CBSE Class 10 Maths Standard 2023 Question Paper (Set 30/2/1) with Solutions

1. Sum of roots $= -\dfrac{b}{a}$.
2. (A) $-\dfrac{-4}{2} = 2$. (B) $-\dfrac{4}{-1} = 4$ ✓. (C) $\dfrac{4/\sqrt{2}}{\sqrt{2}} = 2$. (D) $\dfrac{4}{4} = 1$.

Answer: (B) $-x^2 + 4x + 4 = 0$

1. Arc $= \dfrac{90}{360} \times 2 \times \dfrac{22}{7} \times 14 = \dfrac{1}{4} \times 88 = 22$ cm.

Answer: (A) 22 cm

1. Similar triangles have equal corresponding angles: $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$. So $\angle C = \angle R = 65^\circ$.
2. $\angle B = 180^\circ - 32^\circ - 65^\circ = 83^\circ$.

Answer: (C) $83^\circ$

1. $p = kq$ for some natural number $k$, so $q$ divides both $p$ and $q$. No larger number divides $q$.
2. So $\text{HCF}(p, q) = q$. (Example: $\text{HCF}(12, 4) = 4$.)

Answer: (C) q

1. Three corners are $B(0, 0)$, $C(3, 0)$ and $D(0, 4)$. The sides $BC$ (along the $x$-axis) and $BD$ (along the $y$-axis) meet at a right angle at $B$, so $B$ is a corner between $C$ and $D$.
2. The fourth corner is opposite $B$: it has the $x$-coordinate of $C$ and the $y$-coordinate of $D$, so it is $(3, 4)$. Check: the diagonal from $B$ to $(3, 4)$ has mid-point $(1.5, 2)$, which is also the mid-point of $CD$ ✓.

Answer: (C) $(3, 4)$

Common mistake: CBSE's key gives $(3, 4)$. (The letters $ABCD$ as printed do not go round the rectangle in order, but the fourth vertex is the same.)

1. Coincident lines need $\dfrac{3}{6} = \dfrac{-1}{-r} = \dfrac{8}{16}$.
2. $\dfrac{1}{2} = \dfrac{1}{r}$, so $r = 2$.

Answer: (D) 2

1. Perfect cubes from $1$ to $100$: $1^3 = 1$, $2^3 = 8$, $3^3 = 27$, $4^3 = 64$. That is $4$ cubes ($5^3 = 125$ is too big).
2. $P = \dfrac{4}{100} = \dfrac{1}{25}$.

Answer: (C) $\dfrac{1}{25}$

1. $x = a$ is a vertical line and $y = b$ is a horizontal line, so they are perpendicular and meet at one point.
2. The point has $x = a$ and $y = b$: $(a, b)$.

Answer: (D) intersecting at $(a, b)$

1. Zeroes that are reciprocals have product $1$. Product $= \dfrac{-(k - 2)}{6} = 1$.
2. $-(k - 2) = 6$, so $k - 2 = -6$ and $k = -4$.

Answer: (A) −4

1. Total surface area of a solid hemisphere $= 3\pi r^2$ with $r = \dfrac{d}{2}$.
2. $3\pi \times \dfrac{d^2}{4} = \dfrac{3}{4}\pi d^2$.

Answer: (D) $\dfrac{3}{4}\pi d^2$

Common mistake: Using the radius in the answer. The question asks in terms of the diameter d.

1. At most one tail means $0$ or $1$ tail: $HHH, HHT, HTH, THH$ — $4$ outcomes out of $8$.
2. $P = \dfrac{4}{8}$.

Answer: (B) $\dfrac{4}{8}$

1. $DE \parallel BC$, so $\dfrac{AD}{DB} = \dfrac{AE}{EC}$: $\dfrac{2}{3} = \dfrac{3}{x}$.
2. $x = \dfrac{9}{2}$.

Answer: (D) $\dfrac{9}{2}$

1. Between $7{:}20$ and $7{:}55$ is $35$ minutes. The hour hand moves $30^\circ$ in $60$ minutes, that is $0.5^\circ$ per minute.
2. Angle $= 35 \times \dfrac{1}{2} = \dfrac{35}{2}$ degrees.

Answer: (B) $\left(\dfrac{35}{2}\right)^\circ$

Common mistake: Using 6° per minute (the minute hand). The hour hand turns only half a degree per minute.

1. $x^2 + 4x + 3 = (x + 1)(x + 3)$, so the zeroes are $-1$ and $-3$.

Answer: (D) $-1, -3$

1. Tangents from a point are equal: $PA = PD$ (from $P$) and $CS = DS$ (from $S$).
2. $PA + CS = PD + DS = PS$.

Answer: (C) PS

1. $\alpha + \beta = a$ and $\alpha\beta = -b$.
2. $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = a^2 + 2b$.

Answer: (B) $a^2 + 2b$

1. The line meets the axes at $(a, 0)$ and $(0, b)$. With the origin it forms a right triangle with legs $|a|$ and $|b|$.
2. Area $= \dfrac{1}{2}|ab|$.

Answer: (B) $\dfrac{1}{2}ab$

1. $AB \parallel PQ$, so $\triangle OAB \sim \triangle OQP$ (alternate angles and vertically opposite angles at $O$).
2. $\dfrac{OP}{OB} = \dfrac{PQ}{AB} = \dfrac{2}{6} = \dfrac{1}{3}$, so $OP = \dfrac{3}{3} = 1$ cm.

Answer: (D) 1 cm

1. (A) is the standard tangent–radius theorem. True. (R) is also a standard theorem. True.
2. But the perpendicularity in A is not proved from the equal lengths in R, so R does not explain A.

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

1. (A) $D = 9 - 12 = -3 < 0$, so $p(x)$ has no real zeroes. A is false.
2. (R) A quadratic has at most two real zeroes. True.

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

1. Suppose $2 + \sqrt{3}$ is rational, equal to $\dfrac{p}{q}$ with integers $p, q$ and $q \neq 0$.
2. $\sqrt{3} = \dfrac{p}{q} - 2 = \dfrac{p - 2q}{q}$. As $p, q$ are integers, $\dfrac{p - 2q}{q}$ is rational, so $\sqrt{3}$ would be rational.
3. This contradicts the given fact that $\sqrt{3}$ is irrational. So $2 + \sqrt{3}$ is irrational.

Answer: Proved

1. (a) $4 \times 1 - 4 + \dfrac{3}{4} + p = \dfrac{3}{4}$, so $p = 0$.
2. (b) $\cos A = 1 - \cos^2 A = \sin^2 A$.
3. $\sin^2 A + \sin^4 A = \cos A + \cos^2 A = 1$.

Answer: (a) $p = 0$   (b) $1$

1. Let $A(-2, 3)$, $B(8, 3)$, $C(6, 7)$. $AB^2 = 10^2 = 100$.
2. $BC^2 = (8 - 6)^2 + (3 - 7)^2 = 4 + 16 = 20$ and $AC^2 = (6 + 2)^2 + (7 - 3)^2 = 64 + 16 = 80$.
3. $BC^2 + AC^2 = 100 = AB^2$, so by the converse of Pythagoras the triangle is right-angled at $C$.

Answer: Shown: right angle at C

1. (a) Let the height be $h$, so the shadow is $\sqrt{3}h$. $\tan\theta = \dfrac{h}{\sqrt{3}h} = \dfrac{1}{\sqrt{3}}$, so $\theta = 30^\circ$.
2. (b) $\tan 30^\circ = \dfrac{h}{30}$, so $h = \dfrac{30}{\sqrt{3}} = 10\sqrt{3}$ m.

Answer: (a) $30^\circ$   (b) $10\sqrt{3}$ m

1. $OB \perp AB$ and $OC \perp AC$ (radius and tangent), so $\angle OBA = \angle OCA = 90^\circ$.
2. In quadrilateral $ABOC$: $\angle BAC + \angle BOC = 180^\circ$, so $\angle BOC = 180^\circ - 65^\circ = 115^\circ$.

Answer: $\angle BOC = 115^\circ$

1. (a) $18180 = 2^2 \times 3^2 \times 5 \times 101$ and $7575 = 3 \times 5^2 \times 101$.
2. LCM (greatest powers) $= 2^2 \times 3^2 \times 5^2 \times 101 = 90900$. HCF (smallest powers) $= 3 \times 5 \times 101 = 1515$.
3. (b) They ring together again after the LCM of $6, 12, 18$, which is $36$ minutes.
4. So they ring together next at $6{:}36$ a.m.

Answer: (a) LCM $= 90900$, HCF $= 1515$   (b) $6{:}36$ a.m.

1. LHS $= \dfrac{1 - \cos^2\theta}{\cos\theta} \times \dfrac{1 - \sin^2\theta}{\sin\theta} = \dfrac{\sin^2\theta}{\cos\theta} \times \dfrac{\cos^2\theta}{\sin\theta} = \sin\theta\cos\theta$.
2. RHS $= \dfrac{1}{\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}} = \dfrac{\sin\theta\cos\theta}{\sin^2\theta + \cos^2\theta} = \sin\theta\cos\theta$.
3. LHS = RHS.

Answer: Proved

1. $PQ = QR$, so $PQ^2 = QR^2$: $(5 - 0)^2 + (-3 - 1)^2 = (x - 0)^2 + (6 - 1)^2$.
2. $25 + 16 = x^2 + 25$, so $x^2 = 16$ and $x = 4$ or $x = -4$.

Answer: $x = 4$ or $x = -4$

1. Each blade sweeps a sector of radius $21$ cm and angle $120^\circ$: area $= \dfrac{120}{360} \times \dfrac{22}{7} \times 21 \times 21 = \dfrac{1}{3} \times 1386 = 462$ cm².
2. Two blades (no overlap): $2 \times 462 = 924$ cm².

Answer: $924$ cm²

1. (a) Infinitely many solutions need $\dfrac{2}{2a} = \dfrac{3}{a + b} = \dfrac{7}{28}$.
2. $\dfrac{1}{a} = \dfrac{1}{4}$ gives $a = 4$. $\dfrac{3}{a + b} = \dfrac{1}{4}$ gives $a + b = 12$, so $b = 8$.
3. (b) Add: $348(x + y) = 1740$, so $x + y = 5$. Subtract: $86(x - y) = 86$, so $x - y = 1$.
4. Hence $x = 3$ and $y = 2$.

Answer: (a) $a = 4,\ b = 8$   (b) $x = 3,\ y = 2$

1. $OA = OP$ (radii), so in $\triangle OAP$: $\angle OPA = \angle OAP$. … (i)
2. $OP \perp QR$ (radius and tangent), so $\angle OPA + \angle APR = 90^\circ$. … (ii)
3. $\angle OAP$ is the same as $\angle QAP$ (as $Q$, $O$, $A$ are in a line). So from (i), (ii): $\angle QAP + \angle APR = 90^\circ$.

Answer: Proved

1. $a = 45$ and $d = -6$. $S_n = \dfrac{n}{2}[2 \times 45 + (n - 1)(-6)] = 180$.
2. $\dfrac{n}{2}[96 - 6n] = 180$, so $n(48 - 3n) = 180$, that is $3n^2 - 48n + 180 = 0$ and $n^2 - 16n + 60 = 0$.
3. $(n - 6)(n - 10) = 0$, so $n = 6$ or $n = 10$.
4. Why two answers: the terms are $45, 39, 33, 27, 21, 15, 9, 3, -3, -9, \dots$. The sum of the $7$th, $8$th, $9$th and $10$th terms is $9 + 3 - 3 - 9 = 0$, because some terms are positive and some are negative. So the sum is $180$ after $6$ terms and again after $10$ terms.

Answer: $n = 6$ or $n = 10$ (the terms $7$ to $10$ add up to $0$)

1. (a) Let $PQ = 75$ m be the lighthouse, with ships $R$ ($60^\circ$, nearer) and $S$ ($30^\circ$, farther) on the same side.
2. $\tan 60^\circ = \dfrac{75}{PR}$, so $PR = \dfrac{75}{\sqrt{3}} = 25\sqrt{3}$ m. $\tan 30^\circ = \dfrac{75}{PS}$, so $PS = 75\sqrt{3}$ m.
3. $RS = PS - PR = 75\sqrt{3} - 25\sqrt{3} = 50\sqrt{3} = 50 \times 1.73 = 86.5$ m.
4. (b) Let the building be $AB = 30$ m, the tower $BP = h$, and the point on the ground $C$. $\tan 30^\circ = \dfrac{30}{AC}$, so $AC = 30\sqrt{3}$ m.
5. $\tan 60^\circ = \dfrac{30 + h}{AC} = \sqrt{3}$, so $30 + h = 30\sqrt{3} \times \sqrt{3} = 90$ and $h = 60$ m.

Answer: (a) $86.5$ m   (b) $60$ m

Common mistake: In (a), adding the distances. The ships are on the same side, one behind the other, so subtract.

1. Class marks $x_i$: $5, 15, 25, 35, 45, 55, 65, 75$. $f_i x_i$: $35,\ 210,\ 325,\ 420,\ 900,\ 605,\ 975,\ 600$. $\sum f_i = 100$ and $\sum f_i x_i = 4070$.
2. Mean $= \dfrac{4070}{100} = 40.7$.
3. Cumulative frequencies: $7, 21, 34, 46, 66, 77, 92, 100$. $\dfrac{N}{2} = 50$, so the median class is $40 - 50$: $l = 40$, $cf = 46$, $f = 20$, $h = 10$.
4. Median $= 40 + \dfrac{50 - 46}{20} \times 10 = 40 + 2 = 42$.

Answer: Mean $= 40.7$ and Median $= 42$

1. (a) $D$ and $M$ are mid-points of $BC$ and $QR$, so $BC = 2BD$ and $QR = 2QM$. Given $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$, we get $\dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$.
2. So $\triangle ABD \sim \triangle PQM$ (SSS), giving $\angle B = \angle Q$.
3. In $\triangle ABC$ and $\triangle PQR$: $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ and $\angle B = \angle Q$. So $\triangle ABC \sim \triangle PQR$ (SAS).
4. (b) In $\triangle BMC$ and $\triangle EMD$: $MC = MD$ ($M$ is the mid-point), $\angle CMB = \angle DME$ (vertically opposite) and $\angle MBC = \angle MED$ (alternate angles, $BC \parallel AE$). So $\triangle BMC \cong \triangle EMD$ and $BC = DE$.
5. $AD = BC$ (parallelogram), so $AD = DE$ and $AE = 2BC$.
6. Now $AE \parallel BC$, so $\triangle AEL \sim \triangle CBL$: $\dfrac{EL}{BL} = \dfrac{AE}{BC} = \dfrac{2BC}{BC} = 2$, so $EL = 2BL$.

Answer: (a) $\triangle ABC \sim \triangle PQR$   (b) $EL = 2BL$

1. (i) Quadrant of radius $7$ cm: $\dfrac{1}{4} \times \dfrac{22}{7} \times 7 \times 7 = \dfrac{77}{2} = 38.5$ cm².
2. (ii) $OA = OB = 7 + 3 = 10$ cm with $\angle AOB = 90^\circ$: area $= \dfrac{1}{2} \times 10 \times 10 = 50$ cm².
3. (iii)(a) Shaded area $= 50 - 38.5 = 11.5$ cm². Cost $= 11.5 \times 20 = ₹230$.
4. (iii)(b) Arc $CD = \dfrac{90}{360} \times 2 \times \dfrac{22}{7} \times 7 = 11$ cm.

Answer: (i) $38.5$ cm²   (ii) $50$ cm²   (iii)(a) ₹ $230$   (iii)(b) $11$ cm

1. Cylindrical cup: $r = 3.5$ cm, $h = 14$ cm. Hemispherical cup: $r = 10.5$ cm.
2. (i) Base area $= \pi r^2 = \dfrac{22}{7} \times 3.5 \times 3.5 = 38.5$ cm².
3. (ii)(a) Capacity of the hemisphere $= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3} \times \dfrac{22}{7} \times 10.5^3 = 2425.5$ cm³.
4. (ii)(b) Capacity of the cylinder $= \pi r^2 h = 38.5 \times 14 = 539$ cm³.
5. (iii) Curved surface area $= 2\pi r h = 2 \times \dfrac{22}{7} \times 3.5 \times 14 = 308$ cm².

Answer: (i) $38.5$ cm²   (ii)(a) $2425.5$ cm³   (ii)(b) $539$ cm³   (iii) $308$ cm²

1. Total schools $= 250 + 200 + 290 + 180 + 80 = 1000$.
2. (i) More than $100$ computers: $80$ schools. $P = \dfrac{80}{1000} = 0.08$.
3. (ii)(a) $50$ or fewer: $250 + 200 + 290 = 740$. $P = \dfrac{740}{1000} = 0.74$.
4. (ii)(b) No more than $20$: $250 + 200 = 450$. $P = \dfrac{450}{1000} = 0.45$.
5. (iii) $10$ or less: $250$ schools. $P = \dfrac{250}{1000} = 0.25$.

Answer: (i) $0.08$   (ii)(a) $0.74$   (ii)(b) $0.45$   (iii) $0.25$