Q&A: practice questions with answers

Concept: Swap the signs first, then apply BODMAS.

1. After swapping: $35÷7-2\times 6+13$
2. $=5-12+13$
3. $=6$

Trick: Rewrite the whole expression with new signs before calculating.

Answer: (B) 6

Concept: Replace letters by operators, then use BODMAS.

1. $16+24÷3\times 8-5$
2. $24÷3=8$, $8\times 8=64$
3. $16+64-5=75$

Trick: Division and multiplication are done left to right before + and −.

Answer: (A) 75

Concept: All face north; fix definite positions first, then neighbours.

1. B third from right ⇒ B at 5th (from left).
2. Only one person left of D ⇒ D at 2nd.
3. F right of B, left of G ⇒ F 6th, G 7th.
4. E is neighbour of both C and D ⇒ E 3rd, C 4th, A 1st. Row: A D E C B F G ⇒ 5 people between A and G.

Trick: “E is neighbour of C and D” means E sits between them.

Answer: (B) Five

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

1. A + B: A is sister of B (A is female).
2. B − C: B is son of C.
3. C × D: C is husband of D ⇒ D is mother of A and B.
4. D ÷ E: D is daughter of E ⇒ A is daughter of E’s daughter.

Trick: Mark gender with each code (sister/daughter = female) to fix the final relation.

Answer: (C) Daughter’s daughter

Concept: Letters common to both words share the digits common to both codes.

1. REPS → 9631; EARS → 6914
2. Common letters R, E, S ↔ common digits 9, 6, 1
3. Left over: A ↔ 4 (and P ↔ 3)

Trick: Remove the common part; the leftover letter takes the leftover digit.

Answer: (D) 4

Concept: Draw the Venn diagram; a conclusion follows only if it is true in every possible diagram.

1. Door and Lock are separate; Lock and Key are separate.
2. Door and Key may or may not overlap ⇒ I (No door is a key) is not certain.
3. “No door is a lock” is the same as “No lock is a door” ⇒ II follows.

Yaad rakhein: “No A is B” can always be reversed to “No B is A”. Two negative statements give no definite link between the end terms.

Answer: (B) Only conclusion II follows

Concept: A valid course of action should solve the stated problem practically, without being extreme.

1. Problem: traffic jam due to illegal parking at drop-off time.
2. I: Deploying security to monitor parking and clear jams directly tackles it ⇒ follows.
3. II: Banning all car drops is extreme and impractical; parents can drop children legally ⇒ does not follow.

Trick: Reject actions with “ban/stop all” type extreme words unless the situation demands it.

Answer: (B) Only I follows

Concept: Check each letter position of the clusters separately.

1. 1st letters: O, Q, S, U → +2 → W
2. 2nd letters: R, T, V, X → +2 → Z
3. 3rd letters: N, P, R, T → +2 → V

Trick: Every letter moves +2, so just add 2 to each letter of UXT.

Answer: (B) WZV

Concept: Draw the Coats circle inside the Jackets circle, with Sweaters overlapping Coats.

1. All coats are jackets ⇒ every coat is a jacket, so some jackets (the coats) are coats ⇒ I follows.
2. Some coats are sweaters, and all coats are jackets ⇒ those coats are jackets that are sweaters ⇒ II follows.
3. Hence both conclusions follow.

Yaad rakhein: “All A are B” ⇒ “Some B are A”; “All A are B” + “Some A are C” ⇒ “Some B are C” — both always follow.

Answer: (D) Both conclusions I and II follow

Concept: Check each letter position of the clusters separately.

1. 1st letters: I, L, O, R → +3 → U
2. 2nd letters: M, P, S, V → +3 → Y
3. 3rd letters: H, K, N, Q → +3 → T

Trick: Every letter moves +3, so add 3 to each letter of RVQ.

Answer: (D) UYT

Concept: For a concave mirror, the centre of curvature C lies at $R=2f$ from the pole. An object placed at C gives an image at C itself.

1. Same-size image means magnification $m=-\frac{v}{u}=-1$, so $v=u$.
2. Mirror formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{2}{u}=\frac{1}{f}$.
3. So $u=2f$ (object at C); image is real, inverted and of the same size.

Trick: Object at C → image at C (same size); beyond C → diminished; between F and C → enlarged.

Answer: (B) equal to 2f

Concept: A neuron (nerve cell) is the structural and functional unit of nervous tissue. It has a cell body (cyton), dendrites and an axon.

1. Each neuron has one long axon that carries impulses away from the cell body.
2. It has many short branched dendrites that receive impulses.
3. The cell body contains a single nucleus, not many nuclei – so statement (D) is incorrect.

Yaad rakhein: Neuron = one nucleus, one axon, many dendrites. (Many nuclei is a feature of skeletal muscle fibres.)

Answer: (D) The cell body of a neuron has many nuclei.

Concept: In a food chain or food web, energy flows from the sun → producers → herbivores → carnivores, and never returns to the previous level.

1. Energy captured by autotrophs does not go back to the sun.
2. Energy passed to herbivores does not come back to autotrophs.
3. At each level some energy is lost as heat, so the flow is one-way (unidirectional).

Yaad rakhein: Energy flows (one-way); nutrients cycle (circular).

Answer: (A) Unidirectional

Concept: Evaporation causes cooling. A volatile liquid takes the latent heat of vaporisation from the surface it is on.

1. Acetone is highly volatile and evaporates quickly at room temperature.
2. The energy needed for evaporation is absorbed from the palm.
3. The palm loses heat, so it feels cool.

Yaad rakhein: Acetone, petrol, perfume, spirit on skin → quick evaporation → cooling.

Answer: (B) It will cause the palm to feel cool.

Concept: In respiration, glucose (6-carbon) is broken in the cytoplasm into two molecules of pyruvate (3-carbon). Pyruvate then forms different products.

1. Glucose ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ – 6 carbons ✔
2. Pyruvate – 3 carbons ✔
3. Anaerobic (yeast): ethanol ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{O}\mathrm{H}$ (2 C) + $\mathrm{C}{\mathrm{O}}_2$ ✔
4. Lack of oxygen in muscles: lactic acid ${\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_3$ – 3 carbons, not 4 ✘

Yaad rakhein: Glucose 6 → pyruvate 3 → lactic acid 3 / ethanol 2 + $\mathrm{C}{\mathrm{O}}_2$.

Answer: (A) Lactic acid – 4 carbon molecules

Concept: When copper powder is heated in air, it combines with oxygen to form copper(II) oxide, which is black.

1. Reaction: $2\mathrm{C}\mathrm{u}+{\mathrm{O}}_2\stackrel{\text{heat}}{\to }2\mathrm{C}\mathrm{u}\mathrm{O}$.
2. Copper is brown/reddish; the surface turns black due to $\mathrm{C}\mathrm{u}\mathrm{O}$.
3. Copper(I) oxide $\mathrm{C}{\mathrm{u}}_2\mathrm{O}$ is red, not black.

Yaad rakhein: Heated copper turns black ($\mathrm{C}\mathrm{u}\mathrm{O}$); passing ${\mathrm{H}}_2$ over hot $\mathrm{C}\mathrm{u}\mathrm{O}$ turns it brown again (reduction).

Answer: (A) Black-coloured copper (II) oxide

Concept: Electric current is the rate of flow of charge: $I=\frac{Q}{t}$, so $Q=I\times t$ (time in seconds).

1. $I=1\text{ A}$, $t=5\text{ min}=5\times 60=300\text{ s}$.
2. $Q=I\times t=1\times 300$.
3. $Q=300\text{ C}$.

Trick: Always convert minutes to seconds before using $Q=It$.

Answer: (C) 300 C

Concept: Kinetic energy is the energy possessed by a body due to its motion. It equals the work done to bring the body from rest to velocity $v$.

1. Using $v^2-u^2=2as$ with $u=0$: $s=\frac{v^2}{2a}$.
2. Work $W=F\times s=ma\times \frac{v^2}{2a}$.
3. So $E_k=\frac{1}{2}m{v}^2$.

Yaad rakhein: Double the speed → KE becomes 4 times (since $E_k\propto v^2$).

Answer: (B) $\frac{1}{2}m{v}^2$

Concept: Soaps are sodium or potassium salts of long-chain carboxylic acids (fatty acids), e.g. sodium stearate ${\mathrm{C}}_{17}{\mathrm{H}}_{35}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{N}\mathrm{a}$.

1. Soaps are made by saponification: oil/fat + $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ → soap + glycerol.
2. The ionic end ($-\mathrm{C}\mathrm{O}{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}$) is water-loving; the long hydrocarbon tail is oil-loving.
3. Detergents, in contrast, are sodium salts of sulphonic acids.

Yaad rakhein: Soap → carboxylic acid salt; Detergent → sulphonic acid salt.

Answer: (C) Carboxylic acid

Concept: Acids show acidic behaviour only in water, because water helps them release ${\mathrm{H}}^{+}$ ions.

1. $\mathrm{H}\mathrm{C}\mathrm{l}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_3{\mathrm{O}}^{+}+\mathrm{C}{\mathrm{l}}^{-}$ – without water, ${\mathrm{H}}^{+}$ does not separate from HCl. Statement a is correct.
2. ${\mathrm{H}}^{+}$ cannot exist alone; it combines with water: ${\mathrm{H}}^{+}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_3{\mathrm{O}}^{+}$.
3. So hydrogen ions are always written as ${\mathrm{H}}^{+}(\mathrm{a}\mathrm{q})$ or ${\mathrm{H}}_3{\mathrm{O}}^{+}$. Statement b is correct.

Yaad rakhein: Dry HCl gas does not change the colour of dry blue litmus – no water, no ${\mathrm{H}}^{+}$.

Answer: (C) Both a and b