CBSE Class 10 Maths Standard 2023 Question Paper (Set 30/1/1) with Solutions

1. The zeroes of a polynomial are the $x$-coordinates of the points where its graph meets the $x$-axis.
2. The curve touches the $x$-axis at exactly one point (the top of the arch), so $p(x)$ has $1$ zero.

Answer: (B) 1

1. Write $kx - y - 2 = 0$ and $6x - 2y - 3 = 0$. For infinitely many solutions $\dfrac{k}{6} = \dfrac{-1}{-2} = \dfrac{-2}{-3}$.
2. $\dfrac{k}{6} = \dfrac{1}{2}$ gives $k = 3$, but $\dfrac{1}{2} \neq \dfrac{2}{3}$. The three ratios can never be equal, so no such $k$ exists. (With $k = 3$ the lines are parallel.)

Answer: (B) does not exist

1. In an A.P. the middle term is the mean of the other two, so $2(p + 1) = (p - 1) + (2p + 3)$.
2. $2p + 2 = 3p + 2$, so $p = 0$.

Answer: (C) 0

1. Let the point on the $x$-axis divide $AB$ in the ratio $k : 1$. Its $y$-coordinate is $0$: $\dfrac{-3k + 6}{k + 1} = 0$.
2. $k = 2$, so the ratio is $2 : 1$.

Answer: (D) 2 : 1

1. $OA = OB$, so in $\triangle OAB$: $\angle OBA = \dfrac{180^\circ - 95^\circ}{2} = 42.5^\circ$.
2. $OB \perp PQ$, so $\angle OBQ = 90^\circ$, and $\angle ABQ = 90^\circ - 42.5^\circ = 47.5^\circ$.

Answer: (A) $47.5^\circ$

1. Divide the numerator and denominator by $\cos A$: $\dfrac{4\tan A + 3}{4\tan A - 3}$.
2. $\tan A = \dfrac{3}{2}$, so the value is $\dfrac{6 + 3}{6 - 3} = \dfrac{9}{3} = 3$.

Answer: (C) 3

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

Answer: (A) 1

1. Roots are rational when the discriminant is a perfect square: $D = k^2 + 32$.
2. $k = 2$ gives $D = 36 = 6^2$ ✓. (Smaller positive integers: $k = 1$ gives $33$, not a square.)
3. Roots for $k = 2$: $\dfrac{-2 \pm 6}{4} = 1,\ -2$, both rational.

Answer: (B) 2

1. $\tan^2 30^\circ = \dfrac{1}{3}$, $\sec^2 45^\circ = 2$ and $\sin^2 60^\circ = \dfrac{3}{4}$.
2. $\dfrac{3}{4} \times \dfrac{1}{3} - 2 + \dfrac{3}{4} = \dfrac{1}{4} - 2 + \dfrac{3}{4} = -1$.

Answer: (A) −1

1. $2\pi r h = 94.2$, so $2 \times 3.14 \times r \times 5 = 94.2$ and $31.4\,r = 94.2$.
2. $r = 3$ cm.

Answer: (B) 3 cm

1. Convert the cumulative frequencies to class frequencies by differences: $0$–$10$: $3$; $10$–$20$: $9$; $20$–$30$: $15$; $30$–$40$: $30$; $40$–$50$: $18$; $50$–$60$: $5$.
2. The highest frequency is $30$, so the modal class is $30 - 40$.

Answer: (C) 30 – 40

Common mistake: Reading the cumulative numbers as frequencies. The tallest class is found from the differences.

1. Slant height $l = \sqrt{24^2 + 7^2} = \sqrt{625} = 25$ cm.
2. CSA $= \pi r l = \dfrac{22}{7} \times 7 \times 25 = 550$ cm².

Answer: (C) $550$ cm²

1. Distance $= \sqrt{(2\sqrt{5})^2 + (2\sqrt{5})^2} = \sqrt{20 + 20} = \sqrt{40} = 2\sqrt{10}$ units.

Answer: (A) $2\sqrt{10}$ units

1. Sum of zeroes $= 0$ and product $= -\dfrac{4}{9}$, so the polynomial is $k\left(x^2 - \dfrac{4}{9}\right) = \dfrac{k}{9}(9x^2 - 4)$.
2. Option (D), $5(9x^2 - 4)$, has this form (check: $9x^2 = 4$ gives $x = \pm\dfrac{2}{3}$ ✓).

Answer: (D) $5(9x^2 - 4)$

1. If each of $n$ values grows by $3$, the sum grows by $3n$, and the mean $\dfrac{\text{sum}}{n}$ grows by $3$.

Answer: (B) increases by 3

1. An event either happens or does not happen, so the two probabilities add to $1$: $p + q = 1$.

Answer: (A) $p + q = 1$

1. Probability $= \dfrac{\text{her tickets}}{6000} = 0.08$, so her tickets $= 0.08 \times 6000 = 480$.

Answer: (C) 480

1. People who can swim: $20 - 5 = 15$.
2. $P = \dfrac{15}{20} = \dfrac{3}{4}$.

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

1. (A) On the $y$-axis $x = 0$: $2y = 4$ gives $y = 2$, so the point is $(0, 2)$. True.
2. (R) The distance of $(0, 2)$ from the $x$-axis is $|y| = 2$. True.
3. R does not explain why the line meets the $y$-axis at $(0, 2)$, so R is not the explanation of A.

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

1. (A) $AC = \sqrt{2^2 + 3^2} = \sqrt{13}$, which is irrational. The perimeter $2 + 3 + \sqrt{13}$ is irrational. So A is false.
2. (R) The sum of the squares of two rational numbers is rational. True.

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

1. (a) $x = 3$ is a vertical line through $(3, 0)$ and $y = -4$ is a horizontal line through $(0, -4)$. They meet at $(3, -4)$, so $x = 3$, $y = -4$.
2. (b) $x = 0$ is the $y$-axis and $y = -7$ is a horizontal line. They cross at $(0, -7)$, so the system has a solution: it is consistent.

Answer: (a) $x = 3,\ y = -4$   (b) consistent

1. $XZ \parallel BC$, so $\dfrac{AX}{XB} = \dfrac{AZ}{ZC} = \dfrac{3}{2}$, hence $\dfrac{AX}{AB} = \dfrac{3}{5}$.
2. $XY \parallel BM$, so $\triangle AXY \sim \triangle ABM$: $\dfrac{XY}{BM} = \dfrac{AX}{AB} = \dfrac{3}{5}$.
3. $XY = \dfrac{3}{5} \times 3 = \dfrac{9}{5} = 1.8$ cm.

Answer: $XY = 1.8$ cm

1. (a) Square both sides: $\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 3$, so $1 + 2\sin\theta\cos\theta = 3$.
2. $\sin\theta\cos\theta = 1$.
3. (b) $\mathrm{cosec}\,\alpha = \dfrac{1}{\sin\alpha} = 2$. $\mathrm{cosec}\,\beta = \sqrt{1 + \cot^2\beta} = \sqrt{1 + 3} = 2$.
4. $\mathrm{cosec}\,\alpha + \mathrm{cosec}\,\beta = 2 + 2 = 4$.

Answer: (a) $1$   (b) $4$

Common mistake: (For (a): $\sin\theta + \cos\theta$ is at most $\sqrt{2}$ for any real angle, so $\sqrt{3}$ cannot occur. CBSE's key still gives $1$ by algebra, which is what is shown here.)

1. The number must divide $85 - 1 = 84$ and $72 - 2 = 70$ exactly, and the greatest such number is their HCF.
2. $84 = 2^2 \times 3 \times 7$ and $70 = 2 \times 5 \times 7$, so $\text{HCF} = 2 \times 7 = 14$.

Answer: 14

1. Total balls $= 4 + 3 + 2 = 9$.
2. (i) $P(\text{red}) = \dfrac{4}{9}$. (ii) $P(\text{yellow}) = \dfrac{2}{9}$.

Answer: (i) $\dfrac{4}{9}$   (ii) $\dfrac{2}{9}$

1. Let the numbers be $x$ and $y$ with $x > y$. $\dfrac{1}{2}(x - y) = 2$, so $x - y = 4$. … (i)
2. $x + 2y = 13$. … (ii)
3. Subtract (i) from (ii): $3y = 9$, so $y = 3$ and $x = 7$. (Check: $7 + 6 = 13$ ✓.)

Answer: The numbers are 7 and 3

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

Answer: Proved by contradiction

1. Let $A(-5, 3)$, $B(5, 3)$ and the third vertex $C(x, y)$. $AB = 10$, so $AC = BC = 10$.
2. $AC = BC$: $(x + 5)^2 + (y - 3)^2 = (x - 5)^2 + (y - 3)^2$, giving $20x = 0$ and $x = 0$.
3. $AC^2 = 100$: $25 + (y - 3)^2 = 100$, so $(y - 3)^2 = 75$ and $y - 3 = \pm 5\sqrt{3}$.
4. The origin lies inside the triangle, and $AB$ is at height $3$ above it, so $C$ must be below: $y = 3 - 5\sqrt{3} = 3 - 8.5 = -5.5$.

Answer: $C = (0, -5.5)$

Common mistake: Taking the upper vertex (0, 11.5). The origin must be inside, so the third vertex is below AB.

1. (a) $TP = TQ$ (tangents from $T$), so $\angle TPQ = \angle TQP$. Let $\angle PTQ = \theta$. Then each base angle is $\dfrac{180^\circ - \theta}{2} = 90^\circ - \dfrac{\theta}{2}$.
2. $OP \perp TP$, so $\angle OPT = 90^\circ$ and $\angle OPQ = 90^\circ - \left(90^\circ - \dfrac{\theta}{2}\right) = \dfrac{\theta}{2}$.
3. So $\angle PTQ = \theta = 2\angle OPQ$.
4. (b) Tangents from $D$: $DR = DS = 3$ cm, so $AR = AD - DR = 17 - 3 = 14$ cm.
5. Tangents from $A$: $AQ = AR = 14$ cm, so $QB = AB - AQ = 20 - 14 = 6$ cm.
6. $OQ \perp AB$, $OP \perp BC$ and $\angle B = 90^\circ$, so $BQOP$ is a square: $r = QB = 6$ cm.

Answer: (a) proved   (b) radius $= 6$ cm

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

Answer: Proved

1. (a) Let the radius of the hemisphere be $r$. The height of the cylinder is $h = 2r$.
2. Volume of air $= \pi r^2 h + \dfrac{2}{3}\pi r^3 = 2\pi r^3 + \dfrac{2}{3}\pi r^3 = \dfrac{8}{3}\pi r^3 = \dfrac{1408}{21}$.
3. $\dfrac{8}{3} \times \dfrac{22}{7} \times r^3 = \dfrac{1408}{21}$, so $r^3 = \dfrac{1408}{21} \times \dfrac{21}{176} = 8$ and $r = 2$ m. Then $h = 4$ m.
4. Total height $= h + r = 4 + 2 = 6$ m.
5. (b) Volume of the cone $= \dfrac{1}{3}\pi \times 3^2 \times 12 = 36\pi$ cm³. The filled part is $\dfrac{5}{6}$ of it: $30\pi$ cm³.
6. The hemisphere on top has radius $3$ cm: $\dfrac{2}{3}\pi \times 27 = 18\pi$ cm³.
7. Ice-cream $= 30\pi + 18\pi = 48\pi = 48 \times 3.14 = 150.72$ cm³.

Answer: (a) $6$ m   (b) $150.72$ cm³

1. Given: $\triangle ABC$ with $DE \parallel BC$, $D$ on $AB$ and $E$ on $AC$. To prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
2. Construction: join $BE$ and $CD$; draw $DM \perp AC$ and $EN \perp AB$.
3. $\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)} = \dfrac{\frac{1}{2} \cdot AD \cdot EN}{\frac{1}{2} \cdot DB \cdot EN} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(ADE)}{\text{ar}(CDE)} = \dfrac{\frac{1}{2} \cdot AE \cdot DM}{\frac{1}{2} \cdot EC \cdot DM} = \dfrac{AE}{EC}$.
4. $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels $DE$ and $BC$, so $\text{ar}(BDE) = \text{ar}(CDE)$.
5. Therefore $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

Answer: Basic Proportionality Theorem proved

1. (a) Let $AB = 24$ m be the first tower and $CD = h$ the second, with $BD = x$ between their feet.
2. From the foot $D$ of the second tower, the top $A$ of the first is at $60^\circ$: $\tan 60^\circ = \dfrac{24}{x}$, so $x = \dfrac{24}{\sqrt{3}} = 8\sqrt{3}$ m.
3. From the foot $B$ of the first tower, the top $C$ of the second is at $30^\circ$: $\tan 30^\circ = \dfrac{h}{x}$, so $h = \dfrac{8\sqrt{3}}{\sqrt{3}} = 8$ m.
4. Wire $AC$: the vertical difference of the tops is $24 - 8 = 16$ m. $AC = \sqrt{16^2 + (8\sqrt{3})^2} = \sqrt{256 + 192} = \sqrt{448} = 8\sqrt{7}$ m.
5. (b) The observer is at $B$ and the balloon has centre $O$ and radius $r$. The two tangent lines from $B$ make $60^\circ$, and $BO$ bisects it: the angle between $BO$ and a tangent line is $30^\circ$.
6. The radius to the point of contact is perpendicular to the tangent line, so $\sin 30^\circ = \dfrac{r}{OB}$ and $OB = 2r$. … (i)
7. The centre is at an elevation of $45^\circ$: the height $OA$ of the centre above the ground level satisfies $\sin 45^\circ = \dfrac{OA}{OB}$, so $OA = \dfrac{OB}{\sqrt{2}}$. … (ii)
8. From (i) and (ii): $OA = \dfrac{2r}{\sqrt{2}} = \sqrt{2}\,r$.

Answer: (a) distance $8\sqrt{3}$ m, height $8$ m, wire $8\sqrt{7}$ m   (b) proved

1. Sector area $= \dfrac{60}{360} \times \dfrac{22}{7} \times 14 \times 14 = \dfrac{308}{3}$ cm².
2. The triangle $OAB$ has $OA = OB$ and an angle of $60^\circ$, so it is equilateral with side $14$: area $= \dfrac{\sqrt{3}}{4} \times 14^2 = 49\sqrt{3}$ cm².
3. Minor segment $= \dfrac{308}{3} - 49\sqrt{3}$ cm² $\approx 17.9$ cm².
4. Area of the circle $= \dfrac{22}{7} \times 14 \times 14 = 616$ cm². Major segment $= 616 - \left(\dfrac{308}{3} - 49\sqrt{3}\right) = \dfrac{1540}{3} + 49\sqrt{3}$ cm² $\approx 598.1$ cm².

Answer: Minor segment $\left(\dfrac{308}{3} - 49\sqrt{3}\right) \approx 17.9$ cm²; major segment $\left(\dfrac{1540}{3} + 49\sqrt{3}\right) \approx 598.1$ cm²

1. (a) $\dfrac{a + 10d}{a + 16d} = \dfrac{3}{4}$, so $4a + 40d = 3a + 48d$ and $a = 8d$.
2. $\dfrac{a_5}{a_{21}} = \dfrac{a + 4d}{a + 20d} = \dfrac{12d}{28d} = \dfrac{3}{7}$.
3. $\dfrac{S_5}{S_{21}} = \dfrac{\frac{5}{2}(2a + 4d)}{\frac{21}{2}(2a + 20d)} = \dfrac{5 \times 20d}{21 \times 36d} = \dfrac{25}{189}$.
4. (b) The rows form an A.P.: $22, 21, 20, \dots$ with $a = 22$, $d = -1$ and $S_n = 250$.
5. $\dfrac{n}{2}[44 - (n - 1)] = 250$, so $n(45 - n) = 500$ and $n^2 - 45n + 500 = 0$, giving $(n - 25)(n - 20) = 0$.
6. $n = 25$ would make the top row $22 - 24 = -2$ logs, which is not possible. So $n = 20$ rows and the top row has $a_{20} = 22 - 19 = 3$ logs.

Answer: (a) $3 : 7$ and $25 : 189$   (b) $20$ rows; $3$ logs in the top row

Common mistake: In (b), keeping n = 25. A row cannot have negative logs, so n = 20.

1. (I) New length $= 18 + x$ and new width $= 12 + x$. The area is doubled: $(18 + x)(12 + x) = 2 \times 18 \times 12 = 432$.
2. (II) $216 + 30x + x^2 = 432$, so $x^2 + 30x - 216 = 0$.
3. (III) $(x + 36)(x - 6) = 0$. A length is positive, so $x = 6$. The new dimensions are $24$ cm $\times$ $18$ cm.
4. (III) OR: if $(18 + x)(12 + x) = 220$, then $x^2 + 30x + 216 - 220 = 0$, that is $x^2 + 30x - 4 = 0$.
5. $D = 900 + 16 = 916$, which is not a perfect square, so $x$ cannot be rational. No rational $x$ gives the area $220$ cm².

Answer: (I) $(18 + x)(12 + x) = 432$   (II) $x^2 + 30x - 216 = 0$   (III) $24$ cm $\times$ $18$ cm; OR no rational $x$

1. (I) The highest frequency is $7$, so the modal class is $600 - 800$.
2. (II) Cumulative frequencies: $2, 6, 13, 17, 19, 22, 23, 24$. $N = 24$, $\dfrac{N}{2} = 12$, so the median class is $600 - 800$ ($l = 600$, $cf = 6$, $f = 7$, $h = 200$).
3. Median $= 600 + \dfrac{12 - 6}{7} \times 200 = 600 + \dfrac{1200}{7} = \dfrac{5400}{7} \approx 771.4$ mm.
4. (II) OR: Class marks $300, 500, \dots, 1700$ with frequencies $2, 4, 7, 4, 2, 3, 1, 1$: $\sum f_i x_i = 600 + 2000 + 4900 + 3600 + 2200 + 3900 + 1500 + 1700 = 20400$. Mean $= \dfrac{20400}{24} = 850$ mm.
5. (III) At least $1000$ mm: $2 + 3 + 1 + 1 = 7$ sub-divisions.