Q&A: practice questions with answers

Concept: Find letter-wise shift.

Solution:

1. BIRD → CLSG: +1, +3, +1, +3.
2. RAIN → SDJQ: +1, +3, +1, +3.
3. STAR → T W B U.

Trick: Alternating +1, +3: second letter T + 3 = W leaves TWBU/UWCV; first S + 1 = T → TWBU.

Answer: (B) TWBU

Concept: Sort the word and compare position by position.

Solution:

1. VISCOMETRY → C E I M O R S T V Y.
2. Compare: 5th O = O, 8th T = T, 10th Y = Y.
3. 3 letters unchanged.

Yaad rakhein: Write both words one below the other and tick matches — avoids errors.

Answer: (B) Three

Concept: Look at differences.

Solution:

1. 35 − 24 = 11, 40 − 35 = 5, 51 − 40 = 11, 56 − 51 = 5.
2. Pattern +11, +5 → next 56 + 11 = 67.

Trick: Alternate differences repeat: +11, +5, +11, +5, +11.

Answer: (A) 67

Concept: Between = Total − (left rank + right rank) when they don't overlap.

Solution:

1. Mahesh from left = 59 − 20 + 1 = 40.
2. Kailash = 14th from left.
3. Between = 40 − 14 − 1 = 25.

Trick: Direct: 59 − (20 + 14) = 25.

Answer: (C) 25

Concept: Check gaps between letters.

Solution:

1. DEK: D→E +1, E→K +6.
2. GHN: +1, +6; NOU: +1, +6.
3. NOD: N→O +1 but O→D is −11 — odd one.

Yaad rakhein: In odd-one-out, test the gap pattern on every option, not just the first two letters.

Answer: (D) NOD

Concept: Compare shift from first cluster to second.

Solution:

1. AB−EF, NO−RS, JK−NO: each letter +4.
2. XY−AE: X→A +3, Y→E +6.
3. XY−AE is odd.

Trick: Same-gap pairs: 2nd cluster also consecutive (EF, RS, NO); AE is not consecutive.

Answer: (B) XY−AE

Concept: Find letter-wise shift.

Solution:

1. CMO → DKP: +1, −2, +1.
2. WKL → XIM: +1, −2, +1.
3. OIY → PGZ: +1, −2, +1 ✓.

Trick: First letter must change by +1 — only OIY→PGZ does (others keep it same).

Answer: (A) OIY : PGZ

Concept: Common letter ↔ common digit.

Solution:

1. GODS = 7642, DIVE = 1357.
2. Common letter: D; common digit: 7.
3. So D = 7.

Trick: Ignore order; match the shared letter with the shared digit.

Answer: (B) 7

Concept: Need pattern Vowel–Consonant–Vowel.

Solution:

1. Vowels present: I, U, O only.
2. After I comes K, then Z (consonant); after U comes W, then X; after O comes Y, then X.
3. No consonant has a vowel on both sides → none.

Trick: Only look at the letters right after each vowel — just 3 checks.

Answer: (C) None

Concept: Pattern needed: Odd – Even – Even.

Solution:

1. 5 2 6 ✓ (2 after 5, before 6).
2. 7 8 4 ✓.
3. 3 4 6 ✓ (end).
4. Others (3 4 9, 9 6 7) fail → 3.

Trick: Scan for odd digits and check the next two digits are both even.

Answer: (C) Three

Concept: In a chemical reaction, total mass of reactants = total mass of products.

Solution:

1. CaCO₃ → CaO + CO₂.
2. 10 g = 5.6 g + mass of CO₂.
3. CO₂ = 10 − 5.6 = 4.4 g.

Trick: Molar check: 100 g CaCO₃ → 56 g CaO + 44 g CO₂; so 10 g gives 4.4 g.

Answer: (C) 4.4 g

Concept: Catenation and tetravalency let carbon form millions of compounds.

Solution:

1. Carbon bonds with itself (catenation) forming chains, branches and rings.
2. Valency 4 lets it bond with H, O, N, S, halogens.
3. These form the huge family of organic compounds.

Yaad rakhein: Catenation is strongest in carbon; silicon shows it only to a limited extent.

Answer: (A) Organic compounds

Concept: Turmeric is a natural indicator: yellow in acid/neutral, reddish-brown in base.

1. Turmeric (haldi) turns red/reddish-brown in basic solution — e.g. a haldi stain turns red with soap. So (B) is correct.
2. Red litmus turns blue in bases, not acids — (A) is wrong.
3. Methyl orange is a synthetic indicator, not obtained from turmeric — (C) is wrong.
4. China rose extract turns dark pink (magenta) in acids and green in bases — so 'green in acids' (D) is wrong.

Trick: China rose: acid → magenta, base → green; turmeric: base → red.

Answer: (B) Turmeric turns red in basic solution.

Concept: Echo travels to the surface and back: $v=\frac{2d}{t}$.

Solution:

1. Total distance = $2\times 990=1980$ m.
2. $v=\frac{1980}{6}=330$ m/s.

Trick: Forgetting the 2 gives 165 m/s — a trap option.

Answer: (C) 330 m/s

Concept: Weight F = mg, so F ∝ m on the same planet.

Solution:

1. F = mg = X.
2. Mass tripled → F = 3mg = 3X N.

Yaad rakhein: g does not depend on the body's own mass; only F (weight) changes.

Answer: (D) 3X N

Concept: Object beyond 2F → image between F and 2F, real, inverted, diminished.

Solution:

1. f = 40 cm → 2F = 80 cm; object at 100 cm is beyond 2F.
2. Image forms between F and 2F on the other side: 40–80 cm.
3. (Exact: $\frac{1}{v}=\frac{1}{40}-\frac{1}{100}\to v\approx 66.7$ cm, within range.)

Yaad rakhein: Object at 2F → image at 2F, same size; object between F and 2F → image beyond 2F, enlarged.

Answer: (C) The screen should be placed anywhere between 40 cm and 80 cm from the lens.

Concept: A molecule of two atoms of the same element is a homonuclear diatomic molecule.

Solution:

1. H₂: two H atoms ✓.
2. H₂O and CO contain different elements; O₃ has three atoms (triatomic).

Yaad rakhein: Diatomic elements: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂.

Answer: (A) H₂

Concept: Meristems are growth tissues that keep dividing.

Solution:

1. Cells are small, thin-walled, with dense cytoplasm and prominent nucleus.
2. They lack (or have very small) vacuoles and divide actively.
3. Thick walls and storage are features of permanent tissues.

Yaad rakhein: Types: apical (length), lateral (girth), intercalary (at nodes/leaf base).

Answer: (C) Small, actively dividing cells with thin walls.

Concept: Colloids look homogeneous but are actually heterogeneous.

Solution:

1. 1 ✓: particles are tiny, so they appear uniform.
2. 2 ✗: at molecular level they are heterogeneous (dispersed phase + medium).
3. 3 ✓ Tyndall effect; 4 ✓ milk is an emulsion (colloid).

Trick: Eliminate statement 2 → only option without 2 is '1, 3 and 4 only'.

Answer: (D) 1, 3 and 4 only

Concept: Slope of v–t graph = acceleration.

Solution:

1. Horizontal line → slope 0 → acceleration 0.
2. Velocity stays the same → constant (uniform) velocity.

Yaad rakhein: Horizontal x–t graph = at rest; horizontal v–t graph = constant velocity.

Answer: (B) Constant velocity