Q&A: practice questions with answers

Concept: Remainder on dividing $f(x)$ by $(ax-b)$ = $f\left(\frac{b}{a}\right)$.

1. $3x-6=0\Rightarrow x=2$.
2. $f(2)=4·8+3·4-24+1=32+12-24+1=21$.

Trick: The factor 3 in $(3x-6)$ does not change the remainder – only the root $x=2$ matters.

Answer: (A) 21

Concept: Each triangle centre is the meeting point of a particular set of lines.

1. Altitudes (perpendiculars from vertices) meet at the orthocentre.
2. Medians → centroid; perpendicular bisectors → circumcentre; angle bisectors → incentre.

Yaad rakhein: Mnemonic: Altitude–Ortho, Median–Centroid, Bisector of angle–Incentre, Perp. bisector–Circumcentre.

Answer: (C) Orthocenter

Concept: Combine the fractions using $1-{\cos }^{2}x={\sin }^{2}x$.

1. Sum = $\frac{\sin x[(1+\cos x)+(1-\cos x)]}{1-{\cos }^{2}x}=\frac{2\sin x}{{\sin }^{2}x}=\frac{2}{\sin x}$.
2. $\frac{2}{\sin x}=4\Rightarrow \sin x=\frac{1}{2}\Rightarrow x={30}^{\circ }$.
3. $\sin x+\cos x=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2}$.

Trick: Remember: $\frac{\sin x}{1-\cos x}+\frac{\sin x}{1+\cos x}=2\csc x$ directly.

Answer: (A) $\frac{\sqrt{3}+1}{2}$

Concept: For a circle inside a square, side of square = diameter.

1. $\pi r^2=9856\Rightarrow r^2=9856\times \frac{7}{22}=3136\Rightarrow r=56$ cm.
2. Square side = 112 cm → perimeter 448 cm.
3. Triangle side = $\frac{112}{2}=56$ cm → perimeter 168 cm.
4. Sum = $448+168=616$ cm.

Trick: Perimeters: $4\times 112+3\times 56=56\times (8+3)=56\times 11=616$.

Answer: (A) 616 cm

Concept: Use Pythagoras twice: first for BC, then for AR in ΔABR.

1. $B{C}^2=A{C}^2-A{B}^2=64-48=16\Rightarrow BC=4$ cm.
2. R is the midpoint → $BR=2$ cm.
3. $A{R}^2=A{B}^2+B{R}^2=48+4=52\Rightarrow AR=\sqrt{52}=2\sqrt{13}$ cm.

Trick: $\sqrt{52}$ lies between 7 and 8, and $2\sqrt{13}\approx 7.2$ – the only option in that range.

Answer: (C) $2\sqrt{13}\mathrm{cm}$

Concept: Surface area $\propto r^2$, volume $\propto r^3$.

1. Area ratio 169 : 100 → radius ratio 13 : 10.
2. Volume ratio = ${13}^3:{10}^3$ = 2197 : 1000.
3. Increase = $\frac{1197}{1000}\times 100=119.7\%$.

Trick: +69% area = $(1.3{)}^2$ → radius +30%; volume = $(1.3{)}^3=2.197$ → +119.7%.

Answer: (A) 119.7%

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

1. Ratio A : B = 70000 : 80000 = 7 : 8 (total 15 parts).
2. 1 part = $300000÷15=$ ₹20,000.
3. B's share = $8\times 20000=$ ₹1,60,000.

Trick: Find one part first (total ÷ sum of ratio), then multiply.

Answer: (A) Rs. 160,000

Concept: Convert everything to sin and cos.

1. $\sin x·\sec x=\frac{\sin x}{\cos x}=\tan x$.
2. $\tan x÷\tan x=1$.

Trick: Spot that $\sin x·\sec x$ is itself $\tan x$ – answer in one look.

Answer: (B) 1

Concept: $\sin {30}^{\circ }=\frac{1}{2}$, $\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$.

1. Product = $\frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$.

Trick: $\sin \theta \cos \theta =\frac{1}{2}\sin 2\theta \Rightarrow \frac{1}{2}\sin {60}^{\circ }=\frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$.

Answer: (B) $\frac{\sqrt{3}}{4}$

Concept: Largest equal group that divides every quantity = HCF.

1. $26=2\times 13$, $39=3\times 13$, $52=4\times 13$.
2. Common factor = 13 → HCF = 13 trees per row.

Trick: 'Maximum in each group, equal groups' → HCF; 'minimum common multiple' → LCM.

Answer: (B) 13

Concept: $\text{New average}=(\text{old total}+\text{new weight})÷\text{new count}$; form two equations.

1. Let n students, average A. First joining: $nA+60=(n+1)(A+2)\Rightarrow A+2n=58$.
2. Second joining: $nA+120=(n+2)(A+3.5)\Rightarrow 2A+3.5n=113$.
3. From the first, $A=58-2n$; substitute: $116-4n+3.5n=113\Rightarrow n=6$, $A=46$ kg.
4. Check: $\frac{336}{7}=48(=46+2)$, $\frac{396}{8}=49.5(=46+3.5)$ ✔

Trick: Newcomer's weight = $\text{old average}+\text{rise}\times \text{new count}$: $60=A+2(n+1)$.

Answer: (A) 46 kg

Concept: Total work = LCM; subtract work done, divide the rest by remaining efficiency.

1. $\text{LCM}(12,16,24)=48$ units; A = 4, B = 3, C = 2 units/day.
2. 2 days together: $9\times 2=18$ units → 30 units left.
3. B + C = 5 units/day → $30÷5=6$ days.

Trick: Choose LCM as total work – every efficiency becomes a whole number.

Answer: (A) 6 days

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

1. $R=\frac{25000\times 100}{50000\times 5}$.
2. $=2500000÷250000=10\%$ p.a.

Trick: Interest is half the principal in 5 years → $50\%÷5=10\%$ per year.

Answer: (D) 10%

Concept: $\cot \theta =1$ means $\theta ={45}^{\circ }$.

1. $\theta ={45}^{\circ }\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}$.
2. ${\sin }^2\theta =\frac{1}{2}$.

Trick: Identity way: ${\csc }^2\theta =1+{\cot }^2\theta =2\Rightarrow {\sin }^2\theta =\frac{1}{2}$.

Answer: (C) $\frac{1}{2}$

Concept: $\cot A=\frac{\text{base}}{\text{perpendicular}}$; build a right triangle.

1. Take base = $\sqrt{3}$, perpendicular = 1 → hypotenuse = $\sqrt{3+1}=2$.
2. $\csc A=\frac{2}{1}=2$.
3. $\sec A=\frac{2}{\sqrt{3}}$.

Trick: $\cot A=\sqrt{3}\Rightarrow A={30}^{\circ }$; $\csc {30}^{\circ }=2$, $\sec {30}^{\circ }=\frac{2}{\sqrt{3}}$.

Answer: (D) $\mathrm{cosec}A=2,\sec A=\frac{2}{\sqrt{3}}$

Concept: Identify the tribe tied to medieval Punjab.

1. The Khokhars were a martial tribe of the Punjab (Salt Range–Jhelum area) who often rebelled against the Delhi Sultans in the 13th–14th centuries.
2. Munda and Oraon are tribes of the Chhotanagpur region (Jharkhand); Baiga live in central India (MP/Chhattisgarh).

Yaad rakhein: Jasrat Khokhar troubled the Sayyid rulers in the 15th century – the Khokhars stayed a Punjab power for long.

Answer: (A) The Khokhar

Concept: Check each conquest year against Akbar's chronology.

1. I – Ranthambore fort was taken from Surjan Hada in 1569 (after Chittor, 1568). Correct.
2. II – Gujarat campaign began in 1572; Surat fell in 1573. Correct.

Yaad rakhein: Sequence: Chittor 1568 → Ranthambore & Kalinjar 1569 → Gujarat 1572–73 → Bengal 1576.

Answer: (C) Both I and II

Concept: Moscow Film Festival 1963 – Best Actress honour.

1. Suchitra Sen won the Best Actress (Silver Prize) at Moscow in 1963 for the Bengali film Saat Paake Bandha.
2. She was the first Indian actress to win an international film award.

Yaad rakhein: Suchitra Sen also played the lead in Gulzar's Aandhi (1975).

Answer: (B) Suchitra Sen

Concept: Compare the years of the Indian-linked Nobel prizes.

1. Rabindranath Tagore – Literature, 1913 (Gitanjali) – first Indian and first Asian Nobel laureate.
2. C. V. Raman – Physics 1930; H. G. Khorana – Medicine 1968; V. Ramakrishnan – Chemistry 2009.

Yaad rakhein: 1913 → 1930 → 1968 → 2009: the earliest year wins.

Answer: (B) Rabindranath Tagore

Concept: Match the scheme name with its focus area.

1. Swavalambini ('self-reliant woman') is a women entrepreneurship programme of the Ministry of Skill Development & Entrepreneurship, launched in March 2025.
2. It began with women students of higher-education institutes in the North-East, offering training and mentorship to start ventures.

Trick: Name clue: 'Swavalambi' = self-reliant; the '-ini' ending points to women.

Answer: (B) Women Entrepreneurship Programme