Q&A: practice questions with answers

Concept: Number the letters of the first word 1 2 3 4 and track their order.

1. JAIL (1234) → ALIJ $=2431$ → LIAJ $=4321$ (reverse).
2. ZEAL → ELAZ (2431) → LAEZ (4321) – same.
3. BORE → OERB (2431) → EROB (4321) – matches.

Trick: The third word is just the first word reversed – check that first to eliminate options fast.

Answer: (D) BORE - OERB - EROB

Concept: Two 'Some' statements in a chain give no definite link between the end terms.

1. Rocks–Apples, Apples–Horses, Horses–Nails are each 'Some' links.
2. I: Nails and Apples are linked only through Horses (Some + Some) – no definite conclusion.
3. II: Horses and Rocks are linked only through Apples (Some + Some) – no definite conclusion.

Yaad rakhein: Some + Some = no conclusion. Draw a chain of just-touching circles to see it.

Answer: (C) Neither conclusion (I) nor (II) follows.

Concept: Trace each move on a grid with A at the origin (East = $+x$, North = $+y$).

1. North 7: $(0,7)$; right → East 2: $(2,7)$; right → South 1: $(2,6)$.
2. Left → East 3: $(5,6)$; right → South 3: $(5,3)$; right → West 3: $(2,3)$.
3. Left → South 3: B $=(2,0)$.
4. A $=(0,0)$ lies 2 km to the West of B.

Trick: Only net displacement matters: East $2+3-3=2$, North $7-1-3-3=0$.

Answer: (D) 2 km West

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

1. A + B: A is the mother of B.
2. B − C: B is the husband of C; C + D: C is the mother of D.
3. D × E: D is the sister of E, so E is a child of B and C.
4. B is E's father and A is B's mother – A is E's father's mother.

Trick: Mark gender as soon as a relation reveals it (mother ♀, husband ♂, sister ♀).

Answer: (B) Father’s mother

Concept: Draw the Schools circle inside the Colleges circle, with University completely separate.

1. All schools are colleges ⇒ every school is a college, so some colleges (at least the schools) are schools ⇒ I follows.
2. No college is a university and every school is a college ⇒ no school is a university, so 'Some schools may be universities' is impossible ⇒ II does not follow.

Yaad rakhein: 'All A are B' ⇒ 'Some B are A' always follows; 'All A are B' + 'No B is C' ⇒ 'No A is C', so 'A may be C' is false.

Answer: (C) Only conclusion I follows

Concept: Treat each letter position as its own series.

1. 1st letters: Y, Z, A, B, C ($+1$).
2. 2nd letters: E, G, I, K, M ($+2$).
3. 3rd letters: S, V, Y, B, E ($+3$, wrapping Z → A).
4. Next term: CME.

Trick: Split clusters column-wise; each column usually has a simple constant step.

Answer: (B) CME

Concept: Look at every symbol followed by an odd number, then check what is just before it.

1. Symbols followed by odd numbers: # (1), $ (7), & (3), Ω (5).
2. Before them: # ← 9, $ ← 8, & ← 7, Ω ← 3 – all numbers.
3. No symbol has a symbol before it and an odd number after it.

Trick: Scan for the rarer condition first (odd number after a symbol) – it gives a short list to check.

Answer: (A) None

Concept: All face north, so their left is our left. Fix A and B first.

1. Three people right of A: A at 4. B at 2.
2. F sits between C and G, so C, F, G are consecutive; D is second to the left of C.
3. Only D at 3, C at 5, F at 6, G at 7 fits; E takes seat 1.
4. Between E (1) and D (3): only B – one person.

Trick: 'Immediate neighbour of both C and G' means F sits exactly in the middle of them.

Answer: (C) One

Concept: Letters in a word are coded by digits in jumbled order; compare common letters and digits.

1. TREF → 3528, KERF → 5438.
2. Common letters R, E, F ↔ common digits 3, 5, 8.
3. Remaining: T ↔ 2 and K ↔ 4.

Trick: The letter unique to a word takes the digit unique to its code.

Answer: (C) 4

Concept: Each letter moves back by 11 positions.

1. $V(22)\to K(11)$, $S(19)\to H(8)$, $U(21)\to J(10)$, $R(18)\to G(7)$: all $-11$.
2. Check: $Y(25)\to N(14)$ – also $-11$.
3. SPRO: $S(19)\to H(8)$, $P(16)\to E(5)$, $R(18)\to G(7)$, $O(15)\to D(4)$.

Trick: Shifting all letters by the same amount keeps the pattern – only the 1st letter must be checked to pick an option starting with H, then verify the order.

Answer: (A) HEGD

Concept: Testes are the male reproductive organs; they produce sperms and the hormone testosterone.

1. Testes lie outside the abdominal cavity in the scrotum, because sperm formation needs a temperature lower than body temperature – so (A) and (B) are wrong.
2. Sperms are formed in the testes, not in the seminal vesicles – (D) is wrong.
3. Testes secrete testosterone, which controls sperm formation and male secondary sexual characters – (C) is correct.

Yaad rakhein: Testes = sperm + testosterone; scrotum keeps them cooler than the body.

Answer: (C) Testes secrete testosterone hormone.

Concept: Endoplasmic reticulum (ER) is of two types: rough ER (RER) and smooth ER (SER).

1. RER has ribosomes attached to its surface, which give it a rough, granular look.
2. Ribosomes are the sites of protein synthesis, so RER makes proteins.
3. SER has no ribosomes and helps make fats (lipids).

Trick: Rough = Ribosomes (both start with R); Smooth = no ribosomes, makes lipids.

Answer: (B) ribosomes

Concept: Gravity always acts vertically downward, towards the centre of the Earth.

1. When a body is thrown upward, its motion (velocity) is upward.
2. Gravity acts downward, i.e. opposite to the motion, so the body slows down ($a=-g$).
3. At the highest point $v=0$, then the body falls back.

Trick: Going up → gravity opposes motion; coming down → gravity along motion.

Answer: (B) opposite to the direction of motion

Concept: Metals high in the reactivity series (K, Na, Ca, Mg, Al) cannot be reduced by carbon; they are obtained by electrolytic reduction.

1. Sodium is obtained by electrolysis of molten sodium chloride.
2. At cathode: $\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{e}}^{-}\to \mathrm{N}\mathrm{a}$; at anode chlorine gas is released.
3. Zn and Pb are reduced by carbon; Ag is low in the series.

Yaad rakhein: Top of reactivity series (K, Na, Ca, Mg, Al) → electrolysis of molten salt.

Answer: (C) Na

Concept: Acids are sour, release ${\mathrm{H}}^{+}$ ions in water and have pH less than 7.

1. Acids turn blue litmus red – option (B).
2. Turning red litmus blue, pH greater than 7 and slippery (soapy) feel are properties of bases.
3. So only (B) is a property of acids.

Trick: Acid → Red (blue litmus turns red); Base → Blue.

Answer: (B) They turn blue litmus paper red.

Concept: Distance = total path length; displacement = shortest distance between initial and final positions.

1. Upward journey $=5\text{ m}$, downward journey $=5\text{ m}$.
2. Total distance $=5+5=10\text{ m}$.
3. The body returns to the starting point, so displacement $=0\text{ m}$.

Trick: Back to the starting point → displacement is always zero, distance is not.

Answer: (C) 0m, 10m

Concept: First law of refraction: the incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.

1. Second law (Snell's law): $\frac{\sin i}{\sin r}=\text{constant}$.
2. The blank refers to the ray that strikes the interface – the incident ray.
3. Reflected or emergent rays are not part of this law.

Yaad rakhein: Same-plane law has three members: incident ray, refracted ray, normal.

Answer: (C) Incident

Concept: Metal oxides are basic; they react with acids to form salt and water.

1. Copper oxide is CuO, in which copper is in +2 state.
2. $\mathrm{C}\mathrm{u}\mathrm{O}+2\mathrm{H}\mathrm{C}\mathrm{l}\to \mathrm{C}\mathrm{u}\mathrm{C}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$
3. $\mathrm{C}\mathrm{u}\mathrm{C}{\mathrm{l}}_2$ is copper(II) chloride; the solution turns blue-green.

Yaad rakhein: Metal oxide + acid → salt + water; blue-green colour = $\mathrm{C}\mathrm{u}\mathrm{C}{\mathrm{l}}_2$.

Answer: (D) Copper (II) chloride

Concept: Matter is made of extremely small particles.

1. Particles of matter have spaces between them.
2. They are continuously moving (they have kinetic energy).
3. They attract each other.
4. They are very small, not very large – so (C) is NOT a characteristic.

Yaad rakhein: Three properties: space between, constant motion, mutual attraction – and tiny size.

Answer: (C) They are very large in size.

Concept: For a concave mirror, an object placed between pole P and focus F gives an image behind the mirror.

1. A ray parallel to the axis reflects through F; a ray through C reflects back on itself.
2. The reflected rays diverge, so they meet only when extended behind the mirror.
3. Hence the image is virtual, erect and enlarged (used in shaving/make-up mirrors).

Yaad rakhein: Concave mirror gives a virtual, erect image only when the object is between P and F.

Answer: (C) Virtual and erect