Q&A: practice questions with answers

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

1. U in middle of Q and R; try R to U's left. P is 2nd left of R
2. V and T: T is 2nd to the right of V ⇒ order U, R, T, P, V, S, Q (clockwise)
3. S's neighbours V and Q, not T ✓ (R on U's right fails – P's seat clashes with Q)
4. Third to the left of U: R, T, P ⇒ P

Trick: Fix the 3-person block Q-U-R first, then place others.

Answer: (B) P

Concept: Check the gap between successive letters (cyclic A–Z).

1. GDAX: G→D→A→X, each $-3$
2. KHEB: each $-3$; AXUR: A→X→U→R, each $-3$
3. OKJH: O→K $-4$, K→J $-1$, J→H $-2$ ✗

Trick: Remember A $-3$ wraps to X (26 → 24).

Answer: (D) OKJH

Concept: Each letter position moves $+1$ from one term to the next.

1. HKN → ILO → JMP → KNQ (each letter $+1$)
2. Next: K→L, N→O, Q→R
3. Answer term $=$ LOR

Trick: Inside each term the gap is $+3$: L, O, R keeps it.

Answer: (A) LOR

Concept: Replace letters by signs, then apply BODMAS.

1. A $=÷$, B $=\times$, C $=+$, D $=-$
2. $4+8-9÷3\times 2$
3. $=4+8-3\times 2=4+8-6$
4. $=6$

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

Answer: (B) 6

Concept: Number positions $1$ (top) to $7$ (bottom) and fix the definite clues first.

1. Three boxes above X ⇒ X $=4$; Z at $2$ or $6$
2. A above X and three boxes between Z and A ⇒ Z $=6$, A $=2$; C $=3$
3. Left: B, L, Y at $1,5,7$; B not next to Z ⇒ B $=1$
4. L above Y ⇒ L $=5$, Y $=7$ (lowermost)

Trick: Start with the clue that gives an exact position (X $=4$).

Answer: (A) Y

Concept: Both pairs must follow the same rule; test options.

1. Try rule $\times 4$: $9\times 4=36$ ✓
2. $15\times 4=60$ ⇒ Y $=60$
3. Option (B): X $=9$, Y $=60$ fits; others fail (e.g. $12\times 3=36$ but $15\times 3=45\ne 60$)

Trick: Check each option with one rule for both sides.

Answer: (B) X = 9, Y = 60

Concept: Track net East–West and North–South displacement from A.

1. South $22$ → left (East) $20$ → left (North) $12$ → left (West) $17$ → right (North) $10$
2. East–West: $20-17=3$ km East of A
3. North–South: $-22+12+10=0$
4. P is $3$ km East of A ⇒ drive $3$ km towards West

Trick: N–S cancels to $0$, so only the E–W gap matters.

Answer: (C) 3 km towards West

Concept: Check each even digit: previous digit must be a perfect square ($0,1,4,9$) and next digit odd.

1. …3 4 8 5… : 8 preceded by 4, followed by 5 ✓
2. …2 4 8 9… : 8 preceded by 4, followed by 9 ✓
3. Others fail, e.g. 9 2 2, 1 6 6, 1 6 8

Trick: Scan for 1, 4, 9 first, then check the two digits after it.

Answer: (A) 2

Concept: Facing the centre, left = clockwise. Fix E and place others by left-counts.

1. G immediate left of E; P third to the left of E ⇒ clockwise E, G, _, P
2. Only L between A and P ⇒ L next to P, then A: E, G, _, P, L, A, _
3. B not next to P ⇒ B in last seat, M in the gap: E, G, M, P, L, A, B
4. Fourth to the left of M: P, L, A, B ⇒ B

Trick: In a 7-seat circle, 4th left $=$ 3rd right: M's 3rd right is also B.

Answer: (A) B

Concept: Find the letter-wise shift in the example pairs.

1. SOTQ → PLQN: each letter $-3$
2. OKPM → LHMJ: each letter $-3$ ✓
3. LHMJ $-3$: L→I, H→E, M→J, J→G = IEJG

Trick: The second pair's answer becomes the next question – just apply $-3$ once more.

Answer: (B) IEJG

Concept: Newton's third law: forces always occur in pairs – action and reaction are equal in magnitude, opposite in direction and act on two different bodies.

1. If body A pushes B with force $\overrightarrow{F}$, B pushes A with $-\overrightarrow{F}$.
2. Since they act on different bodies, they never cancel each other.
3. Example: a gun recoils backward when the bullet moves forward.

Yaad rakhein: Equal, opposite, different bodies – that is why a horse can still pull a cart.

Answer: (B) there is an equal and opposite reaction and they act on two different bodies

Concept: Inside a long current-carrying solenoid the magnetic field lines are parallel straight lines, showing a uniform field.

1. Field strength inside: $B={\mu }_{0}nI$ ($n$ = turns per unit length).
2. $B$ does not depend on position inside, so it is the same at all points.
3. Outside, the field resembles that of a bar magnet.

Yaad rakhein: Parallel, equally spaced lines = uniform field. Solenoid inside → uniform.

Answer: (D) Same at all points

Concept: Work is the energy transferred by a force, so work and energy are measured in the same unit, the joule (J).

1. $W=F\times s$ → $\mathrm{N}\mathrm{m}=J$.
2. Energy (kinetic, potential, heat) is also in J.
3. Force → N; inertia is measured by mass → kg. So other pairs differ.

Yaad rakhein: Work = energy spent → same unit joule ($1J=1\mathrm{N}\mathrm{m}$).

Answer: (C) (i) Work and (ii) Energy

Concept: A homologous series is a family of compounds with the same functional group, in which successive members differ by a $\mathrm{C}{\mathrm{H}}_2$ unit.

1. $\mathrm{C}{\mathrm{H}}_3\mathrm{O}\mathrm{H}$, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{O}\mathrm{H}$, ${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{O}\mathrm{H}$, ${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{O}\mathrm{H}$ all have the –OH group (alcohols).
2. Each differs from the next by $\mathrm{C}{\mathrm{H}}_2$ (mass 14 u).
3. Same functional group → similar chemical properties; physical properties change gradually.

Yaad rakhein: Same functional group + difference of $\mathrm{C}{\mathrm{H}}_2$ = homologous series.

Answer: (A) a homologous series

Concept: Plant hormones are either growth promoters or growth inhibitors.

1. Auxin – cell elongation at shoot tip.
2. Gibberellins – stem growth; cytokinins – cell division.
3. Abscisic acid (ABA) – inhibits growth, causes wilting of leaves, so any option with ABA is wrong.

Yaad rakhein: ABA = 'stress hormone' / inhibitor; the other three (A-G-C) promote growth.

Answer: (B) Auxin, Cytokinins and Gibberellins

Concept: Ionic bonds form by complete transfer of electrons from a metal (which loses electrons easily) to a non-metal (which gains electrons).

1. Example: $\mathrm{N}\mathrm{a}\to \mathrm{N}{\mathrm{a}}^{+}+e^{-}$ and $\mathrm{C}\mathrm{l}+e^{-}\to \mathrm{C}{\mathrm{l}}^{-}$.
2. Oppositely charged ions attract → $\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}$.
3. Non-metal + non-metal share electrons (covalent), not transfer.

Yaad rakhein: Metal gives, non-metal takes → ionic bond.

Answer: (A) In an ionic compound, the transfer of electrons takes place from a metal to a non-metal.

Concept: Ohm's law: $V=IR$, so at constant voltage the current is inversely proportional to resistance, $I=\frac{V}{R}$.

1. New resistance $R^{\prime }=\frac{R}{2}$.
2. $I^{\prime }=\frac{V}{R/2}=\frac{2V}{R}=2I$.
3. So the current doubles.

Trick: At fixed $V$: $I\propto \frac{1}{R}$ – halve $R$, double $I$.

Answer: (C) It will be doubled.

Concept: Solids have a definite shape and volume; liquids have definite volume but take the container's shape; gases have neither.

1. Juice, milk, oil, water → liquids; oxygen → gas.
2. Sugar has definite shape (crystals) → solid.
3. Sponge is a solid with pores that hold air; it can be compressed but still keeps its shape.

Yaad rakhein: A sponge is compressible yet still a solid – a common trap.

Answer: (D) Sugar and sponge

Concept: Subatomic particles are particles smaller than an atom that make it up: electrons, protons and neutrons.

1. Electron – negative, outside nucleus.
2. Proton – positive; neutron – neutral; both in nucleus.
3. A molecule is a group of two or more atoms – it is bigger than an atom, not subatomic.

Yaad rakhein: Order of size: subatomic particle $<$ atom $<$ molecule.

Answer: (B) Molecule

Concept: Plaster of Paris (calcium sulphate hemihydrate) is made by heating gypsum at 373 K.

1. $\mathrm{C}\mathrm{a}\mathrm{S}{\mathrm{O}}_4·2{\mathrm{H}}_2\mathrm{O}\stackrel{373K}{\to }\mathrm{C}\mathrm{a}\mathrm{S}{\mathrm{O}}_4·\frac{1}{2}{\mathrm{H}}_2\mathrm{O}+\frac{3}{2}{\mathrm{H}}_2\mathrm{O}$
2. Gypsum loses $1\frac{1}{2}$ water molecules.
3. On adding water it sets back into hard gypsum – used for plasters and casts.

Yaad rakhein: Gypsum has 2 water, POP has $\frac{1}{2}$ – heat removes $1\frac{1}{2}$.

Answer: (B) gypsum is heated at 373 K and it loses water molecules