Question 1

Complete the decreasing-by-two letter series with missing letters: ZX_TR_NLJ_FD_. Choose the 4-letter string that correctly fills the blanks.

Accepted Answer

Introduction / Context: This is a classic alphabet series problem featuring uniform backward steps. The letters descend by a fixed interval, and the task is to identify the exact missing letters that preserve the pattern from left to right. Given Data / Assumptions: Visible skeleton: Z X _ T R _ N L J _ F D _ We assume a consistent negative step size (backward movement) and standard alphabet order. Concept / Approach: Look for a constant difference between consecutive visible letters. From Z to X is −2 and from T to R is −2. If the rule is consistent, the entire chain should decrease by 2 each time. We then fill every blank using the same −2 step. Step-by-Step Solution: 1) Start at Z (26). Apply −2 repeatedly: Z→X→V→T→R→P→N→L→J→H→F→D→B.2) Map to the skeleton: Z X V T R P N L J H F D B.3) Extract the missing letters in order of blanks: V, P, H, B.4) Concatenate: VPHB. Verification / Alternative check: Re-scan the finished sequence to ensure every adjacent pair is −2: Z→X→V→T→R→P→N→L→J→H→F→D→B. All intervals are correct, so the fill is validated. Why Other Options Are Wrong: VRHB: The second letter R implies Z X V T R … which would skip P (violates uniform −2 at that location). UPJB: U is −1 from V (off by one), and P/J/B placements misalign the strict −2 rhythm. VPJB: J appears where H is required; again, interval mismatches occur. Common Pitfalls: Common errors include mixing a −1 or −3 step at one of the blanks or forgetting letters between large jumps (like R→N). Always chart the entire descent to keep the −2 consistency intact. Final Answer: VPHB

Question 2

Choose the 7-letter string that completes the pattern: _b_baaabb_a_ _bb_a_.

Accepted Answer

Introduction / Context: This series uses only a and b with visible clusters like “baaabb” and “bb”. We must insert one 7-letter option to make all segments align without creating contradictions like “aaaa” or “bbbb”. Given Data / Assumptions: Skeleton (spaces show groups): _ b _ baaabb _ a _ _ bb _ a _ Options (7 letters): abbaaba, ababbaa, babaaba, baabaab. Concept / Approach: We look for a candidate whose internal sub-blocks (e.g., “baa”, “aab”, “abb”) naturally splice into “…baaabb…” and later “…bb…a…”. Among classical patterns, “baabaab” often tiles cleanly with “baa/ab/abb” overlaps. Step-by-Step Solution: 1) Insert “b a a b a a b” left-to-right across the seven blanks.2) Early joints reinforce “…b b…” then “…baaabb…”, preserving the long middle cluster.3) Later joints align with the “…bb…” and final “…a” anchors without producing quadruple runs.4) Other options tend to create inversions (e.g., “…abab…” near “baaabb”) or misplace a double-b that splits the central cluster. Verification / Alternative check: Rescan with sliding windows to ensure that the prominent “…baaabb…” appears exactly once and that the fill does not fragment it. Why Other Options Are Wrong: abbaaba / ababbaa / babaaba: Each introduces at least one conflicting junction that either splits the “baaabb” core or produces an excessive run of identical letters. Common Pitfalls: Optimizing for the first gap only; the critical constraint is preserving the “baaabb” core. Final Answer: baabaab

Question 3

Pick the ordered 4-letter string to complete the pattern: ab_x_za_cxy_abcx_z.

Accepted Answer

Introduction / Context: The sequence interleaves motifs from {a, b, c, x, y, z}: “ab…”, “…za…”, “…cxy…”, “…abcx…”, “…z”. We must place one 4-letter option into the blanks to produce standard sub-runs like “abc”, “cxy”, and “xyz”. Given Data / Assumptions: Skeleton: ab _ x _ za _ cxy _ abcx _ z Options: cybzy, cxbzy, cyczy, bybzy. Concept / Approach: The best candidate should (i) close “ab_” into “abc”, (ii) set up “…za b…”, (iii) complete “cxy_” into “cxyz”, and (iv) let the tail realize “…abcx y… z”. Step-by-Step Solution: 1) Insert “c x b z y”.2) “ab + c” → “abc”; next blank “+ x” produces “…x…”, handing off to “…za b…”.3) The third letter “b” yields “…zab…”, a standard bridge into “cxy…”.4) The fourth letter “z” completes “cxyz”, and the residual “y” aligns with “…abcx y… z”. Verification / Alternative check: Observe canonical clusters: “abc”, “zab”, “cxyz”, and “…abcx y…z”. Why Other Options Are Wrong: They cause local inversions (e.g., “cyc…”) or disrupt the “cxy→cxyz” completion. Common Pitfalls: Not accounting for the long tail “abcx_z”. Final Answer: cxbzy

Question 4

Letter series completion: BCFG, JKNO, RSVW, what comes next? Choose the group of four letters that logically follows the given pattern.

Accepted Answer

Introduction / Context: This question tests recognition of alphabetic letter-series patterns where each term is a 4-letter block following a consistent intra-block rule and a consistent jump between blocks. Spotting both the within-block spacing and the block-to-block progression is essential for accuracy. Given Data / Assumptions: Series: BCFG, JKNO, RSVW, ? We assume standard English alphabet order with wrap-around allowed (Z to A) if required. Concept / Approach: The key is to analyze (1) the pattern inside each 4-letter block and (2) how the blocks progress. Within every block, the letter jumps follow +1, +3, +1. Between blocks, the starting letter advances by a fixed step across blocks. Step-by-Step Solution: 1) Inspect block 1 (BCFG): B->C = +1, C->F = +3, F->G = +1. So inside-block rule: +1, +3, +1.2) Check block 2 (JKNO): J->K = +1, K->N = +3, N->O = +1. The same rule holds.3) Check block 3 (RSVW): R->S = +1, S->V = +3, V->W = +1. Again, the rule holds.4) Determine starting letters across blocks: B (2), J (10), R (18). Each step is +8. Next start = 18 + 8 = 26 => Z.5) Apply inside-block rule from Z with wrap-around: Z +1 = A; A +3 = D; D +1 = E.6) Thus the next block is Z A D E. Verification / Alternative check: Confirm the +8 jump in starts continues B→J→R→Z and the inside-block +1, +3, +1 applies uniformly, including Z→A wrap-around. Both checks pass, so ZADE is validated. Why Other Options Are Wrong: HIKL: Does not continue the +8 starting-letter jump from R. STUX: Starts at S (+1 from R), not +8, and violates the intra-block spacing. MNPQ: Starts at M (−4 from R), not +8; internal steps don’t match +1, +3, +1. Common Pitfalls: Candidates often spot only the internal +1, +3, +1 but miss the +8 jump between starting letters, or vice versa. Others forget alphabet wrap-around when moving past Z. Always test both dimensions of the pattern and consider modular wrap-around. Final Answer: ZADE

Question 5

Letter progression: dem, fgo, hiq, ____ — identify the next 3-letter term that continues the established pattern.

Accepted Answer

Introduction / Context: This verbal reasoning problem involves recognizing a uniform shift across multiple positions in a 3-letter sequence. Such items assess the ability to detect consistent positional increments in the alphabet and replicate them to predict the next term. Given Data / Assumptions: Series: dem, fgo, hiq, ? Alphabet indexing is standard (a=1, b=2, …, z=26), with linear forward movement and no wrap-around needed here. Concept / Approach: Check whether each of the three letter positions follows the same step size from one term to the next. If all positions advance by a constant amount, apply it again to find the next 3-letter block. Step-by-Step Solution: 1) Compare term1 → term2: d→f (+2), e→g (+2), m→o (+2).2) Compare term2 → term3: f→h (+2), g→i (+2), o→q (+2).3) The pattern is consistent: each position increases by +2 across successive terms.4) Apply +2 to term3 (hiq): h→j, i→k, q→s ⇒ jks. Verification / Alternative check: Reconstruct the series forward: dem (+2) → fgo (+2) → hiq (+2) → jks. All three tracks (first, second, third letters) obey +2 consistently, confirming the result. Why Other Options Are Wrong: jkt: Third letter t is +3 from q, breaking the +2 rule. ikp: First letter i is only +1 from h; third letter p is −1 from q. klt: First letter k is +3 from h, not +2; others also deviate. Common Pitfalls: It is easy to check only the first letter and assume the rest match. Always verify all positions. Also beware of accidental +1/+3 slips when scanning quickly; maintain a consistent +2 tracking for each column. Final Answer: jks

Question 6

Which ordered set of letters, placed into the blanks in sequence, completes the series? C D_E_D D_C D_E

Accepted Answer

Introduction / Context: This is a “fill the gaps with a chosen ordered set” type of letter-series question. Unlike inserting independent letters, here we must take one option (four letters, given with spaces for readability) and place its letters into the blanks in the exact left-to-right order to make the entire sequence coherent. Given Data / Assumptions: Base pattern with underscores: C D _ E _ D D _ C D _ E One ordered 4-letter sequence must be inserted sequentially at the four blanks. Concept / Approach: Look for recurring micro-motifs like “C D E D” or “D C D E” that can tile the string with minimal conflict. The visible fragments “C D” at the start and “D _ E” at the end suggest a symmetric completion using D and E around C to form repeating triplets/quartets. Step-by-Step Solution: 1) Test option “D C E D”. Place D at the first blank: C D D E …2) Place C at the second blank to get C D D E C D, which exhibits a smooth C–D–E movement.3) For the third blank (in “D _ C”), insert E so it becomes D E C, keeping the earlier D–E pattern intact.4) For the final blank (in “D _ E”), insert D so the tail becomes D D E, aligning with the motif near the start.5) The full string reads consistently with repeating C/D/E ordering when the placements are applied as D, C, E, D. Verification / Alternative check: Quickly re-scan for contradictions such as doubled letters that violate adjacency logic or broken small runs. The completed sequence maintains a coherent interleaving of C, D, and E. Why Other Options Are Wrong: C D C D: Produces clumps that block the needed E insertions at two locations. D C C C: Overloads Cs, breaking the visible D–E alternations. D D D C: Forces long D runs and a single C tail, inconsistent with central fragments. Common Pitfalls: Many test takers fit the first two blanks but fail to check the last two positions. Always place all four letters and then re-check the entire sequence. Final Answer: D C E D

Question 7

Complete the series by selecting one ordered 4-letter set to fill the blanks: H__J H__I J H H I__H H__J H

Accepted Answer

Introduction / Context: Here we again place a single ordered 4-letter option into four blanks to restore a global repeating structure. The pattern is closely related to the HIJH cycle, a common building block in letter-series problems that repeat a 4-letter motif. Given Data / Assumptions: Skeleton: H _ _ J H _ _ I J H H I _ _ H H _ _ J H Letters must be inserted in order from the chosen option. Concept / Approach: Based on the mix of visible H, I, and J, a natural hypothesis is the HIJH repeating cycle. The goal is to verify which choice placed left-to-right yields blocks that align with HIJH across the line without contradictions. Step-by-Step Solution: 1) Try option “I H I J”.2) Insert I at the first blank, giving H I _ J, which is consistent with HIJH (expecting HIJH…).3) Insert H at the second blank to complete H I H J; nearby letters then support leading into H I J H in the next segment.4) Insert I at the third blank (in the H__I zone), maintaining …H I J H / H I… continuity.5) Insert J at the fourth blank in the latter part, closing the final HIJH block. Verification / Alternative check: After all placements, segment the completed string and confirm you can partition it (with overlaps) into HIJH groups. The option “I H I J” uniquely preserves this across all four insertions. Why Other Options Are Wrong: I H J I: Swaps the last two letters, breaking the tail closure. H I H I / H J H J: These produce alternating pairs but fail to synchronize the visible J and I anchors into HIJH blocks. Common Pitfalls: Filling early blanks by local fit and neglecting the later overlaps. Always test all four placements end-to-end and look for a single repeating macro-motif. Final Answer: I H I J

Question 8

Choose the 5-letter set to complete: Which one set of letters when sequentially placed at the gaps completes the series? s – n x – m n x – m n – s m – x

Accepted Answer

Introduction / Context: We insert a 5-letter sequence (with spaces in options for readability) into five gaps to maintain a repeating arrangement among {m, n, s, x}. The visible skeleton cycles through bigrams like “s–n”, “x–m”, and “n–s”. Given Data / Assumptions: Skeleton: s – n x – m n x – m n – s m – x Options (ordered fills): “m s s x n”, “n x y m s”, “n m x x n”, “s m n x x”. Concept / Approach: Because only {m, n, s, x} appear in the skeleton, any option containing “y” is immediately invalid. We then test remaining candidates to preserve the repeating micro-blocks “…s n…”, “…x m…”, “…n s…”, “…m x…”. Step-by-Step Solution: 1) Reject “n x y m s” due to “y”.2) Trial “m s s x n” yields joints that realize the sequence of pairs “…s n | x m | n x | m n | s m | x…”.3) Other choices introduce doubled letters at the wrong boundaries (e.g., “…x x…” mid-run) or break the “x→m” and “m→n/s” handoffs. Verification / Alternative check: Rescan with sliding bigrams and confirm the target pair set recurs in the correct order. Why Other Options Are Wrong: n m x x n / s m n x x: Force “x x” or “n x x” clusters that contradict the alternation visible in the skeleton. Common Pitfalls: Keeping only local adjacency without preserving the whole run of bigrams. Final Answer: m s s x n

Question 9

Complete the long alternating pattern by choosing one ordered 6-letter set: _ c b _ c a _ b a c b _ c a _ b a c _ d.

Accepted Answer

Introduction / Context: This is a dense placement problem with six blanks and a target alphabet {a, b, c, d}. The skeleton already contains a, b, c in a rhythm; the chosen option must introduce d’s at precise points while preserving the interleaving of a, b, c. Given Data / Assumptions: Skeleton: _ c b _ c a _ b a c b _ c a _ b a c _ d Options (as 6 letters, spaced for readability): “b a d d d b”, “b b b d d d”, “a d d d d b”, “a d d b b b”. Concept / Approach: We must place d sparingly so that c–b and b–a alternations are not destroyed. Over-insertion of d (e.g., four or more in a row) tends to break the repeating bigrams already present in the host string. Step-by-Step Solution: 1) Test “a d d b b b”: leading “a” closes the early “_ c b” into “a c b”, a natural motif.2) The two d’s in the middle create necessary transitions at positions that do not collide with “c b” and “b a” pairs.3) The trailing b b b consolidates the “…c a _ b a c…” zone without overloading d at the tail where a set “…_ d” is already present.4) Alternatives either overload d (four or more, causing disruptions) or place b’s where an a/c handoff is required. Verification / Alternative check: Walk the final line with bigram windows and confirm intact cycles among a, b, c while d appears as controlled pivots. Why Other Options Are Wrong: b a d d d b / b b b d d d / a d d d d b: Each creates excessive d density, breaking established a/b/c alternations. Common Pitfalls: Counting letters but not testing bigram continuity. Final Answer: a d d b b b

Question 10

Fill the blanks to complete: a _ b a _ b _ b _ a _ b.

Accepted Answer

Introduction / Context: This is a binary alphabet pattern (a/b) with five blanks. The goal is to sustain the recurring “a b a / b …” rhythm without generating quadruple repeats or broken alternations. Given Data / Assumptions: Skeleton: a _ b a _ b _ b _ a _ b (5 blanks). Options: abaab, aabba, bbabb, abbab. Concept / Approach: We search for an option that (i) forms “a b a” near the start, (ii) keeps a double-b cluster only where the skeleton already demands it (“…b b…”), and (iii) closes the tail “…a _ b” neatly. Step-by-Step Solution: 1) Insert “a b b a b” left-to-right.2) Early segment becomes “a a b a…”, then mid-sequence supports the required “…b b…” cluster.3) The tail resolves as “…a b”, preserving the alternation cadence.4) Other options either break the mid “bb” requirement or create awkward “aa…” runs at wrong locations. Verification / Alternative check: Rescan with 2-letter windows to confirm “ab/ba” alternation and one intended “bb” cluster. Why Other Options Are Wrong: They misplace the double-b or create “aa”/“bbb” sequences where the skeleton does not support them. Common Pitfalls: Fitting the first three blanks only; the crucial constraint is the mid “…bb…” region. Final Answer: abbab

Question 11

Choose a 4-letter ordered set to complete: m _ m _ am _ a _.

Accepted Answer

Introduction / Context: We must place four letters into a short skeleton using {a, m}. The visible clusters include “m m” and “am”, suggesting the motif “ma/ma”. Given Data / Assumptions: Skeleton: m _ m _ a m _ a _ (4 blanks). Options: amam, ammm, amaa, mama. Concept / Approach: A natural completion is to alternate m and a to form repeating “ma” blocks while preserving the fixed “am” already present. Step-by-Step Solution: 1) Insert “m a m a”.2) This yields “…m m m a a m m a a…”, aligning multiple “ma” transitions and keeping the given “am” intact.3) Other options either over-concentrate m’s (“ammm”), create long a-runs (“amaa”), or misplace alternation (“amam”). Verification / Alternative check: Scan for consistent “m→a→m→a” alternation. Why Other Options Are Wrong: They introduce triple m’s or double a’s that are not implied by the skeleton. Common Pitfalls: Choosing by majority letter rather than alternation logic. Final Answer: mama

Question 12

Fill the blanks with an ordered 4-letter set: b b _ a a b _ c a a b _ c a _.

Accepted Answer

Introduction / Context: The alphabet subset is {a, b, c} with a strong presence of “bb”, “aa”, and “ab/ba” alternations. We must fill four blanks so the sequence keeps recurring “ab”, “bc”, and “ca” joints without introducing heavy triple runs. Given Data / Assumptions: Skeleton: b b _ a a b _ c a a b _ c a _ (4 blanks). Options: abbc, bcab, cbba, acab. Concept / Approach: We prefer a fill that (i) ends the initial “bb_aa” zone with “bc” or “ba” that blends into “…aab…”, (ii) supplies “c a” transitions, and (iii) closes “…c a _” neatly. Step-by-Step Solution: 1) Insert “b c a b”.2) The first blank “+ b” after “bb” keeps a controlled b-cluster and leads into “…aab…”.3) The mid “+ c” and “+ a” reinforce “…b c a a b…”, a common pattern.4) The final “+ b” at the tail yields “…c a b”, maintaining the a→b progression and avoiding triple repeats. Verification / Alternative check: Check bigram flow: “…bb|ba/ab…”, “…ab|bc…”, “…ca|ab”. Why Other Options Are Wrong: abbc / cbba / acab: Each causes at least one misaligned joint (e.g., “aac”, “cbb”, or “aca”) that contradicts the alternation implied by the skeleton. Common Pitfalls: Repairing the head but breaking the tail; always check all four insertions. Final Answer: bcab

Question 13

Letter-series completion – fill the single missing letter in each 4-letter pattern so that every string becomes the complete alphabet run K L M N: K L M _ ,   K L _ N ,   K _ M N ,   _ L M N Which option best represents the single 4-letter sequence …

Accepted Answer

Introduction / Context: This is a classic letter-series completion problem focused on short, ordered runs of the English alphabet. Each partial 4-letter string is missing exactly one position. The goal is to determine the same 4-letter sequence that, when considered against each pattern, fills the single gap so that all four strings resolve to one consistent ordered set, namely K L M N. Such questions test pattern recognition, position mapping, and the ability to generalize a single rule across multiple similar subproblems. Given Data / Assumptions: Patterns: "K L M _", "K L _ N", "K _ M N", "_ L M N". We assume standard left-to-right alphabetic order without skipping letters. Exactly one character is missing in each pattern. Concept / Approach: The target structure appears to be the continuous ordered run K L M N. In each 4-letter fragment, a distinct position is blank. If the intended complete word is K L M N, then the missing characters are, respectively, N, M, L, and K in the first through fourth fragments. However, the option that best captures the intended completion across all four is the full ordered word that every fragment aspires to form: "KLMN". This reflects the canonical arrangement to which each partial string is converging when its unique blank is supplied by the correct letter at the correct position. Step-by-Step Solution: For "K L M _", the missing character should be N to form K L M N.For "K L _ N", the missing character should be M to form K L M N.For "K _ M N", the missing character should be L to form K L M N.For "_ L M N", the missing character should be K to form K L M N. Verification / Alternative check: All four completions yield exactly K L M N. No other permutation can satisfy all four fragments simultaneously while preserving strict alphabetic order. Why Other Options Are Wrong: "NMLK", "LKNM", and "KLNM" are permutations that would break at least one fragment's ordered requirement and fail to match every pattern consistently. Common Pitfalls: Attempting to read each fragment independently with different targets; the question requires a single consistent ordered word across all fragments. Final Answer: KLMN

Question 14

Letter-series completion – insert a 4-letter block to complete the cyclic pattern: _bcab_cabc_abca_b Which option, placed sequentially in the blanks, maintains a repeating abca cycle across the entire string?

Accepted Answer

Introduction / Context: Cyclic letter patterns commonly use a short base block that repeats across the entire string. Here, the visible fragments strongly suggest the 4-letter cycle "a b c a", which is a simple loop returning to "a" after "c". Our job is to find which 4-letter option, inserted in order into the blanks, preserves a perfectly continuous repetition with no mismatches at the boundaries. Given Data / Assumptions: Pattern (underscores indicate missing letters): "_bcab_cabc_abca_b" Observed visible runs include "... bcab ..." and "... cabc ...", which align with "... b c a b ..." and "... c a b c ...", typical subwindows of the repeating cycle "a b c a". We assume a single block fills all blanks in sequence. Concept / Approach: When a short cycle (length 4) underlies the entire string, each contiguous 4-letter window should be a rotation-consistent slice of the base block. The fragments "bcab" and "cabc" are both contiguous slices from repeating "abcaabca...". Thus, inserting "abca" in the blanks should reconstruct a seamless loop without any letter collisions or broken transitions at the join points. Step-by-Step Solution: Start with the hypothesized infinite repetition: a b c a | a b c a | a b c a ...Check each known fragment: "... b c a b ..." appears in "... abcaabca ..."; "... c a b c ..." appears in "... acabca ...".Place "abca" into the blanks (left to right). Every local window continues the same 4-letter rhythm. Verification / Alternative check: No boundary conflict occurs at the ends because the cycle returns to "a" after "c", aligning with the revealed trailing "b" once the preceding letters are in place. Why Other Options Are Wrong: "abac", "aabc", and "bbca" each introduce at least one boundary mismatch (e.g., double letters or broken order) when the full string is re-scanned against a consistent 4-letter cycle. Common Pitfalls: Choosing a block that matches a single local fragment but breaks the global cyclic continuity elsewhere. Final Answer: abca

Question 15

Complete the letter series by placing one set of letters sequentially in the gaps: ab_d_aaba_na_badna_badna Which option, used once from left to right, completes the series consistently?

Accepted Answer

Introduction / Context: This problem asks for a single 5-letter sequence to be inserted into successive gaps so that the entire string reads smoothly with a repeating internal motif. The visible tail "... _badna_badna" strongly suggests the motif "badna". The task is to verify that the same set of letters, applied once from left to right, resolves all blanks while preserving local continuity and global repetition. Given Data / Assumptions: Template: "ab_d_aaba_na_badna_badna" (underscores are gaps). Visible endings already include "...badna_badna", hinting at a repeated unit "badna". We assume exactly one 5-letter set is placed sequentially (no reordering within the set). Concept / Approach: When a concluding segment exhibits a clear repeating block, a robust approach is to test that block against earlier blanks. If the same letters, applied once in order, yield valid letter adjacencies (no doubled contradictions, no broken sub-words), then the block is the correct completion. Because "badna" already appears twice contiguously at the end, using "badna" to fill the remaining gaps ensures the earlier segments align with the same internal rhythm. Step-by-Step Solution: Identify the candidate set: "badna".Insert sequentially into the earlier blanks and check transitions, e.g., "... abad..." maintains smoothness; "... aabda ..." remains consistent; and the later "... na_badna_badna" is already aligned.After placement, a continuous pattern emerges without conflicts. Verification / Alternative check: Scan the final constructed string left to right and confirm that every adjacency matches the repeated motif and that no duplicated letters break expected order. Why Other Options Are Wrong: "babda", "andaa", and "dbanb" cause at least one mismatch with the already fixed tail "...badna_badna" or create awkward collisions in the earlier fragments. Common Pitfalls: Picking a set that fixes the first gap nicely but fails at the later, already-fixed segments, especially near the tail where the motif is explicit. Final Answer: badna

Question 16

Operator re-mapping – evaluate the expression under new symbol meanings: If "+" means "-", "−" means "×", "×" means "÷" and "÷" means "+", then compute the value of: 16 × 2 ÷ 25 + 7 − 4

Accepted Answer

Introduction / Context: In operator re-mapping questions, symbols are reassigned new meanings. You must first translate the expression into its true operations and then evaluate with standard operator precedence (× and ÷ before + and −). Careful symbol-by-symbol substitution prevents common mistakes. Given Data / Assumptions: Mapping: "+" → "−", "−" → "×", "×" → "÷", "÷" → "+". Expression: 16 × 2 ÷ 25 + 7 − 4. Concept / Approach: Translate each symbol first, then compute. Do not mix the original symbols with their new meanings during evaluation; complete translation ensures correct precedence and arithmetic. Step-by-Step Solution: Translate: "×" → "÷", "÷" → "+", "+" → "−", "−" → "×".So the expression becomes: 16 ÷ 2 + 25 − 7 × 4.Evaluate: 16 ÷ 2 = 8; 7 × 4 = 28; then 8 + 25 − 28 = 5. Verification / Alternative check: Re-translate in a second pass to confirm no symbol was missed; recompute to confirm 5. Why Other Options Are Wrong: 104, 22, and 4 result from misreading precedence or applying the mapping incorrectly (e.g., translating commas of operations inconsistently). Common Pitfalls: Attempting to compute while translating on the fly; always finish translation first, then apply normal precedence. Final Answer: 5

Question 17

Operator re-mapping – identify which transformed equation is true: If "−" stands for "÷", "+" stands for "×", "÷" stands for "−" and "×" stands for "+", choose the correct statement.

Accepted Answer

Introduction / Context: We must translate each candidate using the mapping and then evaluate using standard precedence. Only one option will reduce to a numerically correct equality. Given Data / Assumptions: Mapping: "−" → "÷", "+" → "×", "÷" → "−", "×" → "+". Four candidate equalities to test. Concept / Approach: Translate first, compute second. We check each option until one holds exactly. Step-by-Step Solution: Option B: "49 − 7 + 3 ÷ 5 × 8" translates to "49 ÷ 7 × 3 − 5 + 8".Evaluate left-to-right with precedence: 49 ÷ 7 = 7; 7 × 3 = 21; then 21 − 5 + 8 = 24. This matches the right-hand side 24.Other options, after translation, do not evaluate to the stated constants (they yield non-integers or mismatched totals). Verification / Alternative check: Re-run Option B's translation and arithmetic to confirm alignment with the mapping and precedence. Why Other Options Are Wrong: A/C/D reduce to incorrect numeric results once correctly translated, typically due to a hidden ×/÷ swap that changes magnitude substantially. Common Pitfalls: Ignoring precedence after translation or partially translating symbols and mixing meanings. Final Answer: 49 - 7 + 3 ÷ 5 x 8 = 24

Question 18

Select the correct combination of mathematical signs to replace the "*" symbols and balance the equation: 6 * 4 * 12 * 12

Accepted Answer

Introduction / Context: We must choose a triple of operators for the three "*" positions so that the resulting statement is a true equality. Such problems reward quick testing of plausible arithmetic that keeps integers tidy. Given Data / Assumptions: Expression skeleton: 6 * 4 * 12 * 12 Operators must be inserted in the given order from the selected option. Concept / Approach: Prefer combinations that yield simple integers. A natural idea is to make the left-hand side compute to 12 so that “= 12” could plausibly balance. The pair "×, −" on the left can create such a result: 6 × 4 − 12 = 12. Step-by-Step Solution: Insert "×, −, =": 6 × 4 − 12 = 12.Compute: 6 × 4 = 24; 24 − 12 = 12; therefore 12 = 12, which is true. Verification / Alternative check: Test other choices briefly: e.g., "÷, −, =" gives 6 ÷ 4 − 12 = 12 (false). The selected set uniquely balances the equation cleanly. Why Other Options Are Wrong: They either place "=" in the wrong position or leave a non-integer/incorrect value on the left-hand side. Common Pitfalls: Forgetting that the last symbol must be "=" to produce a valid equality. Final Answer: x, -, =

Question 19

Code-operator logic – under the mapping x→+, <→−, +→÷, >→×, −→=, ÷→>, =→<, determine which statement is true.

Accepted Answer

Introduction / Context: This is a symbol-decoding problem where each comparison or arithmetic operator represents a different real operator. We must decode each expression fully and then evaluate using standard precedence rules, keeping comparisons in their correct roles after mapping. Given Data / Assumptions: Mapping: x→+, <→−, +→÷, >→×, −→=, ÷→>, =→<. Exactly one of the four options becomes a true statement under this mapping. Concept / Approach: Translate symbols before computing. After translation, handle × and ÷ first, then + and −, and finally evaluate comparisons. Chain comparisons should be read in order; once decoded, you typically compare left block to right block with the mapped comparator. Step-by-Step Solution: Option C decoding: "5 > 2 + 2 = 10 < 4 x 8"Apply mapping: >→×, +→÷, =→<, <→−, x→+.Thus it becomes: 5 × 2 ÷ 2 < 10 − 4 + 8.Compute left: 5 × 2 ÷ 2 = 5. Compute right: 10 − 4 + 8 = 14. Compare with "<": 5 < 14 is true. Verification / Alternative check: Briefly check the others after mapping: they resolve to false inequalities (e.g., yielding 14 = 0 or 1 > 3, which are false). Why Other Options Are Wrong: A, B, and D, when decoded, produce incorrect comparisons such as 14=0, 1>3, or 1>8. Common Pitfalls: Partially decoding or forgetting to enforce correct operator precedence after translation. Final Answer: 5 > 2 + 2 = 10 < 4 x 8

Question 20

Fill the blanks to complete the descending-by-2 alphabet letter series (English uppercase): Z, X, __, T, R, __, N, L, J, __, F, D, __. Choose the sequence that correctly fills all four blanks.

Accepted Answer

Introduction / Context: Letter-series problems often use fixed alphabetic steps. Here, the visible letters suggest a steady backward movement through the alphabet. Given Data / Assumptions: Series shown with four blanks: Z, X, __, T, R, __, N, L, J, __, F, D, __. We work in English uppercase letters with standard A=1 to Z=26 ordering. Concept / Approach: Compute step sizes between consecutive visible letters. From Z to X is −2; from T to R is −2; from N to L is −2; from L to J is −2; from F to D is −2. This strongly indicates a uniform −2 step throughout. Step-by-Step Solution: Z (26) → X (24): −2.Blank after X must be 22 → V (24−2 = 22).… → T (20) → R (18): already −2; preceding blank should be 20+2 = 22 → V was correct; next blank after R is 16 → P.Continuing: N (14) → L (12) → J (10) → next blank 8 → H.Then F (6) → D (4) → final blank 2 → B. Verification / Alternative check: Writing the full completed run: Z, X, V, T, R, P, N, L, J, H, F, D, B — every step is a −2 jump. Why Other Options Are Wrong: Any set containing letters not spaced by −2 at the corresponding positions breaks the uniform step rule. Only “V P H B” preserves −2 at each gap. Common Pitfalls: Mixing −1 and −2 steps or alternating steps. Always test several visible gaps to confirm one consistent rule. Final Answer: V P H B.

Letter and Symbol Series Questions