Q&A: practice questions with answers

Concept: Track each letter position separately.

1. 1st: J, H, F, D → $-2$ → B.
2. 2nd: T, R, P, N → $-2$ → L.
3. 3rd: Q, S, U, W → $+2$ → Y.
4. Next term: BLY.

Trick: Two positions move backward by 2, the last forward by 2.

Answer: (C) BLY

Concept: Draw the minimum diagram; a conclusion follows only if it must be true.

1. Some shirts are pants; some pants are coats – the diagram can be drawn with all pants inside shirts, so (I) “Some pants are not shirts” is not certain.
2. Shirts and coats need not overlap (as drawn), so (II) does not follow.
3. Two “Some” statements give no definite conclusion.

Yaad rakhein: Some + Some = no conclusion.

Answer: (C) Neither conclusion I nor II follows

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

1. Fix Bob. Elma is immediately left of Bob; Ali is second to the left of Bob.
2. Carol is third to the right of Bob (opposite him); Dave is immediately left of Carol.
3. Frank takes the remaining seat, between Dave and Bob.
4. Counting right (anticlockwise) from Elma: Bob, then Frank → one person.

Trick: In a 6-seat table, “third to the left/right” is the seat directly opposite.

Answer: (C) One

Concept: Find the letter gaps inside each pair and between the first letters of pairs.

1. PU-RN-ST: P→U $+5$, R→N $-4$, S→T $+1$; first letters P→R→S: $+2$, $+1$.
2. SX-UQ-VW follows the same: $+5$, $-4$, $+1$; S→U→V.
3. KP-MI-NO: K→P $+5$, M→I $-4$, N→O $+1$; K→M→N ✓.
4. Others fail at the third pair (e.g. HM-JF-KN: K→N $=+3$).

Trick: Check the third pair (gap $+1$) first – it eliminates options fastest.

Answer: (D) KP-MI-NO

Concept: Find each even digit with an odd digit on both sides (odd–even–odd).

1. 1 8 5 ✓ (13th digit), 5 2 7 ✓, 7 8 1 ✓.
2. 7 6 9 ✓, 9 2 9 ✓.
3. Others fail, e.g. 1 6 6, 1 8 4, 8 4 5.
4. Total $=5$.

Trick: Mark only even digits, then glance left and right of each.

Answer: (D) 5

Concept: Compare gaps between consecutive letters (wrap after Z).

1. UZE: U→Z $+5$, Z→E $+5$.
2. BGL: $+5$, $+5$; PUZ: $+5$, $+5$.
3. KUP: K→U $+10$, U→P $-5$ ✗.

Trick: Positions: K = 11, U = 21, P = 16 – gaps not equal.

Answer: (A) KUP

Concept: Replace letters by signs, then apply BODMAS.

1. $14÷2+10-3\times 5$.
2. $=7+10-15$.
3. $=2$.

Trick: Do $÷$ and $\times$ before $+$ and $-$.

Answer: (C) 2

Concept: Find one rule that works for an option on both sides.

1. Try X = 9: $9\times 3+3=30$ ✓.
2. Apply to 12: $12\times 3+3=39$.
3. So X = 9, Y = 39 (option D).

Trick: X = 10 with $\times 3$ would give $Y=36$, but that pair (10, 36) is not an option – so test option pairs, not single values.

Answer: (D) $X=9,Y=39$

Concept: Decode each symbol one by one and draw the family tree.

1. T ÷ U: T is the son of U (T male).
2. U × V: U is the mother of V.
3. V + W: V is the sister of W; so W is also U’s child.
4. W − X: W is the husband of X, so W is male.
5. T and W are both sons of U ⇒ T is W’s brother.

Trick: Mark genders as you decode (T ♂, W ♂) – siblings of the same parent give brother/sister directly.

Answer: (A) Brother

Concept: Find the letter gaps inside each pair and between first letters.

1. EA-GC-IK: E→A $-4$, G→C $-4$, I→K $+2$; E→G→I: $+2$ each.
2. HD-JF-LN: $-4$, $-4$, $+2$; H→J→L ✓ same.
3. KG-MI-OQ: K→G $-4$, M→I $-4$, O→Q $+2$; K→M→O ✓.
4. Others break the pattern (e.g. LG: L→G $=-5$).

Trick: The first-pair gap ($-4$) removes B and C at once; D fails at N→I ($-5$).

Answer: (A) KG-MI-OQ

Concept: Power in a resistor is $P=\frac{V^2}{R}$ and heat produced in time $t$ is $H=P\times t$ (Joule's law of heating).

1. $P=\frac{{10}^2}{10}=\frac{100}{10}=10\text{W}$
2. Time $t=1\text{min}=60\text{s}$
3. $H=10\times 60=600\text{J}$

Trick: Heat in one minute is simply $60\times P$; first find $P$, then multiply by 60.

Answer: (D) $P=10W,H=600J$

Concept: When heat breaks one compound into simpler substances, it is a thermal decomposition reaction.

1. $\mathrm{C}\mathrm{a}\mathrm{C}{\mathrm{O}}_3\stackrel{\text{heat}}{\to }\mathrm{C}\mathrm{a}\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$
2. Limestone ($\mathrm{C}\mathrm{a}\mathrm{C}{\mathrm{O}}_3$) gives quick lime ($\mathrm{C}\mathrm{a}\mathrm{O}$) and carbon dioxide.
3. One reactant → two products using heat, so it is thermal decomposition.

Yaad rakhein: Heat → thermal decomposition; light → photolytic; electricity → electrolytic decomposition.

Answer: (B) Thermal decomposition

Concept: Reduction means loss of oxygen or gain of hydrogen; oxidation is the opposite (gain of oxygen or loss of hydrogen).

1. a) Loss of oxygen → reduction ✔
2. b) Gain of hydrogen → reduction ✔
3. c) and d) describe oxidation ✘

Trick: Reduction = 'O out, H in'. Oxidation = 'O in, H out'.

Answer: (B) Only a and b

Concept: Solids have tightly packed particles with strong forces of attraction, so they have a fixed shape and are almost incompressible.

1. Gases and plasma are highly compressible; their shape changes easily.
2. Liquids are nearly incompressible but take the shape of the container.
3. Solids keep a definite shape — the least change under pressure.

Yaad rakhein: Rigidity and fixed shape: solid > liquid > gas.

Answer: (D) Solid

Concept: Plants do not move, have many dead cells and therefore have low energy needs; so they use relatively slow transport systems (NCERT).

1. Option A wrong: water and minerals move through xylem, not phloem.
2. Option C wrong: food (products of photosynthesis) moves through phloem.
3. Option D wrong: distances are large in tall trees, not in herbs and shrubs.

Yaad rakhein: Xylem → water up (X = 'eXtra water'); Phloem → food (P = 'Products of photosynthesis').

Answer: (B) Plants have low energy needs and use relatively slow transport systems.

Concept: $\text{Mass \% }=\frac{\text{mass of solute}}{\text{mass of solution}}\times 100$, where solution = solute + solvent.

1. Mass of solution $=20+280=300\text{g}$
2. $\frac{20}{300}\times 100=\frac{20}{3}\%$
3. $=6.66\%$

Trick: Never divide by the solvent (280 g) — always by the total solution (300 g).

Answer: (D) 6.66%

Concept: Variations in a population help some individuals survive when the environment changes; this is the basis of natural selection.

1. Global warming raises the water temperature.
2. Bacteria variants that can tolerate heat will survive and reproduce.
3. Cold, water or wind resistance gives no advantage in hotter water.

Trick: Match the variation to the new stress: more heat → heat resistance.

Answer: (A) Resistant to heat

Concept: Ohm's law: at constant temperature, current through a conductor is directly proportional to the potential difference across it: $V\propto I$, i.e. $V=IR$.

1. Joule's law → heat produced $H=I^{2}Rt$.
2. Faraday's law → electromagnetic induction / electrolysis.
3. Newton's laws → motion and force.

Yaad rakhein: Ohm → $V=IR$ (graph of $V$ vs $I$ is a straight line).

Answer: (B) Ohm's law

Concept: Red blood corpuscles contain the iron-containing red pigment haemoglobin, which carries oxygen.

1. Chlorophyll – green pigment of plants.
2. Xanthophyll – yellow pigment of plants.
3. Melanin – skin and hair colour pigment.

Yaad rakhein: Haemoglobin has high affinity for oxygen and contains iron ($\mathrm{F}\mathrm{e}$).

Answer: (C) Haemoglobin

Concept: A seed contains the embryo: plumule (future shoot), radicle (future root) and cotyledons (stored food).

1. Plumule → shoot
2. Radicle → root
3. Cotyledon → food store; sepal and petal are flower parts.

Trick: Radicle sounds like 'radish' — a root!

Answer: (D) radicle