CBSE Class 10 Maths Basic 2024 Question Paper (Set 430/4/1) with Solutions

1. $d = \sqrt{(-1 - 2)^2 + (-5 + 1)^2} = \sqrt{9 + 16} = 5$ units.

Answer: (B) 5 units

1. The $y$-coordinate of the mid-point: $\dfrac{x + 4}{2} = -1$, so $x + 4 = -2$ and $x = -6$. (Check the $x$-coordinate: $\dfrac{4 - 2}{2} = 1$ ✓.)

Answer: (D) −6

1. An event and its complement together make the whole sample space: $P(E) + P(\bar{E}) = 1$.

Answer: (C) $P(E) + P(\bar{E}) = 1$

1. The highest frequency is $14$, so the modal class is $425 - 449$.
2. The classes have gaps ($424$ then $425$), so they are first made continuous by subtracting $0.5$ from each lower limit and adding $0.5$ to each upper limit: the class becomes $424.5 - 449.5$. The lower limit is $424.5$.

Answer: (C) 424.5

1. $\tan\theta = \dfrac{9}{3\sqrt{3}} = \sqrt{3}$, so $\theta = 60^\circ$.

Answer: (A) $60^\circ$

1. $k(1)^2 + 4(1) + k = 0$ gives $2k + 4 = 0$, so $k = -2$.

Answer: (B) −2

1. $k(x + 1)(x - 3)$ has these zeroes for every non-zero constant $k$, so there are infinitely many.

Answer: (D) more than 3

1. $x^2 = 4$ gives $x = 2$ or $x = -2$.

Answer: (B) −2, 2

1. (A) $x^2 - 4x + 5 = 2x - 3$ gives $x^2 - 6x + 8 = 0$: quadratic. (B) $2x^2 - 7x + 3 = x^2 + 4x - 5$ gives $x^2 - 11x + 8 = 0$: quadratic. (D) multiplying by $x$ gives $2x^2 - 5x + 3 = 0$: quadratic.
2. (C) $x^2 + x + 8 = x^2 - 4$ gives $x + 12 = 0$, which is linear. So (C) is not quadratic.

Answer: (C) $x(x + 1) + 8 = (x + 2)(x - 2)$

1. $a_{23} - a_{19} = (23 - 19)d = 4d = 32$, so $d = 8$.

Answer: (A) 8

1. $\dfrac{1}{\tan^2\theta} = \cot^2\theta$ and $\dfrac{1}{\sin^2\theta} = \operatorname{cosec}^2\theta$, so the value is $\cot^2\theta - \operatorname{cosec}^2\theta = -1$.

Answer: (B) −1

1. The region cut off by a chord, between the chord and an arc, is a segment (a sector is bounded by two radii and an arc).

Answer: (C) a segment

1. $1080 = 2^3 \times 3^3 \times 5$, so $x = 3$ and $y = 3$. $x - y = 0$.

Answer: (D) 0

1. $OA \perp PA$ (radius and tangent), so in the right triangle $OAP$: $\angle APO = 180^\circ - 90^\circ - 70^\circ = 20^\circ$.

Answer: (D) $20^\circ$

1. $N = 60$ and $\dfrac{N}{2} = 30$. Cumulative frequencies: $5, 13, 33, \dots$. The first one to reach $30$ is $33$, in the class $20 - 30$.

Answer: (B) 20 – 30

1. Arc $= \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 21 = 22$ cm.

Answer: (D) 22 cm

1. By definition, a tangent meets the circle at exactly one point.

Answer: (A) one point only

1. $\triangle ABC \sim \triangle QPR$ matches $A \leftrightarrow Q$, $B \leftrightarrow P$, $C \leftrightarrow R$. So $AC \leftrightarrow QR$ and $BC \leftrightarrow PR$.
2. $\dfrac{AC}{QR} = \dfrac{BC}{PR}$: $\dfrac{6}{3} = \dfrac{4.6}{x}$, so $x = 2.3$ cm.

Answer: (C) 2.3 cm

1. (R) This is the condition for coincident lines. True.
2. (A) $\dfrac{5}{7}$, $\dfrac{2}{6} = \dfrac{1}{3}$ and $\dfrac{6}{18} = \dfrac{1}{3}$: the first ratio is different, so the lines meet at one point. A has a unique solution, not infinitely many. False.

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

1. (R) An impossible event has no favourable outcomes, so its probability is $0$. True.
2. (A) A die has no face with $8$, so it is an impossible event and $P = 0$. True, and (R) is exactly why.

Answer: (A) Both A and R are true and R is the correct explanation of A

1. $\text{HCF} \times \text{LCM} =$ product of the numbers: $\text{LCM} = \dfrac{306 \times 1314}{18} = 17 \times 1314 = 22338$.

Answer: $22{,}338$

1. Let $O$ be the centre; $XY$ touches the circle at $A$ and $PQ$ at $B$. A tangent is perpendicular to the radius at the point of contact: $\angle OAY = 90^\circ$ and $\angle OBP = 90^\circ$.
2. These are alternate interior angles made by the transversal $AB$ with $XY$ and $PQ$, and they are equal. So $XY \parallel PQ$.

Answer: Proved

1. (a) In $\triangle POQ$ and $\triangle SOR$: $\angle POQ = \angle SOR$ (vertically opposite) and $\angle OPQ = \angle OSR$ (alternate angles, $PQ \parallel SR$). By AA, $\triangle POQ \sim \triangle SOR$.
2. So $\dfrac{OP}{OS} = \dfrac{OQ}{OR}$, that is $OP \times OR = OQ \times OS$.
3. (b) In $\triangle ABC$, $LM \parallel CB$, so $\dfrac{AM}{AB} = \dfrac{AL}{AC}$. (i) In $\triangle ADC$, $LN \parallel CD$, so $\dfrac{AN}{AD} = \dfrac{AL}{AC}$. (ii)
4. From (i) and (ii): $\dfrac{AM}{AB} = \dfrac{AN}{AD}$, hence $\dfrac{AM}{AN} = \dfrac{AB}{AD}$.

Answer: Proved (both parts)

1. (a) $\alpha + \beta = -\dfrac{14}{8} = -\dfrac{7}{4}$ and $\alpha\beta = \dfrac{3}{8}$. $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{-7/4}{3/8} = -\dfrac{14}{3}$.
2. (b) Sum $= -9 + 6 = -3$ and product $= -54$. The polynomial is $x^2 + 3x - 54$ (or $k(x^2 + 3x - 54)$).

Answer: (a) $-\dfrac{14}{3}$   (b) $x^2 + 3x - 54$

1. $\dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} - 1 = \dfrac{1}{4} + \dfrac{3}{4} - 1 = 0$.

Answer: $0$

1. (i) Minor sector $= \dfrac{90}{360} \times 3.14 \times 10 \times 10 = 78.5$ cm².
2. (ii) Area of the circle $= 3.14 \times 100 = 314$ cm². Major sector $= 314 - 78.5 = 235.5$ cm².

Answer: (i) $78.5$ cm²   (ii) $235.5$ cm²

1. (a) Let $ABCD$ circumscribe a circle with centre $O$, touching $AB, BC, CD, DA$ at $P, Q, R, S$. In $\triangle AOP$ and $\triangle AOS$: $OA$ is common, $OP = OS$ (radii) and $AP = AS$ (tangents), so they are congruent (SSS) and $\angle AOP = \angle AOS$. In the same way at $B$, $C$ and $D$.
2. The eight angles at $O$ add up to $360^\circ$ and they come in four equal pairs: $2(\angle AOP + \angle BOP + \angle COQ + \angle DOR) = 360^\circ$. So $\angle AOB + \angle COD = 180^\circ$, and likewise $\angle BOC + \angle AOD = 180^\circ$.
3. (b) $OP \perp PR$, so $\angle OPQ = 90^\circ - 40^\circ = 50^\circ$. $OP = OQ$ (radii), so $\angle OQP = 50^\circ$.
4. In $\triangle POQ$: $\angle POQ = 180^\circ - 50^\circ - 50^\circ = 80^\circ$.

Answer: (a) proved   (b) $80^\circ$

1. (a) $D = 2^2 - 4(2)(9) = 4 - 72 = -68 < 0$, so the equation has no real roots.
2. (b) Equal roots need $D = 4k^2 - 24k = 4k(k - 6) = 0$, so $k = 0$ or $k = 6$. For $k = 0$ the equation is $6 = 0$, not a quadratic, so $k = 6$.
3. Then $6x^2 - 12x + 6 = 0$, that is $x^2 - 2x + 1 = (x - 1)^2 = 0$ and the root is $x = 1$ (twice).

Answer: (a) no real roots   (b) $k = 6$; roots $1, 1$

1. (i) There are $2$ red kings: $P = \dfrac{2}{52} = \dfrac{1}{26}$.
2. (ii) Not black means red: $26$ cards. $P = \dfrac{26}{52} = \dfrac{1}{2}$.
3. (iii) There is one ace of hearts: $P = \dfrac{1}{52}$.

Answer: (i) $\dfrac{1}{26}$   (ii) $\dfrac{1}{2}$   (iii) $\dfrac{1}{52}$

1. Suppose $x = 2 + 5\sqrt{3}$ is rational. Then $\sqrt{3} = \dfrac{x - 2}{5}$, which is rational.
2. This contradicts the given fact that $\sqrt{3}$ is irrational, so $2 + 5\sqrt{3}$ is irrational.

Answer: Proved

1. LHS $= \tan^2 A + 2\tan A\sec A + \sec^2 A + \tan^2 A - 2\tan A\sec A + \sec^2 A = 2\tan^2 A + 2\sec^2 A$.
2. $= 2\left(\dfrac{\sin^2 A}{\cos^2 A} + \dfrac{1}{\cos^2 A}\right) = 2\,\dfrac{1 + \sin^2 A}{\cos^2 A} = 2\,\dfrac{1 + \sin^2 A}{1 - \sin^2 A} =$ RHS.

Answer: Proved

1. (a) Each cube has side $\sqrt[3]{125} = 5$ cm. The cuboid is $10 \times 5 \times 5$ cm. Volume $= 2 \times 125 = 250$ cm³.
2. Surface area $= 2(lb + bh + hl) = 2(10 \times 5 + 5 \times 5 + 5 \times 10) = 2 \times 125 = 250$ cm².
3. (b) $r = \dfrac{7}{2}$ cm and the cone height $h = \dfrac{7}{2}$ cm. Volume $= \dfrac{2}{3}\pi r^3 + \dfrac{1}{3}\pi r^2 h = \pi r^3$ (since $h = r$).
4. $= \dfrac{22}{7} \times \dfrac{343}{8} = \dfrac{539}{4} = 134.75$ cm³.

Answer: (a) $250$ cm³ and $250$ cm²   (b) $\dfrac{539}{4} = 134.75$ cm³

1. Younger children: in the right triangle with height $2$ m and angle $30^\circ$: $\sin 30^\circ = \dfrac{2}{AC} = \dfrac{1}{2}$, so $AC = 4$ m.
2. Older children: height $4$ m and angle $60^\circ$: $\sin 60^\circ = \dfrac{4}{PR} = \dfrac{\sqrt{3}}{2}$, so $PR = \dfrac{8}{\sqrt{3}} = \dfrac{8\sqrt{3}}{3} \approx 4.62$ m.

Answer: $4$ m and $\dfrac{8\sqrt{3}}{3} \approx 4.62$ m

1. (a) $\triangle ABC \sim \triangle PQR$ gives $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$ and $\angle A = \angle P$. $BD$ and $QM$ are medians, so $AC = 2AD$ and $PR = 2PM$: $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$.
2. In $\triangle ABD$ and $\triangle PQM$: $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$ and $\angle A = \angle P$. By SAS, $\triangle ABD \sim \triangle PQM$, so $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$.
3. (b)(i) $\triangle ABC \sim \triangle FEG$ gives $\angle ACB = \angle FGE$, and halving, $\angle 1 = \angle 3$ and $\angle 2 = \angle 4$ (the bisected halves). Also $\angle A = \angle F$. In $\triangle ACD$ and $\triangle FGH$: $\angle A = \angle F$ and $\angle 1 = \angle 3$, so $\triangle ACD \sim \triangle FGH$ (AA) and $\dfrac{CD}{GH} = \dfrac{AC}{FG}$.
4. (ii) In $\triangle DCB$ and $\triangle HGE$: $\angle B = \angle E$ (from the similarity) and $\angle 2 = \angle 4$. By AA, $\triangle DCB \sim \triangle HGE$.

Answer: Proved (both parts)

1. Let the first year’s production be $a$ and the yearly increase $d$. $a_4 = a + 3d = 720$ (i) and $a_8 = a + 7d = 880$ (ii).
2. (ii) $-$ (i): $4d = 160$, so $d = 40$ and $a = 600$.
3. Production in the $10^{\text{th}}$ year $= a_{10} = 600 + 9 \times 40 = 960$.
4. Total in the first $7$ years $= S_7 = \dfrac{7}{2}[2(600) + 6(40)] = \dfrac{7}{2} \times 1440 = 5040$.

Answer: $960$ TV sets in the $10^{\text{th}}$ year; $5040$ in the first $7$ years

1. (i) The highest frequency is $22$, so the modal class is $10 - 15$ and its upper limit is $15$.
2. (ii) $N = 80$ and $\dfrac{N}{2} = 40$. Cumulative frequencies: $13, 29, 51, 69, 80$. The median class is $10 - 15$.
3. (iii)(a) $l = 10$, $f_1 = 22$, $f_0 = 16$, $f_2 = 18$, $h = 5$. Mode $= 10 + \dfrac{22 - 16}{44 - 16 - 18} \times 5 = 10 + \dfrac{6}{10} \times 5 = 13$.
4. (iii)(b) Median $= l + \dfrac{\frac{N}{2} - cf}{f} \times h = 10 + \dfrac{40 - 29}{22} \times 5 = 10 + 2.5 = 12.5$.

Answer: (i) $15$   (ii) $10 - 15$   (iii)(a) $13$   (iii)(b) $12.5$

1. (i) Reading the grid, $C = (5, 4)$.
2. (ii) $B = (6, 6)$, so $BO = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}$ units.
3. (iii)(a) $A(2, 7)$, $B(6, 6)$, $C(5, 4)$. In the parallelogram $ABCD$ the diagonals bisect each other: the mid-point of $AC$ equals the mid-point of $BD$. $\left(\dfrac{2 + 5}{2}, \dfrac{7 + 4}{2}\right) = \left(\dfrac{6 + x}{2}, \dfrac{6 + y}{2}\right)$, so $x = 1$ and $y = 5$. $D = (1, 5)$.
4. (iii)(b) $AC = \sqrt{(5 - 2)^2 + (4 - 7)^2} = \sqrt{9 + 9} = 3\sqrt{2}$ units.

Answer: (i) $(5, 4)$   (ii) $6\sqrt{2}$   (iii)(a) $D(1, 5)$   (iii)(b) $3\sqrt{2}$

1. (i) $3x + 2y = 80$ and $4x + 3y = 110$.
2. (ii) Multiply the first by $3$ and the second by $2$: $9x + 6y = 240$ and $8x + 6y = 220$. Subtracting, $x = 20$. Then $2y = 80 - 60 = 20$ and $y = 10$.
3. (a) A notebook costs ₹ $20$. (b) A pen costs ₹ $10$.
4. (iii) $6 \times 20 + 3 \times 10 = 120 + 30 = ₹150$.

Answer: (i) $3x + 2y = 80$, $4x + 3y = 110$   (ii)(a) ₹ $20$   (ii)(b) ₹ $10$   (iii) ₹ $150$