Q&A: practice questions with answers

Concept: A 20-letter string usually hides 5 blocks of 4 letters that change by a fixed step.

1. Split: _BFK | G_HM | IF_O | KHL_ | MJN_.
2. 1st letters: _, G, I, K, M → step +2, so blank = E.
3. 2nd letters: B, _, F, H, J → blank = D; 3rd letters: F, H, _, L, N → blank = J.
4. 4th letters: K, M, O, _, _ → Q, S.
5. Blanks in order: E D J Q S.

Trick: Test only the first blank: block-1 must start with E (…E, G, I…). That alone removes (b) and (d); the second letter D then removes (a).

Answer: (C) EDJQS

Concept: Letters common to both words share the common digits; the leftover letter takes the leftover digit.

1. RATE → 7532, PART → 6735.
2. Common letters R, A, T ↔ common digits 7, 3, 5.
3. RATE has extra E with extra digit 2 → E = 2.
4. PART has extra P with extra digit 6 → P = 6.

Trick: Cancel the shared digits (7, 3, 5) from the PART code – whatever remains is P.

Answer: (B) 6

Concept: If four different letters are all adjacent to one face, the sixth letter must be opposite it.

1. F appears in both positions of the dice.
2. Its neighbours are E, D (first view) and A, B (second view) – four different letters.
3. Only C is never seen next to F, so C is opposite F.

Trick: Common face + list its 4 neighbours; the missing sixth letter is its opposite.

Answer: (C) F

Concept: Successive differences grow by 1 each time.

1. 460 − 451 = 9, 470 − 460 = 10.
2. Next difference = 11 → 470 + 11 = 481.
3. Check: 481 + 12 = 493, 493 + 13 = 506 ✓

Trick: Work backwards from the end too: 506 − 13 = 493, 493 − 12 = 481 – both sides agree.

Answer: (A) 481

Concept: Treat each position of the 3-letter clusters as its own series.

1. 1st letters E → I → M → Q: +4 each → U.
2. 2nd letters I → G → E → C: −2 each → A.
3. 3rd letters K → J → I → H: −1 each → G.

Trick: Options (b) and (d) both start UA; just check the last letter: H − 1 = G.

Answer: (D) UAG

Concept: Facing the centre: left = clockwise, right = anticlockwise.

1. Fix D. Q is D's immediate left, S is D's immediate right → clockwise order S, D, Q.
2. From Q: 2 seats left gives E; 3 seats right gives F (seat opposite Q).
3. Remaining seat → R. Clockwise: D, Q, R, E, F, S.
4. From F going left (clockwise): S, D, Q, then R → 3 persons between.

Trick: In a 6-seat circle, 'third to the right/left' is always the seat directly opposite.

Answer: (A) Three

Concept: Find the change in each part separately and apply it to the new term.

1. NT6 → QR1: N+3 = Q, T−2 = R, 6−5 = 1.
2. FN8 → IL3 fits the same: F+3 = I, N−2 = L, 8−5 = 3.
3. JL16 → J+3 = M, L−2 = J, 16−5 = 11 → MJ 11.

Trick: The number rule (−5) gives 11, leaving (a) and (d); first letter J+3 = M fits both, second letter L−2 = J picks (d).

Answer: (D) MJ 11

Concept: If A is wholly inside B and B is separate from C, then A is separate from C.

1. Onion ⊂ Potato; Potato has no link with Tomato or Corn.
2. (I) Onion lies inside Potato, which never touches Tomato → no onion is a tomato ✓.
3. (II) Corn and Tomato are both just outside Potato; their meeting is only possible, not certain ✗.

Trick: Two 'No' statements about a common class (Potato) never give a definite relation between the other two.

Answer: (C) Only conclusion (I) follows.

Concept: Rearrange from Z to A and compare position by position.

1. Original: A C T I V I T Y
2. Z→A order: Y V T T I I C A
3. Same place: 3rd (T) and 6th (I) only.

Trick: Write the two rows one under the other and tick matching columns – takes 10 seconds.

Answer: (D) Two

Concept: Differences form consecutive odd numbers.

1. 570−567 = 3, 575−570 = 5, 582−575 = 7, 591−582 = 9.
2. Next difference = 11 → 591 + 11 = 602.

Trick: Odd-number differences mean term = 566 + n²: 566 + 6² = 566 + 36 = 602.

Answer: (D) 602

Concept: The given shape must be traced inside an option in the same orientation (no rotation).

1. Figure (X): two vertical lines joined at the top, crossed by one straight line that extends outside on both sides.
2. In option (c) the middle column gives the two verticals and the top edge closes them.
3. The lower horizontal lines on the left and right sit at the same level, forming the crossing line.
4. Other options lack a crossing line at one common height.

Trick: Pick the most unusual part of (X) – here the straight cut sticking out on both sides – and search only for that.

Answer: (C)

Concept: Facing the centre: left = clockwise, right = anticlockwise.

1. Seat E. G is placed so that E is second to G's left; F sits just left of G (G is on F's right).
2. B is third to E's right; A is B's immediate right; C is E's immediate left.
3. Clockwise order: E, C, D, A, B, G, F (D takes the last seat).
4. From D moving left: A, then B → 1 person between.

Trick: Start with the person mentioned most often (E) and fix the 'third/second' clues first; one-seat clues come last.

Answer: (D) One

Concept: A word present in both sentences gets the code present in both coded forms.

1. 'I eat breakfast' → hk mr It; 'breakfast is healthy' → xz hk bd.
2. Common word: breakfast. Common code: hk.
3. So breakfast = hk.

Trick: Don't decode everything – just circle the repeated word and the repeated code.

Answer: (B) hk

Concept: Track the movement of the symbols and the position of the block separately.

1. Six symbols sit in two columns; each step they shift one cell in a clockwise loop.
2. From figure 4 (left 7, 9, 5; right M, K, T) the next step gives left 9, 5, T and right 7, M, K.
3. The block alternates sides, so in figure 5 it returns to the right with Ω on the left.
4. Only option (c) satisfies both changes.

Trick: First use the block's side to cut two options, then check just one symbol (T moving to the left column).

Answer: (C)

Concept: Each fold doubles every cut as a mirror image about the fold line.

1. Two folds (horizontal, then vertical) → each cut appears 4 times.
2. The circle at the folded corner shows up at all four corners.
3. The circle near the open corner becomes four circles around the centre; both triangles also appear four times.
4. Only option (a) has corner circles and a ring of four central circles.

Trick: Count shapes first: 2 folds → each shape × 4. Any option with a different count is out.

Answer: (A)

Concept: Every letter is moved forward by the same number of places.

1. UHAR → XKDU: U+3 = X, H+3 = K, A+3 = D, R+3 = U.
2. SDWC → VGZF confirms +3.
3. Shift IMOS by three: (I, M, O, S) → (L, P, R, V).

Trick: Only the first letter is needed: I+3 = L, and only option (c) begins with L.

Answer: (C) LPRV

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

1. 51 ÷ 6 + 9 × 84 − 4 becomes 51 − 6 × 9 + 84 ÷ 4.
2. 6 × 9 = 54; 84 ÷ 4 = 21.
3. 51 − 54 + 21 = 18.

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

Answer: (C) 18

Concept: Two unrelated members of one larger class are drawn as two separate circles inside a big circle.

1. Tigers and monkeys are both animals → both inside 'Animals'.
2. A tiger can never be a monkey → their circles do not touch.
3. Option (b) shows exactly this.

Yaad rakhein: Similar pattern: Rose, Lily, Flowers or Iron, Copper, Metals – same diagram.

Answer: (B)

Concept: Compare the 2nd and 3rd letters with the 1st letter of each cluster.

1. Pattern (−2, +1) from the first letter: B→Z & C, N→L & O, P→N & Q.
2. QPR gives (−1, +1): P is only one step behind Q, so it breaks the pattern.

Trick: The 3rd letter is +1 in all four, so look only at the middle letter – the one with a gap of 1 instead of 2 is odd.

Answer: (C) QPR

Concept: Decode each symbol pair one by one and build a small family tree.

1. P + Q: P is Q's daughter. Q × R: Q is R's wife → R is P's father.
2. R − S: R is S's brother. S ÷ T: S is T's father.
3. So R is T's father's brother, and P (R's daughter) is T's father's brother's daughter.

Trick: P is female (daughter) and of T's generation – that removes both 'wife' options; R and S are brothers on the father's side, removing 'mother's'.

Answer: (C) Father's brother's daughter