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

Alphabet Test Questions