Q&A: practice questions with answers

Concept: Fix certain positions (bottom = 1) and fill the rest.

Solution:

1. D = 5th from bottom; five between E and B → they are at 1 and 7
2. G below B, above D → B = 7, G = 6, E = 1
3. F just below A, C above A → A = 3, F = 2, C = 4
4. Between C (4) and G (6): D → 1 box

Trick: 'Five between' in 7 boxes means one is top and the other bottom.

Answer: (C) One

Concept: Sort the letters and compare position by position.

Solution:

1. Original: A L T E R I N G
2. Sorted: A E G I L N R T
3. Only A (position 1) stays the same → 1

Trick: Write both rows one under the other and tick the matches.

Answer: (B) One

Concept: Total = rank from top + rank from bottom − 1.

Solution:

1. Total = 37 + 155 − 1
2. = 191

Yaad rakhein: Subtract 1 because the person is counted in both ranks.

Answer: (C) 191

Concept: Find the operation linking first and second number.

Solution:

1. $29\times 2+1=59$
2. $35\times 2+1=71$
3. $41\times 2+1=83$ ✓

Trick: Second number is always odd and just above double — 46,91 / 56,112 / 53,105 all fail.

Answer: (C) 41, 83

Concept: The common word gets the common code.

Solution:

1. 'lamp' is in both sentences
2. Common code in 'nh rz vk' and 'rz bw mt' = rz
3. lamp = rz

Trick: Match repeated word with repeated code.

Answer: (D) rz

Concept: Track x (east) and y (north) coordinates.

Solution:

1. Start (0, 0); 7 S → (0, −7); left (E) 2 → (2, −7)
2. Right (S) 4 → (2, −11); left (E) 6 → (8, −11)
3. Left (N) 13 → (8, 2); left (W) 8 → (0, 2) = M
4. M is 2 km north of L → drive 2 km south

Trick: East total 2 + 6 − 8 = 0; North total −7 − 4 + 13 = +2 → go 2 km South.

Answer: (B) 2 km towards South

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

Solution:

1. (5 + 3) × 2 = 16
2. (6 + 3) × 2 = 18
3. (8 + 12) × 2 = 40 ✓; others: (7+4)×2 = 22, (4+2)×2 = 12, (6+4)×2 = 20 ✗

Trick: Middle number must be even (it is a double) — this removes (4, 13, 2) and (6, 19, 4) at once.

Answer: (D) (8, 40, 12)

Concept: Track each letter position separately.

Solution:

1. 1st letters: M, U, C, K → +8 each → S
2. 2nd letters: F, N, V, D → +8 → L
3. 3rd letters: Y, G, O, W → +8 → E
4. Answer = SLE

Trick: All three positions move +8 (wrap after Z), so find one and match.

Answer: (D) SLE

Concept: Number 7 seats, counting rightwards from Aw (Fe in the question means Fn).

Solution:

1. Aw = 0, Bt = 2 (2nd right), Ck = 4 (3rd left = 4th right)
2. Free seats 1, 3, 5, 6: Gy, Er = Gy + 1, Dv = Er + 2 → only Gy = 5, Er = 6, Dv = 1
3. Fn = 3
4. Fn is 4th to the right of Er = 3rd to the left

Trick: In 7 seats, k-th to the right = (7 − k)-th to the left.

Answer: (A) Third to the left

Concept: Each position moves by a fixed step (wrap after Z).

Solution:

1. 1st: Z, O, D, S → +15 → S + 15 = H
2. 2nd: J, Y, N, C → +15 → R
3. 3rd: F, U, J, Y → +15 → N
4. Answer = HRN

Trick: +15 is the same as −11: S − 11 = H, C − 11 = R, Y − 11 = N.

Answer: (A) HRN

Concept: Only about 10% of energy passes from one trophic level to the next (Lindeman's 10% law).

Solution:

1. About 90% energy is lost as heat, respiration and waste at each level
2. After 4 levels only ~0.1% of producer energy remains — too little to support another level
3. Producers are not rare; decomposers act at all levels, not just top

Yaad rakhein: Energy flow in an ecosystem is unidirectional; the energy pyramid is always upright.

Answer: (B) Only 10% of energy passes to the next level

Concept: $F=ma$, where $a=\frac{v-u}{t}$.

Solution:

1. $a=\frac{5-0}{10}=0.5{\text{m/s}}^2$
2. $m=\frac{F}{a}=\frac{100}{0.5}$
3. $=200$ kg

Trick: $m=\frac{Ft}{v}=\frac{100\times 10}{5}=200$ kg (impulse = change in momentum).

Answer: (B) 200 kg

Concept: Galvanisation = coating iron/steel with a thin layer of zinc to prevent rusting.

Solution:

1. Zinc layer blocks air and moisture
2. Zinc is more reactive, so it corrodes first and protects iron even if scratched
3. Tinning uses tin; electroplating is a general method using electricity

Yaad rakhein: G for Galvanise, Z for Zinc — 'GZ' pair; tin-coating is for food cans.

Answer: (A) Galvanisation

Concept: Xylem has tracheids, vessels, xylem parenchyma and xylem fibres, each with a role.

Solution:

1. Xylem parenchyma (only living part) stores food
2. Xylem fibres are dead, thick-walled cells → mechanical support
3. Water/mineral transport is done mainly by tracheids and vessels

Yaad rakhein: Of the 4 xylem elements, only xylem parenchyma is living.

Answer: (B) Supportive function

Concept: Cell shape and size are adapted to the specific job the cell does.

Solution:

1. Nerve cells are long and branched to carry messages
2. RBCs are biconcave discs to carry oxygen
3. All body cells of one organism have the same genome — so the last option is wrong

Yaad rakhein: Division of labour in multicellular organisms leads to specialised cells.

Answer: (A) Their structure is adapted to perform specific functions.

Concept: Metal loses electrons, non-metal gains them; ions are held by electrostatic attraction.

Solution:

1. Na (2,8,1) loses 1 e⁻ → Na⁺; Cl (2,8,7) gains it → Cl⁻
2. Both get noble-gas configuration
3. Strong attraction between Na⁺ and Cl⁻ makes NaCl stable (ionic, not covalent)

Trick: Metal + non-metal → transfer (ionic); non-metal + non-metal → sharing (covalent).

Answer: (A) Sodium donates one electron to chlorine, forming oppositely charged ions held by strong electrostatic forces.

Concept: Rays from infinity are parallel; a concave lens diverges them so they appear to come from the focus.

Solution:

1. Object at infinity → image at focus F₁ (same side)
2. Image is virtual and erect
3. Image is highly diminished (point-sized)

Yaad rakhein: A concave lens always gives a virtual, erect, diminished image for any object position.

Answer: (B) Virtual, erect, and highly diminished at the focus

Concept: After fertilisation, ovary → fruit and ovule → seed.

Solution:

1. Zygote inside the ovule forms the embryo; ovule becomes seed
2. Ovary ripens into fruit
3. Sepals, petals, stamens, stigma and style usually wither and fall

Trick: 'Ovary is bigger → fruit (bigger); ovule is smaller → seed (smaller)'.

Answer: (C) The ovary develops into a fruit while the ovule becomes a seed

Concept: Object between F₁ and 2F₁ of a convex lens → image beyond 2F₂.

Solution:

1. Image forms beyond 2F₂ on the other side
2. It is real and inverted
3. It is enlarged

Trick: Object between F and 2F ↔ image beyond 2F (they swap with object beyond 2F). Closer object (but beyond F) → bigger image.

Answer: (A) Enlarged, real, inverted

Concept: Striated (skeletal) muscles are voluntary and attached to bones.

Solution:

1. Striated muscles: long, cylindrical, multinucleate, light-dark bands, voluntary
2. Smooth muscle: involuntary (gut, blood vessels)
3. Cardiac: involuntary, heart only; tendon is connective tissue, not muscle

Yaad rakhein: Tendon joins muscle to bone; ligament joins bone to bone.

Answer: (C) Striated (skeletal) muscle