Q&A: practice questions with answers

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. Profit = $960-800=$ ₹160.
2. $\frac{160}{800}\times 100=20\%$.

Trick: $\frac{\text{SP}}{\text{CP}}=\frac{960}{800}=1.2$ → 20% profit in one step.

Answer: (C) $20\%$

Concept: Equivalent discount = $a+b-\frac{ab}{100}$.

1. $=30+20-\frac{30\times 20}{100}$
2. $=50-6=44\%$.

Trick: Multiply the 'paid' parts: $0.7\times 0.8=0.56$ → paid 56%, discount 44%.

Answer: (D) 44%

Concept: Cheaper : Dearer = (Dearer − Mean) : (Mean − Cheaper).

1. 20% and 60%, mean 50%.
2. 20% part : 60% part = (60 − 50) : (50 − 20) = 10 : 30 = 1 : 3.

Trick: The mean (50) is closer to 60, so more of the 60% solution is needed – only (D) fits.

Answer: (D) $1:3$

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

1. $R=\frac{2805\times 100}{8500\times 4}=\frac{280500}{34000}=8.25\%$.
2. Check: $\frac{8500\times 4\times 8.25}{100}=$ ₹2,805 ✔

Trick: SI per 1% = $\frac{8500\times 4}{100}=$ ₹340 → $\frac{2805}{340}=8.25$.

Answer: (B) $8.25\%$

Concept: Outer surface of a hemispherical bowl = curved surface = $2\pi r^2$.

1. $2\times \frac{22}{7}\times 7\times 7=308{\text{cm}}^2$.
2. Cost = $308\times 0.50=$ ₹154.

Trick: With $r=7$: $\pi r^2=154$, so $2\pi r^2=308$ – and ₹0.50 just halves it back to 154.

Answer: (A) ₹154

Concept: Sum of interior angles = $(n-2)\times {180}^{\circ }$.

1. $n=10$.
2. $(10-2)\times {180}^{\circ }=8\times {180}^{\circ }={1440}^{\circ }$.

Trick: Each extra side adds 180°: pentagon 540°, hexagon 720° … decagon 1440°.

Answer: (C) 1440°

Concept: Exterior angle = ${180}^{\circ }-\text{interior}$; $n={360}^{\circ }÷\text{exterior}$.

1. Exterior = ${180}^{\circ }-{160}^{\circ }={20}^{\circ }$.
2. $n=360÷20=18$.

Trick: Always go via the exterior angle – one subtraction and one division.

Answer: (C) 18

Concept: Equal material → equal volumes.

1. $\frac{1}{3}\pi r^{2}h=\frac{4}{3}\pi r^3$.
2. Cancel $\frac{1}{3}\pi r^2$: $h=4r$.

Trick: Cone : sphere with same r → $\frac{h}{3}=\frac{4r}{3}$ → $h=4r$. Learn it as a ready result.

Answer: (C) $4r$

Concept: Volume of big sphere = total volume of small spheres.

1. $\frac{4}{3}\pi (10{)}^3=125\times \frac{4}{3}\pi r^3$.
2. $1000=125r^3\to r^3=8\to r=2$ m.

Trick: Radius ratio = cube root of count: $\frac{R}{r}=\sqrt[3]{125}=5\to r=\frac{10}{5}=2$.

Answer: (B) 2 m

Concept: Pure recurring decimal = repeating digits ÷ as many 9s.

1. $0.272727\dots =\frac{27}{99}$.
2. $\frac{27}{99}=\frac{3}{11}$ (divide by 9).

Trick: Remember $\frac{1}{11}=0.0909\dots$, so $\frac{3}{11}=0.2727\dots$.

Answer: (C) $\frac{3}{11}$

Concept: Find total fare slab-wise, then split in the given ratio.

1. First 50 km: $50\times 18=$ ₹900.
2. Next 75 km: $75\times 14=$ ₹1,050.
3. Total = ₹1,950; $B=\frac{3}{5}\times 1950=$ ₹1,170.

Trick: $1950÷5=390$ per part → B's 3 parts = 1170.

Answer: (C) ₹1170

Concept: $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$.

1. $\tan \theta =\frac{5}{12}$.
2. $\sin 2\theta =\frac{2\times \frac{5}{12}}{1+\frac{25}{144}}=\frac{\frac{10}{12}}{\frac{169}{144}}$.
3. $=\frac{10}{12}\times \frac{144}{169}=\frac{120}{169}$.

Trick: Triplet 5-12-13: $\sin 2\theta =\frac{2\times 5\times 12}{{13}^2}=\frac{120}{169}$. Any sine must be $\le 1$, so $\frac{26}{10}$ and $\frac{169}{120}$ are out at once.

Answer: (D) $\frac{120}{169}$

Concept: Equalise the common term y.

1. x : y = 5 : 6, y : z = 3 : 7 → ×2 → 6 : 14.
2. x : z = 5 : 14.

Trick: $\frac{x}{z}=\frac{x}{y}\times \frac{y}{z}=\frac{5}{6}\times \frac{3}{7}=\frac{15}{42}=\frac{5}{14}$.

Answer: (B) 5 : 14

Concept: A sum doubles when SI = P, i.e. $R\times T=100$.

1. $R\times 12=100$.
2. $R=\frac{100}{12}=8.33\%\left(8\frac{1}{3}\%\right)$.

Trick: Doubling at SI: $R\times T=100$; tripling: $R\times T=200$.

Answer: (B) 8.33%

Concept: ${\sin }^{2}x+{\cos }^{2}x=1$, with $x=2A$.

1. ${\sin }^{2}2A=1-{\left(\frac{12}{13}\right)}^2=1-\frac{144}{169}=\frac{25}{169}$.
2. $\sin 2A=\frac{5}{13}$.

Trick: 12 and 13 → the third side of the 5-12-13 triangle is 5, so $\sin =\frac{5}{13}$. Values above 1 ($\frac{169}{25}$, $\frac{13}{5}$) are impossible.

Answer: (D) $\frac{5}{13}$

Concept: Revenue reforms of Akbar's reign.

1. Raja Todar Mal was Akbar's finance/revenue minister (Diwan) and one of his Navratnas.
2. He introduced land measurement and cash assessment – the Zabti / Dahsala system (1580).

Yaad rakhein: Akbar's 'Navratnas' also include Birbal, Tansen, Abul Fazl and Man Singh.

Answer: (C) Akbar

Concept: The SCO Charter fixes its official languages.

1. Official working languages: Russian and Chinese.
2. SCO was set up in 2001 (Shanghai); India became a full member in 2017.
3. English/Hindi are not official SCO languages.

Yaad rakhein: SCO Secretariat: Beijing; RATS (anti-terror body): Tashkent.

Answer: (B) Russian and Chinese

Concept: First EAS venue.

1. The 1st East Asia Summit was held in December 2005 at Kuala Lumpur, Malaysia.
2. It began with 16 members (ASEAN-10 + India, China, Japan, South Korea, Australia, New Zealand); USA and Russia joined in 2011.

Yaad rakhein: EAS is always hosted by the country chairing ASEAN that year.

Answer: (C) Kuala Lumpur

Concept: Match the book with its author.

1. The Jungle Book (1894) – Rudyard Kipling (born in Bombay, 1865); hero Mowgli.
2. Kipling won the Nobel Prize in Literature in 1907; he also wrote 'Kim'.
3. Mark Twain – Tom Sawyer; H. G. Wells – The Time Machine; Dickens – Oliver Twist.

Trick: Kipling + India + Mowgli – think 'K for Kim, K for Kipling'.

Answer: (B) Rudyard Kipling

Concept: Current affairs: appointment in Ministry of Defence (May 2025).

1. Shailendra Nath Gupta (IDES, 1990 batch) took charge as Director General Defence Estates on 31 May 2025.
2. DGDE works under the Ministry of Defence and oversees defence land and cantonment boards.

Yaad rakhein: DGDE headquarters: New Delhi.

Answer: (C) Shailendra Nath Gupta