Q&A: practice questions with answers

Concept: Each letter of the first word is moved back by a fixed pattern of places.

1. ERHT → BLEN: E−3=B, R−6=L, H−3=E, T−6=N. Pattern: −3, −6, −3, −6.
2. The pair XSNR → UMKL obeys the same rule ✔.
3. DIOQ → D(4)−3=A(1), I(9)−6=C(3), O(15)−3=L(12), Q(17)−6=K(11).

Trick: Find only the first letter: D−3 = A. Only option (D) starts with A.

Answer: (D) ACLK

Concept: Sort the letters A→Z, pick the two positions, then count the alphabet letters strictly between them.

1. KNIFERS in order: E F I K N R S.
2. 2nd from left = F (6); 3rd from right = N (14).
3. Letters between = 14 − 6 − 1 = 7 (G up to M).

Trick: Letters strictly between two letters = difference of positions − 1.

Answer: (A) 7

Concept: Swap the signs first, then solve with BODMAS.

1. Swap: $+\leftrightarrow ÷$ and $-\leftrightarrow \times$.
2. $9-4\times 58+2÷22$ becomes $9\times 4-58÷2+22$.
3. $=36-29+22=29$.

Trick: Rewrite the whole expression once with new signs before calculating — most mistakes come from swapping in your head.

Answer: (A) 29

Concept: Treat 1st, 2nd and 3rd letters of each group as three separate series.

1. 1st letters: R, T, V, X → +2 → Z.
2. 2nd letters: F, H, J, L → +2 → N.
3. 3rd letters: G, I, K, M → +2 → O. Next group = ZNO.

Trick: In every group the last two letters are consecutive (FG, HI…). So after N must come O — only ZNO fits.

Answer: (D) ZNO

Concept: Decode each symbol and move from T back to P.

1. S ÷ T: S is father of T.
2. R × S: R is wife of S → R is T's mother.
3. Q − R: Q is brother of R → Q is T's maternal uncle (mother's brother).
4. P + Q: P is daughter of Q → P is T's mother's brother's daughter.

Trick: Read the chain from the end (T) towards P; each link adds one word to the relation.

Answer: (B) Mother's brother's daughter

Concept: Find the gap pattern between consecutive letters of each cluster.

1. CEF: +2, +1; RTU: +2, +1; GIJ: +2, +1.
2. UWY: U→W +2, W→Y +2 (not +1).
3. So UWY breaks the pattern.

Trick: In three clusters the last two letters are neighbours (EF, TU, IJ). Not so in WY.

Answer: (B) UWY

Concept: Decide which pairs can share members and which cannot.

1. Sister and Brother: no person is both → two separate circles.
2. Friend: some friends are sisters, some are brothers → Friend overlaps both.
3. Required figure: one circle cutting two separate circles — option (C).

Trick: Gender-based classes (sister/brother) never overlap; a role like 'friend' crosses both.

Answer: (C)The figure where Friend overlaps both separate circles

Concept: Check the differences between terms.

1. Differences: 1, 3, 5, 7 (odd numbers).
2. Next difference = 9.
3. $895+9=904$.

Trick: Differences 1, 3, 5, 7 ⇒ terms are $879+n^2$ (0, 1, 4, 9, 16, 25) → $879+25=904$.

Answer: (C) 904

Concept: A face touches four faces; the sixth one is its opposite.

1. View 1: W is beside U and V.
2. View 2: W is beside Y and Z.
3. W touches U, V, Y, Z → the remaining letter X is opposite W.

Trick: Common face in both views → list its four neighbours; the leftover letter is opposite.

Answer: (A) W

Concept: A word appearing in two sentences gets the code appearing in both codes.

1. Common word in 'here I come' and 'I am going' = I.
2. Common code in 'Is rc tp' and 'rc as bo' = rc.
3. So I = rc.

Trick: Don't be fooled by 'Is' looking like 'I' — codes need not resemble words.

Answer: (A) rc

Concept: Count the triangles and see which one disappears each step.

1. Triangles: 9 → 8 → 7, so next figure has 6.
2. Right and middle columns stay the same; the left column loses its lowest triangle each time.
3. Only the row-2 left triangle remains; it disappears next — figure (A).

Trick: First count shapes (9, 8, 7 → 6) and reject options with a wrong count.

Answer: (A)Figure (a)

Concept: Find the shift for each position and apply it to the new word.

1. NSBX → QQZA: N+3=Q, S−2=Q, B−2=Z, X+3=A (wrap around).
2. OKIR → RIGU confirms +3, −2, −2, +3.
3. DTAF → D+3=G, T−2=R, A−2=Y, F+3=I = GRYI.

Trick: Use wrap-around: after Z comes A, before A comes Z.

Answer: (D) GRYI

Concept: Two 'some' statements in a chain never give a definite link between the end classes.

1. Actors are inside Writers; Writers–Dancers and Dancers–Stars are only partial overlaps.
2. Dancers may overlap only the non-actor part of Writers → 'Some dancers are actors' not certain.
3. Stars link only to Dancers → 'Some stars are actors' not certain.

Trick: Some + Some = no conclusion. Draw the possible diagram where conclusions fail.

Answer: (B) Neither conclusion (I) nor (II) follows.

Concept: Split into 4-letter groups; each position moves +3 from group to group.

1. Groups: _GCE, G_FH, JM_K, MPL_, PSO_.
2. 1st letters: D, G, J, M, P; 2nd: G, J, M, P, S; 3rd: C, F, I, L, O; 4th: E, H, K, N, Q.
3. Blanks: D, J, I, N, Q → DJINQ.

Trick: First blank must be G−3 = D → options (A) or (C); second blank G+3 = J → only (C).

Answer: (C) DJINQ

Concept: Unfold in reverse order; each unfold mirrors the cut across the fold line.

1. Fold 1: top half down. Fold 2: right half onto left. Cut made in the final quarter.
2. Open fold 2 → cut copied left–right into the other bottom quarter.
3. Open fold 1 → both cuts mirrored up–down → 4 cuts, one per quarter, mirror pairs — figure (A).

Trick: 2 folds = 4 copies of the cut; each pair across a fold line is a mirror image.

Answer: (A)Figure (a)

Concept: Fix the end persons first; remember south-facing people's right is towards the viewer's left.

1. Number seats 1–4. Row 2 (north): S at extreme right = seat 4; A two seats away = seat 2.
2. Row 1 (south): N at its extreme right = seat 1; O second to left of T → T seat 2, O seat 4; E seat 3.
3. M should not face E (seat 3) → M seat 1, D seat 3.
4. Seat 4: O faces S.

Trick: Ends first: 'No one to the right of X' always puts X at an end — place S and N first.

Answer: (D) O and S

Concept: Number seats 1–7 left to right and fix fixed positions first.

1. N at seat 7 (right end). Three between N and X → X at 3.
2. Two between X and W → W at 6.
3. M third left of L, Y just right of L → M 1, L 4, Y 5; O gets 2.
4. Between O (2) and W (6): X, L, Y = 3.

Trick: People between seats a and b = |a − b| − 1 = 6 − 2 − 1 = 3.

Answer: (B) Three

Concept: If differences grow by a fixed amount, add the next difference.

1. Differences: 6, 10, 14, 18 (each +4).
2. Next difference = 22.
3. 76 + 22 = 98.

Trick: Next term = last term + last difference + 4 = 76 + 18 + 4 = 98 — one line, no table needed.

Answer: (D) 98

Concept: A vertical mirror swaps left and right; top and bottom stay as they are.

1. Thick shaded strip on the right edge → moves to the left edge.
2. Shaded block at top-left → goes to top-right; bottom shaded strip now starts from the left.
3. Slanting lines that drop from left to right now drop from right to left — matches figure (B).

Trick: Pick one corner feature (top-left box) and check it lands at top-right without moving up or down.

Answer: (B)Figure (b)

Concept: Letters common to both words use the digits common to both codes; the leftovers pair up.

1. FILE = 6312, LIVE = 3241.
2. Common letters I, L, E ↔ common digits 3, 1, 2.
3. Leftover in LIVE = V; leftover digit in 3241 = 4 → V = 4.

Trick: Strike out what is common; the one letter and one digit left over belong together.

Answer: (A) 4