Q&A: practice questions with answers

Concept: Marble Palace is a 19th-century mansion in north Kolkata (Muktaram Babu Street).

1. It was built in 1835 by Raja Rajendra Mullick, a wealthy Bengali merchant and art collector; it houses paintings by Rubens and Reynolds, marble statues and a small zoo.
2. The other three were social reformers/writers, not builders of this mansion.

Yaad rakhein: Marble Palace (1835) – Rajendra Mullick; his descendants still live there.

Answer: (D) Raja Rajendra Mullick

Concept: Coal India's inter-company sports meets are hosted by its subsidiaries in rotation.

1. The 2025-26 Powerlifting, Weightlifting and Bodybuilding Championship was hosted by Western Coalfields Limited at Nagpur in October 2025, and the host WCL team won the overall championship.
2. SECL, MCL and SCCL (Singareni) fielded teams but did not take the overall title.

Yaad rakhein: WCL is headquartered in Nagpur – host and overall champion 2025-26.

Answer: (C) WCL

Concept: Post-1991 financial sector reforms were based on the Narasimham Committee reports.

1. Narasimham Committee I (1991) recommended deregulating interest rates (moving from administered to market-determined rates), lowering SLR/CRR and phasing out directed credit.
2. Chelliah (1991) dealt with tax reforms, Kelkar (2002) with direct/indirect taxes, and Rangarajan (1997) with disinvestment/balance of payments matters.

Yaad rakhein: Banking reforms → Narasimham (I: 1991, II: 1998); tax reforms → Chelliah, Kelkar.

Answer: (A) Narasimham Committee I (1991)

Concept: Magnification $m=\frac{v}{u}$ with the Cartesian sign convention (distances against the incident light are negative).

1. Object in front: $u=-10$ cm. Image on the same side (virtual): $v=-30$ cm.
2. $m=\frac{v}{u}=\frac{-30}{-10}=+3$.
3. Positive $m$ ⇒ erect, magnified virtual image (a convex lens used as a magnifying glass).

Trick: Image on the same side as the object ⇒ virtual and erect ⇒ $m$ is positive; magnitude $=\frac{30}{10}=3$.

Answer: (D) 3

Concept: The SHANTI Act, 2025 (Sustainable Harnessing and Advancement of Nuclear Energy for Transforming India) replaces the Atomic Energy Act, 1962 and opens the sector to private participation.

1. It gives a statutory basis to the Atomic Energy Regulatory Board (AERB) as the independent nuclear safety regulator, in place of its earlier executive-order status.
2. The other three names are not bodies created by this Act.

Yaad rakhein: SHANTI Act 2025 – private players in nuclear power + statutory AERB; target 100 GW nuclear by 2047.

Answer: (D) Atomic Energy Regulatory Board

Concept: Net force $=$ mass $\times$ acceleration (Newton's 2nd law); zero net force ⇒ no change in motion (1st law).

1. Push 30 N right and friction 30 N left ⇒ net force $=0$ ⇒ acceleration $=0$.
2. The box was initially at rest, so it stays at rest (static friction balances the push).

Yaad rakhein: Balanced forces keep the state unchanged – rest stays rest, uniform motion stays uniform.

Answer: (A) The box remains at rest.

Concept: Match protected areas with their flagship species.

1. Kaziranga National Park (Assam, UNESCO World Heritage Site) holds about two-thirds of the world's greater one-horned rhinoceros – a conservation success story.
2. Gir – Asiatic lion; Periyar – elephants/tigers; Silent Valley – lion-tailed macaque and rainforest.

Yaad rakhein: Rhino → Kaziranga (also Pobitora, Manas, Orang in Assam; Dudhwa in UP).

Answer: (D) Kaziranga National Park

Concept: Paragraph alignment (left/right/centre/justify) works horizontally; vertical placement inside a table cell is a cell property.

1. Select the cells → right-click → Table Properties → Cell tab → Vertical alignment: Center (or use Layout tab → Alignment group → Align Center).
2. Align Left/Right/Justify only move text sideways, not up or down.

Yaad rakhein: Home tab = horizontal alignment; Table Properties/Layout tab = vertical alignment in cells.

Answer: (C) Right-click selection > Table Properties > Cell tab > Vertical alignment: Center

Concept: The Congress split at Surat (1907) over methods of struggle.

1. Moderates (Gokhale, Naoroji) believed in petitions, prayers and constitutional means; Extremists (Tilak, Lajpat Rai, Bipin Chandra Pal) demanded Swaraj through mass participation, swadeshi, boycott and direct political agitation.
2. Options A, C and D describe moderate or irrelevant positions.

Yaad rakhein: Lal-Bal-Pal = Extremists = “Swaraj is my birthright” + boycott & mass agitation.

Answer: (B) Emphasised mass participation and direct political agitation

Concept: The Industrial Policy Resolution of 30 April 1956 is called the “economic constitution of India”.

1. Its declared objective was to accelerate growth and establish a socialist pattern of society (adopted by the Avadi session of Congress, 1955), giving the public sector the commanding heights of heavy industry.
2. It classified industries into Schedules A, B and C – the opposite of reducing the public sector or privatising heavy industry.

Yaad rakhein: IPR 1948 – mixed economy; IPR 1956 – socialist pattern, Mahalanobis heavy-industry model (2nd Plan).

Answer: (A) Establishing a socialist pattern of society

Concept: Angle bisector theorem: the bisector of $\angle X$ divides the opposite side in the ratio of the other two sides, $\frac{XY}{XZ}=\frac{YT}{TZ}$.

1. $TZ=YZ-YT=6-3.2=2.8$ cm.
2. $\frac{5.6}{XZ}=\frac{3.2}{2.8}$
3. $XZ=\frac{5.6\times 2.8}{3.2}=\frac{15.68}{3.2}=4.9$ cm.

Trick: $3.2:2.8=8:7$, so $XY:XZ=8:7$. One part $=5.6÷8=0.7$, hence $XZ=7\times 0.7=4.9$ – no long division needed.

Answer: (D) 4.9 cm

Concept: Mixed recurring decimal $0.a\bar{b}$: write all digits minus the non-recurring digits over 9s followed by 0s. Here it is faster to subtract first.

1. $3x=5.4\bar{6}-2.3\bar{6}$. The recurring parts are identical, so they cancel: $3x=5.4-2.3=3.1$.
2. $x=\frac{3.1}{3}=\frac{31}{30}$.
3. Check by fractions: $2.3\bar{6}=\frac{236-23}{90}=\frac{71}{30}$, $5.4\bar{6}=\frac{546-54}{90}=\frac{164}{30}$; difference $\frac{93}{30}$, so $x=\frac{31}{30}$ ✓.

Trick: When both sides have the same recurring tail (…666), subtract directly – the tails vanish and you avoid the 90-denominator work.

Answer: (D) $\frac{31}{30}$

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

1. $4500=\frac{30000\times x\times 3}{100}=900x$.
2. $x=\frac{4500}{900}=5$.

Trick: Interest per year $=4500÷3=1500$; $1500$ is $5\%$ of $30000$ – done mentally.

Answer: (A) 5

Concept: Under compound interest, if a sum becomes $n$ times in $t$ years, it becomes $n^k$ times in $kt$ years.

1. $3$ times in $7$ years.
2. $27=3^3$, so the time is $3\times 7=21$ years.

Trick: Only in CI do multiples "stack": $3\to 9\to 27$ takes $7+7+7$ years. Under SI you would add equal interests instead.

Answer: (D) 21 years

Concept: A radius is perpendicular to the tangent at the point of contact, and the two tangents from an external point are equal.

1. In right $△OPQ$: $PQ=\sqrt{O{Q}^2-O{P}^2}=\sqrt{{41}^2-9^2}=\sqrt{1681-81}=\sqrt{1600}=40$ cm.
2. $QR=PQ=40$ cm, so $△PQR$ is isosceles with apex angle ${60}^{\circ }$ → the base angles are also ${60}^{\circ }$ → equilateral, $PR=40$ cm.
3. Perimeter $=3\times 40=120$ cm.

Trick: $(9,40,41)$ is a Pythagorean triplet; and "equal tangents + ${60}^{\circ }$ between them" always makes an equilateral triangle.

Answer: (C) 120 cm

Concept: A number leaving the same remainder $r$ on division by several divisors = (multiple of their LCM) $+r$.

1. $\text{LCM}(8,7,11)=616$.
2. Largest 4-digit multiple of 616: $9999÷616=16$ remainder $143$, so $616\times 16=9856$.
3. Required number $=9856+4=9860$.

Trick: Check options quickly: subtract 4 and test divisibility by 8 and 11 – $9856$ is $8\times 1232$ and $11\times 896$ ✓.

Answer: (C) 9860

Concept: $\text{M.D.}=\frac{\sum |x_i-\overline{x}|}{n}$.

1. Mean $\overline{x}=\frac{5+10+15+20+25}{5}=15$.
2. Absolute deviations: $10,5,0,5,10$; sum $=30$.
3. $\text{M.D.}=\frac{30}{5}=6$.

Trick: For an AP the mean is the middle term; deviations are symmetric, so just double the one-sided sum ($10+5=15$) and divide by $n$.

Answer: (B) 6

Concept: A path of width $w$ outside a rectangle adds $2w$ to both length and breadth, so the perimeter increases by $8w$ regardless of the field's size.

1. Outer dimensions: $(68.5+4.8)$ m by $(45.6+4.8)$ m $=73.3$ m $\times 50.4$ m.
2. Outer perimeter $=2(73.3+50.4)=247.4$ m; original perimeter $=2(68.5+45.6)=228.2$ m.
3. Difference $=247.4-228.2=19.2$ m $=8\times 2.4$ ✓.

Trick: Perimeter difference $=8w$ for an outside path (and the same for an inside path): $8\times 2.4=19.2$ – the field dimensions are a distraction.

Answer: (C) 19.2 m

Concept: The perpendicular from the centre bisects a chord. Let OM ⊥ AB with OM = 3 cm; M is the midpoint of both AB and CD.

1. Outer circle: $AM=\sqrt{{13}^2-3^2}=\sqrt{160}=4\sqrt{10}$ cm.
2. Inner circle: $CM=\sqrt{7^2-3^2}=\sqrt{40}=2\sqrt{10}$ cm.
3. $AC=AM-CM=4\sqrt{10}-2\sqrt{10}=2\sqrt{10}$ cm.

Trick: For concentric circles, $AC=BD$ always, and $AC=\frac{AB-CD}{2}$ – one subtraction of half-chords.

Answer: (A) $2\sqrt{10}$

Concept: Successive changes of $a\%$ and $b\%$ give a net change of $a+b+\frac{ab}{100}\%$.

1. $a=+20$, $b=-20$: net $=20-20+\frac{(20)(-20)}{100}=-4\%$.
2. Check with ₹100: $100\to 120\to 120\times 0.8=96$, a $4\%$ decrease.

Trick: Equal increase and decrease of $x\%$ always gives a net decrease of $\frac{x^2}{100}\%$: here $\frac{400}{100}=4\%$.

Answer: (B) 4% decrease