Q&A: practice questions with answers

Concept: Apply +1 to odd digits and −2 to even digits, then compare.

1. 7452183 → 7→8, 4→2, 5→6, 2→0, 1→2, 8→6, 3→4.
2. New number: 8 2 6 0 2 6 4.
3. Highest 8 − lowest 0 = 8.

Trick: The 2 becomes 0 and 7 becomes 8 — the extremes are clear at once.

Answer: (C) 8

Concept: Track each position separately.

1. 1st letters: F, D, B, Z, X (−2 each, wrapping B → Z).
2. 2nd letters: K, I, G, E, C (−2).
3. 3rd letters: P, N, L, J, H (−2). Answer: XCH.

Trick: Every letter moves 2 steps back; wrap around after A.

Answer: (A) XCH

Concept: 'Some A are B' can always be converted to 'Some B are A'.

1. Some houses are small → Some small are houses: I follows.
2. Some small are forks → Some forks are small: II follows.
3. 'All forks are knives' is not needed here.

Trick: Conclusions that are just converses of 'Some' statements always follow.

Answer: (C) Both conclusions (I) and (II) follow

Concept: Find the change in each letter and in the number.

1. JR 11 → MV 2: J+3 = M, R+4 = V, 11 − 9 = 2.
2. KL 7 → NP −2: K+3 = N, L+4 = P, 7 − 9 = −2 ✓.
3. WO 2 → W+3 = Z, O+4 = S, 2 − 9 = −7 → ZS −7.

Trick: Rule: (+3, +4, −9). Check the number first: only A and D have −7, then O+4 = S picks D.

Answer: (D) ZS -7

Concept: Decode each symbol from left to right and draw the family tree.

1. M ÷ N: M is father of N. N + O: N is sister of O.
2. O − P: O is brother of P, so P is also M's child.
3. P × Q: P is wife of Q → P is M's daughter.
4. M is the father of Q's wife.

Trick: Chain: M → daughter P → her husband Q ⇒ wife's father (father-in-law).

Answer: (D) Wife's Father

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

1. After interchange: $116\times 5+423÷3-13$
2. $=580+141-13$
3. $=708$

Trick: Rewrite the whole expression with new signs before calculating — never mix old and new signs.

Answer: (C) 708

Concept: When two words share a letter and their codes share a digit, that digit is the code of the common letter.

1. FIRE and ACID have only one common letter: I.
2. 5371 and 8247 have only one common digit: 7.
3. So I = 7.

Trick: Match common letters with common digits — no need to decode the full words.

Answer: (D) 7

Concept: Compare the alphabet positions of corresponding letters in each pair.

1. BK – IF: $B(2)+7=I(9)$, $K(11)-5=F(6)$; similarly HQ – OL and PY – WT follow $+7,-5$.
2. TC – AY: $T+7=A$ (cyclic) but $C(3)\to Y(25)$ is $-4$ (cyclic), not $-5$.
3. So TC – AY is the odd one.

Trick: Check the second letter only: $-5$ from C gives X, not Y.

Answer: (C) TC – AY

Concept: Fix the definite clue first, then place others around it (Mon = 1 … Sun = 7).

1. U = Tue (2). Two people between P and U ⇒ P = Fri (5).
2. T after P and before Q ⇒ T = Sat, Q = Sun.
3. Left: Mon, Wed, Thu. V just before R (V not Sat) ⇒ V = Wed, R = Thu; S = Mon.
4. Between S (Mon) and Q (Sun): U, V, R, P, T = 5.

Trick: Between Mon and Sun there are always 5 days.

Answer: (D) Five

Concept: Find the rule from both given pairs and apply it.

1. MS 14 → OV 8: $M+2=O$, $S+3=V$, $14-6=8$
2. NN 5 → PQ −1: $N+2=P$, $N+3=Q$, $5-6=-1$
3. SR 6 → $S+2=U$, $R+3=U$, $6-6=0$ ⇒ UU 0

Trick: Rule: $+2$, $+3$, $-6$.

Answer: (B) UU 0

Concept: Check every number: the element just before it and just after it must both be symbols.

1. Qualifying numbers: * 6 @, @ 1 £, $ 7 #, # 6 %, * 5 €.
2. 8, 6 (after 8), 1, 9 fail because a neighbour is a number.
3. Count = 5.

Trick: Scan in groups of three (symbol–number–symbol) from left to right.

Answer: (B) 5

Concept: Facing the centre, 'left' means moving clockwise.

1. Place D. G is immediate left of D; A is 3rd to the left of D (D → G → _ → A).
2. Only B sits between H and A ⇒ B next to A on the far side, then H.
3. Remaining seats: one next to A and one next to D. C is not A's neighbour ⇒ C next to D, I next to A.
4. Third to the left of I: A, B, H ⇒ H.

Trick: Draw the circle and always move clockwise for 'left' when everyone faces the centre.

Answer: (D) H

Concept: Use the 'exactly 4 between' clue to fix Q and V (Mon = 1 … Sun = 7).

1. 4 between Q and V ⇒ (Mon, Sat) or (Tue, Sun). Neither on Monday ⇒ Q, V are Tue and Sun.
2. If Q = Sun, T = Sat; left Mon, Thu, Fri — no room for three consecutive P, S, R. ✗
3. So Q = Tue, T = Mon, V = Sun; P, S, R = Thu, Fri, Sat; U = Wed.
4. Before Q: only T ⇒ 1.

Trick: A block of three consecutive days (P-S-R) quickly rules out the wrong case.

Answer: (A) 1

Concept: Look at the differences between consecutive terms.

1. $872-2=870$, $870-4=866$, $866-6=860$, $860-8=852$
2. Next: $852-10=842$

Trick: Differences are consecutive even numbers 2, 4, 6, 8, 10.

Answer: (B) 842

Concept: Track each letter position of the clusters separately.

1. 1st letters: O, M, K, I → G (−2 each)
2. 2nd letters: Q, O, M, K → I (−2 each)
3. 3rd letters: J, H, F, D → B (−2 each)

Trick: All three positions move back by 2 together.

Answer: (A) GIB

Concept: Odd digit → +1, even digit → −2, then look for repeats.

1. 6318574 → 6−2, 3+1, 1+1, 8−2, 5+1, 7+1, 4−2
2. New number: 4426682
3. Repeated digits: 4, 2, 6 ⇒ three.

Trick: Write the new digits directly under the old ones to avoid mistakes.

Answer: (A) Three

Concept: The pair rule must be the same on both sides: # → USO and LJF → %.

1. Test option (C): LJF → IGC: L−3 = I, J−3 = G, F−3 = C.
2. Same rule: XVR → USO: X−3 = U, V−3 = S, R−3 = O. ✓
3. Other options fail (e.g., LJF → IZC is not a uniform shift).

Trick: Find the rule from the right side (LJF → %) and check the left side.

Answer: (C) # = XVR, % = IGC

Concept: Find how the middle number comes from the outer two.

1. $(14,66,19)$: $(14+19)\times 2=66$
2. $(43,134,24)$: $(43+24)\times 2=134$
3. $(6,42,15)$: $(6+15)\times 2=42$ ✓; others give $36\ne 35$, $98\ne 97$, $82\ne 81$

Trick: Middle = 2 × (first + third).

Answer: (C) (6, 42, 15)

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

1. 'relax' is the only word in both sentences.
2. 'pu' is the only code in both 'ha mi pu' and 'pu co lz'.
3. So relax = pu.

Trick: Common word ↔ common code.

Answer: (B) pu

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

1. Fix C; B is 3rd to the left of C.
2. F is 2nd to the left of E and E is not next to B ⇒ E sits immediately right of C, F immediately left of C.
3. D is immediate right of A ⇒ A and D take the two seats after B; G takes the seat between F and B.
4. Moving clockwise from G, the next person is B ⇒ B is to the immediate left of G.

Trick: After fixing the two 'fixed-gap' pairs, only one seat is left — fill it and read the answer.

Answer: (B) To the immediate left