Q&A: practice questions with answers

Concept: .ppsx is the PowerPoint Show format.

1. A .ppsx file opens directly in Slide Show mode instead of the editing view.
2. .pptx = normal editable presentation; .potx = template; read-only/compression are not the purpose of .ppsx.

Yaad rakhein: ppSx = Show; poTx = Template.

Answer: (B) To save the presentation in a format that opens directly in Slide Show mode

Concept: The 'Dare Square' motorcycle expedition (January 2025) was an Indian Navy outreach rally.

1. It was flagged off by the Eastern Naval Command as part of INS Ranvijay's 37th anniversary – statement 1 correct.
2. It promoted mental-health awareness, road safety and maritime heritage among youth – statement 2 correct; it ended at the Dogra Regimental Centre, Ayodhya, reflecting the ship's affiliation with the Dogra Regiment – statement 4 correct.
3. It was not an Indian Air Force collaboration covering the north-eastern states – statement 3 incorrect.

Yaad rakhein: Navy rally → Army regiment link (Dogra, Ayodhya); no Air Force role.

Answer: (D) Only 1, 2 and 4

Concept: Dividend Distribution Tax (DDT) was paid by a company on the dividends it distributed to shareholders.

1. Statement 1 is correct.
2. DDT was abolished in Budget 2020-21 (from 1 April 2020), not 2025-26 – statement 2 is wrong.
3. Dividends are now taxed in the hands of shareholders; removal ended double taxation and made Indian equity more attractive – statement 3 correct.

Yaad rakhein: DDT abolished – Budget 2020 (Nirmala Sitharaman).

Answer: (C) Only 1 and 3

Concept: The National Conference on 'Environment – 2025' was organised by the National Green Tribunal (NGT) in New Delhi.

1. President Droupadi Murmu inaugurated it on 29 March 2025.
2. She stressed that environmental protection is a shared responsibility of citizens and institutions.

Yaad rakhein: Environment – 2025 conference: NGT organiser, President Murmu inaugurated.

Answer: (C) Droupadi Murmu

Concept: The Indian High Courts Act, 1861 empowered the Crown (Her Majesty) to issue Letters Patent establishing High Courts.

1. Under it, High Courts were set up at Calcutta, Bombay and Madras in 1862, replacing the Supreme Courts and Sadar Adalats.
2. Regulating Act 1773 set up the Supreme Court at Calcutta; GoI Act 1935 created the Federal Court.

Yaad rakhein: 1861 Act → 1862 High Courts (Calcutta, Bombay, Madras).

Answer: (A) Indian High Courts Act, 1861

Concept: 'Parvat Shiromani Bhagat Singh Koshyari' is a biography of the former Uttarakhand Chief Minister and Maharashtra Governor.

1. It was written by Madan Mohan Sati, a journalist-writer from Uttarakhand.
2. Arun Shourie, PS Raman and Gopichand P Hinduja are not its authors.

Yaad rakhein: Koshyari's biography – by Madan Mohan Sati.

Answer: (A) Madan Mohan Sati

Concept: Justice Syed Shah Mohammed Quadri classified judicial review in India into three categories.

1. (i) Judicial review of constitutional amendments.
2. (ii) Judicial review of legislation of Parliament, State Legislatures and subordinate legislation; (iii) judicial review of administrative action of the Union, States and authorities under the State.
3. Review of judicial appointments is not one of his categories.

Yaad rakhein: Quadri's three: Amendments – Legislation – Administrative action.

Answer: (D) Judicial review of judicial appointments to the Supreme Court and High Courts

Concept: Average of the first $n$ natural numbers $=\frac{n+1}{2}$.

1. Here $n=17$.
2. Average $=\frac{17+1}{2}=\frac{18}{2}=9$.
3. Check: sum $=\frac{17\times 18}{2}=153$, and $153÷17=9$.

Trick: For consecutive numbers, average = middle term; the middle of 1 to 17 is 9.

Answer: (A) 9

Concept: Take CP $=100$; then MP and SP follow from the given percentages.

1. MP is 28% above CP: MP $=128$.
2. Profit 3%: SP $=103$.
3. Discount $=128-103=25$.
4. Discount% $=\frac{25}{128}\times 100=19.53\%$.

Trick: Discount% $=\frac{\text{markup}-\text{profit}}{100+\text{markup}}\times 100=\frac{28-3}{128}\times 100$. The MP ₹397 is not even needed.

Answer: (D) $19.53\%$

Concept: For $n$ (odd) observations in ascending order, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ term.

1. Ascending: 11, 13, 21, 34, 58, 64, 65, 70, 73, 78, 88.
2. $n=11$, so median $=$ the $\frac{11+1}{2}=6^{\text{th}}$ term.
3. 6th term $=64$.

Trick: Always sort first – the 6th number in the given (unsorted) list is 73, a trap.

Answer: (A) 64

Concept: Use $\sec \theta =\frac{1}{\cos \theta }$, $\text{cosec}\theta =\frac{1}{\sin \theta }$ and ${\sin }^2\theta +{\cos }^2\theta =1$.

1. $\sec \theta -\cos \theta =\frac{1-{\cos }^2\theta }{\cos \theta }=\frac{{\sin }^2\theta }{\cos \theta }$.
2. $\text{cosec}\theta -\sin \theta =\frac{{\cos }^2\theta }{\sin \theta }$.
3. $\sin \theta -\text{cosec}\theta =-\frac{{\cos }^2\theta }{\sin \theta }$.
4. Product $=\frac{{\sin }^2\theta }{\cos \theta }\times \frac{{\cos }^2\theta }{\sin \theta }\times (-\frac{{\cos }^2\theta }{\sin \theta })=-{\cos }^3\theta$.

Trick: The last bracket is negative (since $\sin \theta <\text{cosec}\theta$), so the answer must be negative; checking with $\theta ={30}^{\circ }$ gives $-{\cos }^3\theta$.

Answer: (D) $-{\cos }^3\theta$

Concept: Equal final amounts: $P_1(1+r{)}^{n_1}=P_2(1+r{)}^{n_2}$.

1. $P\times (1.1{)}^{11}=M\times (1.1{)}^{12}$ ⇒ $P=1.1M$, i.e. $P:M=11:10$.
2. $11+10=21$ parts $=$ ₹1617 ⇒ 1 part $=77$.
3. Piyush $=11\times 77=$ ₹847.

Trick: Shares are in ratio $(1+\frac{r}{100}{)}^{\text{gap}}:1=11:10$ for a 1-year gap at 10%.

Answer: (B) ₹847

Concept: $(a\pm b{)}^2=a^2+b^2\pm 2ab$.

1. $(a-b{)}^2=168-2\times 27=114$.
2. $(a+b{)}^2=168+2\times 27=222$.
3. $\frac{a-b}{a+b}=\sqrt{\frac{114}{222}}=\sqrt{\frac{57}{111}}$ (positive since $a>b$).

Trick: ${\left(\frac{a-b}{a+b}\right)}^2=\frac{S-2P}{S+2P}$ directly.

Answer: (A) $\sqrt{\frac{57}{111}}$

Concept: If $a:b::b:x$, then $b$ is the mean proportional and $x=\frac{b^2}{a}$.

1. $x=\frac{96.9\times 96.9}{5.1}$.
2. $\frac{96.9}{5.1}=19$.
3. $x=19\times 96.9=1841.1$.

Trick: Spot that $96.9=19\times 5.1$; the ratio is $1:19$, so $x=96.9\times 19$.

Answer: (D) 1841.1

Concept: Take total work = LCM of the days and find each person's daily work.

1. Total work $=\text{LCM}(11,33)=33$ units; A $=3$ units/day, B $=1$ unit/day.
2. In 3 days together: $3\times (3+1)=12$ units.
3. Remaining $=33-12=21$ units.
4. A alone: $\frac{21}{3}=7$ days.

Trick: Unit method avoids fractions completely.

Answer: (A) 7

Concept: $\sigma =\sqrt{\frac{\sum (x-\overline{x}{)}^2}{n}}$.

1. Sum $=552$, $n=8$, mean $\overline{x}=69$.
2. Deviations: $-4,-22,-10,6,13,22,-14,9$.
3. Squares: $16+484+100+36+169+484+196+81=1566$.
4. $\sigma =\sqrt{\frac{1566}{8}}=\sqrt{195.75}\approx 13.99$.

Trick: ${14}^2=196$, and $195.75$ is just below it, so $\sigma$ is just below 14.

Answer: (B) 13.99

Concept: $x^2+2xy+y^2=(x+y{)}^2$.

1. $100a^2=(10a{)}^2$ and $225b^2=(15b{)}^2$.
2. Middle term check: $2\times 10a\times 15b=300ab$ ✔ (positive).
3. So the expression $=(10a+15b{)}^2$.

Trick: All terms positive → '+' inside; $a^2$ has coefficient 100, so $10a$ comes with $a$.

Answer: (C) $(10a+15b{)}^2$

Concept: Upstream speed $=B-S$, downstream speed $=B+S$.

1. Upstream speed $=\frac{62.3}{75/60}=\frac{62.3}{1.25}=49.84\text{km/h}$.
2. $B:S=3:2$ ⇒ $3k-2k=k=49.84$; downstream $=5k=249.2\text{km/h}$.
3. Time $=\frac{67.6}{49.84}+\frac{35.8}{249.2}=1.3563+0.1437=1.5$ h.

Trick: Downstream speed is 5 times upstream, so total time $=\frac{67.6+\frac{35.8}{5}}{49.84}=\frac{74.76}{49.84}=1.5$.

Answer: (B) 1.5

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

1. $P=$ ₹2,00,000, $R=8\%$, $T=4$ years.
2. $\text{SI}=\frac{200000\times 8\times 4}{100}=64000$.

Trick: $8\%\times 4=32\%$ of 2,00,000 $=$ 64,000. Option (D) 2,64,000 is the amount, not the interest.

Answer: (B) 64,000

Concept: Volume of metal in a shell $=\frac{4}{3}\pi (R^3-r^3)$; mass $=$ density $\times$ volume.

1. $R^3-r^3=216-27=189$.
2. $V=\frac{4}{3}\times \frac{22}{7}\times 189=4\times 22\times 9=792{\text{cm}}^3$.
3. Mass $=792\times 11=8712$ g $=8.712$ kg.

Trick: $189$ is divisible by both 3 and 7 ($189=3\times 7\times 9$), so cancel early.

Answer: (A) 8.712