Q&A: practice questions with answers

Concept: Track the positions in both steps.

1. LIES → ILSE: positions 1-2-3-4 become 2-1-4-3 (swap letters in pairs). FROG → RFGO ✓.
2. ILSE → LSIE: positions become 2-3-1-4 (1st letter moves to 3rd place). RFGO → FGRO ✓.
3. CUPS → UCSP (pairs swapped ✓) → CSUP (U-C-S-P → C-S-U-P: 2-3-1-4 ✓).
4. PALE → ALEP, SCAM → MSCA, LUCK → UCLK do not follow the pair-swap in step 1.

Trick: Step 1: swap 1↔2 and 3↔4; Step 2: order 2-3-1-4.

Answer: (D) च

Concept: Decode each symbol and build the tree from the right end.

1. L x M: L is the father of M. E ≠ L: E is the son of L → E and M are siblings.
2. D ₹ E: D is the sister of E → D is also a sister of M.
3. C ÷ D: C is the daughter of D → C is the daughter of M's sister, i.e. M's niece.

Trick: Children of the same father are siblings; a sibling's daughter = niece ("sister's daughter").

Answer: (C) ए

Concept: Take A as origin; a left turn rotates the direction anticlockwise: W → S → E → N → W.

1. West 20 km: $(-20,0)$. Left (South) 18 km: $(-20,-18)$. Left (East) 21 km: $(1,-18)$.
2. Left (North) 28 km: $(1,10)$. Left (West) 1 km: P $=(0,10)$.
3. P is directly 10 km north of A, so to return he must drive 10 km towards the South.

Trick: Net east–west: $-20+21-1=0$; net north–south: $-18+28=+10$ (north).

Answer: (C) क

Concept: Replace each symbol by its meaning, then apply BODMAS.

1. Given: + → −, − → ×, × → ÷, ÷ → +.
2. $21-4÷10+36\times 4$ becomes $21\times 4+10-36÷4$.
3. $=84+10-9=85$.

Trick: Do ÷ and × before + and − even after substitution.

Answer: (B) प

Concept: Number the seats 1 to 7 from left to right; all face north.

1. No one is to the right of M → M at 7. Exactly three between O and M → O at 3. Exactly one between M and L → L at 5.
2. H is immediately left of G: of the free seats 1, 2, 4, 6 only 1–2 are adjacent → H at 1, G at 2.
3. Q is not next to M → Q not at 6 → Q at 4, F at 6.
4. Row: H, G, O, Q, L, F, M. Between F (6) and H (1): G, O, Q, L = four.

Trick: Use the "immediately left/right" pair to fill the only adjacent free seats.

Answer: (D) "

Concept: Positions 1 (bottom) to 7 (top).

1. Only four boxes above U → U at 3. Only one above R → R at 6.
2. Exactly one box between R and A → A at 4 (position 8 does not exist).
3. S immediately above T: free positions 1, 2, 5, 7 → T at 1, S at 2.
4. P is below B → P at 5, B at 7. Stack: B, R, P, A, U, S, T. Between B (7) and S (2): R, P, A, U = four.

Trick: "Only n boxes above" ⇒ position $=7-n$.

Answer: (C) "

Concept: Mark each vowel and check its immediate neighbours.

1. Series: K D A Z Z X L E L A R A E P E Y U Z F U Q P.
2. A(3): D–Z ✓; E(8): L–L ✓; A(10): L–R ✓; A(12): R–E ✗ (vowel after); E(13): A before ✗; E(15): P–Y ✓; U(17): Y–Z ✓; U(20): F–Q ✓.
3. Count = 6.

Trick: Y is a consonant here; two adjacent vowels (A E) disqualify each other.

Answer: (D) Y

Concept: Look at the differences between known consecutive terms.

1. $81-63=18$, $105-81=24$, $135-105=30$ → differences increase by 6.
2. Working backwards the differences are 6, 12, 18, 24, 30: $45+6=51$ and $51+12=63$ ✓.

Trick: Differences in A.P. (6, 12, 18, …) → fill the gap from the known pairs.

Answer: (B) अ

Concept: Apply the rule digit by digit, keeping the order.

1. 7281654: 7 (odd) → 9, 2 (even) → 1, 8 → 7, 1 → 3, 6 → 5, 5 → 7, 4 → 3.
2. New number: 9173573. Third digit from the left = 7.

Trick: Only the third digit matters: 8 is even → $8-1=7$.

Answer: (C) क

Concept: Compare the gap inside each pair and between the two pairs.

1. HL–ZS: H→L $+4$, Z→S $-7$. QU–IB: Q→U $+4$, I→B $-7$. WA–OH: W→A $+4$ (wrap), O→H $-7$.
2. GK–YP: G→K $+4$ but Y→P $=16-25=-9$, not $-7$.
3. Hence GK–YP is the odd one out.

Trick: First letter $+18$ gives the third letter in every option; only the last gap (−7 vs −9) breaks in D.

Answer: (D) ह

Concept: Substitute the letters by operators and use BODMAS.

1. P = ÷, Q = ×, R = +, S = −.
2. 40 R 16 P 4 S 3 Q 5 $=40+16÷4-3\times 5$.
3. $=40+4-15=29$.

Trick: Division and multiplication first: $16÷4=4$, $3\times 5=15$.

Answer: (A) प

Concept: Find the differences between consecutive terms.

1. $109-92=17$, $143-109=34$, $194-143=51$, $262-194=68$.
2. Differences are multiples of 17: 17, 34, 51, 68 → next 85.
3. $262+85=347$.

Trick: Differences $=17\times 1,2,3,4,5$.

Answer: (C) अ

Concept: Each letter of the group moves by the same step with wrap-around.

1. 1st letters: R(18) → K(11) → D(4) → W(23) → P(16): $-7$ each time ($4-7=-3\to 23$).
2. 2nd letters: N(14) → G(7) → Z(26) → S(19) → L(12): $-7$.
3. 3rd letters: T(20) → M(13) → F(6) → Y(25) → R(18): $-7$.
4. Next group: PLR.

Trick: Every letter $-7$; below 1 add 26.

Answer: (D) ह

Concept: Positions 1 (bottom) to 7 (top).

1. Only two boxes below E → E at 3. Only one above R → R at 6.
2. Exactly one box between R and H → H at 4.
3. G immediately above J: free positions 1, 2, 5, 7 → J at 1, G at 2.
4. F is below I → F at 5, I at 7. Stack: I, R, F, H, E, G, J. Between I (7) and F (5): only R = one.

Trick: Fix the exact-position clues, then place the "immediately above" pair in the only adjacent gap.

Answer: (B) प

Concept: Try reverse alphabetical positions (A = 26 … Z = 1) when the normal sum does not match.

1. NHJCR: reverse positions 13 + 19 + 17 + 24 + 9 $=82$; code 77 $=82-5$.
2. QETLO: 10 + 22 + 7 + 15 + 12 $=66$; code 61 $=66-5$ ✓.
3. PHASJ: 11 + 19 + 26 + 8 + 17 $=81$; code $=81-5=76$.

Trick: Code = (sum of reverse positions) − 5; reverse position $=27-$ normal position.

Answer: (C) क

Concept: The program counter (PC) is a CPU register that holds the memory address of the next instruction to be executed.

1. During the fetch cycle the CPU reads the address from the PC, fetches that instruction and increments the PC.
2. Secondary memory (hard disk) is too slow and is not accessed directly for instruction fetch; output registers and input buffers hold data, not addresses.

Yaad rakhein: PC = "pointer to the next instruction"; IR (instruction register) = holds the current instruction.

Answer: (C) System program counter

Concept: After Raja Ram Mohan Roy died in 1833, Debendranath Tagore (father of Rabindranath) revived and led the Brahmo Samaj; he had founded the Tattvabodhini Sabha in 1839 and joined the Samaj in 1843.

1. Keshub Chandra Sen joined later (1857) and split away in 1866 to form the Brahmo Samaj of India.
2. Dwarkanath Tagore was a business associate of Roy, not the Samaj's leader; Vivekananda belonged to the Ramakrishna movement.

Yaad rakhein: Brahmo leaders in order: Ram Mohan Roy → Debendranath Tagore → Keshub Chandra Sen.

Answer: (A) Debendra Nath Tagore

Concept: The Bastar Olympics is a mass sports event of the Bastar division of Chhattisgarh, started in 2024 to connect youth of the Maoist-affected region with the mainstream; its second edition was launched in October 2025.

1. Bastar (Jagdalpur) is in southern Chhattisgarh; the games cover village, block, district and divisional levels.
2. Jharkhand, Madhya Pradesh and Odisha have no "Bastar" region.

Yaad rakhein: Bastar = Chhattisgarh (Dussehra of Bastar, Chitrakote falls, Bastar Olympics).

Answer: (A) Chhattisgarh

Concept: Speed of sound in a gas $v\propto \sqrt{T}$ (absolute temperature).

1. At higher temperature molecules move faster and pass on the vibration more quickly, so sound travels faster – roughly $0.6\text{m/s}$ more per ${}^{\circ }\text{C}$ rise in air.
2. Density falls with heating, but pressure falls in the same proportion, so density alone does not slow sound; option (C) is wrong.

Yaad rakhein: Sound in air: $\approx 331\text{m/s}$ at $0^{\circ }\text{C}$, $\approx 343\text{m/s}$ at ${20}^{\circ }\text{C}$.

Answer: (D) It increases, because the air molecules move faster

Concept: The Hornbill Festival, the "Festival of Festivals", is held every year from 1 to 10 December at Kisama Heritage Village near Kohima, Nagaland.

1. It brings together all the Naga tribes with traditional dance, music, indigenous games and food; named after the hornbill bird revered in Naga folklore.
2. Assam has Bihu, Manipur has Sangai Festival, Meghalaya has Wangala/Cherry Blossom festivals.

Yaad rakhein: Hornbill – Nagaland; Sangai – Manipur; Wangala – Meghalaya; Bihu – Assam.

Answer: (D) Nagaland