Q&A: practice questions with answers

Concept: Decode the symbols one by one and draw the tree.

1. H + I → H is the wife of I.
2. I * O → I is the father of O, so H is the mother of O.
3. O × D → O is the mother of D; D – J → D is the brother of J, so O is also J's mother.
4. H is the mother of J's mother → mother's mother.

Trick: Two generations up on the mother's side = mother's mother (maternal grandmother).

Answer: (B) Mother's mother

Concept: Look at the differences between consecutive terms.

1. Differences: 9, 14, 25, 30, 41.
2. Pattern of these differences: $+5,+11,+5,+11$ → next is $+5$.
3. Next difference $=41+5=46$, so $?=125+46=171$.

Trick: When the differences are irregular, take differences once more – an alternating pattern often appears.

Answer: (D) 171

Concept: Track each letter position and the number separately.

1. First letters: B, F, J, N → $+4$ each → next R.
2. Second letters: D, H, L, P → T; third letters: F, J, N, R → V.
3. Numbers: 2, 5, 8, 11 → $+3$ → 14.
4. Missing term = RTV 14.

Trick: Each cluster has letters 2 apart (B_D_F); just add 4 to the first letter and fill the rest.

Answer: (A) RTV 14

Concept: Draw the Venn diagram with the minimum overlap allowed by the statements.

1. Some pens are markers → Pens and Markers overlap.
2. All markers are stands → Markers circle lies inside Stands.
3. The pens that are markers are therefore also stands → Conclusion I follows.
4. Stands may contain many things that are not markers → II (All stands are markers) does not follow.

Trick: Some A are B + All B are C ⇒ Some A are C. ‘All B are C’ never reverses to ‘All C are B’.

Answer: (B) Only conclusion (I) follows

Concept: Take A as the origin, with east and north positive, and follow each move.

1. 23 km north → $(0,23)$; left turn = west 8 km → $(-8,23)$.
2. Left = south 28 km → $(-8,-5)$; left = east 17 km → $(9,-5)$.
3. Left = north 5 km → P at $(9,0)$ – on the same horizontal line as A.
4. P is 9 km east of A, so he must drive 9 km to the west.

Trick: East–west net: $17-8=9$ km east; north–south net: $23+5-28=0$.

Answer: (B) 9 km to the west

Concept: Track each letter position with its alphabet number (A = 1 … Z = 26) and the number separately.

1. First letters: B(2) → W(23) → R(18) → M(13): each step $-5$ (B − 5 wraps to W) → next $13-5=8$ = H.
2. Second letters: V → Q → L → G ($-5$) → B; third letters: Y → T → O → J ($-5$) → E.
3. Numbers: 32, 16, 8, 4 → $÷2$ → 2.
4. Missing term = HBE 2.

Trick: All three letters move back 5 together, the number halves.

Answer: (D) HBE 2

Concept: Write the original number and the ascending-order number one below the other and compare position by position.

1. Original: 3 7 4 9 6 8 5.
2. Ascending: 3 4 5 6 7 8 9.
3. Same digit at the same place: 1st (3) and 6th (8).
4. So two digits keep their positions.

Trick: Compare column by column – never count by memory.

Answer: (C) Two

Concept: Compare each letter of the first cluster with the matching letter of the second (A = 1 … Z = 26).

1. VB-TN: $V-2=T$, $B+12=N$.
2. SY-QK: $S-2=Q$, $Y+12=K$ (25 + 12 = 37, $37-26=11$). KQ-IC: $K-2=I$, $Q+12=C$.
3. PV-MH: $P-3=M$ (not $-2$), $V+12=H$ → the first rule breaks.
4. So PV – MH is the odd one.

Trick: Check the simple shift (first letters) first – the odd pair often fails there.

Answer: (D) PV – MH

Concept: A word common to two sentences gets the code common to both coded sentences.

1. ‘tale jaw bow’ → vl ud np.
2. ‘law bow fresh’ → pq re ud.
3. Common word = bow; common code = ud → bow = ud.

Trick: Match the common word with the common code – order of codes does not matter.

Answer: (A) ud

Concept: Find how each letter of the first cluster changes into the second (A = 1 … Z = 26).

1. VUT – PON: $V(22)-6=P(16)$, $U-6=O$, $T-6=N$; also each cluster is three consecutive letters going backward.
2. TSR – NML: again $-6$ for every letter.
3. YXW: consecutive backward ✓; $Y(25)-6=S(19)$, $X-6=R$, $W-6=Q$ → SRQ.
4. YWU is not consecutive; RRP and RQP give a wrong shift.

Trick: First-letter check: Y − 6 = S, so only options starting with S survive.

Answer: (C) YXW−SRQ

Concept: Vowels move one letter forward; consonants move one letter back.

1. P → O, E → F, R → Q, B → A, L → K, A → B, D → C.
2. New group: O F Q A K B C.
3. All seven letters are different, so no letter appears twice → zero.

Trick: Write the new letters in alphabetical order (A B C F K O Q) – repeats become obvious at once.

Answer: (D) Zero

Concept: Try each option, then solve using BODMAS.

1. As given: $(6\times 14)÷2+18\times 4-28÷7+34=42+72-4+34=144\ne 87$.
2. Interchange 4 and 2: $(6\times 14)÷4+18\times 2-28÷7+34$.
3. $=21+36-4+34=87$ ✓.

Trick: The value must fall from 144 to 87 – swap numbers so that a big term (18 × 4) shrinks.

Answer: (D) 4 and 2

Concept: All face north, so left and right are the solver's left and right. Fix the sure positions first.

1. Only K is to the right of E → K at the right end (7), E at 6.
2. One person between I and E → I at 4. Two people between C and I → C at 1 (7 is taken).
3. R, S, T fill 2, 3, 5 in the order R – S – T (S between R and T) → R 2, S 3, T 5.
4. Row: C R S I T E K → third from the right is T.

Trick: Clues with ‘only’ (only K to the right) lock the end seats – start there.

Answer: (D) T

Concept: Facing north, left–right is the same as ours. Place the people fixed by count clues first.

1. Five people to the right of D → D is 2nd from the left. Four people to the left of H → H is 5th.
2. R is the neighbour of both A and D → R at 3 and A at 4 (R at 1 would leave no seat for A).
3. Seats 1, 6, 7 remain; J is immediately right of B → B 6, J 7, so U 1.
4. Row: U D R A H B J → U and D are adjacent, no one between them.

Trick: ‘n people to the right’ in a row of 7 means position $7-n$ from the left.

Answer: (C) None

Concept: Number the positions 1 (top) to 7 (bottom) and place the fixed boxes first.

1. Three boxes above C → C at 4. One box between D and C → D at 2 or 6.
2. Three boxes between D and E, and E above C → D at 6, E at 2. G just below E → G at 3.
3. Positions 1, 5, 7 remain for A, B, F. A not next to D (5 or 7) → A at 1; F above B → F 5, B 7.
4. Stack: A, E, G, C, F, D, B → box just above F is C.

Trick: Test both cases of an ‘either–or’ clue; the one that breaks a later rule is dropped.

Answer: (D) C

Concept: Follow each letter position and the number separately.

1. First letters: A, E, I, M → $+4$ → Q.
2. Second letters: D, H, L, P → T; third letters: G, K, O, S → W.
3. Numbers: 3, 6, 9, 12 → $+3$ → 15.
4. Missing term = QTW 15.

Trick: Inside each cluster letters are 3 apart (A_ _D_ _G) – Q, T, W keeps this gap.

Answer: (C) QTW 15

Concept: Number positions 1 (top) to 6 (bottom).

1. No box above G → G at 1. Three boxes below A → A at 3.
2. Two boxes between A and E → E at 6.
3. Positions 2, 4, 5 remain; F above H above D → F 2, H 4, D 5.
4. H and D are adjacent → zero boxes between them.

Trick: ‘No box above X’ means X is on top – the easiest anchor.

Answer: (A) Zero

Concept: Find the operation that turns the first number into the second in both given pairs.

1. $104-89=15$ and $172-157=15$ → rule: add 15.
2. $138+15=153$ ✓.
3. Others: $56+15=71\ne 72$, $97+15=112\ne 111$, $39+15=54\ne 64$.

Trick: Small, equal differences → simple addition rule; check the difference first.

Answer: (D) 138, 153

Concept: A word common to two sentences has the code common to both coded sentences.

1. ‘bat view month’ → nd wz nh.
2. ‘oven month gum’ → on ju wz.
3. Common word = month; common code = wz → month = wz.

Trick: Common word ↔ common code – ignore positions.

Answer: (C) wz

Concept: Number the positions 1 (top) to 7 (bottom).

1. Only F above G → F at 1, G at 2. Three boxes below D → D at 4.
2. Two boxes between D and C → C at 1 or 7; 1 is F, so C at 7.
3. Positions 3, 5, 6 remain; A above B above E → A 3, B 5, E 6.
4. Stack: F, G, A, D, B, E, C → box just above B is D.

Trick: ‘Only F above G’ fixes the top two boxes at once.

Answer: (D) D