Q&A: practice questions with answers

Concept: In acid, ${\text{MnO}}_4^{-}$ (Mn $+7$) gains 5 electrons to become ${\text{Mn}}^{2+}$; each ${\text{Fe}}^{2+}$ loses 1 electron.

1. Electron balance: one ${\text{MnO}}_4^{-}$ needs 5 ${\text{Fe}}^{2+}$.
2. Balance O with $4{\text{H}}_2\text{O}$ and H with $8{\text{H}}^{+}$.
3. ${\text{MnO}}_4^{-}+5{\text{Fe}}^{2+}+8{\text{H}}^{+}\to {\text{Mn}}^{2+}+5{\text{Fe}}^{3+}+4{\text{H}}_2\text{O}$ (charge $+17=+17$).

Yaad rakhein: Acidic ${\text{KMnO}}_4$: 5 electrons, 8 ${\text{H}}^{+}$, 4 ${\text{H}}_2\text{O}$. ${\text{MnO}}_2$ forms only in neutral medium.

Answer: (B) ${\text{MnO}}_4^{-}+5{\text{Fe}}^{2+}+8{\text{H}}^{+}\to {\text{Mn}}^{2+}+5{\text{Fe}}^{3+}+4{\text{H}}_2\text{O}$

Concept: Tau proteins stabilise microtubules in neurons; abnormal tau clumps (tangles) are a hallmark of Alzheimer's disease.

1. In February 2025, University of Cologne researchers reported that the tau isoform '1N4R' is especially toxic to nerve cells.
2. Targeting 1N4R tau is being explored as a new approach to treating Alzheimer's disease.

Yaad rakhein: Tau + amyloid-beta = Alzheimer's markers.

Answer: (C) Treating Alzheimer’s disease

Concept: Lok Adalats work under the Legal Services Authorities Act, 1987.

1. Their award is final and binding, treated as a civil court decree; cases settle fast and no court fee is charged (fee paid earlier is refunded).
2. No appeal lies against a Lok Adalat award – so 'right to appeal' is not an advantage.

Yaad rakhein: Lok Adalat: fast, free, final – no appeal.

Answer: (D) Right to appeal

Concept: Dr Manmohan Singh (1932–2024), economist, died on 26 December 2024.

1. He first became Prime Minister on 22 May 2004 (UPA-I) and served two full terms till 2014.
2. 1991 was when he became Finance Minister under P. V. Narasimha Rao and launched economic reforms.

Yaad rakhein: 1991 – Finance Minister (LPG reforms); 2004–2014 – Prime Minister.

Answer: (A) 2004

Concept: If $a:b=b:c$, then $c$ is the third proportional of $a$ and $b$, so $c=\frac{b^2}{a}$.

1. First three even natural numbers: $2+4+6=12$, so $B=12$.
2. $C=\frac{{12}^2}{29}=\frac{144}{29}$.
3. $\frac{144}{29}\approx 4.9655\approx 4.97$.

Trick: Third proportional = (middle term)${}^2$ ÷ first term.

Answer: (D) $4.97$

Concept: A number divisible by both 6 and 31 (co-prime) must be divisible by $6\times 31=186$.

1. $3720÷186=20$ exactly.
2. Check others: 2132, 4916, 2858 are not divisible by 3 (digit sums 8, 20, 23), so not by 6.

Trick: First test divisibility by 6 (even + digit sum divisible by 3): only 3720 (digit sum 12) passes.

Answer: (A) $3720$

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$. $D<0$ ⇒ no real roots.

1. $a=2,b=4,c=8$.
2. $D=4^2-4\times 2\times 8=16-64=-48$.
3. $D<0$, so the roots are imaginary (no real roots).

Trick: If $a$ and $c$ have the same sign and $b^2<4ac$, roots are never real.

Answer: (D) no real roots

Concept: Pipe A runs for the full 42 minutes; Pipe B must fill whatever A cannot.

1. A in 42 min fills $\frac{42}{84}=\frac{1}{2}$ of the cistern.
2. Remaining $\frac{1}{2}$ is filled by B.
3. Time for B $=\frac{1}{2}\times 36=18$ min.

Trick: 42 is exactly half of 84, so B only needs half of its 36 minutes.

Answer: (A) 18 minutes

Concept: Mode = the value that occurs most often.

1. Count repeats: 74 appears 4 times; 69, 77, 80, 87, 89 appear twice each; others once.
2. Highest frequency = 4 for 74.

Trick: Tick values that repeat while scanning once; only the repeating ones matter.

Answer: (A) 74

Concept: $A=P(1+\frac{RT}{100})$ for simple interest.

1. $RT=4\times 5=20\%$, so $A=1.2P$.
2. $1.2P=720$ ⇒ $P=\frac{720}{1.2}=600$.

Trick: 20% SI ⇒ Amount : Principal = 6 : 5, so P = ₹$720\times \frac{5}{6}$ = ₹600.

Answer: (A) ₹600

Concept: Dividend = Divisor × Quotient + Remainder; HCF by Euclid's method.

1. Let smaller = $S$: $L=5S+178$ and $L-S=2858$.
2. $4S+178=2858$ ⇒ $S=670$, $L=3528$.
3. $\text{HCF}(3528,3828)$: $3828-3528=300$; $3528=11\times 300+228$; $300-228=72$; $228=3\times 72+12$; $72=6\times 12$.
4. HCF = 12.

Trick: HCF of two numbers also divides their difference (300) – check options: 12 divides 300, 3528 and 3828.

Answer: (C) 12

Concept: $\text{SP}=\text{MP}\times \frac{100-d}{100}$ and $\text{CP}=\text{SP}\times \frac{100}{100+g}$.

1. SP = ₹$5960\times \frac{91}{100}$ = ₹5423.60.
2. CP = ₹$5423.6\times \frac{100}{130}$ = ₹4172.

Trick: One line: CP = ₹$\frac{5960\times 91}{130}=\frac{5960\times 7}{10}$ = ₹4172 (since $\frac{91}{130}=\frac{7}{10}$).

Answer: (C) ₹4,172

Concept: In a right triangle, $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. Base $=15\text{m}$, $\theta ={45}^{\circ }$.
2. $\tan {45}^{\circ }=\frac{h}{15}=1$.
3. $h=15\text{m}$.

Trick: At ${45}^{\circ }$ elevation, height = distance from the base.

Answer: (C) $15\text{m}$

Concept: If $x(1+r{)}^{12}=y(1+r{)}^{13}$, then $x=y(1+r)$.

1. Arjun gets $x$, Kartik gets $y$: $x(1.1{)}^{12}=y(1.1{)}^{13}$ ⇒ $x=1.1y$.
2. $x+y=1218$ ⇒ $2.1y=1218$.
3. $y=\frac{1218}{2.1}=580$.

Trick: Shares are in ratio $11:10$ (one extra year at 10%); Kartik = ₹$\frac{10}{21}\times 1218$ = ₹580.

Answer: (B) ₹580

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $61.6=3M-2\times 47.4=3M-94.8$.
2. $3M=156.4$.
3. $M=\frac{156.4}{3}=52.13\approx 52.1$.

Trick: $\text{Median}=\frac{\text{Mode}+2\text{Mean}}{3}=\frac{61.6+94.8}{3}$.

Answer: (B) $52.1$

Concept: Angle sum of a quadrilateral is ${360}^{\circ }$. If two adjacent angles add to ${180}^{\circ }$, the sides opposite their common side are parallel.

1. $2+8+3+7=20$ parts ⇒ 1 part $={18}^{\circ }$: $\angle P={36}^{\circ }$, $\angle Q={144}^{\circ }$, $\angle R={54}^{\circ }$, $\angle S={126}^{\circ }$.
2. $\angle P+\angle Q={180}^{\circ }$ ⇒ $PS\parallel QR$.
3. $\angle Q+\angle R={198}^{\circ }\ne {180}^{\circ }$ ⇒ $PQ$ not parallel to $SR$.
4. Exactly one pair of parallel sides ⇒ trapezium.

Trick: Ratio terms $2+8=3+7=10$ (half of 20) ⇒ co-interior pair ⇒ trapezium; all angles unequal rules out rhombus/square.

Answer: (A) Trapezium

Concept: After paying 40% rent, the remaining 60% of salary covers the other expenses and savings.

1. Other expenses + savings = ₹$442+9115+946$ = ₹10,503.
2. $60\%$ of salary = ₹10,503.
3. Salary = ₹$\frac{10503\times 100}{60}$ = ₹17,505.

Trick: Salary must be a multiple of 5 such that $0.6\times$ salary = 10503 – only 17,505 works ($17505\times 0.6=10503$).

Answer: (C) ₹17,505

Concept: $V=\frac{1}{3}\pi r^{2}h$, $l=\sqrt{r^2+h^2}$, $\text{TSA}=\pi r(r+l)$.

1. $\frac{1}{3}\times \frac{22}{7}\times r^2\times 21=1078$ ⇒ $22r^2=1078$ ⇒ $r^2=49$, $r=7$.
2. $l=\sqrt{49+441}=\sqrt{490}=7\sqrt{10}$.
3. $\text{TSA}=\frac{22}{7}\times 7\times (7+7\sqrt{10})=22\times 7(1+\sqrt{10})=154(1+\sqrt{10}){\text{cm}}^2$.

Trick: With $h=21$, $\frac{1}{3}\pi h=22$, so $r^2=\frac{1078}{22}=49$ instantly.

Answer: (C) $154(1+\sqrt{10}){\text{cm}}^2$

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

1. Total work = LCM(2, 52) = 52 units; A = 26 units/day, B = 1 unit/day.
2. B in 26 days: $26\times 1=26$ units; remaining $52-26=26$ units.
3. A's time $=\frac{26}{26}=1$ day.

Trick: B did half the work (26 of 52 days), so A needs half of 2 days = 1 day.

Answer: (C) 1

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

1. $R^3-r^3=729-512=217$.
2. Volume $=\frac{4}{3}\times \frac{22}{7}\times 217=\frac{2728}{3}{\text{cm}}^3$.
3. Mass $=21\times \frac{2728}{3}=19096\text{g}=19.096\text{kg}$.

Trick: $217=7\times 31$ cancels the 7 of $\pi$ and $21$ cancels the 3 → $4\times 22\times 31\times 7=19096$.

Answer: (B) 19.096