Q&A: practice questions with answers

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

1. Fix M; N is 3rd to M's left. M's neighbours are E and L.
2. If L were on M's left, G (two places right of L) would clash with E; so L is on M's right, E on M's left, G two right of L.
3. D, F fill the two gaps; D not beside N ⇒ order (clockwise): M, E, F, N, G, D, L.
4. From F going right: E, M, then L ⇒ 2 people between.

Trick: Start with the person mentioned most (M) and test both sides for the neighbours.

Answer: (B) Two

Concept: Drop all symbols and keep only the digits in order.

1. Digits left: 3, 2, 1, 4, 6, 8, 7, 9, 1, 5.
2. 7th from left $=7$.

Trick: Write only the digits in a line, then count – don't count symbols by mistake.

Answer: (D) 7

Concept: Check the gaps inside each cluster (positions: A=1 … Z=26).

1. TQ–OG: T$-3=$Q, O$-8=$G. ZW–UM: Z$-3=$W, U$-8=$M.
2. BY–WO: B$-3=$Y (cyclic), W$-8=$O.
3. GK–PW: G$+4=$K, P$+7=$W ✗.

Trick: Backward counting wraps around: B → A → Z → Y is $-3$.

Answer: (C) GK – PW

Concept: Vowels → previous letter; consonants → next letter.

1. S→T, H→I, O→N, W→X, I→H, N→O, G→H ⇒ TINXHOH.
2. Original letters S, H, O, W, I, N, G: H, O, I, N appear in the new group; S, W, G do not.

Trick: Vowel O→N and I→H create letters N and H; consonants N→O and H→I bring O and I back.

Answer: (C) 4

Concept: Facing north: left end $=$ position 1, right end $=$ position 7.

1. Only B is left of A ⇒ B $=1$, A $=2$.
2. Four between B and Q ⇒ Q $=6$.
3. R-E-P together with E in middle, R not beside Q ⇒ R $=3$, E $=4$, P $=5$.
4. F takes the last seat, 7 (extreme right).

Trick: Fix 'only' clues first – they lock exact positions.

Answer: (D) F

Concept: Common letters get common digits; the unique letter gets the unique digit.

1. POKED → 8, 5, 2, 6, 3; TOKED → 3, 6, 5, 8, 4.
2. Common letters O, K, E, D ↔ common digits 3, 5, 6, 8.
3. P ↔ 2 (only in first), T ↔ 4 (only in second).

Trick: Cancel common digits in both codes; what remains pairs with the odd letters.

Answer: (D) 4

Concept: Each letter of a group moves one step forward in the next group.

1. P→Q→R→S→T; R→S→T→U→V; T→U→V→W→X.
2. Next group $=$ TVX.

Trick: Inside each group the letters skip one (P_R_T), so the answer must also skip one: TVX.

Answer: (D) TVX

Concept: Track net East–West and North–South movement (South first, then left turns from South face East).

1. 3 km South, left → 2 km East, left → 2 km North, right → 3 km East, left → 1 km North.
2. North–South: $-3+2+1=0$.
3. East: $2+3=5$ km ⇒ B is 5 km East of A.

Trick: Net vertical $=0$ means A and B are on the same line; just add the East moves.

Answer: (D) 5 km West

Concept: Common letters get common digits; the unique letter gets the unique digit.

1. REST → 4, 5, 2, 9; HERS → 5, 2, 7, 9.
2. Common letters R, E, S ↔ common digits 2, 5, 9.
3. T ↔ 4 (only in REST), H ↔ 7 (only in HERS).

Trick: Cancel common digits in both codes; the leftover digit belongs to the odd letter.

Answer: (A) 7

Concept: Arrange the letters alphabetically and compare position by position.

1. Original: D R A W I N G.
2. Sorted: A D G I N R W.
3. Compare: D/A, R/D, A/G, W/I, I/N, N/R, G/W – no letter stays in place.

Trick: Write both words one under the other and tick matching columns.

Answer: (A) 0

Concept: A convex mirror always forms a virtual, erect and diminished image behind the mirror, between P and F.

1. As the object moves farther away, the image shrinks and moves towards F.
2. For an object at infinity, parallel rays reflect and appear to diverge from the focus F.
3. So the image is highly diminished, point-sized and virtual, located at F behind the mirror.

Trick: Convex mirror: object at $\infty$ → image at F (behind); any other position → between P and F.

Answer: (B) behind the mirror at the focus F

Concept: The plasma membrane is the selectively permeable outer boundary of every cell.

1. It is a lipid bilayer (mainly phospholipids).
2. Proteins are embedded in or attached to this bilayer (fluid mosaic model).
3. Cellulose is found in the plant cell wall, not in the membrane.

Yaad rakhein: Membrane = lipids + proteins (fluid mosaic); cell wall = cellulose.

Answer: (C) Both proteins and lipids

Concept: Variations arise mainly because DNA copying during reproduction is never 100% accurate.

1. Before cell division, the cell makes a copy of its DNA.
2. Small errors in this copying create slightly different DNA in offspring.
3. These variations help some individuals survive changing environments, aiding the species over time.

Yaad rakhein: Inaccurate DNA copying → variation → survival of species (NCERT Class 10).

Answer: (B) DNA copying

Concept: Oxidising agents add oxygen to alcohols and convert them into carboxylic acids.

1. Alkaline $\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4$ (or acidified ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$) is a strong oxidising agent.
2. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{O}\mathrm{H}\stackrel{\mathrm{a}\mathrm{l}\mathrm{k}.\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4+\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{t}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}$
3. Pd and Ni are hydrogenation catalysts; Zn is a reducing metal.

Trick: Purple $\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4$ loses its colour while it oxidises the alcohol.

Answer: (C) Alkaline potassium permanganate

Concept: The number of heart chambers differs among vertebrate groups.

1. Fishes: 2 chambers (1 atrium, 1 ventricle).
2. Amphibians and most reptiles: 3 chambers (2 atria, 1 ventricle), so some mixing of blood occurs.
3. Birds, mammals and crocodiles: 4 chambers.

Trick: Fish 2 → Frog 3 → Bird/Man 4.

Answer: (D) Three

Concept: Members of a homologous series have the same functional group and differ by a $-\mathrm{C}{\mathrm{H}}_2-$ unit.

1. $\mathrm{C}{\mathrm{H}}_3\mathrm{O}\mathrm{H}$, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{O}\mathrm{H}$, ${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{O}\mathrm{H}$ fit ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+1}\mathrm{O}\mathrm{H}$ — alcohols, each differing by $\mathrm{C}{\mathrm{H}}_2$.
2. ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{O}\mathrm{H}$ (phenol) has –OH on a benzene ring, not on an alkyl chain.
3. It does not fit the general formula, so it is the odd one out.

Trick: Check the formula ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+1}\mathrm{O}\mathrm{H}$: for $n=6$ it should be ${\mathrm{C}}_6{\mathrm{H}}_{13}\mathrm{O}\mathrm{H}$, not ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{O}\mathrm{H}$.

Answer: (D) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{O}\mathrm{H}$

Concept: Starlight passes through atmospheric layers of continuously changing density and refractive index.

1. Light from a star bends (refracts) repeatedly on entering the atmosphere.
2. Since the air is always moving, the amount of bending keeps changing.
3. So the amount of light reaching the eye flickers — the star twinkles.

Yaad rakhein: Twinkling, advance sunrise, delayed sunset → atmospheric refraction. Blue sky → scattering.

Answer: (B) the atmospheric refraction of light from the star

Concept: Voltage is work done per unit charge: $V=\frac{W}{Q}$.

1. Charge $Q=I\times t$.
2. Substitute: $V=\frac{W}{Q}=\frac{W}{I\times t}$.
3. So voltage $=\frac{\text{Work done}}{\text{Current}\times \text{Time}}$.

Trick: Units check: $\frac{\text{J}}{\text{A}·\text{s}}=\frac{\text{J}}{\text{C}}=\text{V}$.

Answer: (D) $\frac{\text{Work done}}{\text{Current}\times \text{Time}}$

Concept: Paramecium is a unicellular protozoan whose body is covered with cilia used for movement and feeding.

1. Amoeba is unicellular but moves by pseudopodia, not cilia.
2. Ticks (arthropods) and leeches (annelids) are multicellular.
3. So the unicellular ciliated organism is Paramecium.

Trick: Paramecium → cilia; Amoeba → pseudopodia; Euglena → flagellum.

Answer: (A) Paramecium

Concept: Evaporation is a surface phenomenon — only surface particles escape into vapour.

1. Larger surface area exposes more particles to the air.
2. More particles escape per second, so evaporation becomes faster.
3. That is why clothes are spread out to dry.

Yaad rakhein: Evaporation ↑ with surface area, temperature and wind speed; ↓ with humidity.

Answer: (B) increased