CBSE Class 10 Maths Standard 2023 Question Paper (Set 30/6/1) with Solutions

1. $p^2 = \dfrac{32}{50} = \dfrac{16}{25}$, so $p = \dfrac{4}{5}$ (or $-\dfrac{4}{5}$), which is a ratio of two integers: a rational number.

Answer: (C) rational number

1. The distance from the $x$-axis is the absolute value of the $y$-coordinate: $|8| = 8$ units. (The $6$ would be the distance from the $y$-axis.)

Answer: (C) 8 units

1. $k(x + 5)(x + 3)$ has the zeroes $-5$ and $-3$ for every non-zero constant $k$, so there are infinitely many such polynomials.

Answer: (D) more than 3

1. On the $y$-axis $x = 0$, so $-y = 3$ and $y = -3$. The point is $(0, -3)$.

Answer: (A) $(0, -3)$

1. Circumference $= 2\pi r$ is proportional to $r$, so the radii are in the same ratio, $4 : 5$. (The areas would be in the ratio $16 : 25$.)

Answer: (D) 4 : 5

1. $x^2 - 1 = (x - 1)(x + 1)$, so the zeroes are $1$ and $-1$ and $\alpha + \beta = 0$. (Also, the sum of zeroes is $-\dfrac{b}{a} = 0$.)

Answer: (D) 0

1. $\dfrac{\cos^2\theta - 1}{\sin^2\theta} = \dfrac{-\sin^2\theta}{\sin^2\theta} = -1$.

Answer: (D) −1

1. The perimeters of similar triangles are in the ratio of the corresponding sides: $\dfrac{\text{perimeter } PQR}{\text{perimeter } ABC} = \dfrac{PQ}{AB} = \dfrac{6}{8}$.
2. Perimeter of $\triangle PQR = 36 \times \dfrac{3}{4} = 27$ cm.

Answer: (B) 27 cm

1. Equal roots need the discriminant to be zero: $b^2 - 4ac = 0$, so $c = \dfrac{b^2}{4a}$.

Answer: (D) $\dfrac{b^2}{4a}$

1. $DB = AB - AD = 7 - 3 = 4$ cm. By the basic proportionality theorem ($DE \parallel BC$): $\dfrac{AE}{EC} = \dfrac{AD}{DB} = \dfrac{3}{4}$.
2. $AE = 3 \times \dfrac{3}{4} = 2.25$ cm.

Answer: (B) 2.25 cm

1. Total $= 5 + 8 + 7 = 20$ balls. "Neither blue nor pink" means yellow: $P = \dfrac{7}{20}$.

Answer: (C) $\dfrac{7}{20}$

1. $V = \dfrac{1}{3} \times (\text{base area}) \times h = \dfrac{1}{3} \times 156 \times 8 = 52 \times 8 = 416$ cm³.

Answer: (D) $416$ cm³

1. $3$ chairs cost $3x$ and $1$ table costs $y$: $3x + y = 900$.
2. $5$ chairs cost $5x$ and $3$ tables cost $3y$: $5x + 3y = 2100$.

Answer: (C) $3x + y = 900,\ 5x + 3y = 2100$

1. The radii $CA$ and $CB$ are perpendicular to the tangents, so in the quadrilateral $PACB$: $\angle ACB = 360^\circ - 90^\circ - 90^\circ - 55^\circ = 125^\circ$.
2. The angle at the centre is twice the angle at the circle on the same arc: $\angle AQB = \dfrac{125^\circ}{2} = 62\tfrac{1}{2}^\circ$.

Answer: (A) $62\tfrac{1}{2}^\circ$

Common mistake: Choosing 125° (the angle at the centre) or 55° (the angle at P) instead of half the central angle.

1. A deck has $3$ face cards (J, Q, K) in each of $4$ suits: $12$ cards. $P = \dfrac{12}{52} = \dfrac{3}{13}$.

Answer: (B) $\dfrac{3}{13}$

1. (A) $\sin\theta \cdot \dfrac{\cos\theta}{\sin\theta} = \cos\theta$ ✓. (B) $\cos\theta \cdot \dfrac{\sin\theta}{\cos\theta} = \sin\theta$ ✓. (C) is a standard identity ✓.
2. (D) The identity is $\sec^2\theta - \tan^2\theta = 1$, so $\tan^2\theta - \sec^2\theta = -1$, not $1$. This one is not true.

Answer: (D) $\tan^2\theta - \sec^2\theta = 1$

1. Sum of zeroes $= -(a + 1) = 2 + (-3) = -1$, so $a + 1 = 1$ and $a = 0$.
2. Product of zeroes $= b = 2 \times (-3) = -6$.

Answer: (D) $a = 0,\ b = -6$

1. The first term is the sum of the first one term: $a = S_1 = 3(1)^2 + 1 = 4$.
2. (Check: $S_2 = 14$, so $a_2 = 10$ and $d = 10 - 4 = 6$ ✓.)

Answer: (D) 4

1. (R) For a quadratic with rational coefficients, $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, so an irrational root $p + \sqrt{q}$ always comes with $p - \sqrt{q}$. True.
2. (A) is exactly that rule applied to $5 + \sqrt{7}$, so it is true and R explains it.

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

1. (R) is a standard identity, so it is true.
2. (A) $(\operatorname{cosec}\theta - \cot\theta)(\operatorname{cosec}\theta + \cot\theta) = \operatorname{cosec}^2\theta - \cot^2\theta = 1$, so the two expressions multiply to $1$: they are reciprocals. R is exactly why.

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

1. (a) A number ends in $0$ only if it is divisible by $5$. $6^n = (2 \times 3)^n = 2^n \times 3^n$, and by the fundamental theorem of arithmetic its only prime factors are $2$ and $3$. There is no $5$ in it, so $6^n$ never ends in $0$.
2. (b) $72 = 2^3 \times 3^2$ and $120 = 2^3 \times 3 \times 5$. HCF $= 2^3 \times 3 = 24$ and LCM $= 2^3 \times 3^2 \times 5 = 360$.
3. (Check: $24 \times 360 = 8640 = 72 \times 120$ ✓.)

Answer: (a) shown   (b) HCF $= 24$, LCM $= 360$

1. $P$ is on the $y$-axis, so $P = (0, y)$. $Q$ is on the $x$-axis, so $Q = (x, 0)$.
2. Mid-point formula: $\dfrac{0 + x}{2} = 2$ gives $x = 4$, and $\dfrac{y + 0}{2} = 5$ gives $y = 10$.

Answer: $P(0, 10)$ and $Q(4, 0)$

1. Let the pole be $AB = 18$ m and the shadow $AP$. In the right triangle $PAB$: $\tan\theta = \dfrac{AB}{AP} = \dfrac{18}{AP}$.
2. $\dfrac{6}{7} = \dfrac{18}{AP}$, so $AP = \dfrac{18 \times 7}{6} = 21$ m.

Answer: $21$ m

1. $B$, $O$, $C$ are on one straight line, so $\angle AOB = 180^\circ - 130^\circ = 50^\circ$.
2. A tangent is perpendicular to the radius at the point of contact, so $\angle OAP = 90^\circ$.
3. In $\triangle AOP$: $\angle APB = 180^\circ - (50^\circ + 90^\circ) = 40^\circ$.

Answer: $\angle APB = 40^\circ$

1. (a) $DE \parallel BC$, so $\dfrac{AD}{DB} = \dfrac{AE}{EC}$: $\dfrac{x}{x - 2} = \dfrac{x + 2}{x - 1}$.
2. $x(x - 1) = (x + 2)(x - 2)$, so $x^2 - x = x^2 - 4$ and $x = 4$.
3. (b) In $\triangle AOB$ and $\triangle COD$: $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ (alternate angles, $AB \parallel DC$).
4. So $\triangle AOB \sim \triangle COD$ by AA, and the corresponding sides are proportional: $\dfrac{OA}{OC} = \dfrac{OB}{OD}$.

Answer: (a) $x = 4$   (b) shown

1. A point on the $x$-axis has $y = 0$. Let $P$ divide $AB$ in the ratio $k : 1$.
2. $y = \dfrac{k(-5) + 1(3)}{k + 1} = 0$, so $-5k + 3 = 0$ and $k = \dfrac{3}{5}$.
3. The ratio is $\dfrac{3}{5} : 1 = 3 : 5$. (So $P = \left(\dfrac{3(-2) + 5(6)}{8}, 0\right) = (3, 0)$.)

Answer: $3 : 5$

1. (a) $26 = 2 \times 13$, $65 = 5 \times 13$, $117 = 3^2 \times 13$.
2. HCF $= 13$ (the only common prime). LCM $= 2 \times 3^2 \times 5 \times 13 = 1170$.
3. (b) Suppose $\sqrt{2}$ is rational: $\sqrt{2} = \dfrac{p}{q}$ with $p, q$ co-prime integers and $q \neq 0$.
4. Squaring, $2q^2 = p^2$, so $p^2$ is even and therefore $p$ is even: $p = 2a$.
5. Then $4a^2 = 2q^2$, so $q^2 = 2a^2$ is even and $q$ is even: $q = 2b$.
6. Now $2$ divides both $p$ and $q$, which contradicts that they are co-prime. So $\sqrt{2}$ is irrational.

Answer: (a) HCF $= 13$, LCM $= 1170$   (b) proved

1. $AB = AC$, so the angles opposite them are equal: $\angle ABC = \angle ACB$. Since $E$ is on $CB$ produced and $F$ is on $AC$, $\angle ECF$ is the same angle as $\angle ACB$. So $\angle ABD = \angle ECF$.
2. $\angle ADB = \angle EFC = 90^\circ$ (given: $AD \perp BC$, $EF \perp AC$).
3. In $\triangle ABD$ and $\triangle ECF$ two angles are equal, so $\triangle ABD \sim \triangle ECF$ by AA.

Answer: Proved

1. (a) Let the numbers be $x$ and $15 - x$. $\dfrac{1}{x} + \dfrac{1}{15 - x} = \dfrac{3}{10}$, so $\dfrac{15}{x(15 - x)} = \dfrac{3}{10}$.
2. $150 = 3x(15 - x)$, so $3x^2 - 45x + 150 = 0$, that is $x^2 - 15x + 50 = 0$ and $(x - 10)(x - 5) = 0$. The numbers are $10$ and $5$.
3. (b) $\alpha + \beta = 7$ and $\alpha\beta = 10$. $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 49 - 20 = 29$, and $\alpha^2\beta^2 = 100$.
4. The equation is $x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2 = 0$, that is $x^2 - 29x + 100 = 0$.

Answer: (a) $10$ and $5$   (b) $x^2 - 29x + 100 = 0$

1. LHS $= \dfrac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = \dfrac{\cos A + 1}{\cos A} \times \cos A = 1 + \cos A$.
2. Multiply and divide by $(1 - \cos A)$: $1 + \cos A = \dfrac{(1 + \cos A)(1 - \cos A)}{1 - \cos A} = \dfrac{1 - \cos^2 A}{1 - \cos A}$.
3. $1 - \cos^2 A = \sin^2 A$, so LHS $= \dfrac{\sin^2 A}{1 - \cos A} =$ RHS.

Answer: Proved

1. Area of the sector $= \dfrac{60}{360} \times \pi r^2 = \dfrac{1}{6} \times \dfrac{22}{7} \times 21 \times 21 = 231$ cm².
2. Length of the arc $= \dfrac{60}{360} \times 2\pi r = \dfrac{1}{6} \times 2 \times \dfrac{22}{7} \times 21 = 22$ cm.

Answer: Area $= 231$ cm²; arc length $= 22$ cm

1. (a) Tangents from an external point are equal, so $TP = TQ$ and $\angle TPQ = \angle TQP$. Let $\angle PTQ = \theta$. Then $\angle TPQ = \dfrac{180^\circ - \theta}{2} = 90^\circ - \dfrac{\theta}{2}$.
2. $OP \perp TP$, so $\angle OPT = 90^\circ$. Hence $\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \dfrac{\theta}{2}\right) = \dfrac{\theta}{2}$.
3. So $\angle PTQ = \theta = 2\angle OPQ$.
4. (b) Tangents from a point are equal: $AQ = AR$, $BQ = BP$ and $CR = CP$.
5. $2AQ = AQ + AR = (AB + BQ) + (AC + CR) = AB + AC + (BP + CP) = AB + AC + BC$.
6. So $AQ = \dfrac{1}{2}(AB + AC + BC) = \dfrac{1}{2}(\text{Perimeter of } \triangle ABC)$.

Answer: (a) proved   (b) proved

1. Radius of both $= 7$ cm. The height of the cone equals the diameter, so $h = 14$ cm.
2. Volume of the hemisphere $= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi \times 343 = \dfrac{686\pi}{3}$ cm³.
3. Volume of the cone $= \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \times 49 \times 14 = \dfrac{686\pi}{3}$.
4. Total volume $= \dfrac{1372\pi}{3} = \dfrac{1}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 28 = \dfrac{4312}{3} \approx 1437.33$ cm³.

Answer: $\dfrac{4312}{3} \approx 1437.33$ cm³

1. (a) $\dfrac{a + 10d}{a + 17d} = \dfrac{2}{3}$, so $3a + 30d = 2a + 34d$ and $a = 4d$.
2. $\dfrac{a_5}{a_{21}} = \dfrac{a + 4d}{a + 20d} = \dfrac{8d}{24d} = \dfrac{1}{3}$.
3. $\dfrac{S_5}{S_{21}} = \dfrac{\frac{5}{2}(2a + 4d)}{\frac{21}{2}(2a + 20d)} = \dfrac{5 \times 12d}{21 \times 28d} = \dfrac{5}{49}$.
4. (b) $S_6 = 36$: $3(2a + 5d) = 36$, so $2a + 5d = 12$. $S_{16} = 256$: $8(2a + 15d) = 256$, so $2a + 15d = 32$.
5. Subtracting, $10d = 20$, so $d = 2$ and $a = 1$.
6. $S_{10} = \dfrac{10}{2}[2(1) + 9(2)] = 5 \times 20 = 100$.

Answer: (a) $a_5 : a_{21} = 1 : 3$ and $S_5 : S_{21} = 5 : 49$   (b) $100$

1. (i) The total is $250$: $20 + 60 + 70 + x + 60 = 210 + x = 250$, so $x = 40$.
2. Class marks: $90, 110, 130, 150, 170$. $\sum f_ix_i = 20(90) + 60(110) + 70(130) + 40(150) + 60(170) = 1800 + 6600 + 9100 + 6000 + 10200 = 33700$.
3. Mean $= \dfrac{33700}{250} = 134.8$ g.
4. (ii) The highest frequency is $70$, so the modal class is $120 - 140$: $l = 120$, $f_1 = 70$, $f_0 = 60$, $f_2 = 40$, $h = 20$.
5. Mode $= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 120 + \dfrac{10}{140 - 60 - 40} \times 20 = 120 + \dfrac{10}{40} \times 20 = 125$ g.

Answer: $x = 40$; mean $= 134.8$ g; modal mass $= 125$ g

1. (i) Batch I: $20x + 5y = 9000$. Batch II: $5x + 25y = 26000$.
2. (ii) Divide the first by $5$: $4x + y = 1800$, so $y = 1800 - 4x$. Divide the second by $5$: $x + 5y = 5200$.
3. Substitute: $x + 5(1800 - 4x) = 5200$, so $x + 9000 - 20x = 5200$ and $19x = 3800$. $x = 200$ and $y = 1800 - 800 = 1000$.
4. (a) A poor child pays ₹ $200$ a month. (b) The difference is $1000 - 200 = ₹800$.
5. (iii) $10x + 20y = 10(200) + 20(1000) = 2000 + 20000 = ₹22000$.

Answer: (i) $20x + 5y = 9000,\ 5x + 25y = 26000$   (ii)(a) ₹ $200$   (ii)(b) ₹ $800$   (iii) ₹ $22000$

1. (i) In the right triangle $OPB$: $\cos 30^\circ = \dfrac{OP}{OB}$, so $\dfrac{\sqrt{3}}{2} = \dfrac{36}{OB}$ and $OB = \dfrac{72}{\sqrt{3}} = 24\sqrt{3}$ cm.
2. (ii)(a) $\tan 30^\circ = \dfrac{PB}{36}$, so $PB = \dfrac{36}{\sqrt{3}} = 12\sqrt{3}$ cm. $\tan 45^\circ = \dfrac{AP}{36}$, so $AP = 36$ cm.
3. $AB = AP - PB = 36 - 12\sqrt{3} = 12(3 - \sqrt{3})$ cm.
4. (ii)(b) Area of $\triangle OPB = \dfrac{1}{2} \times OP \times PB = \dfrac{1}{2} \times 36 \times 12\sqrt{3} = 216\sqrt{3}$ cm².
5. (iii) The height of Section A from the base is $AP = 36$ cm.