Q&A: practice questions with answers

Concept: When everyone faces the centre, 'left' = next seat clockwise, 'right' = next seat anticlockwise.

1. Left-chain: B, then G (left of B), then E, then C → clockwise B–G–E–C.
2. D is right of B and F is right of D, A is left of F → clockwise A–F–D–B.
3. Join: A–F–D–B–G–E–C, and after C the circle returns to A.

Trick: Build one long clockwise chain from all 'immediate left' clues; the seat after the last name closes the circle.

Answer: (C) A

Concept: Classes with no common member are drawn as separate, non-touching circles.

1. Shoe is footwear, jacket is clothing, bow is a weapon/ribbon – none is a type of another.
2. No item can be two of these at once, so there is zero overlap.
3. Three separate circles = figure (a).

Trick: Ask 'can one object be both?' for each pair – three 'no' answers means three disjoint circles.

Answer: (A)Three separate circles

Concept: A vertical mirror swaps left and right only; top and bottom stay unchanged.

1. Filled dots at top-right and bottom-left shift to top-left and bottom-right.
2. The diagonal that rose to the right must now rise to the left (slope reverses).
3. Option (b) keeps the old slope; only (c) has both changes.

Trick: Check one slanted line first – in a vertical mirror its slope must reverse; this alone removes most options.

Answer: (C)Figure (c)

Concept: Faces adjacent to a common face cannot be opposite to it; the left-over number is opposite.

1. 6 is visible in both positions.
2. Neighbours of 6: 5, 4 (first view) and 1, 3 (second view).
3. Only 2 is not a neighbour, so 2 and 6 are opposite.

Trick: Common face + list its 4 neighbours → the missing sixth number is its opposite.

Answer: (C) 6

Concept: A word repeated in two sentences gets the code repeated in both code groups.

1. 'do the math' → ab cd ef; 'math is fun' → ef gh ij.
2. Common word: math; common code: ef.
3. So math = ef.

Trick: Circle the repeated word and the repeated code – match them directly.

Answer: (C) ef

Concept: The symbol moves one corner clockwise each step and a new symbol appears each time.

1. Positions: top-right → bottom-right → bottom-left → top-left.
2. Next position: top-right again.
3. (b), (c) fail the position; (d) repeats the square; (a) has a new star at top-right.

Trick: Track only the corner first, then remove options that repeat an old symbol.

Answer: (A)Figure (a)

Concept: Swap the signs first, then simplify using BODMAS.

1. After swapping: $69+2\times 7-35÷7$.
2. $2\times 7=14,35÷7=5$.
3. $69+14-5=78$.

Trick: Rewrite the whole expression with new signs before doing any arithmetic – most errors come from swapping mid-way.

Answer: (A) 78

Concept: Each letter shifts back by a fixed number of places.

1. PQTU − 4 each → LMPQ ✓; EHLN − 4 each → ADHJ ✓.
2. Apply to FMRW: F(6)−4 = B(2), M(13)−4 = I(9), R(18)−4 = N(14), W(23)−4 = S(19).
3. Result: BINS.

Trick: Check only the first letter of each option: F−4 = B leaves only option (b).

Answer: (B) BINS

Concept: Fix M, then place others using the 'one/two between' clues; facing centre, right = anticlockwise.

1. Seats 1–7 clockwise, M = 1. One between M and L from M's right → L = 6.
2. H = 3 (one between M and H); J = 5 (one between J and H).
3. I = 4 (one between L and I); K = 7 (two between K and I); G = 2.
4. Right of G = seat 1 (M), left of G = seat 3 (H).

Trick: Place the person mentioned in the most clues (H) early; the last empty seat goes to the person never mentioned (G).

Answer: (A) M and H

Concept: Decode each sign, then trace the chain from C to F.

1. C − D: C is D's wife. D × E: D is E's brother.
2. E ÷ F: E is F's daughter, so brother D is F's son.
3. F + G: F is father of G (not needed).
4. F is the father of C's husband.

Trick: Only the links connecting C to F matter – skip the rest (F + G).

Answer: (A) Husband's father

Concept: Subtracted values increase by 2 each step.

1. 410 − 2 = 408, 408 − 4 = 404, 404 − 6 = 398.
2. Next: 398 − 8 = 390.
3. Check: 390 − 10 = 380 ✓.

Trick: Even-number differences 2, 4, 6, 8, 10 – always verify with the last given term.

Answer: (D) 390

Concept: Write the letters in Z→A order and compare place by place.

1. IDENTITY = I D E N T I T Y.
2. Reverse order: Y T T N I I E D.
3. Match at 4th place (N) and 6th place (I) → two letters.

Trick: Write both rows one under the other and tick matches – avoid doing it mentally.

Answer: (B) Two

Concept: Find a block that repeats and fill blanks from it.

1. Given letters fit the 5-letter block GPLMD.
2. Series: GPLMD GPLMD GPLM.
3. Blanks (1,2,5,7,9,11,14) = G P D P M G M.

Trick: Test the option's letters in the blanks – (b) makes 'GPLMD' repeat perfectly.

Answer: (B) GPDPMGM

Concept: Codes are jumbled; remove common letters and common digits to pair the leftovers.

1. DENS = 8976, SENT = 6739.
2. Common letters E, N, S ↔ common digits 9, 7, 6.
3. Left-over: D = 8 and T = 3.

Trick: Odd letter out ↔ odd digit out – no need to decode E, N, S.

Answer: (D) 3

Concept: A conclusion follows only if it is true in every possible diagram.

1. Pants ⊂ Trousers; Shirts ⊂ Coats; Pants and Shirts disjoint.
2. (I) 'No coat is a trouser' – coats may or may not overlap trousers → not certain.
3. (II) 'Some coats are pants' – no link forces it → not certain.

Trick: If two classes are never linked by any statement, any definite conclusion about them fails.

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

Concept: Find the step pattern inside each cluster.

1. LMJ: +1, −3; EFC: +1, −3; FGD: +1, −3.
2. ONL: −1, −2 – different pattern.
3. So ONL is odd.

Trick: First two letters consecutive forward (LM, EF, FG) – ON goes backward, spotted instantly.

Answer: (B) ONL

Concept: Every letter moves 2 places forward (with wrap-around after Z).

1. OQUT + 2 each → QSWV ✓; IKON + 2 each → KMQP ✓.
2. ZBFE: Z→B, B→D, F→H, E→G.
3. Result: BDHG.

Trick: Third letter decides: F+2 = H appears only in option (a).

Answer: (A) BDHG

Concept: Constant difference series (AP).

1. $19-11=8,27-19=8$, … each step +8.
2. $43+8=51$.

Trick: All terms are 11 more than multiples of 8 → only 51 ($=8\times 5+11$) fits.

Answer: (D) 51

Concept: Each fold doubles the layers, so the number of holes doubles on unfolding.

1. Two folds → 4 layers; one punch cuts 4 holes.
2. Unfolding reflects the hole across each fold line.
3. Result: one hole in each quarter, symmetric about both centre lines.

Trick: Holes on unfolding = holes punched × 2^{number of folds} $=1\times 2^2=4$.

Answer: (A)Figure (a)

Concept: Each group is 3 consecutive letters in reverse; the first letter drops by 6.

1. First letters: R(18) → L(12) → F(6).
2. 6 − 6 = 0 → wraps to 26 = Z, so group = ZYX.
3. Check: Z(26) − 6 = 20 = T → TSR ✓.

Trick: For backward wrap, add 26: 0 + 26 = 26 = Z.

Answer: (A) ZYX