Q&A: practice questions with answers

Concept: Find which class contains the others and which classes only partly share members.

1. Any father or brother is a male child of someone, so both are subsets of Son (big outer circle).
2. A man may be a father and a brother at once, but not always – so Father and Brother circles partly overlap.
3. Required picture: two overlapping circles inside one big circle.

Trick: Ask 'Is every X a Y?' for each pair – a 'yes' means draw X inside Y.

Answer: (A)the figure with two intersecting circles inside a larger circle

Concept: When a sheet is folded once, every cut appears twice on opening – mirrored about the fold line.

1. The bottom half is lifted onto the top half, so the crease is the horizontal middle line.
2. The four small holes cut in a row stay clear of the crease → it appears in the top half and again, mirrored, in the bottom half.
3. The triangle touches the crease → its two halves join into one rhombus in the centre.

Trick: A shape cut on the fold opens as one doubled shape; a shape cut away from the fold opens as two copies.

Answer: (A)the figure with a row of 4 holes at the top, a matching row at the bottom and one rhombus at the centre

Concept: Check the gap between 1st→2nd and 2nd→3rd letters of each cluster.

1. ZCA: Z(26)+3 = C (wrap), C−2 = A ✔
2. GJH: G+3 = J, J−2 = H ✔; LOM: L+3 = O, O−2 = M ✔
3. MPD: M+3 = P, but P(16) → D(4) is −12 ✘

Trick: Once the first step (+3) matches everywhere, test only the last letter: it must be 1 letter after the first letter (Z→A, G→H, L→M). M→D fails instantly.

Answer: (C) MPD

Concept: Each position has its own fixed shift, which is the same for every word.

1. MARVEL → SXROCP: shifts are +6, −3, 0, −7, −2, +4 (STRING → YQRBLK confirms).
2. CIRCLE: Apply to CIRCLE → I (C+6), F (I−3), R (no change), V (C−7, wrapping), J (L−2), I (E+4).
3. Code = IFRVJI.

Trick: Find only the 1st and last letters first: C+6 = I and E+4 = I → only IFRVJI ends and starts with I.

Answer: (D) IFRVJI

Concept: A long series with blanks is usually one block repeated; find the block and fill the gaps.

1. 24 places → block of 8: tarivolv × 3.
2. Fill: tarivolv | tarivolv | tarivolv
3. Missing letters in order: i v a o l i o v.

Trick: Spot the repeated start 'tar' – the distance between two 'tar's (8) gives the block length.

Answer: (B) ivaoliov

Concept: Separate the changing element from the fixed one.

1. The two small squares stay the same in every box – fixed part.
2. The letter goes back one place each time: Q → P → O → N → M.
3. So the next box has M with two small squares.

Trick: Eliminate options that alter the fixed part (number of squares) first, then check the letter.

Answer: (A)the figure showing M with two small squares

Concept: The code depends only on how many letters the word has.

1. SBZFPA: 6 letters → $6\times 3=18$ ✔
2. GNPQ: 4 letters → $4\times 3=12$ ✔
3. LCEZRDX: 7 letters → $7\times 3=$ 21

Trick: When codes are small multiples like 18, 12, divide by the letter count first – a constant ratio (3) cracks it.

Answer: (A) 21

Concept: Number seats 1–4 from the left of the drawing; row 2 faces north so its 'right' = drawing's right.

1. Two people between L and M → they occupy ends 1 and 4 of row 1.
2. W is just right of U and U faces L → U = 1, W = 2, L = 1, M = 4.
3. X not at an end → X = 3, V = 4.
4. Person facing N is just left of V → X (3), so N = 3, O = 2. Seat 2 of row 2 is W.

Trick: 'Only two between' in a row of 4 always means the two ends – fix them first.

Answer: (D) W

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

1. Black block at top-right → moves to top-left.
2. Black bar at bottom (left of centre) → stays at bottom, now right of centre.
3. The slanting line also flips its direction; only option B shows all three changes.

Trick: Track one corner element only: the side flips, the level (top/bottom) never changes.

Answer: (B)the second figure

Concept: Replace the signs first, then solve using BODMAS.

1. Swap − ↔ ÷ and × ↔ +: $24\times 25÷5+2-9$
2. $24\times 25=600$; $600÷5=120$
3. $120+2-9=$ 113

Trick: Do $24\times (25÷5)=24\times 5=120$ – smaller numbers, faster.

Answer: (C) 113

Concept: Handle the letters and the number separately.

1. Letters: each moves +1 (B→C, D→E; S→T, T→U).
2. Number: ×3 (11→33, 10→30).
3. UB 2 → VC and $2\times 3=6$ → VC 6.

Trick: Number part alone leaves UC 6 / VC 6; the U→V shift then picks VC 6.

Answer: (C) VC 6

Concept: Use fixed clues first; seats 1–7 from left (all face north).

1. Only L is left of X → L = 1, X = 2.
2. Four between L and N → N = 6.
3. O-M-W need 3 adjacent seats without seat 6 → seats 3, 4, 5 with M = 4.
4. O can't be next to N → O = 3, W = 5; remaining Y = 7 (extreme right).

Trick: 'Only one person to the left of X' = X is 2nd from the left – a direct fix.

Answer: (B) Y

Concept: Sort the letters Z→A and compare each position with the original word.

1. Arranged Z→A, DETECTOR becomes TTROEEDC.
2. Place-by-place, the original DETECTOR and the new TTROEEDC never share a letter at the same position → zero.

Trick: First letter after sorting must be the 'largest' letter (T); D ≠ T – then just scan the rest quickly.

Answer: (C) Zero

Concept: Split the digit string into 2-digit numbers and look at the differences.

1. Numbers: 53, 42, 49, 38, 45, 34, ?
2. Differences: −11, +7, −11, +7, −11, …
3. Next: 34 + 7 = 41

Trick: Alternate terms fall by 4 each (53, 49, 45 and 42, 38, 34) – so the next after 45 is 41.

Answer: (A) 41

Concept: Read 3-digit numbers; differences rise by 2.

1. Terms: 123, 121, 117, 111, ?, 93
2. Differences: −2, −4, −6, −8, −10
3. 111 − 8 = 103; check 103 − 10 = 93 ✔

Trick: Always verify with the last given term (93) – it confirms the pattern in one step.

Answer: (A) 103

Concept: Track each position of the three-letter terms separately.

1. 1st letters: L, J, H, F → D (−2)
2. 2nd letters: Y, W, U, S → Q (−2)
3. 3rd letters: G, E, C, A → Y (A−2 wraps to Y)

Trick: The first letter alone (D) eliminates all other options.

Answer: (A) DQY

Concept: Letters common to two words share the common digits; the leftover letter gets the leftover digit.

1. COWS = 5691, COST = 1679.
2. Common letters C, O, S ↔ common digits 1, 6, 9.
3. Extra in COST is T; extra digit in 1679 is 7 → T = 7 (and W = 5).

Trick: Cancel common letters and common digits – what remains pairs up directly.

Answer: (B) 7

Concept: Decode the chain pair by pair; keep only links that connect M to K.

1. K − L: K is daughter of L → L is K's parent.
2. L × M: L is sister of M → L is female, so L = K's mother.
3. M is L's brother → M is K's mother's brother. (M + N, N ÷ O are not needed.)

Trick: Stop decoding once both names in the question are linked – the rest is filler.

Answer: (B) Mother's brother

Concept: 'All A are B' + 'No B is C' ⇒ 'No A is C'; two classes both outside a third class have no definite relation.

1. All handles ⊂ locks, and locks have no shelf → no handle is a shelf. (I) follows ✔
2. Cupboards and shelves are each only known to be outside locks – their link is unknown. (II) is just a possibility ✘

Trick: All + No = No (for the subset). 'Some' between two outsiders never follows.

Answer: (D) Only conclusion (I) follows

Concept: On unfolding, each cut is reflected across the fold line.

1. Bottom half folded up → fold line = horizontal centre line.
2. Two triangles away from the fold → two near the top corners plus their upside-down mirrors near the bottom corners (4 triangles).
3. Semicircle on the fold → opens as one full circle in the centre.

Trick: Count holes: off-fold cuts double (2 → 4 triangles), on-fold cut becomes one whole shape (1 circle).

Answer: (D)the figure with four triangles near the corners and one circle in the centre