CBSE Class 10 Maths Standard 2024 Question Paper (Set 30/3/1) with Solutions

1. Write the second as $3x + 6y - 1 = 0$. Then $\dfrac{a_1}{a_2} = \dfrac{1}{3}$, $\dfrac{b_1}{b_2} = \dfrac{2}{6} = \dfrac{1}{3}$ and $\dfrac{c_1}{c_2} = \dfrac{5}{-1} = -5$.
2. $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, so the lines are parallel: no solution.

Answer: (D) no solution

1. $d = \dfrac{1 - 4x}{2x} - \dfrac{1}{2x} = \dfrac{-4x}{2x} = -2$.
2. Check: $\dfrac{1 - 8x}{2x} - \dfrac{1 - 4x}{2x} = \dfrac{-4x}{2x} = -2$ ✓.

Answer: (B) −2

1. Total outcomes $= 36$. Same numbers: $(1,1), (2,2), \dots, (6,6)$, so $6$ outcomes.
2. $P(\text{different}) = 1 - \dfrac{6}{36} = \dfrac{30}{36} = \dfrac{5}{6}$.

Answer: (B) $\dfrac{5}{6}$

1. $P(\text{correct}) = 1 - \dfrac{2}{3} = \dfrac{1}{3}$.
2. $\dfrac{x}{6} = \dfrac{1}{3}$, so $x = 2$.

Answer: (A) 2

1. $3780 = 2^2 \times 3^3 \times 5 \times 7$.
2. LCM takes the greatest power of each prime, so the power of $3$ in the LCM is $\max(x, 1) = 3$, giving $x = 3$.

Answer: (C) 3

1. $2x^2 - 3x - 9 = 2x^2 - 6x + 3x - 9 = 2x(x - 3) + 3(x - 3) = (2x + 3)(x - 3)$.
2. The zeroes are $x = 3$ and $x = -\dfrac{3}{2}$.

Answer: (A) $3,\ -\dfrac{3}{2}$

1. $\tan 60^\circ = \dfrac{h}{30}$, so $h = 30\sqrt{3}$ m.

Answer: (B) $30\sqrt{3}$

1. $\cos\theta = \dfrac{\sqrt{3}}{2}$ gives $\theta = 30^\circ$, and $\sin\phi = \dfrac{1}{2}$ gives $\phi = 30^\circ$.
2. $\tan(\theta + \phi) = \tan 60^\circ = \sqrt{3}$.

Answer: (A) $\sqrt{3}$

1. Two circles that do not meet have $4$ common tangents; touching externally, $3$; intersecting at two points, $2$ (the two outer ones); touching internally, $1$.
2. So the maximum is $2$.

Answer: (C) 2

1. $OT \perp PT$ (radius perpendicular to the tangent), so in $\triangle OTP$: $\angle TOP = 180^\circ - 90^\circ - 35^\circ = 55^\circ$.
2. The diameter through $T$ is a straight line, so the angle $x$ on the other side of $OP$ is $180^\circ - 55^\circ = 125^\circ$.

Answer: (D) $125^\circ$

Common mistake: Stopping at $55^\circ$. The angle marked $x$ is the supplementary angle on the other side of $OP$.

1. If the diagonals $AC$ and $BD$ meet at $O$ with $\dfrac{OA}{OC} = \dfrac{OB}{OD}$, then $\triangle OAB \sim \triangle OCD$, so $\angle OAB = \angle OCD$ and $AB \parallel CD$.
2. Only one pair of opposite sides is guaranteed parallel, so the quadrilateral is a trapezium. (In a parallelogram the diagonals bisect each other, which is a special case of this ratio.)

Answer: (D) trapezium

1. $DE \parallel BC$, so $\triangle ADE \sim \triangle ABC$: $\dfrac{DE}{BC} = \dfrac{AD}{AB}$.
2. $AB = AD + DB = 2 + 3 = 5$. $\dfrac{DE}{7.5} = \dfrac{2}{5}$, so $DE = 3$ cm.

Answer: (B) 3

1. $\text{HCF} \times \text{LCM} = 2520 \times 6600$, so $\text{LCM} = \dfrac{2520 \times 6600}{40} = 2520 \times 165 = 415800$.
2. $415800 = 252 \times 1650$, so $k = 1650$.

Answer: (A) 1650

1. (A) $\sqrt{16} = 4$ and $\sqrt{4} = 2$ are rational, so it is not a pair of irrationals. (D) $\sqrt{36} = 6$ is rational.
2. (B) $\sqrt{5} \times \sqrt{2} = \sqrt{10}$ is irrational.
3. (C) $\sqrt{3}$ and $\sqrt{27} = 3\sqrt{3}$ are both irrational, and $\sqrt{3} \times 3\sqrt{3} = 9$, rational. ✓

Answer: (C) $(\sqrt{3}, \sqrt{27})$

1. Odd prime digits among $1$ to $9$: $3, 5, 7$. That is $3$ out of $9$.
2. $P = \dfrac{3}{9} = \dfrac{1}{3}$. (The prime $2$ is even.)

Answer: (A) $\dfrac{1}{3}$

1. Sum of all five $= 5 \times 15 = 75$. Sum of the first three $= 3 \times 14 = 42$. Sum of the last three $= 3 \times 17 = 51$.
2. The third observation is counted in both sums, so $42 + 51 - 75 = 18$.

Answer: (C) 18

1. Arc length $= \dfrac{90}{360} \times 2\pi r = \dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7 = 11$ cm.
2. Perimeter $=$ arc $+ 2r = 11 + 14 = 25$ cm.

Answer: (D) 25 cm

Common mistake: Stopping at the arc length (11 cm). The perimeter of a sector also includes the two radii.

1. The angle between a tangent and a chord is $70^\circ$, so the angle in the alternate segment is $70^\circ$.
2. The angle at the centre is twice the angle at the circumference on the same arc: $\angle MON = 2 \times 70^\circ = 140^\circ$.

Answer: (B) $140^\circ$

1. (R) This is the section formula, which is correct.
2. (A) With $m_1 : m_2 = 1 : 2$: $x = \dfrac{1 \times (-1) + 2 \times 1}{3} = \dfrac{1}{3}$ and $y = \dfrac{1 \times 1 + 2 \times 2}{3} = \dfrac{5}{3}$. The point is $\left(\dfrac{1}{3}, \dfrac{5}{3}\right)$, not $\left(-\dfrac{1}{3}, \dfrac{5}{3}\right)$. So A is false.

Answer: (D) A is false, R is true

1. $P(\text{boundary}) = \dfrac{9}{45} = \dfrac{1}{5}$, so $P(\text{no boundary}) = 1 - \dfrac{1}{5} = \dfrac{4}{5}$. A is true.
2. R, $P(E) + P(\text{not } E) = 1$, is true and is exactly the rule used to get A. So R explains A.

Answer: (A) Both true, and R explains A

1. Total outcomes $= 52$.
2. (i) There is $1$ queen of hearts: $P = \dfrac{1}{52}$.
3. (ii) There are $4$ jacks, so $52 - 4 = 48$ cards are not jacks. $P = \dfrac{48}{52} = \dfrac{12}{13}$.

Answer: (i) $\dfrac{1}{52}$   (ii) $\dfrac{12}{13}$

1. (a) Add the equations: $6x = 30$, so $x = 5$. Then $y = 13 - 10 = 3$. $x - y = 5 - 3 = 2$.
2. (b) Let the numbers be $x$ and $y$ with $x > y$: $x + y = 105$ and $x - y = 45$. Add: $2x = 150$, so $x = 75$ and $y = 30$.

Answer: (a) $2$   (b) $75$ and $30$

1. (a) $PA = PB$, so $PA^2 = PB^2$: $(x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$.
2. $-14x + 49 - 2y + 1 = -6x + 9 - 10y + 25$, so $-8x + 8y + 16 = 0$, that is $x - y - 2 = 0$.
3. (b) $O$ is the mid-point of the diameter $AB$: $\dfrac{y + 7}{2} = -3y$, so $y + 7 = -6y$ and $y = -1$.
4. Then $O = (2, 3)$ and $B = (5, 7)$. Radius $= OB = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{9 + 16} = 5$.

Answer: (a) $x - y - 2 = 0$   (b) $y = -1$, radius $= 5$

1. In $\triangle EAB$ and $\triangle ECD$: $\dfrac{EA}{EC} = \dfrac{EB}{ED}$ (given).
2. $\angle AEB = \angle CED$ (vertically opposite angles).
3. So $\triangle EAB \sim \triangle ECD$ by SAS similarity.

Answer: $\triangle EAB \sim \triangle ECD$

1. $\cos 45^\circ = \dfrac{1}{\sqrt{2}}$, $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$, $\sec 30^\circ = \dfrac{2}{\sqrt{3}}$, $\mathrm{cosec}\,30^\circ = 2$.
2. Numerator $= \dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2} = \dfrac{2 + \sqrt{6}}{2\sqrt{2}}$. Denominator $= \dfrac{2}{\sqrt{3}} + 2 = \dfrac{2(1 + \sqrt{3})}{\sqrt{3}}$.
3. Value $= \dfrac{2 + \sqrt{6}}{2\sqrt{2}} \times \dfrac{\sqrt{3}}{2(1 + \sqrt{3})} = \dfrac{2\sqrt{3} + 3\sqrt{2}}{4\sqrt{2}(1 + \sqrt{3})}$.

Answer: $\dfrac{2\sqrt{3} + 3\sqrt{2}}{4\sqrt{2}(1 + \sqrt{3})}$ (about $0.4987$)

1. (a) $S_7 = \dfrac{7}{2}(2a + 6d) = 49$, so $a + 3d = 7$. … (i) $S_{17} = \dfrac{17}{2}(2a + 16d) = 289$, so $a + 8d = 17$. … (ii)
2. Subtract: $5d = 10$, so $d = 2$ and $a = 1$. $S_{20} = \dfrac{20}{2}[2 \times 1 + 19 \times 2] = 10 \times 40 = 400$.
3. (b) $\dfrac{a + 9d}{a + 29d} = \dfrac{1}{3}$: $3a + 27d = a + 29d$, so $a = d$. … (i)
4. $S_6 = \dfrac{6}{2}(2a + 5d) = 42$, so $2a + 5d = 14$. … (ii) Put $a = d$: $7d = 14$, so $d = 2$ and $a = 2$.

Answer: (a) $S_{20} = 400$   (b) $a = 2$, $d = 2$

1. $x^2 - 15 = (x - \sqrt{15})(x + \sqrt{15})$, so the zeroes are $\sqrt{15}$ and $-\sqrt{15}$.
2. Sum of zeroes $= \sqrt{15} - \sqrt{15} = 0$. From the polynomial: $-\dfrac{\text{coeff of } x}{\text{coeff of } x^2} = -\dfrac{0}{1} = 0$ ✓.
3. Product of zeroes $= \sqrt{15} \times (-\sqrt{15}) = -15$. From the polynomial: $\dfrac{\text{constant}}{\text{coeff of } x^2} = \dfrac{-15}{1} = -15$ ✓.

Answer: Zeroes $\pm\sqrt{15}$; relationships verified

1. First line: $y = x + 1$. Points: $(0, 1)$, $(2, 3)$, $(4, 5)$.
2. Second line: $y = 5 - x$. Points: $(0, 5)$, $(2, 3)$, $(5, 0)$.
3. Plot both lines. They meet at $(2, 3)$, so $x = 2$ and $y = 3$.

Answer: $x = 2,\ y = 3$

1. (a) The $Y$-axis has $x = 0$. Let the segment from $(5, 3)$ to $(-1, 6)$ be divided in the ratio $k : 1$: $x = \dfrac{-k + 5}{k + 1} = 0$.
2. $k = 5$. The ratio is $5 : 1$.
3. (b) $PR = 2QR$ means $PR : RQ = 2 : 1$. $x = \dfrac{2 \times 3 + 1 \times (-2)}{3} = \dfrac{4}{3}$ and $y = \dfrac{2 \times 2 + 1 \times 5}{3} = 3$.

Answer: (a) $5 : 1$   (b) $R = \left(\dfrac{4}{3}, 3\right)$

1. LHS $= \dfrac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta(2\cos^2\theta - 1)} = \tan\theta \cdot \dfrac{1 - 2\sin^2\theta}{2\cos^2\theta - 1}$.
2. $2\cos^2\theta - 1 = 2(1 - \sin^2\theta) - 1 = 1 - 2\sin^2\theta$.
3. So LHS $= \tan\theta \cdot \dfrac{1 - 2\sin^2\theta}{1 - 2\sin^2\theta} = \tan\theta =$ RHS.

Answer: Proved

1. Let $AB$ be a chord of a circle with centre $O$, and let the tangents at $A$ and $B$ meet at $P$. We prove $\angle PAB = \angle PBA$.
2. Tangents from an external point are equal: $PA = PB$.
3. In $\triangle PAB$, $PA = PB$, so the angles opposite them are equal: $\angle PAB = \angle PBA$.
4. So the two tangents make equal angles with the chord.

Answer: Proved

1. (a) Let the original speed be $x$ km/h. Original time $= \dfrac{2800}{x}$ and new time $= \dfrac{2800}{x - 100}$, which is $\dfrac{1}{2}$ hour more.
2. $\dfrac{2800}{x - 100} - \dfrac{2800}{x} = \dfrac{1}{2}$, so $\dfrac{2800 \times 100}{x(x - 100)} = \dfrac{1}{2}$ and $x^2 - 100x - 560000 = 0$.
3. $(x - 800)(x + 700) = 0$. Speed is positive, so $x = 800$ km/h. Original duration $= \dfrac{2800}{800} = 3.5$ hours $= 3$ h $30$ min.
4. (b) Let the numerator be $x$, so the denominator is $2x + 1$ and the fraction is $\dfrac{x}{2x + 1}$. The sum with its reciprocal is $\dfrac{x}{2x + 1} + \dfrac{2x + 1}{x} = \dfrac{58}{21}$.
5. $21[x^2 + (2x + 1)^2] = 58x(2x + 1)$, so $21(5x^2 + 4x + 1) = 116x^2 + 58x$, giving $105x^2 + 84x + 21 = 116x^2 + 58x$ and $11x^2 - 26x - 21 = 0$.
6. $(x - 3)(11x + 7) = 0$. A numerator is positive here, so $x = 3$ and the fraction is $\dfrac{3}{7}$.

Answer: (a) $3.5$ hours ($3$ h $30$ min)   (b) $\dfrac{3}{7}$

Common mistake: In (a), answering the original speed (800 km/h). The question asks for the original duration of the flight.

1. Statement: if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
2. Given: $\triangle ABC$ with $DE \parallel BC$, where $D$ is on $AB$ and $E$ is on $AC$. To prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
3. Construction: join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.
4. $\text{ar}(\triangle ADE) = \dfrac{1}{2} \times AD \times EN$ and $\text{ar}(\triangle BDE) = \dfrac{1}{2} \times DB \times EN$, so $\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)} = \dfrac{AD}{DB}$. … (i)
5. $\text{ar}(\triangle ADE) = \dfrac{1}{2} \times AE \times DM$ and $\text{ar}(\triangle CDE) = \dfrac{1}{2} \times EC \times DM$, so $\dfrac{\text{ar}(ADE)}{\text{ar}(CDE)} = \dfrac{AE}{EC}$. … (ii)
6. $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels $DE$ and $BC$, so $\text{ar}(BDE) = \text{ar}(CDE)$. … (iii)
7. From (i), (ii) and (iii): $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. Hence proved.

Answer: Basic Proportionality Theorem proved

1. Let the point be $P$, at a distance $x$ from the foot $F$ of the building. The building has height $FT = 20$ m and the tower is $TU = h$ m.
2. Bottom of the tower (top of the building) at $45^\circ$: $\tan 45^\circ = \dfrac{20}{x} = 1$, so $x = 20$ m. … (i)
3. Top of the tower at $60^\circ$: $\tan 60^\circ = \dfrac{h + 20}{x} = \sqrt{3}$, so $h + 20 = x\sqrt{3}$. … (ii)
4. Put $x = 20$: $h + 20 = 20\sqrt{3}$, so $h = 20(\sqrt{3} - 1)$ m $\approx 14.64$ m.

Answer: Height of the tower $= 20(\sqrt{3} - 1)$ m $\approx 14.64$ m

Common mistake: Using $h$ alone with $\tan 60^\circ$. The $60^\circ$ is measured to the top of the tower, at height $h + 20$.

1. (a) Lower cylinder: $r = 14$ cm, $h = 200$ cm. Upper cylinder: $r = 7$ cm, $h = 50$ cm.
2. Volume $= \pi \times 14^2 \times 200 + \pi \times 7^2 \times 50 = \dfrac{22}{7} \times 196 \times 200 + \dfrac{22}{7} \times 49 \times 50 = 123200 + 7700 = 130900$ cm³.
3. Mass $= 130900 \times 8$ g $= 1047200$ g $= 1047.2$ kg.
4. (b) Radius of each hemisphere $=$ radius of the cylinder $= 2$ mm. Length of the cylindrical part $= 14 - 2 - 2 = 10$ mm.
5. Surface area $=$ CSA of the cylinder $+ 2 \times$ CSA of a hemisphere $= 2\pi r h + 2 \times 2\pi r^2 = 2 \times \dfrac{22}{7} \times 2 \times 10 + 4 \times \dfrac{22}{7} \times 4 = \dfrac{880 + 352}{7} = \dfrac{1232}{7} = 176$ mm².
6. Volume $= \pi r^2 h + 2 \times \dfrac{2}{3}\pi r^3 = \dfrac{22}{7} \times 4 \times 10 + \dfrac{4}{3} \times \dfrac{22}{7} \times 8 = \dfrac{880}{7} + \dfrac{704}{21} = \dfrac{3344}{21} \approx 159.24$ mm³.

Answer: (a) $1047.2$ kg   (b) surface area $176$ mm², volume $\dfrac{3344}{21} \approx 159.24$ mm³

Common mistake: In (b), using $14$ mm as the length of the cylinder. The two hemispheres take $2 + 2 = 4$ mm, leaving $10$ mm.

1. (i) Area of the square field $= 20 \times 20 = 400$ m².
2. (ii)(a) Each horse grazes a quarter circle of radius $7$ m, and the four quarter circles together make a full circle: area $= \pi r^2 = \dfrac{22}{7} \times 7 \times 7 = 154$ m².
3. (ii)(b) With a $10$ m rope, one horse grazes a quarter circle: $\dfrac{1}{4} \times 3.14 \times 10 \times 10 = 78.5$ m².
4. (iii) With a rope of length $7$ cm, the four quarter circles together have area $\dfrac{22}{7} \times 7 \times 7 = 154$ cm² $= 0.0154$ m². The area left ungrazed $= 400 - 0.0154 = 399.9846$ m².

Answer: (i) $400$ m²   (ii)(a) $154$ m²   (ii)(b) $78.5$ m²   (iii) $399.9846$ m²

Common mistake: In (iii), subtracting 154 from 400. The rope is 7 cm, not 7 m, so the grazed area is 154 cm², which is only 0.0154 m².

1. The ages are in inclusive classes, so convert them to continuous classes by subtracting $0.5$ and adding $0.5$: $14.5 - 19.5,\ 19.5 - 24.5,\ \dots,\ 49.5 - 54.5$.
2. (i) The highest frequency is $132$, so the modal class is $19.5 - 24.5$ and its lower limit is $19.5$.
3. (ii)(a) Cumulative frequencies: $62,\ 194,\ 290,\ 327,\ 340,\ 351,\ 361,\ 365$. $N = 365$ and $\dfrac{N}{2} = 182.5$, which first falls within $194$, so the median class is $19.5 - 24.5$.
4. (ii)(b) Participants younger than $50$: all classes except $50 - 54$: $62 + 132 + 96 + 37 + 13 + 11 + 10 = 361$.
5. (iii) $3\,\text{Median} = \text{Mode} + 2\,\text{Mean}$.

Answer: (i) $19.5$   (ii)(a) $19.5 - 24.5$   (ii)(b) $361$   (iii) $3\,\text{Median} = \text{Mode} + 2\,\text{Mean}$