Q&A: practice questions with answers

Concept: Every fold doubles each cut; two folds mean each cut appears 4 times, mirrored across the fold lines.

1. The paper is folded twice → it opens into 4 layers, so each mark is repeated 4 times.
2. Cuts made: 1 square + 1 circle → open sheet shows 4 squares + 4 circles (8 holes).
3. Unfold the last fold first, then the earlier one, mirroring the holes each time. In each column the order reads circle–square–square–circle.
4. Only option (C) matches this mirrored pattern.

Trick: Count holes first: $(\text{cuts})\times 2^{\text{folds}}=2\times 4=8$. Reject options with the wrong count, then check the mirror symmetry.

Answer: (C) figure (c)

Concept: Track each letter position of the clusters separately.

1. 1st letters: K, M, O, Q → +2 each → S.
2. 2nd letters: R, T, V, X → +2 → Z.
3. 3rd letters: B, D, F, H → +2 → J.
4. Next cluster = SZJ.

Trick: Check just the first letter (Q+2 = S) to cut options to SZK/SZJ, then the last letter H+2 = J settles it.

Answer: (C) SZJ

Concept: Sort the letters A→Z and compare position by position.

1. Original word: ROTCHES; arranged A→Z: CEHORST.
2. Check place by place – every place differs (e.g. 1st: R vs C, 4th: C vs O, 7th: S vs T).
3. Unchanged letters = 0.

Trick: The first letter of the sorted word must be the smallest letter (C); if the word does not start with it, position 1 is already changed – go on checking only likely spots.

Answer: (B) Zero

Concept: Swap the signs as told, then solve using BODMAS.

1. Swap $+\leftrightarrow -$ and $\times \leftrightarrow ÷$: $48÷3+16-6\times 3$
2. Division/multiplication first: $16+16-18$
3. = $32-18=14$

Trick: Rewrite the whole expression with new signs before calculating – most mistakes come from swapping signs mid-way.

Answer: (B) 14

Concept: Split the 20-letter string into 5 groups of 4; each column moves +1 from group to group.

1. Groups: _KHL | J_IM | KM_N | LNK_ | MOL_
2. Column 1: _, J, K, L, M → I; Column 2: K, _, M, N, O → L
3. Column 3: H, I, _, K, L → J; Column 4: L, M, N, _, _ → O, P
4. Blanks in order: I, L, J, O, P → ILJOP.

Trick: Last two blanks fill column 4 as L, M, N, O, P – so the answer must end in “OP”; only ILJOP does.

Answer: (C) ILJOP

Concept: Follow how symbols enter, slide and move up in each step.

1. At every step a new symbol enters at the bottom-right.
2. The bottom row shifts one place left; the symbol pushed out moves up to the left of the row above.
3. Applying this to the 4th figure: f enters, bottom row reads dot, H, f; the star goes up; π stays on top.
4. This arrangement is option (C).

Trick: Track only the newest symbol's place and the top element – two checks usually remove 3 options.

Answer: (C) figure (c)

Concept: Find figure X in an option with the same size and orientation (no rotation).

1. X = a circle sitting on the top vertex of an upright triangle with a horizontal base.
2. (B): two slanting lines meeting just below a circle plus the bottom edge form this triangle, with the circle on its apex. ✔
3. (A) circle is inside the triangle; (C) triangle points down; (D) circle is separate.

Trick: Search for the most distinctive part first (circle touching an apex), then confirm the rest of the shape around it.

Answer: (B) figure (b)

Concept: Look at consecutive differences.

1. Differences: 37, 34, 31, 28 → each 3 less.
2. Next difference = 28 − 3 = 25.
3. 148 + 25 = 173.

Trick: Differences shrink, so the next jump must be smaller than 28 → answer lies below 148 + 28 = 176; only 173 and 168 qualify, and 168 needs a jump of 20.

Answer: (C) 173

Concept: North-facing row: left = west; south-facing row: left = east.

1. Number seats 1–4 from west. Row 2 (north): no one left of S → S = 1; one between T and S → T = 3.
2. Row 1 (south): no one left of L → L = 4 (east end).
3. Second to the right of L (right = west) = seat 2 → E faces it, so E = 2, B = 4.
4. O is second to the right of G → G = 3, O = 1, F = 2.
5. Ends: row 1 → O, L; row 2 → S, B. Option with both at ends: O and B.

Trick: Fix the “no one to the left” persons first – they give two end seats instantly.

Answer: (B) O and B

Concept: When codes are jumbled, match letters common to both words with symbols common to both codes.

1. TRIP → #, @, &, $; PART → &, ^, @, #.
2. Common letters T, R, P ↔ common symbols #, @, &.
3. Leftover in PART: A; leftover in its code: ^ → A = ^.

Trick: Uncommon letter ↔ uncommon symbol. Likewise I = $.

Answer: (C) ^

Concept: Decode only the part of the chain linking the two persons asked.

1. E + F → E is F's daughter.
2. F × G → F is G's sister, so F is female → F is E's mother.
3. G − H → G is a brother (male).
4. G is the brother of E's mother → mother's brother.

Trick: Stop as soon as both names (E and G) are connected – the rest (H, I) is noise.

Answer: (D) Mother's brother

Concept: Fix the definite positions first (ends), then fill the rest.

1. Seats 1–7 from left. No one right of M → M = 7; two between M and F → F = 4.
2. F immediately left of L → L = 5; three between E and L → E = 1.
3. Free seats 2, 3, 6; G–H adjacent → G = 2, H = 3; K = 6.
4. Between K (6) and H (3): seats 4, 5 → 2 persons.

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

Answer: (C) Two

Concept: If one face stays fixed, the dice has rotated about that axis; the four side faces move in a cycle.

1. 8 is on the left face in both views → rotation about the 8-axis.
2. 9 moved from the right face to the top, so every face moved one step: bottom → right.
3. 12 appears on the right in view II → it was at the bottom in view I, below the top face 7.
4. Top and bottom are opposite → 12 is opposite 7.

Yaad rakhein: A face visible together with another in any view can never be its opposite – use this to eliminate options quickly.

Answer: (C) 7

Concept: Find the uniform shift between the pairs.

1. D→A, G→D, H→E, K→H: each −3. MQTU → JNQR: also −3.
2. HLQR − 3: H→E, L→I, Q→N, R→O.
3. Result: EINO.

Trick: Only the first letter is needed: H − 3 = E, and only one option starts with E.

Answer: (C) EINO

Concept: Compare gaps between consecutive letters.

1. BDG: B→D +2, D→G +3.
2. EGJ: +2, +3; QSV: +2, +3.
3. MOQ: +2, +2 → different.

Trick: The first gap is +2 in all four, so compare only the last gap: 3, 2, 3, 3 – the odd value marks the answer.

Answer: (B) MOQ

Concept: Consecutive differences decrease by 1.

1. Differences: +10, +9, +8, +7.
2. Next difference = +6.
3. 684 + 6 = 690.

Trick: Decreasing-by-1 pattern → just add (last difference − 1) to the last term.

Answer: (D) 690

Concept: Decide overlap/separation for each pair of classes.

1. Car–Petrol: some cars run on petrol, but petrol is used elsewhere too → partial overlap.
2. Car–Electric: some cars are electric, and electricity runs other things → partial overlap.
3. Petrol–Electric: nothing common → separate.
4. Picture: Car in the middle cutting two separate circles → option (B).

Trick: Two classes with nothing in common + one class linking both = “chain” of three circles.

Answer: (B) figure (b)

Concept: Some + All = Some; All + No = No.

1. Some balls are toys + All toys are plastic → Some balls are plastic. (I) ✔
2. All toys are plastic + No plastic is water → No toy is water. (II) ✔
3. Both follow.

Trick: Rule pairs: I + A = I, A + E = E. Apply directly without drawing.

Answer: (A) Both conclusions (I) and (II) follow.

Concept: $\text{Code}=2\times (\text{sum of alphabet positions})$.

1. NBSF: $14+2+19+6=41$; $41\times 2=82$ ✔
2. IMSR: $9+13+19+18=59$; $59\times 2=118$ ✔
3. AYLC: $1+25+12+3=41$; $41\times 2=82$.

Trick: AYLC has the same letter-sum (41) as NBSF, so it must get the same code, 82.

Answer: (D) 82

Concept: Each letter moves one step back (A wraps to Z).

1. GFBE → FEAD and AZVY → ZYUX: every letter −1.
2. CBXA − 1: C→B, B→A, X→W, A→Z.
3. Result: BAWZ.

Trick: Check the third letter only: X − 1 = W – only BAWZ has W.

Answer: (D) BAWZ