Q&A: practice questions with answers

Concept: ${\sin }^2+{\cos }^2=1$; required expression = $(\sin \theta -\cos \theta {)}^2=1-2\sin \theta \cos \theta$.

1. Square the given: $1+2\sin \theta \cos \theta =\frac{36}{25}\to 2\sin \theta \cos \theta =\frac{11}{25}$.
2. Required = $1-\frac{11}{25}=$ $\frac{14}{25}$.

Trick: $(s+c{)}^2+(s-c{)}^2=2\to (s-c{)}^2=2-\frac{36}{25}=\frac{14}{25}$.

Answer: (B) $\frac{14}{25}$

Concept: Angle subtended by a diameter on the circle = 90°.

1. ∠ACB = 90°.
2. ∠CBA = 180° − 90° − 35° = 55°.

Trick: The other two angles are complementary: 90° − 35° = 55°.

Answer: (D) 55°

Concept: The centre must be equidistant from all three points.

1. Perpendicular bisectors of two joining segments meet at one unique point (the circumcentre).
2. That fixes one centre and one radius → exactly one circle.

Yaad rakhein: 3 non-collinear points → 1 circle; 3 collinear points → none; 1 or 2 points → infinitely many.

Answer: (A) 1

Concept: $\text{Chord}=2r\sin \left(\frac{\theta }{2}\right)$.

1. $AB=2\times 15\times \sin {30}^{\circ }=30\times \frac{1}{2}=$ 15 cm.

Trick: 60° at the centre → triangle OAB is equilateral → chord = radius.

Answer: (A) 15 cm

Concept: $\text{Perimeter}=2(l+b)$; convert the ratio to an l–b relation.

1. $l:2(l+b)=1:3\to 3l=2l+2b\to l=2b$.
2. $\text{Area}=2b^2=200\to b^2=100\to b=$ 10 cm.

Trick: Option check: b = 10 → l = 20 → area 200 ✓, perimeter $60=3\times 20$ ✓.

Answer: (A) 10 cm

Concept: $\text{Area}=ab\sin \theta$; $\text{height on a side}=\text{area}÷\text{that side}$.

1. $\text{Area}=24\times 16\times \sin {60}^{\circ }=384\times \frac{\sqrt{3}}{2}=$ $192\sqrt{3}{\text{cm}}^2$.
2. Height on 24 cm side $=192\sqrt{3}÷24=$ $8\sqrt{3}$ cm.

Trick: $\text{Height on the longer side}=\text{shorter side}\times \sin \theta =16\times \frac{\sqrt{3}}{2}=8\sqrt{3}$ – no area needed.

Answer: (B) $192\sqrt{3}{\mathrm{cm}}^2,8\sqrt{3}\mathrm{cm}$

Concept: Add all three sums to get x + y + z, then subtract each pair.

1. Sum: 2(x + y + z) = 24k → x + y + z = 12k.
2. x = 12k − 9k = 3k, y = 12k − 7k = 5k, z = 12k − 8k = 4k.
3. x : y : z = 3 : 5 : 4.

Trick: Check options: 3 + 5 = 8, 5 + 4 = 9, 4 + 3 = 7 ✓.

Answer: (B) $3:5:4$

Concept: $\text{Common term}=(\text{sum of first group}+\text{sum of last group})-\text{total sum}$.

1. Total = $11\times 50=550$.
2. First 6 = 288, last 6 = 330 → together 618.
3. 6th number = $618-550=$ 68.

Trick: Deviation method: 6th = $50+6\times (-2)+6\times (+5)=50-12+30=68$.

Answer: (D) 68

Concept: Extra interest comes only from the extra rate: $\Delta \text{SI}=\frac{P\times \Delta r\times T}{100}$.

1. $\Delta \text{SI}=3120-2400=720$.
2. $720=\frac{P\times 3\times 4}{100}\to P=\frac{720\times 100}{12}=$ ₹6,000.

Trick: Extra 3% for 4 years = 12% of P $=720\to P=6000$.

Answer: (C) Rs. 6000

Concept: Same period → profit shared in ratio of investments.

1. 50000 : 75000 = 2 : 3.
2. Bhavna = $\frac{3}{5}\times 90000=$ ₹54,000.

Trick: 1 part $=\frac{90000}{5}=18000$; Bhavna = $3\times 18000$.

Answer: (C) Rs. 54,000

Concept: $\text{Required value}=\text{target total}-\text{existing total}$.

1. Target = $5\times 150=750$.
2. Made = $140+160+135+145=580$.
3. 5th day = $750-580=$ 170.

Trick: Deviations from 150: $-10,+10,-15,-5=-20$ → 5th = $150+20=170$.

Answer: (A) 170

Concept: Parallel lines have equal slopes; for $ax+by=c$, slope $=-\frac{a}{b}$.

1. Given slope $=-\frac{5}{4}$.
2. Only $5x+4y=35$ keeps the same x and y coefficients → slope $-\frac{5}{4}$.
3. Different constant ($35\ne 20$) → parallel, not the same line.

Trick: Parallel to $ax+by=c$ is always $ax+by=k$ – just look for the same left side.

Answer: (C) $5x+4y=35$

Concept: 'x% more than' = multiply by $\frac{100+x}{100}$.

1. $C=200\times 1.5=300\to C+D=500$.
2. $A+B=500\times 3=1500$ (200% more = 3 times).
3. $A:B=3:2\to A=\frac{3}{5}\times 1500=$ 900.

Trick: '200% more' means 3 times, not 2 times – the usual trap.

Answer: (B) 900

Concept: $T=\frac{\text{SI}\times 100}{P\times R}$.

1. $\text{SI}=18500-12500=6000$.
2. $T=\frac{6000\times 100}{12500\times 8}=\frac{600000}{100000}=$ 6 years.

Trick: 8% of 12500 = 1000 per year → $\frac{6000}{1000}=6$ years.

Answer: (C) 6 years

Concept: Find work done in one 2-day cycle using LCM units.

1. Work = LCM(12, 24) = 24 units; A = 2/day, B = 1/day.
2. One cycle (A + B, 2 days) = 3 units.
3. $\frac{24}{3}=8$ cycles = 16 days (ends exactly, no remainder).

Trick: If total ÷ cycle work is exact, answer = cycles × 2; otherwise handle the leftover day by day.

Answer: (B) 16

Concept: Contemporary chronicle of the early Delhi Sultanate.

1. Minhaj-us-Siraj wrote Tabaqat-i-Nasiri (dedicated to Sultan Nasiruddin Mahmud).
2. He noted that Raziyya (r. 1236–40) was more capable than her brothers.
3. Amir Khusrau – poet of the Khalji era; Ibn Khaldun and Al-Masudi – Arab/North African scholars, not Delhi chroniclers.

Trick: Minhaj → Tabaqat-i-Nasiri → Nasiruddin; it covers Raziyya's reign.

Answer: (B) Minhaj-us-Siraj

Concept: 'Banda' = slave; 'Bandagan' = plural.

1. Bandagan were slaves bought by the Sultan and trained for military and administrative service.
2. They depended wholly on the Sultan, so they were trusted with high posts.
3. Iltutmish's group of elite slave-officers was known as the 'Chalisa' (Forty).

Trick: Bandagan = 'bande' of the Sultan → purchased slave soldiers.

Answer: (D) Slave soldiers purchased by the Sultan

Concept: High Courts are covered in Part VI, Articles 214–231.

1. Art. 214 provides that every State is to have its own High Court.
2. Art. 224 – additional/acting judges; Art. 226 – writ jurisdiction of HCs.
3. Art. 114 – Appropriation Bills (Parliament).

Yaad rakhein: Art. 231 lets Parliament set up a common High Court for two or more States.

Answer: (A) Article 214

Concept: NDA admitted women for the first time in 2022 (after a Supreme Court order).

1. The NDA course is 3 years long.
2. So the first women's batch passed out in 2025 (148th course).
3. NDA is at Khadakwasla, Pune – the world's first tri-service academy.

Trick: Entry year + 3-year course = passing-out year: 2022 + 3 = 2025.

Answer: (C) 2025

Concept: ODF Plus = village keeps ODF status and also manages solid and liquid waste.

1. Under Swachh Bharat Mission (Grameen) Phase II, about 5.61 lakh villages were ODF Plus by December 2024.
2. ODF Plus stages: Aspiring → Rising → Model.

Yaad rakhein: SBM was launched on 2 October 2014; Phase II (ODF Plus focus) began in 2020-21.

Answer: (A) 5.61 lakh