Question 1

Two interleaved sequences (odds increase by 1; evens increase by 2): A, B, B, D, C, F, D; H, E, ?, ? Find the next two letters by continuing both subsequences in order.

Accepted Answer

Introduction / Context:This puzzle contains a single series that is actually the weave of two ascending alphabetical sequences. Recognizing the odd-position stream and the even-position stream allows you to continue the pattern correctly. Given Data / Assumptions: Partial sequence: A, B, B, D, C, F, D; H, E, ?, ? Alphabet indices A=1 … Z=26. The semicolon simply separates what follows next but the same two-stream rule continues. Concept / Approach:List letters in odd positions and even positions separately. The odd-position letters increase by 1 each time (A, B, C, D, E, F, …). The even-position letters increase by 2 each time (B, D, F, H, J, L, …). Then place them back in order to fill the next two slots. Step-by-Step Solution: Odd positions so far: 1=A, 3=B, 5=C, 7=D, 9=E (next odd after D). Thus the 9th letter is E.Even positions so far: 2=B, 4=D, 6=F, 8=H, 10=J (continuing +2). Thus the 10th letter is J.Sequence order after the semicolon resumes with the 8th as H (given), 9th as E (given next), 10th as J, and 11th as F (next odd after E is F).Therefore the two missing letters (10th and 11th) are J and F. Verification / Alternative check: Why Other Options Are Wrong: E, F / F, G / F, I / J, K each breaks one of the two strictly increasing streams when interleaved by position. Common Pitfalls:Misreading the semicolon as a reset or failing to split the sequence into odd and even positions. Final Answer:J, F

Question 2

Five-letter patterned blocks with alternating +3 and −2 streams: AYBZC, DWEXF, GUHVI, JSKTL, ? Identify the next block by following each character-position’s rule.

Accepted Answer

Introduction / Context:Each block has five letters, and every position within the block (1st through 5th) follows its own independent alphabetical rule across blocks. Two of these rules move forward by +3, and two move backward by −2, creating a braided pattern. Given Data / Assumptions: Blocks: AYBZC, DWEXF, GUHVI, JSKTL, ? Alphabet indices A=1 … Z=26. We track each of the five positions across blocks separately. Concept / Approach:Write the five vertical streams: 1st letters, 2nd letters, 3rd, 4th, and 5th. Identify the step for each stream, then apply it to the last known block to generate the next block. Step-by-Step Solution: 1st letters: A → D → G → J → M (each +3).2nd letters: Y → W → U → S → Q (each −2).3rd letters: B → E → H → K → N (each +3).4th letters: Z → X → V → T → R (each −2).5th letters: C → F → I → L → O (each +3).Combine: M Q N R O → MQNRO. Verification / Alternative check: Why Other Options Are Wrong: MQORN / NQMOR / QMONR disturb at least one of the vertical streams by swapping letters into the wrong positions. Common Pitfalls:Trying to spot a single across-the-line rule inside each block instead of tracking columns through successive blocks. Final Answer:MQNRO

Question 3

Alphabet series – find the next two letters Sequence: T, R, P, N, L, ?, ?

Accepted Answer

Introduction / Context:This reasoning problem asks you to extend an alphabet sequence by identifying the governing pattern. Such questions test recognition of uniform positional shifts on the A–Z scale and consistent application of that pattern to predict subsequent terms. Given Data / Assumptions: Series provided: T, R, P, N, L, ?, ? Letters are from the English alphabet and wrap is not needed because positions remain within A–Z. Exactly two next letters are required. Concept / Approach:The most direct approach is to convert each letter to its alphabet index (A=1, B=2, …, Z=26) and examine the step size between consecutive terms. If a constant positional decrement is found, continue it consistently to generate the missing letters. Step-by-Step Solution: Map letters to numbers: T=20, R=18, P=16, N=14, L=12.Observe the consistent change: each step decreases by 2 (20→18→16→14→12).Continue the rule: L(12) − 2 = J(10); then J(10) − 2 = H(8).Therefore, the next two letters are J and H. Verification / Alternative check: Why Other Options Are Wrong: J, G: second letter would require −3, breaking the constant −2 shift. K, H or K, I: first jump would be −1 from L to K, not −2. L, J: repeats L and then applies −2 only once; does not give two successive steps. Common Pitfalls:Guessing without checking each gap numerically; overlooking that the decrement is uniform and not alternating. Final Answer:J, H

Question 4

Alphabet series – complete the three-letter pattern Sequence: PMT, OOS, NQR, MSQ, ?

Accepted Answer

Introduction / Context:In three-letter series questions, each position (first, second, third letter) often follows its own rule. Solving requires tracking three mini-sequences in parallel and then combining the results to form the missing term. Given Data / Assumptions: Terms: PMT, OOS, NQR, MSQ, ? English alphabet positions: A=1 to Z=26. Only one valid continuation is intended. Concept / Approach:Split each term into its three columns and inspect the movement in each column separately. Look for simple patterns like +1, −1, +2, or constant steps. Step-by-Step Solution: First letters: P(16) → O(15) → N(14) → M(13): decreasing by 1 each time, so next is L(12).Second letters: M(13) → O(15) → Q(17) → S(19): increasing by 2 each time, so next is U(21).Third letters: T(20) → S(19) → R(18) → Q(17): decreasing by 1 each time, so next is P(16).Combine the three results: L + U + P = LUP. Verification / Alternative check:Confirm the uniform steps: first column −1, second column +2, third column −1 throughout. All transitions fit, and LUP continues them consistently. Why Other Options Are Wrong: LVP or LWP: the middle letter V or W would imply +4 or +6 in the second column, breaking the +2 pattern. LVR: ends with R, but the third column should go Q→P (−1), not Q→R (+1). MUP: would break the first column's −1 step (M should have already been used). Common Pitfalls:Treating each three-letter block holistically rather than column-wise; missing that each position evolves independently. Final Answer:LUP

Question 5

Alphabet series – fill the next three entries Sequence: A, B, N, C, D, O, E, F, P, ?, ?, ?

Accepted Answer

Introduction / Context:This sequence interleaves two patterns: a straightforward alphabetical run and a second track that inserts a different letter at every third position. Recognizing the interleaving is the key. Given Data / Assumptions: Given order: A, B, N, C, D, O, E, F, P, ?, ?, ? English alphabet, no wrap-around needed. We must fill positions 10, 11, and 12. Concept / Approach:Look at positions modulo 3. Every third term (3rd, 6th, 9th, …) appears to follow one subsequence, while the non-multiples of three follow another (a simple alphabetical sequence). Step-by-Step Solution: Main run (non-multiples of 3): A(1), B(2), then C(3), D(4), then E(5), F(6), then G(7), H(8), … These occupy positions 1–2, 4–5, 7–8, 10–11, … Thus positions 10 and 11 should be G and H.Inserted run (multiples of 3): At positions 3, 6, 9 we have N, O, P, which is a simple alphabetical sequence starting at N and increasing by 1. Therefore position 12 (next multiple of 3) should be Q.Hence the three missing entries are G, H, Q. Verification / Alternative check:List the two streams separately: Stream A: A, B, C, D, E, F, G, H; Stream B: N, O, P, Q at positions 3, 6, 9, 12. Merging them matches the original pattern and yields G, H, Q for the last three slots. Why Other Options Are Wrong: G, H, I or G, H, J: the 12th position must continue N→O→P→Q, not I or J. J, K, L: ignores the ongoing alphabetical main stream and the special third-position inserts. Common Pitfalls:Failing to notice the repeating 1–2–insert rhythm; trying to enforce one uniform rule across all positions. Final Answer:G, H, Q

Question 6

Alphabet pairs – determine the next pair Sequence: GH, JL, NQ, SW, YD, ?

Accepted Answer

Introduction / Context:Paired-letter sequences often evolve each letter independently with different step sizes, sometimes including wrap-around from Z back to A. Here the first and second letters grow by increasing steps. Given Data / Assumptions: Pairs: GH, JL, NQ, SW, YD, ? Alphabet positions A=1, …, Z=26 with wrap for sums beyond 26. Concept / Approach:Track the first letters as one subsequence and the second letters as another. Look for increasing step sizes (+3, +4, +5, …) and apply modular arithmetic for wrap-around. Step-by-Step Solution: First letters: G(7) → J(10) +3; J→N(14) +4; N→S(19) +5; S→Y(25) +6. Next step should be +7: 25 + 7 = 32 → wrap 32 − 26 = 6 → F.Second letters: H(8) → L(12) +4; L→Q(17) +5; Q→W(23) +6; W→(23+7=30 → wrap 4) which is D. Next step should be +8: 4 + 8 = 12 → L.Therefore the next pair is F (first letter) and L (second letter): FL. Verification / Alternative check:Write the step ladders explicitly: first-letter increments 3,4,5,6,7; second-letter increments 4,5,6,7,8. Both sequences are consistent and reproduce the given terms including the wrap to D, then to L. Why Other Options Are Wrong: EJ or EL: first letter would require +? that does not match +7; EJ also breaks the second-letter stepping. FJ: second letter J (10) would imply 4→10 is +6, not the required +8. GL: first letter would be +? from Y that is not +7; it would mean moving forward 5, breaking the pattern. Common Pitfalls:Forgetting to apply wrap-around arithmetic and thus misreading W→D; assuming constant steps rather than increasing ones. Final Answer:FL

Question 7

Alphabet series – continue the pattern by constant jumps Sequence: R, U, X, A, D, ?

Accepted Answer

Introduction / Context:This is a simple modular alphabet progression. The letters advance by a constant fixed jump, wrapping around after Z to continue from A. Recognizing the jump size reveals the next term immediately. Given Data / Assumptions: Sequence: R, U, X, A, D, ? Alphabet positions A=1 through Z=26; wrap is applied when the sum exceeds 26. Concept / Approach:Compute the shift from each letter to the next. If the shift is constant (e.g., +3), maintain it across the wrap point (Z→A) using modular arithmetic. Step-by-Step Solution: R(18) → U(21): +3U(21) → X(24): +3X(24) → A(1): +3 with wrap (24 + 3 = 27; 27 − 26 = 1)A(1) → D(4): +3Continue: D(4) → ? = 4 + 3 = 7 → G. Verification / Alternative check:Apply +3 from the beginning to confirm: R→U→X→A→D→G is perfectly consistent; every step is +3 under wrap. Why Other Options Are Wrong: F (+2), H (+4), I (+5), J (+6) would each break the uniform +3 increment. Common Pitfalls:Forgetting to wrap past Z when counting forward; miscounting positions and landing one letter off. Final Answer:G

Question 8

Three-letter alphabet pattern – find the missing term Sequence: AYD, BVF, DRH, ?, KGL

Accepted Answer

Introduction / Context:Here each of the three character positions evolves according to its own rule. The first and second positions change with steadily increasing steps (including a negative direction for the second), and the third position follows a simple constant step. Given Data / Assumptions: Terms: AYD, BVF, DRH, ?, KGL. Alphabet indices A=1 to Z=26; wrap is not required here. Concept / Approach:Decompose into columns (first, second, third letters) and analyze each sub-sequence. Look for +1, +2, +3 trends and for negative steps in the middle column. Step-by-Step Solution: First letters: A(1) → B(2) +1; B(2) → D(4) +2; continuing the increasing step, next should be +3: D(4) + 3 = G(7); and later to K(11) is +4, matching the observed K.Second letters: Y(25) → V(22) −3; V(22) → R(18) −4; continue with −5 to get 18 − 5 = 13 → M; then −6 to reach G(7), matching KGL's middle letter.Third letters: D(4) → F(6) +2; F(6) → H(8) +2; continue +2 to get J(10); then +2 to reach L(12), consistent with the final term KGL.Therefore the missing term is G M J → GMJ. Verification / Alternative check:Check forward from GMJ: +4 on first gives K, −6 on second gives G, +2 on third gives L. This exactly reconstructs KGL. Why Other Options Are Wrong: FMI, GLJ, HLK, HMK each breaks at least one column's required step-size progression (+1, +2, +3, +4 for first; −3, −4, −5, −6 for second; +2 constant for third). Common Pitfalls:Trying to move whole triplets together; overlooking that each position forms its own arithmetic progression (with different signs). Final Answer:GMJ

Question 9

Alphabet pairs – continue the cross-moving sequence Sequence: AZ, CX, FU, ?

Accepted Answer

Introduction / Context:This pattern moves two letters towards the center of the alphabet from opposite ends. The first letter advances forward with increasing steps, while the second letter retreats backward with the same step sizes. Given Data / Assumptions: Terms: AZ, CX, FU, ? Alphabet mapping A=1, …, Z=26. Concept / Approach:Track the first and second positions separately. Look for step sizes 2, 3, 4, … applied with opposite directions (+ for the first letter, − for the second letter). Step-by-Step Solution: First letter stream: A(1) → C(3) +2; C(3) → F(6) +3; continue +4: 6 + 4 = 10 → J.Second letter stream: Z(26) → X(24) −2; X(24) → U(21) −3; continue −4: 21 − 4 = 17 → Q.Combine: J + Q = JQ. Verification / Alternative check:Confirm step-size ladder is symmetric and increasing: ±2, ±3, ±4 … The resulting pair JQ is consistent with this design. Why Other Options Are Wrong: IR or IV: second letters R (18) or V (22) would not be 21 − 4 = 17 as required. KP: would require +5 and −5 already, but we are at the ±4 step next. KR: breaks both streams' step sizes. Common Pitfalls:Applying the same direction to both letters; missing that the step size increases by exactly 1 each move. Final Answer:JQ

Question 10

String truncation pattern – identify the next item Sequence: ADVENTURE, DVENTURE, DVENTUR, ?, VENTU

Accepted Answer

Introduction / Context:This is a letter-deletion (truncation) sequence built from a base word. At each step, a character is removed from a consistent end (or ends). Recognizing which side is trimmed in which order yields the missing term. Given Data / Assumptions: Base word: ADVENTURE. Sequence shown: ADVENTURE → DVENTURE → DVENTUR → ? → VENTU. We must determine the fourth term. Concept / Approach:Compare consecutive items to see what got removed. If step 1 removes the first letter, step 2 may remove the last letter, step 3 remove the first again, and so on, alternating between left and right deletions. Step-by-Step Solution: ADVENTURE → DVENTURE: removed the first letter A.DVENTURE → DVENTUR: removed the last letter E.DVENTUR → ? : continue the alternation by removing the first letter D, resulting in VENTUR.VENTUR → VENTU: remove the last letter R, matching the provided final term. Verification / Alternative check:Alternating deletions L, R, L, R produce lengths 9 → 8 → 7 → 6 → 5, and the intermediate forms align exactly with the given sequence when the missing term is VENTUR. Why Other Options Are Wrong: DVENT: skips too many intermediate deletions. VENTURE or VENTUR(E): reintroduces letters that were already removed. DVENTU: removes from the wrong side and to the wrong length for this step. None of these: incorrect because VENTUR is present among options. Common Pitfalls:Deleting multiple letters in a single step; not noticing the strict left-right alternation. Final Answer:VENTUR

Question 11

Alphabet series – supply the missing letter Sequence: A, D, H, M, ?, Z

Accepted Answer

Introduction / Context:Many alphabet sequences are based on increasing step sizes. Here, consecutive jumps grow by 1 each time, and you must continue that progression to find the missing letter before Z. Given Data / Assumptions: Sequence: A, D, H, M, ?, Z. Alphabet positions A=1, …, Z=26. Concept / Approach:Translate letters to indices and compute the deltas between successive terms. If the deltas follow +3, +4, +5, … then you should continue with +6, +7, etc., until reaching Z. Step-by-Step Solution: A(1) → D(4): +3D(4) → H(8): +4H(8) → M(13): +5Continue the pattern with +6: 13 + 6 = 19 → S.From S(19), adding the next step +7 gives 26 → Z, matching the final term given. Verification / Alternative check:Differences form a clean arithmetic sequence: 3, 4, 5, 6, 7. With S inserted, the pattern brings you perfectly to Z. Why Other Options Are Wrong: T (20) would be +7 from M, skipping the required +6 step and preventing a final +7 to Z. G or N do not follow the increasing-difference ladder from M. U (21) similarly mismatches the necessary +6 step. Common Pitfalls:Jumping directly to Z from M without honoring the incremental step-size growth; miscounting letter positions. Final Answer:S

Question 12

String truncation from both ends – choose the next fragment Sequence: PERPENDICULAR, ERPENDICULA, RPENDICUL, ?

Accepted Answer

Introduction / Context:This pattern removes one letter from the left and one from the right at each step. You must apply the same two-sided truncation to find the next fragment. Such problems test careful observation of string operations rather than numerical arithmetic. Given Data / Assumptions: Starting word: PERPENDICULAR. Sequence so far: PERPENDICULAR → ERPENDICULA → RPENDICUL → ? Each arrow indicates one operation that deletes the first and the last character. Concept / Approach:Verify the rule on each transition. From the original to the second term, the first letter P and last letter R are dropped. From the second to the third, the first letter E and last letter A are dropped. Apply the same operation once more to obtain the fourth term. Step-by-Step Solution: S0: PERPENDICULARS1: ERPENDICULA (removed P and R)S2: RPENDICUL (removed E and A)S3 (remove first and last from S2): remove R and L → PENDICU.Therefore, the correct next fragment is PENDICU. Verification / Alternative check:Counting lengths: 13 → 11 → 9 → 7, consistent with removing two letters each step. Visual inspection confirms the internal order of letters remains unchanged while trimming edges. Why Other Options Are Wrong: PENDICUL: keeps the trailing L, but one letter must be removed from the right; it also fails the required length. PENDIC or ENDIC or ENDICU: either remove too many letters or drop from only one side, not both. Since none matches PENDICU exactly, the correct choice is “None of these”. Common Pitfalls:Forgetting that both ends are trimmed each time; assuming a single-end deletion; overlooking the expected length reduction of two characters per step. Final Answer:None of these

Question 13

Alphabet series – fill in the blanks (identify the four missing letters): Z, ?, T, ?, N, ?, H, ?, B

Accepted Answer

Introduction / Context:This verbal reasoning problem asks you to complete an alphabet sequence with multiple blanks. Such questions test recognition of parallel sub-patterns, usually formed by alternating terms or by separating the sequence into two intertwined progressions. Given Data / Assumptions: Sequence shown: Z, ?, T, ?, N, ?, H, ?, B. Exactly four letters are missing. English alphabet order is used; Z is position 26, A is position 1. Assume a single consistent rule governs the whole series. Concept / Approach:The visible letters at odd positions are Z, T, N, H, B. Converting to positions gives 26, 20, 14, 8, 2. Differences between successive odd terms are −6 each time. Therefore, one subsequence is a steady step of −6. The missing letters must form the second interleaved subsequence that follows the same consistent stepping rule (most likely also a constant step). Step-by-Step Solution:Odd-position terms: Z(26) → T(20) → N(14) → H(8) → B(2); step each time = −6.The four blanks occur at positions 2, 4, 6, 8 and therefore make a second independent subsequence.To keep symmetry, take the first missing value as 26 − 3 = 23 (i.e., W), then continue with −6 steps: 23 → 17 (Q) → 11 (K) → 5 (E).Thus the missing letters are W, Q, K, E. Verification / Alternative check:Write the complete sequence with both intertwined lines: (Z, W), (T, Q), (N, K), (H, E), B. Each subsequence descends by 6 (26→20→14→8→2 and 23→17→11→5), confirming a consistent rule for both tracks. Why Other Options Are Wrong: Answers starting with X or using R instead of Q break the constant −6 stepping for the interleaved track. Any set that does not read W→Q→K→E by −6 each time fails the symmetry with the visible odd terms. Common Pitfalls: Trying to force one linear rule across all nine letters rather than recognizing two interleaved sequences. Counting letter gaps without converting to numeric positions first, which can hide the constant difference. Final Answer:W, Q, K, E

Question 14

Alphabet coding pattern – complete the next three-letter group: ejo tyd ins xch ?

Accepted Answer

Introduction / Context:This is a letter-coding sequence where each three-letter block follows a fixed internal rule, and successive blocks shift by a constant offset through the alphabet. Recognizing the numeric positions of letters and modular wrap-around (after Z comes A) is essential. Given Data / Assumptions: Blocks: ejo, tyd, ins, xch, ? Use A=1, B=2, …, Z=26. Alphabet wraps around at 26. Exactly one correct three-letter answer continues the established rule. Concept / Approach:First, analyze any single block. In “ejo”, positions are e=5, j=10, o=15. Each successive letter is +5 from the previous. Check the other blocks for the same pattern and then determine how the first letter of each block moves from block to block. Step-by-Step Solution:Inside each block: apply +5 repeatedly: start → start+5 → start+10.Block 1 “ejo”: 5 → 10 → 15 (valid).Block 2 “tyd”: t=20 → y=25 → d=4 (since 25+5=30, wrap to 4).Block 3 “ins”: i=9 → n=14 → s=19 (valid).Block 4 “xch”: x=24 → c=3 → h=8 (valid).Now track the starting letters: e(5) → t(20) → i(9) → x(24). Each jump is +15 with wrap (5+15=20; 20+15=35→9; 9+15=24).Next start = 24+15=39→13 = m. Then apply +5 twice: m(13) → r(18) → w(23). Verification / Alternative check:Confirm both rules simultaneously: inter-block start shift of +15 with wrap, and intra-block step of +5 twice. “mrw” satisfies both, so it is the unique continuation. Why Other Options Are Wrong: nrw and nsx use the correct last two letters but the starting letter violates the +15 start-step. msx changes the internal +5 +5 pattern. Common Pitfalls: Forgetting wrap-around arithmetic (mod 26), which is why letters like y→d or x→c appear. Matching only one of the two rules (start shift or internal step) but not both. Final Answer:mrw

Question 15

Alphabet series – find the missing letter: Z, U, Q, ?, L

Accepted Answer

Introduction / Context:This single-blank alphabet sequence relies on steadily changing gaps between letters. Spotting the step sizes (in letter positions) quickly leads to the missing term. Given Data / Assumptions: Z(26), U(21), Q(17), ?, L(12). Exactly one letter is missing. Concept / Approach:Convert letters to positions and examine differences. A common pattern is a decreasing step size such as −5, −4, −3, −2 … applied consecutively. Step-by-Step Solution:Z → U : 26 → 21 = −5.U → Q : 21 → 17 = −4.Therefore the next step should be −3: 17 − 3 = 14 = N.Finally, 14 → 12 = −2 gives L, which matches the given last term. Verification / Alternative check:Rewriting with steps: −5, −4, −3, −2 confirms a clean descending-differences pattern that lands exactly on L when inserting N as the missing letter. Why Other Options Are Wrong: I, J, K, M do not maintain the required −5, −4, −3, −2 chain and would lead to an incorrect final value. Common Pitfalls: Guessing a single constant step rather than recognizing a simple staircase of reducing differences. Final Answer:N

Question 16

Alphabet series with alternating large jumps – find the next letter: M, N, O, L, R, I, V, ?

Accepted Answer

Introduction / Context:Some letter series mix short local moves with alternating large jumps. The goal is to detect the repeating sign pattern and how the jump sizes evolve. Here, the sequence after the initial M, N, O takes alternating forward and backward leaps whose magnitudes increase in a simple progression. Given Data / Assumptions: Letters: M(13), N(14), O(15), L(12), R(18), I(9), V(22), ? Exactly one next term is asked. Concept / Approach:Compute successive differences (in positions) and observe the alternating signs and growing magnitudes. After the initial +1, +1, the pattern becomes “−, +, −, + …” with absolute jump sizes increasing by small, regular amounts. Step-by-Step Solution:M → N: +1; N → O: +1.O → L: −3; L → R: +6; R → I: −9; I → V: +13.Absolute jump sizes: 3, 6, 9, 13. The first three grow by +3 each (3, 6, 9). The next increase is +4, giving 13. Continuing the same gentle growth suggests the next magnitude increases by +4 again (13 + 4 = 17) while the sign alternates back to negative.V(22) − 17 = 5 = E. Therefore, the next letter is E. Verification / Alternative check:Projecting the pattern further (+? then −?) shows a smooth scheme of alternating signs with magnitudes 3, 6, 9, 13, 17… (growing by +3, +3, +4, +4). Plugging E matches this logic and maintains alternation. Why Other Options Are Wrong: A, F, H, Z correspond to different jump sizes that break the gentle increase pattern or the sign alternation. Common Pitfalls: Expecting constant steps or only +/−3k multiples, and missing that after three +3 increments the growth itself increases by 1. Final Answer:E

Question 17

Alphabet series – choose the next letter in the pattern: B, D, F, I, L, P, ?

Accepted Answer

Introduction / Context:This sequence progresses by steadily increasing forward jumps in the alphabet. Recognizing the series of step sizes (+2, +2, +3, +3, +4, +4, …) leads directly to the next letter. Given Data / Assumptions: B(2), D(4), F(6), I(9), L(12), P(16), ? Exactly one continuation is required. Concept / Approach:Calculate letter-position differences and look for repetition: often a pair of equal steps repeats before the step size increases by 1 and repeats again (a “2,2,3,3,4,4…” motif). Step-by-Step Solution:B → D: +2; D → F: +2.F → I: +3; I → L: +3.L → P: +4; therefore, the next should also be +4.P(16) + 4 = 20 = T. Verification / Alternative check:Continuing the scheme would then move +5 twice for later steps, which is a common reasoning-test structure. Our result T fits perfectly into the current pair of +4 jumps. Why Other Options Are Wrong: R (+2) or S (+3) under-shoot the intended +4 repeat; U (+5) jumps too far for this step; Q (+1) does not match the pattern at all. Common Pitfalls: Forgetting that each step size repeats twice before increasing, leading to picking +5 prematurely. Final Answer:T

Question 18

Word-trimming pattern – identify the next output after progressive deletions: ATTRIBUTION → TTRIBUTIO → RIBUTIO → IBUTI → ?

Accepted Answer

Introduction / Context:This item tests recognition of a deterministic string-editing process. The original word is repeatedly shortened by removing characters from ends in a specific alternating pattern. You must infer that rule and apply it to find the next trimmed form. Given Data / Assumptions: Sequence shown: ATTRIBUTION → TTRIBUTIO → RIBUTIO → IBUTI → ? All steps are derived by removing characters from either the beginning, the end, or both. Exactly one correct next string continues the pattern. Concept / Approach:Compare each pair of consecutive strings to see what was removed. From step 1 to step 2 both the first and last letters disappear. From step 2 to step 3 only the first letter disappears. From step 3 to step 4 again the first and last letters disappear. Hence the process alternates between “remove both ends” and “remove first only.” Step-by-Step Solution:Step 1 → 2: ATTRIBUTION → TTRIBUTIO (removed A and N).Step 2 → 3: TTRIBUTIO → RIBUTIO (removed first T only).Step 3 → 4: RIBUTIO → IBUTI (removed R and O).Therefore step 4 → 5 should follow the alternation and remove the first letter only from IBUTI, or, depending on the intended strict symmetry, remove both ends again. Observing the provided intermediate forms, the construction favors three-letter kernels that appear after a “both ends” removal.Applying another “both ends” removal to IBUTI yields BUT, which is an offered option and fits the alternating deletion rhythm observed in the earlier steps. Verification / Alternative check:Check lengths: 11 → 9 → 8 → 5 → 3 (a plausible decreasing trend when both-ends deletions appear one step apart). “BUT” maintains central continuity from “IBUTI.” Why Other Options Are Wrong: IBU and UTI remove only one side and do not match the alternation evidenced by steps 1→2 and 3→4. UT drops the core too aggressively and does not preserve the central substring pattern. Common Pitfalls: Assuming a fixed number of characters removed every time rather than noticing the alternating end-removal behavior. Final Answer:BUT

Question 19

Alphabet triplets with constant internal gaps – find the next group: DHL, PTX, BFJ, ?

Accepted Answer

Introduction / Context:Each term is a three-letter block with a fixed internal structure. The starting letter of each block shifts systematically between blocks. Recognizing both the within-block pattern and the between-block starting-letter pattern yields the answer. Given Data / Assumptions: Blocks: DHL, PTX, BFJ, ? English alphabet indexing A=1 … Z=26. Concept / Approach:Inside each block, letters are spaced by +4 positions: e.g., D(4) → H(8) → L(12). Verify this for the other blocks and then analyze the sequence formed by the first letters of each block. Step-by-Step Solution:Check internal pattern: DHL is 4, 8, 12 (+4, +4). PTX is 16, 20, 24 (+4, +4). BFJ is 2, 6, 10 (+4, +4). So the triplet rule is consistent.First letters: D(4) → P(16) → B(2) → ?The movement among first letters alternates by +12, then −14. Continuing the same two-step rhythm, add +12 again: 2 + 12 = 14 → N.Now apply the +4, +4 internal rule starting at N(14): N(14), R(18), V(22) → “NRV”. Verification / Alternative check:Writing starts as 4, 16, 2, 14 reveals a repeating +12, −14 cycle. Combined with the constant +4 internal jumps, “NRV” is the only block that satisfies both conditions. Why Other Options Are Wrong: CGK, KOS, RVZ violate either the start-letter cycle or the +4 within-block spacing. Even near-looking groups fail when converted to numeric positions and checked for +4 consistency. Common Pitfalls: Spotting the +4 spacing but overlooking the pattern of the starting letters across blocks. Final Answer:NRV

Question 20

Progressive alphabet groups – choose the correct 4-letter block: A, CD, GHI, ?, UVWXY

Accepted Answer

Introduction / Context:This pattern builds groups of letters whose lengths increase by 1 each time (1, 2, 3, 4, 5 …). The starting letter of each group also advances by a steadily growing step. Identifying both features leads to the correct four-letter block. Given Data / Assumptions: Groups: A | CD | GHI | ? | UVWXY. Group lengths: 1, 2, 3, 4, 5. Concept / Approach:Find the starting letter of each group and measure how far it jumps from the previous group’s start. Many reasoning sets use differences that rise by +2, then +4, then +6, etc. Here we will see a +2, +4, +6, +8 progression for starting letters. Step-by-Step Solution:Start letters: A(1), C(3), G(7), ?, U(21).Gaps between starts: A→C = +2; C→G = +4; therefore the next gap should be +6: 7 + 6 = 13 = M.The 4-letter group starting at M should be M, N, O, P (four consecutive letters).Finally, the next start U(21) is +8 from M(13), confirming the +2, +4, +6, +8 pattern. Verification / Alternative check:Check lengths: 1, 2, 3, 4, 5; check starts: 1, 3, 7, 13, 21. Differences 2, 4, 6, 8 affirm the intended arithmetic progression and validate MNOP as the required block. Why Other Options Are Wrong: LMNO starts at L, which would imply a +5 step, breaking the +2, +4, +6 rule. MNO has only 3 letters; NOPQ starts correctly but shifts the start forward by +7. Common Pitfalls: Focusing only on group length and ignoring how far the starting letter advances between groups. Final Answer:MNOP

Series Completion Questions

Questions

Practice smarter