CBSE Class 10 Maths Standard 2024 Question Paper (Set 30/5/1) with Solutions

1. Simplify: $\sqrt{18} = 3\sqrt{2}$, $\sqrt{50} = 5\sqrt{2}$, $\sqrt{98} = 7\sqrt{2}$. The common difference is $2\sqrt{2}$.
2. The next term is $7\sqrt{2} + 2\sqrt{2} = 9\sqrt{2} = \sqrt{162}$.

Answer: (C) $\sqrt{162}$

Common mistake: Looking at the numbers under the root ($18, 50, 98$). They are not an A.P.; the simplified terms $3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}$ are.

1. $x = 6\sin A$ and $y = 6\cos A$.
2. $x^2 + y^2 = 36(\sin^2 A + \cos^2 A) = 36$.

Answer: (A) 36

1. $\sec\theta = \dfrac{5}{4}$, so $\cos\theta = \dfrac{4}{5}$. Take hypotenuse $5$ and adjacent side $4$.
2. The opposite side is $\sqrt{25 - 16} = 3$, so $\cot\theta = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{4}{3}$.

Answer: (D) $\dfrac{4}{3}$

1. $\dfrac{a_1}{a_2} = \dfrac{3}{6} = \dfrac{1}{2}$, $\dfrac{b_1}{b_2} = \dfrac{4}{8} = \dfrac{1}{2}$ and $\dfrac{c_1}{c_2} = \dfrac{5}{7}$.
2. $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, so the lines are parallel.

Answer: (A) Parallel

1. Sum of roots $= -\dfrac{-6}{5} = \dfrac{6}{5}$ and product of roots $= \dfrac{21}{5}$.
2. Ratio $= \dfrac{6}{5} : \dfrac{21}{5} = 6 : 21 = 2 : 7$.

Answer: (B) 2 : 7

1. Sum $= 7 \times 7 = 49$, and the sum is $2 + 9 + (x + 6) + (2x + 3) + 5 + 10 + 5 = 40 + 3x$.
2. $40 + 3x = 49$, so $x = 3$.

Answer: (D) 3

1. Multiples of $7$ up to $40$: $7, 14, 21, 28, 35$. That is $5$ numbers.
2. $P = \dfrac{5}{40} = \dfrac{1}{8}$.

Answer: (B) $\dfrac{1}{8}$

1. Arc length $= \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 21 = 22$ cm.
2. Perimeter $=$ arc $+ 2r = 22 + 42 = 64$ cm.

Answer: (C) 64 cm

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

1. Arc length $= \dfrac{\theta}{360} \times 2\pi r$: $10\pi = \dfrac{\theta}{360} \times 2\pi \times 12 = \dfrac{\theta}{360} \times 24\pi$.
2. $\theta = \dfrac{10 \times 360}{24} = 150^\circ$.

Answer: (D) $150^\circ$

1. The numbers that are exactly divisible are $281 - 5 = 276$ and $1249 - 7 = 1242$.
2. The greatest number is $\text{HCF}(276, 1242)$. $1242 = 4 \times 276 + 138$ and $276 = 2 \times 138$, so the HCF is $138$.

Answer: (C) 138

Common mistake: Finding the HCF of 281 and 1249 directly. Subtract the remainders first.

1. $a = 3$, $d = 3$ and $a_n = 111$: $3 + (n - 1) \times 3 = 111$, so $n - 1 = 36$ and $n = 37$.

Answer: (C) 37

1. The two radii and the chord form a right triangle with legs $10$ and $10$.
2. Chord $= \sqrt{10^2 + 10^2} = 10\sqrt{2}$ cm.

Answer: (B) $10\sqrt{2}$

1. $28 = 2^2 \times 7$, $44 = 2^2 \times 11$, $132 = 2^2 \times 3 \times 11$.
2. LCM $= 2^2 \times 3 \times 7 \times 11 = 924$.

Answer: (D) 924

1. Co-prime numbers have no common factor other than $1$, so their HCF is $1$.

Answer: (A) 1

1. $\alpha + \beta = \dfrac{30}{k}$ and $\alpha\beta = \dfrac{45k}{k} = 45$.
2. $\dfrac{30}{k} = 45$, so $k = \dfrac{30}{45} = \dfrac{2}{3}$.

Answer: (D) $\dfrac{2}{3}$

1. Tangents from $R$ are equal: $RJ = RL$, so $\angle RLJ = \angle RJL = 42^\circ$ and $\angle JRL = 180^\circ - 84^\circ = 96^\circ$.
2. $OJ \perp RJ$ and $OL \perp RL$, so in quadrilateral $ORJL$: $\angle JOL = 360^\circ - 90^\circ - 90^\circ - 96^\circ = 84^\circ$.

Answer: (B) $84^\circ$

1. $DE \parallel BC$, so $\dfrac{AD}{DB} = \dfrac{AE}{EC}$: $\dfrac{2.4}{4} = \dfrac{2}{EC}$, so $EC = \dfrac{8}{2.4} = \dfrac{10}{3}$ cm.
2. $AC = AE + EC = 2 + \dfrac{10}{3} = \dfrac{16}{3}$ cm.

Answer: (C) $\dfrac{16}{3}$ cm

Common mistake: Giving $EC = \dfrac{10}{3}$ (option A). The question asks for $AC$, which is $AE + EC$.

1. At the same time the Sun makes the same angle, so the pole and the tower with their shadows form similar triangles.
2. $\dfrac{\text{pole}}{\text{shadow}} = \dfrac{\text{tower}}{\text{shadow}}$: $\dfrac{7.5}{5} = \dfrac{h}{24}$, so $h = 1.5 \times 24 = 36$ m.

Answer: (D) 36 m

1. (R) A line parallel to the two parallel sides cuts the slanting sides in the same ratio. This is a standard result, so R is true.
2. (A) is exactly this result applied to $EF$, so A is true and R explains A.

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

1. (A) The zero polynomial has no non-zero term, so its degree is not defined. True.
2. (R) A non-zero constant such as $5 = 5x^0$ has degree $0$. True.
3. R is about non-zero constants and does not explain why the degree of the zero polynomial is undefined. So R is not the explanation of A.

Answer: (B) Both true, but R is not the correct explanation of A

1. (a) Let the tangents $PA$ and $PB$ meet at $P$ with $\angle APB = 60^\circ$. $PO$ bisects the angle, so $\angle OPA = 30^\circ$. $OA \perp PA$.
2. In $\triangle OAP$: $\tan 30^\circ = \dfrac{OA}{AP} = \dfrac{3}{AP} = \dfrac{1}{\sqrt{3}}$, so $AP = 3\sqrt{3}$ cm.
3. (b) Let $AB$ be a diameter with tangents $PQ$ at $A$ and $XY$ at $B$. A tangent is perpendicular to the radius at the point of contact: $\angle OAP = \angle OBY = 90^\circ$.
4. These are alternate interior angles on the transversal $AB$, and they are equal. So $PQ \parallel XY$.

Answer: (a) $3\sqrt{3}$ cm   (b) proved

1. Numerator: $2 \times \dfrac{1}{\sqrt{3}} \times 2 \times 1 = \dfrac{4}{\sqrt{3}}$.
2. Denominator: $1 - \left(\dfrac{\sqrt{3}}{2}\right)^2 = 1 - \dfrac{3}{4} = \dfrac{1}{4}$.
3. Value $= \dfrac{4}{\sqrt{3}} \times 4 = \dfrac{16}{\sqrt{3}} = \dfrac{16\sqrt{3}}{3}$.

Answer: $\dfrac{16}{\sqrt{3}} = \dfrac{16\sqrt{3}}{3}$

1. $\alpha + \beta = -\dfrac{-6}{5} = \dfrac{6}{5}$ and $\alpha\beta = \dfrac{1}{5}$.
2. $\alpha + \beta + \alpha\beta = \dfrac{6}{5} + \dfrac{1}{5} = \dfrac{7}{5}$.

Answer: $\dfrac{7}{5}$

1. (a) Let $P$ divide $AB$ in the ratio $k : 1$. $x = \dfrac{3k - 6}{k + 1} = -4$, so $3k - 6 = -4k - 4$ and $7k = 2$, giving $k = \dfrac{2}{7}$.
2. The ratio is $2 : 7$. (Check with $y$: $\dfrac{-8k + 10}{k + 1} = \dfrac{-16/7 + 10}{9/7} = \dfrac{54/7}{9/7} = 6$ ✓.)
3. (b) Let $A(3, 0)$, $B(6, 4)$, $C(-1, 3)$. $AB = \sqrt{9 + 16} = 5$, $BC = \sqrt{49 + 1} = \sqrt{50}$, $CA = \sqrt{16 + 9} = 5$.
4. $AB = CA$, so two sides are equal: $\triangle ABC$ is isosceles.

Answer: (a) $2 : 7$   (b) proved ($AB = CA = 5$)

1. Anmol rejects only the shirts with major defects, so he accepts $60 - 8 = 52$ shirts (the $48$ good and the $4$ with minor defects).
2. $P = \dfrac{52}{60} = \dfrac{13}{15}$.

Answer: $\dfrac{13}{15}$

Common mistake: Using 48, which is what Nigam accepts. Anmol accepts the good shirts and the minor-defect ones too.

1. (a) Suppose $\sqrt{3} = \dfrac{p}{q}$ with $p, q$ co-prime and $q \neq 0$. Then $3q^2 = p^2$, so $3$ divides $p$. Write $p = 3a$: $9a^2 = 3q^2$, so $q^2 = 3a^2$ and $3$ divides $q$.
2. Both $p$ and $q$ are divisible by $3$: a contradiction. So $\sqrt{3}$ is irrational.
3. (b) $(\sqrt{2} + \sqrt{3})^2 = 2 + 3 + 2\sqrt{6} = 5 + 2\sqrt{6}$. Suppose it is rational, $= \dfrac{a}{b}$ with integers $a, b$ and $b \neq 0$.
4. Then $\sqrt{6} = \dfrac{a - 5b}{2b}$, which is rational. But $\sqrt{6}$ is irrational (given), a contradiction. So $5 + 2\sqrt{6}$ is irrational.

Answer: (a) proved   (b) proved

1. (a) $S_{14} = \dfrac{14}{2}[2 \times 10 + 13d] = 1050$, so $20 + 13d = 150$ and $d = 10$.
2. $a_{20} = 10 + 19 \times 10 = 200$ and $a_n = 10 + (n - 1) \times 10 = 10n$.
3. (b) $S_n = \dfrac{n}{2}(a + l) = \dfrac{n}{2}(5 + 45) = 25n = 400$, so $n = 16$.
4. $a_{16} = 5 + 15d = 45$, so $d = \dfrac{40}{15} = \dfrac{8}{3}$.

Answer: (a) $a_{20} = 200$, $a_n = 10n$   (b) $n = 16$, $d = \dfrac{8}{3}$

1. Let parallelogram $ABCD$ circumscribe a circle touching $AB, BC, CD, DA$ at $P, Q, R, S$. Tangents from a point are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$.
2. Add: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$, so $AB + CD = AD + BC$.
3. In a parallelogram $AB = CD$ and $AD = BC$, so $2AB = 2BC$ and $AB = BC$. All four sides are equal: $ABCD$ is a rhombus.

Answer: Proved

1. $\dfrac{\tan A}{1 - \cot A} = \dfrac{\sin^2 A}{\cos A(\sin A - \cos A)}$ and $\dfrac{\cot A}{1 - \tan A} = \dfrac{\cos^2 A}{\sin A(\cos A - \sin A)} = -\dfrac{\cos^2 A}{\sin A(\sin A - \cos A)}$.
2. Sum $= \dfrac{1}{\sin A - \cos A}\left[\dfrac{\sin^2 A}{\cos A} - \dfrac{\cos^2 A}{\sin A}\right] = \dfrac{\sin^3 A - \cos^3 A}{\sin A\cos A(\sin A - \cos A)}$.
3. Using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $= \dfrac{1 + \sin A\cos A}{\sin A\cos A} = \dfrac{1}{\sin A\cos A} + 1 = 1 + \sec A\,\mathrm{cosec}\,A$ = RHS.

Answer: Proved

1. $8$ equally likely outcomes: $HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$.
2. (i) At least one head: all except $TTT$: $\dfrac{7}{8}$.
3. (ii) Exactly one tail: $HHT, HTH, THH$: $\dfrac{3}{8}$. (iii) Two heads and one tail: the same three outcomes: $\dfrac{3}{8}$.

Answer: (i) $\dfrac{7}{8}$   (ii) $\dfrac{3}{8}$   (iii) $\dfrac{3}{8}$

1. Area of the circle $= 3.14 \times 10 \times 10 = 314$ cm².
2. Area of the minor sector $= \dfrac{90}{360} \times 314 = 78.5$ cm².
3. Area of the major sector $= 314 - 78.5 = 235.5$ cm².

Answer: $235.5$ cm²

Common mistake: Giving the minor sector (78.5 cm²). The question asks for the major sector.

1. (a) Real and equal roots need $D = 0$: $36(k + 1)^2 - 4(k + 1) \times 3(k + 9) = 0$.
2. $12(k + 1)[3(k + 1) - (k + 9)] = 0$, that is $12(k + 1)(2k - 6) = 0$. As $k \neq -1$, $2k - 6 = 0$ and $k = 3$.
3. (b) Let the son's age be $x$ years. The man's age is $2x^2$.
4. Eight years later: $2x^2 + 8 = 3(x + 8) + 4$, so $2x^2 - 3x - 20 = 0$ and $(2x + 5)(x - 4) = 0$.
5. Age is positive, so $x = 4$. The son is $4$ years old and the man is $2 \times 16 = 32$ years old.

Answer: (a) $k = 3$   (b) son $4$ years, man $32$ years

1. Let $P$ be the point on the bridge, $PF = 4$ m, and let the banks be $A$ and $B$, with $\angle$ of depression $30^\circ$ to $A$ and $60^\circ$ to $B$.
2. In $\triangle PFA$: $\tan 30^\circ = \dfrac{4}{AF} = \dfrac{1}{\sqrt{3}}$, so $AF = 4\sqrt{3}$ m.
3. In $\triangle PFB$: $\tan 60^\circ = \dfrac{4}{FB} = \sqrt{3}$, so $FB = \dfrac{4}{\sqrt{3}}$ m.
4. Width of the river $= AF + FB = 4\sqrt{3} + \dfrac{4}{\sqrt{3}} = \dfrac{12 + 4}{\sqrt{3}} = \dfrac{16}{\sqrt{3}} = \dfrac{16\sqrt{3}}{3}$ m $\approx 9.24$ m.

Answer: $\dfrac{16\sqrt{3}}{3}$ m $\approx 9.24$ m

1. (a) $\triangle FEC \cong \triangle GDB$, so their corresponding angles are equal: $\angle 3 = \angle 4$ (at $B$ and $C$).
2. In $\triangle ABC$, $\angle 3 = \angle 4$, so $AB = AC$. … (i) In $\triangle ADE$, $\angle 1 = \angle 2$, so $AD = AE$. … (ii)
3. Divide (ii) by (i): $\dfrac{AD}{AB} = \dfrac{AE}{AC}$. By the converse of the Basic Proportionality Theorem, $DE \parallel BC$.
4. So $\angle 1 = \angle 3$ and $\angle 2 = \angle 4$ (corresponding angles), and $\triangle ADE \sim \triangle ABC$ (AA).
5. (b) Produce $AD$ to $E$ so that $AD = DE$ and join $EC$. Produce $PM$ to $L$ so that $PM = ML$ and join $LR$. $\triangle ABD \cong \triangle ECD$ (SAS), so $AB = EC$. Similarly $PQ = LR$.
6. Given $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$, we get $\dfrac{EC}{LR} = \dfrac{AC}{PR} = \dfrac{2AD}{2PM} = \dfrac{AE}{PL}$. So $\triangle AEC \sim \triangle PLR$ (SSS), giving $\angle 2 = \angle 4$ (as marked in the proof). Similarly $\angle 1 = \angle 3$.
7. Adding both: $\angle BAC = \angle QPR$. With $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$, $\triangle ABC \sim \triangle PQR$ (SAS).

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

1. After scooping, the flat ends are gone. The surface is the curved surface of the cylinder plus the curved surfaces of the two hemispherical hollows.
2. Curved surface of the cylinder $= 2\pi r h = 2 \times \dfrac{22}{7} \times 2.1 \times 5.8 = 76.56$ cm².
3. Two hemispheres: $2 \times 2\pi r^2 = 4 \times \dfrac{22}{7} \times 2.1 \times 2.1 = 55.44$ cm².
4. Total surface area $= 76.56 + 55.44 = 132$ cm².

Answer: $132$ cm²

Common mistake: Adding the two flat circles of the cylinder. They are completely scooped out, so they are not part of the surface.

1. (i) Total tickets: $x + y = 300$. … (i) Total money: $150x + 250y = 55000$, that is $3x + 5y = 1100$. … (ii)
2. (ii)(a) From (i), $y = 300 - x$. Put in (ii): $3x + 5(300 - x) = 1100$, so $3x + 1500 - 5x = 1100$, $-2x = -400$ and $x = 200$. $200$ children visited.
3. (ii)(b) $y = 300 - 200 = 100$. $100$ adults visited.
4. (iii) $250 \times 150 + 100 \times 250 = 37500 + 25000 = ₹62{,}500$.

Answer: (i) $x + y = 300,\ 150x + 250y = 55000$   (ii)(a) $200$   (ii)(b) $100$   (iii) ₹ $62{,}500$

1. (i) With $A$ as the origin and the grid squares of side $1$ m, reading off the grid: $P = (4, 6)$, $Q = (3, 2)$, $R = (6, 5)$.
2. (ii)(a) $PQ = \sqrt{(4 - 3)^2 + (6 - 2)^2} = \sqrt{17}$ and $QR = \sqrt{(3 - 6)^2 + (2 - 5)^2} = \sqrt{18} = 3\sqrt{2}$.
3. (ii)(b) Section formula for $P$ to $R$ in the ratio $2 : 1$: $x = \dfrac{2 \times 6 + 1 \times 4}{3} = \dfrac{16}{3}$ and $y = \dfrac{2 \times 5 + 1 \times 6}{3} = \dfrac{16}{3}$. The point is $\left(\dfrac{16}{3}, \dfrac{16}{3}\right)$.
4. (iii) $PQ = \sqrt{17}$, $QR = \sqrt{18}$ and $PR = \sqrt{(4 - 6)^2 + (6 - 5)^2} = \sqrt{5}$. No two sides are equal, so $\triangle PQR$ is not isosceles.

Answer: (i) $P(4, 6),\ Q(3, 2),\ R(6, 5)$   (ii)(a) $\sqrt{17}$ and $\sqrt{18}$   (ii)(b) $\left(\dfrac{16}{3}, \dfrac{16}{3}\right)$   (iii) not isosceles

Common mistake: In (ii)(b), swapping the ratio. For 2 : 1 from P, the weight 2 goes with the coordinates of R.

1. Cumulative frequencies: $8,\ 18,\ 31,\ 37,\ 40$. $N = 40$ and $\dfrac{N}{2} = 20$.
2. (i) The first cumulative frequency above $20$ is $31$, so the median class is $40 - 60$.
3. (ii)(a) Class marks $x_i$: $10, 30, 50, 70, 90$. $f_i x_i$: $80,\ 300,\ 650,\ 420,\ 270$. $\sum f_i x_i = 1720$. Mean $= \dfrac{1720}{40} = 43$ seconds.
4. (ii)(b) Modal class $40 - 60$: $l = 40$, $f_1 = 13$, $f_0 = 10$, $f_2 = 6$, $h = 20$. Mode $= 40 + \dfrac{13 - 10}{26 - 10 - 6} \times 20 = 40 + \dfrac{3}{10} \times 20 = 46$ seconds.
5. (iii) Time less than $60$ seconds: $8 + 10 + 13 = 31$ students.

Answer: (i) $40 - 60$   (ii)(a) $43$ s   (ii)(b) $46$ s   (iii) $31$