the alphabetical positions: C=3, A=1, S=19, T=20, R=18, P=16, H=8, O=15, N=14, E=5.Step 3: Examine relevant pairs and compare gaps. For example, pair S and T at positions 3 and 4 has zero letters between them in the word, and in the alphabet S and T are consecutive, so there are zero letters between them there as well. Therefore S–T is a valid pair.Step 4: Continue systematically and you will find six valid pairs: S–T, T–O, T–N, R–P, H–E, and O–N. In each of these cases, the number of letters between the two in the word matches the number of letters between them in alphabetical order.Step 5: For all other pairs, either the gap in the word and the gap in the alphabet do not match, or the letters are repeated in such a way that the counts do not line up. After checking all pairs, we confirm that there are exactly six qualifying pairs. Verification / Alternative check: A careful numerical check for each of the identified pairs confirms correctness. For example, T is the twentieth letter and O is the fifteenth, so they are separated by four letters in the alphabet (one less than the difference in their positions). In the word, T and O occur at positions 4 and 9, with four letters between them as well. Similar calculations verify that T–N, R–P, H–E, and O–N satisfy the condition. No additional pair satisfies both gap conditions, which confirms the count of six pairs. Why Other Options Are Wrong: Options a, b, and c (3, 4, or 5 pairs) underestimate the number of qualifying pairs and typically result from incomplete checking or early stopping. Option e (2) is far too small and would mean that several valid pairs have been overlooked. Common Pitfalls: Because there are many possible pairs, students often lose track while counting and either skip some pairs or double count. Another common error is to forget to subtract one when counting letters between two positions in the word or in the alphabet, which leads to off by one mistakes. Writing positions clearly, proceeding in a fixed order, and comparing both word gaps and alphabet gaps carefully helps avoid such errors and leads to the correct answer of six pairs. Final Answer: There are 6 such pairs of letters in the word CASTRAPHONE.

Question 15

In the English alphabet written in order from A to Z, first find the letter that is the tenth letter to the left of the second last letter of the alphabet, and then move eight letters to the right from that letter. Which letter do you reach?

Accepted Answer

Introduction / Context: This logical reasoning question tests your understanding of positions in the English alphabet and your ability to apply multi step directional instructions such as moving a certain number of letters to the left or right. Such questions are very common in banking exams, SSC, and other aptitude tests, so learning a systematic approach is extremely useful. Given Data / Assumptions: • The English alphabet is written in a straight line from left to right as A, B, C, D, ..., X, Y, Z. • The last letter of the alphabet is Z and the second last letter is Y. • Left means moving towards A and right means moving towards Z. • We must first find the letter that is the tenth to the left of Y, and then from that letter move eight positions to the right. Concept / Approach: The core concept is ordinal positions in a fixed sequence. Every letter occupies a definite position from the left, from 1 to 26. Moving a certain number of places to the left means subtracting that count from the position, while moving to the right means adding that count. By converting letters to position numbers, doing simple addition or subtraction, and then converting back to letters, we can answer reliably without confusion. Step-by-Step Solution: Step 1: Write positions of letters. A is position 1, B is 2, ..., Y is 25, Z is 26.Step 2: Identify the second last letter of the alphabet. The last letter is Z (26), so the second last letter is Y (25).Step 3: Find the letter that is the tenth to the left of Y. Moving left means subtracting 10 from the position of Y.Step 4: Position of Y is 25, so the required position is 25 - 10 = 15.Step 5: The 15th letter of the alphabet is O.Step 6: Now move eight letters to the right of O. Moving right means adding 8 to the position of O.Step 7: Position of O is 15, so new position = 15 + 8 = 23.Step 8: The 23rd letter of the alphabet is W, so W is the final answer. Verification / Alternative check: You can also count directly on the alphabet line. From Y, move left one by one to reach O as the tenth letter to the left, and then from O move right one by one until you count eight steps reaching W. Doing this slow manual count confirms that the computed answer W is correct and removes any doubt about off by one errors. Why Other Options Are Wrong: • X: This is one step to the right of W and would correspond to moving nine letters instead of eight from O, so it does not satisfy the given condition. • I: I is much earlier in the alphabet and does not lie at the computed final position after performing both left and right movements correctly. • H: H lies near the beginning of the alphabet and cannot be reached by the described two stage movement starting from Y. It results from incorrect counting or misunderstanding of directions. • T: T is at position 20, which would correspond to moving only five steps right from O. This again does not match the required eight steps to the right. Common Pitfalls: Many learners confuse whether to include the starting letter when counting steps. Remember that when the question says tenth to the left, you move one letter at a time and count moves, not letters including the starting position. Another frequent mistake is mixing up the second last letter (Y) with the last letter (Z). Some students also subtract or add one extra step, which shifts the final letter to X instead of W. Using position numbers systematically helps avoid all these mistakes. Final Answer: The letter that you reach after performing the given two step movement in the alphabet is W.

Question 16

In the word "DISTURBANCE", the first letter is interchanged with the last letter, the second letter is interchanged with the tenth letter, the third letter is interchanged with the ninth letter, and so on from both ends towards the centre. In the ne…

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Introduction / Context: This question checks your ability to follow a pattern of letter interchanges within a word and then identify the position of a specific letter in the transformed word. Such word manipulation questions are common in alphabet test and reasoning sections of competitive exams, where accuracy in following instructions is more important than mere speed. Given Data / Assumptions: • The original word is "DISTURBANCE". • The first letter is swapped with the last letter, the second letter with the tenth letter, the third letter with the ninth letter, the fourth letter with the eighth letter, and the fifth letter with the seventh letter. • The middle letter (sixth position) remains as it is, because there is no partner letter left to swap with from the opposite end. • We must find which letter appears immediately after the letter T in the new word obtained after all these swaps. Concept / Approach: The concept is symmetric swapping from both ends toward the centre of the word. Instead of trying to imagine all changes in your head, it is better to write the positions and letters clearly, then perform the swaps step by step. After that, you examine the transformed word to locate the position of T and the letter that comes immediately after it. Step-by-Step Solution: Step 1: Write the original word with positions: 1 D, 2 I, 3 S, 4 T, 5 U, 6 R, 7 B, 8 A, 9 N, 10 C, 11 E.Step 2: Swap position 1 with 11: D and E interchange, giving E at 1 and D at 11.Step 3: Swap position 2 with 10: I and C interchange, so C goes to 2 and I goes to 10.Step 4: Swap position 3 with 9: S and N interchange, so N goes to 3 and S goes to 9.Step 5: Swap position 4 with 8: T and A interchange, so A goes to 4 and T goes to 8.Step 6: Swap position 5 with 7: U and B interchange, so B goes to 5 and U goes to 7. Position 6 (R) remains unchanged.Step 7: Now write the final arrangement: 1 E, 2 C, 3 N, 4 A, 5 B, 6 R, 7 U, 8 T, 9 S, 10 I, 11 D. So the new word is "ECNABRUTSID".Step 8: In this new word, the letter T appears at position 8, and the letter immediately after T is at position 9, which is S. Verification / Alternative check: You can verify by performing the swaps directly in the string "DISTURBANCE" without positions, but it is safer to keep positions to avoid missing any pair. Another quick check is to count the number of swaps: there are five swaps for an eleven letter word, leaving the sixth letter fixed. Repeating the process once again in reverse should bring you back to the original, confirming that the swap logic is correct and that T really moved to position 8 with S after it. Why Other Options Are Wrong: • I: The letter I appears later in the word at position 10 in the transformed arrangement, not immediately after T at position 8. • T: T itself is the reference letter; the question asks for the letter that comes after T, so choosing T means ignoring the instruction. • N: N appears at position 3 in the final word and has no relation to the immediate neighbour of T. • R: R appears in the middle at position 6 but is not adjacent to T in the new arrangement. Common Pitfalls: Candidates sometimes swap incorrectly by mixing positions or by forgetting that the middle letter of an odd length word remains fixed. Another common error is to stop after writing the swapped word partially and guessing the answer without carefully checking the final arrangement. A disciplined, position based approach prevents these typical mistakes and leads to the correct neighbour of T. Final Answer: In the newly formed word, the letter that comes immediately after T is S.

Question 17

Using each letter only once, how many distinct meaningful English words (using common everyday vocabulary) can be formed from the letters E, S, R and O?

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Introduction / Context: This is a classic alphabet test question on word formation. You are given a small set of letters and asked how many meaningful English words can be made from them when each letter is used exactly once. Such questions check your vocabulary, your ability to permute letters, and your discipline in counting distinct words without repetition or overcounting. Given Data / Assumptions: • The given letters are E, S, R and O. • Each letter can be used at most once in any word we form. • We only count meaningful English words that are commonly used, such as basic nouns or adjectives that a general exam taker is expected to know. • We ignore very rare or highly technical dictionary forms in order to match typical exam expectations. Concept / Approach: The approach is to think of all possible permutations of the four letters and then filter them by meaning. Because four letters can create many jumbled arrangements, a helpful trick is to look for naturally familiar patterns such as names of flowers, body parts, or basic adjectives. You do not need to list all 24 permutations; instead, you search systematically for common words and then check whether there are more that you might have missed. Step-by-Step Solution: Step 1: List the letters: E, S, R, O.Step 2: Try to form common words you know. One obvious and very familiar word is "ROSE" using R O S E. This is a common flower and an accepted English word.Step 3: Rearrange to look for an adjective. From the same letters we can form "SORE" using S O R E, which is a common word meaning painful.Step 4: Check if any other very common four letter words appear. Arrangements like "ORES" and "ROES" are legitimate dictionary words, but they are much less frequent and often considered beyond the expected level in many aptitude exams.Step 5: Following typical exam convention, we count "ROSE" and "SORE" as the main commonly used words.Step 6: Therefore, the number of meaningful English words formed in the intended sense is two. Verification / Alternative check: If you wish, you can systematically generate all permutations of the four letters and then check each one against your vocabulary or a dictionary. Doing so confirms that "ROSE" and "SORE" are the two most standard four letter words. The existence of rarer forms such as "ORES" and "ROES" does not usually affect exam keys, which focus on the everyday vocabulary that test takers are expected to know. Why Other Options Are Wrong: • None: This is wrong because at least "ROSE" is a very familiar word that clearly can be formed from the letters given. • One: This would be correct only if there were exactly one standard word, but both "ROSE" and "SORE" are basic words, so the count is higher than one. • Three: A count of three would require another equally common four letter word, which is not present at the same everyday level as "ROSE" and "SORE". • Four: To reach four, we would need several more widely used words. That goes beyond standard expectations for this puzzle and does not match typical answer keys. Common Pitfalls: Some students forget that each letter can be used only once and may try forms like "ROSEE". Others count rare or technical words and end up with a number that does not match the official answer. A balanced strategy is to first find obvious common words, then think for a few more seconds for any additional very familiar words, and stop when you have convincingly exhausted that pool. In this case, the reliable count is two. Final Answer: The number of meaningful English words that can be formed from E, S, R and O, under standard exam assumptions, is two.

Question 18

Study the following letter sequence: A A B A B C A B C D A B C D E A B C D __. Which letter should come next in place of the blank to continue the pattern correctly?

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Introduction / Context: This is a letter sequence question where you must recognise the underlying pattern and then extend it correctly. Alphabet test problems like this are designed to check how quickly you can observe repetition, grouping, and gradual growth in patterns, which is a common skill required in competitive examinations. Given Data / Assumptions: • The given sequence is: A A B A B C A B C D A B C D E A B C D __. • All letters are from the English alphabet and appear in increasing order within certain groups. • We need to identify the logical rule that generates this sequence and find the missing letter that should follow D at the end. Concept / Approach: The key idea is to look for grouping in the sequence rather than treating all letters as one long stream. Many such questions are constructed using blocks that repeat with a slight extension each time. By splitting the sequence into possible segments like A, A B, A B C, A B C D, and A B C D E, we can see an increasing pattern where each new group adds one more letter to the previous block. Step-by-Step Solution: Step 1: Rewrite the sequence with visual grouping: A / A B / A B C / A B C D / A B C D E / A B C D __.Step 2: Observe the pattern of groups: the first group has only A, the second group has A B, the third group has A B C, the fourth has A B C D, and the fifth has A B C D E.Step 3: Notice that each new group starts with A and adds one more letter in alphabetical order compared with the previous group.Step 4: The sequence shown near the end is A B C D forming the beginning of the next group. According to the pattern, this group should ultimately be A B C D E.Step 5: Therefore, the letter that must come after this final D is E, to complete the block as A B C D E. Verification / Alternative check: You can verify the pattern by counting group sizes. The group lengths are 1, 2, 3, 4, and 5. The last incomplete group has length 4 and obviously should become the next group of size 5. Since all such groups follow alphabetical order starting from A, the missing fifth letter in the last group must be E. This confirms E as the only consistent answer. Why Other Options Are Wrong: • A: If A were placed, the pattern of steadily growing groups would be broken because A B C D A would not fit the A to E progression. • C: Repeating C at that point does not match the rule of adding the next new letter in the alphabet, which must be E. • B: Adding B again would create A B C D B, which neither forms a standard alphabetical block nor matches the earlier structure. • D: Repeating D would give two consecutive Ds and disturb the regular pattern of increasing groups, so it cannot be correct. Common Pitfalls: Many students try to see the pattern in consecutive letters only and miss out on the grouping idea. Others may think that the sequence is repeating A B C D, but that fails when you notice the earlier A B C D E block. A reliable strategy is to mark commas or slashes and ask yourself whether there is a progressive structure in the lengths of these blocks. Once that is done, such problems become straightforward. Final Answer: The letter that should come next to continue the given pattern correctly is E.

Question 19

In the word "DEPRESSION", the first and second letters are interchanged, the third and fourth letters are interchanged, the fifth and sixth letters are interchanged, and so on for each pair. After all such adjacent swaps, which letter will be the se…

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Introduction / Context: This question combines word manipulation with position counting. You must perform systematic pairwise swaps of letters in the word "DEPRESSION" and then identify the letter at a specified position from the right side. These kinds of problems test attention to detail and your ability to follow a structured transformation before counting positions. Given Data / Assumptions: • The original word is "DEPRESSION". • We interchange letters in adjacent pairs: positions (1,2), (3,4), (5,6), (7,8) and (9,10). • The length of the word is ten letters, so all letters participate in a pair. • After forming the new word through these swaps, we must find the seventh letter from the right end. Concept / Approach: The problem involves two key concepts: adjacent swapping and position counting from the right. It is safer to work from left to right when computing the new arrangement, and then convert the requested right side position into the corresponding left side index. The seventh from the right in a ten letter word is the fourth from the left, because position from left equals total length minus position from right plus one. Step-by-Step Solution: Step 1: Write the original word with index positions: 1 D, 2 E, 3 P, 4 R, 5 E, 6 S, 7 S, 8 I, 9 O, 10 N.Step 2: Swap positions 1 and 2: E, D, P, R, E, S, S, I, O, N.Step 3: Swap positions 3 and 4: E, D, R, P, E, S, S, I, O, N.Step 4: Swap positions 5 and 6: E, D, R, P, S, E, S, I, O, N.Step 5: Swap positions 7 and 8: E, D, R, P, S, E, I, S, O, N.Step 6: Swap positions 9 and 10: E, D, R, P, S, E, I, S, N, O. The final word is "EDRPSEISNO".Step 7: We require the seventh letter from the right. The word has 10 letters, so position from left = 10 - 7 + 1 = 4. The fourth letter from the left in "EDRPSEISNO" is P. Verification / Alternative check: An alternative is to list the letters from the right after forming the new word. Starting from the right, the letters are O (1st), N (2nd), S (3rd), I (4th), E (5th), S (6th), P (7th), R (8th), D (9th), E (10th). Clearly, the seventh letter in this right side count is P, confirming the previous computation. Why Other Options Are Wrong: • R: R is the fourth from the right and eighth from the left in the new word, so it does not match the requested seventh from the right position. • O: O becomes the first letter from the right, so it lies at the extreme end rather than at the seventh place. • S: Two S letters appear, but neither occupies the seventh from right position; they are located closer to the middle. • D: D appears at the second position from the left and therefore is the ninth from the right, which is not the required position. Common Pitfalls: Students frequently miscount the pairings or forget one of the swaps. Another very common mistake is to treat seventh from the right as seventh from the left, which leads to a completely different letter. Always remember to convert positions carefully when counting from the right by using the relation left index = total length minus right index plus one. Final Answer: After performing all the adjacent swaps, the seventh letter from the right end of the new arrangement is P.

Question 20

In the sequence S L U A Y J V E I O N Q G Z B D R H, certain letters are used to form the pairs LA, UJ, YI and EG in a specific pattern. Following the same pattern, which pair of letters should come next in the series: LA, UJ, YI, EG, __?

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Introduction / Context: This is a letter series question where the given long sequence S L U A Y J V E I O N Q G Z B D R H acts as a source for forming smaller two letter pairs. The task is to identify how the example pairs LA, UJ, YI and EG are created from this source sequence and then predict the next pair correctly. Such questions examine pattern recognition and careful position tracking. Given Data / Assumptions: • The base sequence is: S L U A Y J V E I O N Q G Z B D R H (positions 1 to 18 from left). • The pairs we already have are: LA, UJ, YI and EG. • Each pair is formed by selecting two letters from specific positions in the base sequence, and this selection follows a consistent numerical pattern that we must uncover. Concept / Approach: The key idea is to map each pair back to the positions of its letters in the base sequence and then look for a pattern in those positions. Once we see how the starting and second positions of each pair are chosen, we can extend the sequence to find the positions of the next pair. The pattern in position differences often follows a simple arithmetic progression that makes prediction easier. Step-by-Step Solution: Step 1: Label the base sequence with positions: 1 S, 2 L, 3 U, 4 A, 5 Y, 6 J, 7 V, 8 E, 9 I, 10 O, 11 N, 12 Q, 13 G, 14 Z, 15 B, 16 D, 17 R, 18 H.Step 2: Identify LA. L is at position 2 and A is at position 4. So first pair uses positions (2, 4) with a gap of 2.Step 3: Identify UJ. U is at position 3 and J at position 6, so second pair uses (3, 6) with a gap of 3.Step 4: Identify YI. Y is at position 5 and I is at position 9, so third pair uses (5, 9) with a gap of 4.Step 5: Identify EG. E is at position 8 and G is at position 13, so fourth pair uses (8, 13) with a gap of 5.Step 6: Now observe the starting positions: 2, 3, 5, 8. The increments are +1, +2, +3. Following this pattern, the next increment should be +4, so the next starting position is 8 + 4 = 12.Step 7: The gaps for the second letters are 2, 3, 4, 5, so the next gap should be 6. Therefore, the second position is 12 + 6 = 18.Step 8: At position 12 we have Q, and at position 18 we have H. So the next pair in the series is QH. Verification / Alternative check: You can verify by writing the positions in pairs: (2,4), (3,6), (5,9), (8,13), and (12,18). The differences between first elements are 1, 2 and 3, and between second elements they increase by 1 repeatedly as 2, 3, 4, 5. This consistent pattern confirms that (12,18) is the logical continuation, resulting in pair QH. Why Other Options Are Wrong: • ZH: Z is at position 14 and H at 18, which does not follow the specific starting position and gap progression we discovered. • IB: I is at position 9 and B at 15, again not matching the required pair of starting and ending positions. • NR: N is at position 11 and R at 17, which breaks both the starting index pattern and the increasing gap rule. • VG: V is at position 7 and G at 13, so the pair would be (7,13), which again does not align with the sequence of starting indices 2, 3, 5, 8, 12. Common Pitfalls: Students may try to see a direct alphabetic pattern in the letters L, A, U, J, Y, I, E and G rather than working with positions in the base sequence. Without mapping positions carefully, it is very hard to detect the arithmetic pattern in the index differences. Another typical mistake is to miscount gaps or mix up which position is associated with which letter, leading to wrong conclusions. Final Answer: Following the established positional pattern, the next pair of letters in the series is QH.

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