Q&A: practice questions with answers

Concept: Net displacement = combine moves on each axis.

1. East 10 and West 10 cancel → E–W = 0.
2. North = 17 km.
3. Distance from start = 17 km.

Trick: Same-length opposite legs → answer is simply the middle leg.

Answer: (B) 17 km

Concept: Fill fixed facts first, then use 'does not' clues to settle the rest.

1. Fixed: Ankush–Cricket, Sita–Hockey, Deepak–Football.
2. Left: Badminton, Tennis for Meena and Mohan.
3. Meena does not play Tennis → Meena–Badminton, Mohan–Tennis.

Trick: Direct clues remove 3 of 5 items; one negative clue finishes the last pair.

Answer: (A) Meena

Concept: Place fixed people, then use 'before/after' chains to fix the rest.

1. P = Tuesday. S cannot be Wed, Fri, Sat → S is Mon or Thu.
2. Q before R before S → S needs two days before it → S = Thursday; Q = Monday, R = Wednesday.
3. Fri, Sat left for T and U; T not Friday → T = Saturday, U = Friday.

Trick: Chains like Q→R→S push the last person late — test the latest allowed day first.

Answer: (B) T - Saturday

Concept: Find the common link in most pairs, then spot the pair that breaks it.

1. Sun–Shine, Moon–Glow, Star–Twinkle: second word = how the body gives light.
2. A cloud gives no light, and 'fall' describes rain, not the cloud.
3. Odd pair: Cloud – Fall.

Trick: Turn each pair into a sentence: 'The ___ ___s.' — 'The cloud falls' sounds wrong.

Answer: (D) Cloud - Fall

Concept: Check the gap between letters in each option.

1. A-D-G-J, B-E-H-K, C-F-I-L: each step +3.
2. D-G-I-L: +3, +2, +3 → breaks the rule.

Trick: Use position values (D4 G7 I9 L12) — the uneven jump shows immediately.

Answer: (D) $D-G-I-L$

Concept: Find the shared relation, then pick the pair with a different relation.

1. School–Students, Hospital–Patients, Market–Shopkeeper: place → people found there.
2. Book–Page: whole → part (no people).
3. So Book : Page does not conform.

Trick: Ask 'Who is found at the ___?' — it fails only for Book.

Answer: (D) Book : Page

Concept: A compass shows only horizontal directions: 4 cardinal + 4 intermediate.

1. North-West, South-East, South-West are intermediate (corner) directions.
2. 'Downwards' is a vertical direction — not on a compass.
3. Inconsistent item: Downwards.

Yaad rakhein: Compass = 8 flat directions (N, S, E, W, NE, NW, SE, SW); up/down never appear.

Answer: (C) Downwards

Concept: Each term should be double the previous term.

1. 10-20-40, 4-8-16, 8-16-32 → $\times 2$ each step.
2. 6-12-23: $12\times 2=24$, not 23.
3. Distinct pattern: 6-12-23.

Trick: Odd last term in a doubling pattern? Doubling always gives an even number → 23 is impossible.

Answer: (C) 6-12-23

Concept: Without evidence, no one can be declared guilty or innocent.

1. A past record only creates suspicion, not proof.
2. Jailing both or punishing the old offender punishes without evidence.
3. Trusting the new person clears him without evidence → best is to wait for evidence.

Trick: In decision questions, reject extreme actions (punish/trust blindly) and choose the fair, evidence-based step.

Answer: (C) Wait until evidence is found

Concept: Write the word and its sorted form one below the other and compare positions.

1. Original: EDUCATION.
2. Sorted: ACDEINOTU.
3. No position has the same letter → zero unchanged.

Trick: Check vowels/ends first: first letter E ≠ A, last N ≠ U — then scan the middle quickly.

Answer: (C) Zero

Concept: Natural numbers are counting numbers: 1, 2, 3, …

1. 7, 6 and 4 are counting numbers → natural.
2. 0 is not used for counting → it is a whole number, not natural.

Yaad rakhein: Whole numbers = Natural numbers + 0. So 0 is the only whole number that is not natural.

Answer: (C) 0

Concept: Every natural number is also an integer (N ⊂ W ⊂ Z).

1. Integers include negatives and zero, so −8 and 0 qualify only as integers.
2. 8 counts things → natural, and therefore also an integer.

Yaad rakhein: Any positive whole number like 8 sits in N, W, Z, Q and R all at once.

Answer: (B) 8

Concept: Transport/packing expenses are added to the cost price.

1. $\text{CP}=220\times 160+1600=35200+1600=$ ₹36,800.
2. $\text{SP}=120\times 180+100\times 170=21600+17000=$ ₹38,600.
3. Profit = ₹1,800 → Profit % = $\frac{1800}{36800}\times 100=\frac{225}{46}=$ $4\frac{41}{46}\%$ ($\approx 4.89\%$).

Trick: 1% of 36800 = 368; $1800÷368$ is a little under 5 → only (D) lies between 4.8 and 5.

Answer: (D) $4\frac{41}{46}\%$

Concept: For incentre O: $\angle QOR={90}^{\circ }+\frac{\angle P}{2}$.

1. ${130}^{\circ }={90}^{\circ }+\frac{\angle P}{2}$.
2. $\frac{\angle P}{2}={40}^{\circ }$ → $\angle QPR=$ ${80}^{\circ }$.

Trick: Incentre: $90+\frac{A}{2}$; Excentre: $90-\frac{A}{2}$; Circumcentre: 2A; Orthocentre: $180-A$.

Answer: (C) 80°

Concept: For $A+B={90}^{\circ }$: $\csc B=\sec A,\sec B=\csc A,\tan B=\cot A$.

1. $\frac{{\tan }^2{40}^{\circ }}{{\csc }^2{50}^{\circ }}=\frac{{\tan }^2{40}^{\circ }}{{\sec }^2{40}^{\circ }}={\sin }^2{40}^{\circ }$.
2. $\frac{{\cot }^2{40}^{\circ }}{{\sec }^2{50}^{\circ }}=\frac{{\cot }^2{40}^{\circ }}{{\csc }^2{40}^{\circ }}={\cos }^2{40}^{\circ }$.
3. Sum = ${\sin }^2{40}^{\circ }+{\cos }^2{40}^{\circ }=1$; and $2·\tan {35}^{\circ }·\tan {45}^{\circ }·\tan {55}^{\circ }=2\times 1\times 1=2$.
4. Total = $1+2=$ 3.

Trick: $\tan \theta ·\tan ({90}^{\circ }-\theta )=1$ — pairs like ${35}^{\circ }$–${55}^{\circ }$ vanish to 1 instantly.

Answer: (B) 3

Concept: $\cos ({90}^{\circ }-p)=\sin p$.

1. $2\sin p·\sin p=1$ → ${\sin }^{2}p=\frac{1}{2}$ → $\sin p=\frac{1}{\sqrt{2}}$.
2. $p={45}^{\circ }$ → $2p={90}^{\circ }$.
3. $\sin {90}^{\circ }+\cos {90}^{\circ }=1+0=$ 1.

Trick: $2{\sin }^{2}p=1$ is the standard signal for $p={45}^{\circ }$.

Answer: (C) 1

Concept: Under SI, interest per year is constant; difference of amounts = interest for the gap years.

1. Interest for 4 years = $13384-10504=$ ₹2,880 → 1 year = ₹720.
2. Principal = $10504-2880=$ ₹7,624.
3. $R=\frac{720}{7624}\times 100\approx$ 9.44%.

Trick: Since the gap (4 yrs) equals the first period (4 yrs), $P=2\times A_1-A_2=21008-13384=7624$ directly.

Answer: (C) $9.44\%$

Concept: Amount = 340% of P → SI = 240% of P.

1. Rate = $240\%÷8=30\%$ per year.
2. SI on ₹4000 for 3 years = $\frac{4000\times 30\times 3}{100}=$ ₹3,600.

Trick: 3 years $\times 30\%=90\%$ → 90% of 4000 = 3600 in one line.

Answer: (A) Rs. 3600

Concept: $\text{Interest}=\frac{\text{Share}\times \text{Rate}\times \text{Time}}{100}$; compare the two interests.

1. Shared money = $80\%$ of $x=\frac{4x}{5}$; Saroj = $\frac{7}{16}$ of it $=\frac{7x}{20}$, Sita $=\frac{9x}{20}$.
2. Saroj's $\text{SI}=\frac{7x}{20}\times 35\%=\frac{49x}{400}$; Sita's $\text{SI}=\frac{9x}{20}\times 55\%=\frac{99x}{400}$.
3. Difference = $\frac{50x}{400}=\frac{x}{8}=3304$ → $x=$ ₹26,432.

Trick: Answer must be a multiple of 8 ($\frac{x}{8}=3304$) — only $26432=8\times 3304$.

Answer: (B) Rs. 26,432

Concept: Stick = hypotenuse; height on wall = hypotenuse × sin(angle with floor).

1. Angle with the floor $={30}^{\circ }$, so height $=38\times \sin {30}^{\circ }$.
2. $=38\times \frac{1}{2}=$ 19 cm.

Trick: The side opposite ${30}^{\circ }$ in a right triangle is half the hypotenuse.

Answer: (C) 19 cm