CBSE Class 10 Maths Standard 2024 Question Paper (Set 30/2/1) with Solutions

1. Infinitely many solutions need $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.
2. $\dfrac{3}{6} = \dfrac{-1}{-k} = \dfrac{8}{16}$. So $\dfrac{1}{2} = \dfrac{1}{k}$, giving $k = 2$.

Answer: (B) 2

1. Section formula with $m : n = 5 : 2$: $x = \dfrac{5 \times 1 + 2 \times 4}{7} = \dfrac{13}{7}$.
2. $y = \dfrac{5 \times 2 + 2 \times (-5)}{7} = 0$. So $P = \left(\dfrac{13}{7}, 0\right)$.

Answer: (C) $\left(\dfrac{13}{7}, 0\right)$

1. $a_{15} - a_{11} = (a + 14d) - (a + 10d) = 4d$.
2. $4d = 48$, so $d = 12$.

Answer: (A) 12

1. $D = b^2 - 4ac = 1 - 4 = -3 < 0$.
2. A negative discriminant means there are no real roots: the roots are not real.

Answer: (D) not-real

1. $\text{HCF} \times \text{LCM} =$ product of the two numbers: $40 \times \text{LCM} = 2520 \times 6600$.
2. $\text{LCM} = \dfrac{2520 \times 6600}{40} = 2520 \times 165 = 415800$.
3. $415800 = 252 \times 1650$, so $k = 1650$.

Answer: (A) 1650

1. $DE \parallel BC$, so $\triangle ADE \sim \triangle ABC$ and $\dfrac{DE}{BC} = \dfrac{AD}{AB}$.
2. $AB = AD + DB = 5 + 2.5 = 7.5$ cm. $\dfrac{DE}{12} = \dfrac{5}{7.5} = \dfrac{2}{3}$.
3. $DE = 8$ cm.

Answer: (C) 8 cm

Common mistake: Using $\dfrac{AD}{DB}$ as the ratio of $DE : BC$. The ratio of $DE : BC$ is $AD : AB$.

1. $\sec\theta \cdot \sin\theta = \dfrac{1}{\cos\theta} \cdot \sin\theta = \tan\theta$.
2. $\sin\theta = \cos\theta$ gives $\tan\theta = 1$ (and $\theta = 45^\circ$). So the value is $1$.

Answer: (C) 1

1. Total outcomes $= 36$. A sum more than $10$ means $11$ or $12$.
2. Sum $11$: $(5, 6), (6, 5)$. Sum $12$: $(6, 6)$. That is $3$ outcomes.
3. $P = \dfrac{3}{36} = \dfrac{1}{12}$.

Answer: (D) $\dfrac{1}{12}$

1. $\alpha + \beta = -\dfrac{3}{5}$ and $\alpha\beta = -\dfrac{7}{5}$.
2. $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{-3/5}{-7/5} = \dfrac{3}{7}$.

Answer: (C) $\dfrac{3}{7}$

1. For similar triangles the ratio of perimeters equals the ratio of corresponding sides.
2. $\dfrac{PQ}{AB} = \dfrac{\text{perimeter of } PQR}{\text{perimeter of } ABC} = \dfrac{48}{56} = \dfrac{6}{7}$.

Answer: (B) $\dfrac{6}{7}$

Common mistake: Inverting the ratio. The question asks $PQ / AB$, with $PQR$ on top.

1. In $\triangle ADP$ and $\triangle CBP$: $\angle DAP = \angle DAB$ and $\angle BCP = \angle BCD$ are angles on the same arc $DB$, so they are equal.
2. $\angle APD = \angle CPB$ (vertically opposite angles).
3. So $\triangle ADP \sim \triangle CBP$ (AA), with $A \leftrightarrow C$, $D \leftrightarrow B$, $P \leftrightarrow P$.

Answer: (D) $\triangle ADP \sim \triangle CBP$

Common mistake: Writing the vertices in the wrong order. The order of letters shows which angles match.

1. Adding $2$ to every observation does not change their order, so the middle observation is still the same one, now $2$ larger.
2. So the median increases by $2$. (The mean also increases by $2$.)

Answer: (A) increases by 2

1. Cards $6$ to $55$: $55 - 6 + 1 = 50$ cards.
2. Perfect squares in this range: $9, 16, 25, 36, 49$ ($4$ is below $6$ and $64$ is above $55$). That is $5$ squares.
3. $P = \dfrac{5}{50} = \dfrac{1}{10}$.

Answer: (C) $\dfrac{1}{10}$

Common mistake: Taking the total as 55. The cards start at 6, so there are 50 of them.

1. Tangents from $P$ are equal: $PB = PA = 5$ cm. $\angle P = 90^\circ$.
2. In right $\triangle APB$: $AB = \sqrt{PA^2 + PB^2} = \sqrt{25 + 25} = 5\sqrt{2}$ cm.

Answer: (B) $5\sqrt{2}$ cm

1. The diagonals of a rectangle are equal. Use the diagonal $XY$: $XY = \sqrt{(0 + 3)^2 + (4 - 0)^2} = \sqrt{9 + 16} = 5$ units.

Answer: (A) 5 units

1. $a = -29$ and $d = -26 - (-29) = 3$. $a_n = a + (n - 1)d = 16$.
2. $-29 + 3(n - 1) = 16$, so $3(n - 1) = 45$ and $n - 1 = 15$, so $n = 16$.

Answer: (B) $16^{\text{th}}$

1. The angle between a tangent and a chord equals the angle in the alternate segment.
2. $\angle CAT = \angle CBA = 40^\circ$.

Answer: (D) $40^\circ$

Common mistake: Using the angle in a semicircle ($90^\circ$) to get $50^\circ$. The tangent-chord angle theorem gives the answer directly.

1. The value obtained by the largest number of students is the most frequent value: the mode.

Answer: (B) mode

1. (R) $\sin^2\theta + \cos^2\theta = 1$ is true for every angle.
2. (A) $\cos^2 A = 1 - \dfrac{1}{9} = \dfrac{8}{9}$, so $\cos A = \dfrac{2\sqrt{2}}{3}$ (positive for an acute angle). True.
3. The value of $\cos A$ in (A) was found using (R), so R explains A.

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

1. (R) Each face of the cube is $10 \times 10 = 100$ cm². True.
2. (A) Two cubes have $12$ faces, but the $2$ joined faces are hidden, so the cuboid has $10$ faces: $10 \times 100 = 1000$ cm², not $1200$ cm². (Check: the cuboid is $20 \times 10 \times 10$ and $2(200 + 200 + 100) = 1000$.) So A is false.

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

Common mistake: Counting all 12 faces. When the cubes are joined, two faces disappear.

1. $15^n = (3 \times 5)^n = 3^n \times 5^n$. Its only prime factors are $3$ and $5$.
2. A number ends with $0$ only if it is divisible by $10 = 2 \times 5$, so it needs both $2$ and $5$ as prime factors. $15^n$ has no factor $2$, so it can never end with $0$.

Answer: No, $15^n$ cannot end with $0$

1. $AB = \sqrt{(-5 - 1)^2 + 0^2} = 6$.
2. $BC = \sqrt{(-5 + 2)^2 + (0 - 5)^2} = \sqrt{9 + 25} = \sqrt{34}$ and $CA = \sqrt{(1 + 2)^2 + (0 - 5)^2} = \sqrt{34}$.
3. $BC = CA$ and $AB$ is different, so two sides are equal. It is not right-angled because $BC^2 + CA^2 = 68 \neq 36 = AB^2$. So $\triangle ABC$ is isosceles.

Answer: $\triangle ABC$ is an isosceles triangle

1. (a) $2 \times \left(\dfrac{1}{2}\right)^2 \times 2 + (\sqrt{3})^2 = 2 \times \dfrac{1}{4} \times 2 + 3 = 1 + 3 = 4$.
2. (b) $\sin(A + B) = \dfrac{\sqrt{3}}{2}$ gives $A + B = 60^\circ$. $\cos(A - B) = 1$ gives $A - B = 0^\circ$.
3. Adding: $2A = 60^\circ$, so $A = 30^\circ$ and $B = 30^\circ$.

Answer: (a) $4$   (b) $A = B = 30^\circ$

1. Join $OA$ and $OC$. $OA = OC$ (radii), so $\angle OAC = \angle OCA$.
2. A tangent is perpendicular to the radius at the point of contact: $\angle OAB = \angle OCD = 90^\circ$.
3. Subtract: $\angle OAB - \angle OAC = \angle OCD - \angle OCA$, that is $\angle BAC = \angle DCA$.
4. Yes, $\angle BAC = \angle DCA$.

Answer: Yes, $\angle BAC = \angle DCA$

1. (a) Let the point on $y = x$ divide the segment in the ratio $k : 1$. It is $\left(\dfrac{-k + 3}{k + 1},\ \dfrac{6k - 5}{k + 1}\right)$.
2. It lies on $y = x$, so $-k + 3 = 6k - 5$, giving $7k = 8$ and $k = \dfrac{8}{7}$. The ratio is $8 : 7$.
3. (b) $E$ is the mid-point of $AC$: $E = \left(\dfrac{3 - 1}{2},\ \dfrac{0 + 3}{2}\right) = \left(1, \dfrac{3}{2}\right)$.
4. $BE = \sqrt{(6 - 1)^2 + \left(4 - \dfrac{3}{2}\right)^2} = \sqrt{25 + \dfrac{25}{4}} = \sqrt{\dfrac{125}{4}} = \dfrac{5\sqrt{5}}{2}$.

Answer: (a) $8 : 7$   (b) $\dfrac{5\sqrt{5}}{2}$

1. (a) $S_m = S_n$: $\dfrac{m}{2}[2a + (m - 1)d] = \dfrac{n}{2}[2a + (n - 1)d]$.
2. $2a(m - n) = d[(n^2 - n) - (m^2 - m)] = d(n - m)(n + m - 1)$. Divide by $(m - n)$: $2a = -d(m + n - 1)$, that is $2a + (m + n - 1)d = 0$.
3. $S_{m+n} = \dfrac{m + n}{2}[2a + (m + n - 1)d] = \dfrac{m + n}{2} \times 0 = 0$.
4. (b) Let the three numbers be $a - d$, $a$, $a + d$. Sum: $3a = 24$, so $a = 8$.
5. Squares: $(8 - d)^2 + 64 + (8 + d)^2 = 194$, so $2d^2 + 192 = 194$ and $d^2 = 1$, $d = \pm 1$.
6. The numbers are $7, 8, 9$ (or $9, 8, 7$).

Answer: (a) $S_{m+n} = 0$   (b) $7, 8, 9$

1. Suppose $\sqrt{5} = \dfrac{p}{q}$ with $p, q$ co-prime and $q \neq 0$. Then $5q^2 = p^2$, so $5$ divides $p^2$ and hence $5$ divides $p$. Write $p = 5a$.
2. $25a^2 = 5q^2$, so $q^2 = 5a^2$, so $5$ divides $q^2$ and hence $q$.
3. $5$ is a common factor of $p$ and $q$, contradicting that they are co-prime. So $\sqrt{5}$ is irrational.

Answer: Proved by contradiction

1. (a) Join $BQ$. $AB$ is a diameter, so $\angle AQB = 90^\circ$ (angle in a semicircle).
2. In $\triangle AQB$: $\angle ABQ = 180^\circ - 90^\circ - 30^\circ = 60^\circ$, so the exterior angle $\angle QBP = 180^\circ - 60^\circ = 120^\circ$.
3. By the tangent-chord theorem $\angle BQP = \angle BAQ = 30^\circ$. In $\triangle BQP$: $\angle BPQ = 180^\circ - 120^\circ - 30^\circ = 30^\circ$.
4. $\angle BPQ = \angle BQP$, so the sides opposite them are equal: $BP = BQ$.
5. (b) Let the circle touch $AB, BC, CD, DA$ at $P, Q, R, S$. Join $OP, OQ, OR, OS$. Tangents from a point are equal and the radii are equal, so $\triangle POB \cong \triangle QOB$ and $\angle 1 = \angle 2$. Similarly the other three pairs give $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$.
6. The angles around $O$ add up to $360^\circ$: $2(\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^\circ$, so $\angle 1 + \angle 8 + \angle 4 + \angle 5 = 180^\circ$.
7. $\angle AOB = \angle 1 + \angle 8$ and $\angle COD = \angle 4 + \angle 5$, so $\angle AOB + \angle COD = 180^\circ$.

Answer: (a) $BP = BQ$   (b) $\angle AOB + \angle COD = 180^\circ$

1. Write $1 = \sec^2\theta - \tan^2\theta$ in the numerator: $(\sec^2\theta - \tan^2\theta) + (\sec\theta - \tan\theta)$.
2. $= (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) + (\sec\theta - \tan\theta) = (\sec\theta - \tan\theta)(\sec\theta + \tan\theta + 1)$.
3. So LHS $= \dfrac{(\sec\theta - \tan\theta)(\sec\theta + \tan\theta + 1)}{1 + \sec\theta + \tan\theta} = \sec\theta - \tan\theta$.
4. $= \dfrac{1}{\cos\theta} - \dfrac{\sin\theta}{\cos\theta} = \dfrac{1 - \sin\theta}{\cos\theta} =$ RHS.

Answer: Proved

1. Class marks $x_i$: $5, 15, 25, 35, 45$ and frequencies $f_i$: $12, 23, 34, 25, 6$ with $\sum f_i = 100$.
2. $f_i x_i$: $60,\ 345,\ 850,\ 875,\ 270$. Sum $\sum f_i x_i = 2400$.
3. Mean $= \dfrac{2400}{100} = 24$.

Answer: Mean marks = 24

1. Let the digit at the ten's place be $x$. The unit's digit is $x - 5$.
2. The product is $36$: $x(x - 5) = 36$, so $x^2 - 5x - 36 = 0$.
3. $(x - 9)(x + 4) = 0$. A digit cannot be negative, so $x = 9$ and the unit's digit is $4$.
4. The number is $94$. (Check: $9 \times 4 = 36$ ✓ and $9 - 4 = 5$ ✓.)

Answer: The number is 94

1. (a) First line: $y = -3x - 4$. Points: $(0, -4)$, $(-1, -1)$, $(-2, 2)$.
2. Second line: $y = 3x + 2$. Points: $(0, 2)$, $(-1, -1)$, $(1, 5)$.
3. Plot both lines. They cross at $(-1, -1)$, so $x = -1$ and $y = -1$.
4. (b) Let the number of correct answers be $x$ and the number of wrong answers be $y$. $3x - y = 40$. … (i) $4x - 2y = 50$, that is $2x - y = 25$. … (ii)
5. Subtract (ii) from (i): $x = 15$. Then $y = 3 \times 15 - 40 = 5$.
6. Total number of questions $= 15 + 5 = 20$.

Answer: (a) $x = -1,\ y = -1$   (b) $20$ questions

1. (a) Given: $\triangle ABC$ with $DE \parallel BC$. To prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. Join $BE$ and $CD$; draw $EF \perp AB$ and $DG \perp AC$.
2. $\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)} = \dfrac{\frac{1}{2} \cdot AD \cdot EF}{\frac{1}{2} \cdot DB \cdot EF} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(ADE)}{\text{ar}(CDE)} = \dfrac{AE}{EC}$.
3. $\text{ar}(BDE) = \text{ar}(CDE)$ (same base $DE$, same parallels), so $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
4. (b) Produce $AD$ to $E$ so that $AD = DE$ and join $EC$. Produce $PM$ to $N$ so that $PM = MN$ and join $NR$.
5. $\triangle ADB \cong \triangle EDC$ (SAS), so $AB = EC$. Similarly $PQ = NR$. Given $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$, we get $\dfrac{EC}{NR} = \dfrac{AC}{PR} = \dfrac{AE}{PN}$ (as $AE = 2AD$ and $PN = 2PM$).
6. So $\triangle AEC \sim \triangle PNR$ (SSS), giving $\angle 1 = \angle 2$. Similarly $\angle 3 = \angle 4$, so $\angle 1 + \angle 3 = \angle 2 + \angle 4$, that is $\angle A = \angle P$.
7. With $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$ and $\angle A = \angle P$: $\triangle ABC \sim \triangle PQR$ (SAS).

Answer: (a) BPT proved   (b) $\triangle ABC \sim \triangle PQR$

1. Let $AB = 45$ m be the lighthouse and let the ships be at $P$ ($30^\circ$) and $Q$ ($60^\circ$). Angle of depression equals angle of elevation from the ship.
2. For the ship at $60^\circ$: $\tan 60^\circ = \sqrt{3} = \dfrac{45}{BQ}$, so $BQ = \dfrac{45}{\sqrt{3}} = 15\sqrt{3}$ m.
3. For the ship at $30^\circ$: $\tan 30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{45}{BP}$, so $BP = 45\sqrt{3}$ m.
4. The ships are on opposite sides, so $PQ = BP + BQ = 45\sqrt{3} + 15\sqrt{3} = 60\sqrt{3} = 60 \times 1.73 = 103.8$ m.

Answer: $60\sqrt{3} \approx 103.8$ m

Common mistake: Subtracting the two distances. The ships are on opposite sides of the lighthouse, so the distances add.

1. Perimeter of a sector $= 2r + \dfrac{\theta}{360} \times 2\pi r = 20$.
2. $2 \times 5.6 + \dfrac{\theta}{360} \times 2 \times \dfrac{22}{7} \times 5.6 = 20$, so $11.2 + \dfrac{\theta}{360} \times 35.2 = 20$.
3. $\dfrac{\theta}{360} \times 35.2 = 8.8$, so $\dfrac{\theta}{360} = \dfrac{1}{4}$ and $\theta = 90^\circ$.
4. Area $= \dfrac{\theta}{360} \times \pi r^2 = \dfrac{1}{4} \times \dfrac{22}{7} \times 5.6 \times 5.6 = 24.64$ m².