CBSE Class 10 Maths Standard 2023 Question Paper (Set 30/4/1) with Solutions

1. The least composite number is $4$ and the least prime is $2$.
2. $\text{HCF}(4, 2) = 2$ and $\text{LCM}(4, 2) = 4$. The ratio is $2 : 4 = 1 : 2$.

Answer: (A) 1 : 2

1. $x^2 + 5x - 2x - 10 = x(x + 5) - 2(x + 5) = (x - 2)(x + 5) = 0$, so the roots are $2$ and $-5$.

Answer: (A) 2, −5

1. Simplify: $\sqrt{6}$, $\sqrt{24} = 2\sqrt{6}$, $\sqrt{54} = 3\sqrt{6}$. The common difference is $\sqrt{6}$.
2. The next term is $4\sqrt{6} = \sqrt{96}$.

Answer: (B) $\sqrt{96}$

1. The distance from the $x$-axis is the absolute value of the $y$-coordinate: $|7| = 7$.

Answer: (B) 7

1. Radius $= \dfrac{d}{2}$. Area $= \dfrac{1}{2}\pi\left(\dfrac{d}{2}\right)^2 = \dfrac{1}{8}\pi d^2$.

Answer: (C) $\dfrac{1}{8}\pi d^2$

1. The empirical relation is $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$.

Answer: (A) Mode = 3 Median − 2 Mean

1. Write $2x - 5y - 6 = 0$ and $6x - 15y - 18 = 0$. $\dfrac{2}{6} = \dfrac{-5}{-15} = \dfrac{-6}{-18} = \dfrac{1}{3}$.
2. All three ratios are equal, so the lines are coincident.

Answer: (C) coincident

1. $x^2 - 1 = (x - 1)(x + 1)$ has zeroes $1$ and $-1$. Their sum is $0$ (also $-\dfrac{b}{a} = 0$).

Answer: (D) 0

1. $\tan\theta = \dfrac{\text{pole}}{\text{shadow}} = \dfrac{6}{2\sqrt{3}} = \sqrt{3}$.
2. So $\theta = 60^\circ$.

Answer: (A) $60^\circ$

1. $\sec\theta = \dfrac{1}{\cos\theta} = \dfrac{\sin\theta}{\cos\theta} \times \dfrac{1}{\sin\theta} = \dfrac{\mathrm{cosec}\,\theta}{\cot\theta}$.
2. $\mathrm{cosec}\,\theta = \sqrt{1 + \cot^2\theta}$, so $\sec\theta = \dfrac{\sqrt{1 + \cot^2\theta}}{\cot\theta}$.

Answer: (C) $\dfrac{\sqrt{1 + \cot^2\theta}}{\cot\theta}$

1. Total outcomes $= 36$. Difference $3$: $(1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3)$ — $6$ outcomes.
2. $P = \dfrac{6}{36} = \dfrac{1}{6}$.

Answer: (C) $\dfrac{1}{6}$

1. $\triangle ABC \sim \triangle QPR$ means $A \leftrightarrow Q$, $B \leftrightarrow P$, $C \leftrightarrow R$. So $AC \leftrightarrow QR$ and $BC \leftrightarrow PR$.
2. $\dfrac{AC}{QR} = \dfrac{BC}{PR}$: $\dfrac{6}{3} = \dfrac{5}{x}$, so $x = 2.5$ cm.

Answer: (B) 2.5 cm

1. Distance $= \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = 10$.

Answer: (D) 10

1. $PQ$ is a tangent at $Q$, so $OQ \perp PQ$ and $\angle OQP = 90^\circ$.
2. In $\triangle POQ$: $x + y + 90^\circ = 180^\circ$, so $x + y = 90^\circ$.

Answer: (B) $90^\circ$

1. $OA \perp TA$ (radius and tangent), so $\triangle OAT$ is right-angled at $A$.
2. $\cos 30^\circ = \dfrac{TA}{OT}$, so $TA = 4 \times \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$ cm. (Check: $OA = 4\sin 30^\circ = 2$ and $TA^2 = 16 - 4 = 12$.)

Answer: (A) $2\sqrt{3}$ cm

1. $PQ \parallel BC$, so $\dfrac{AP}{PB} = \dfrac{AQ}{QC}$: $\dfrac{4}{6} = \dfrac{8}{QC}$, so $QC = 12$ cm.
2. $AC = AQ + QC = 8 + 12 = 20$ cm.

Answer: (B) 20 cm

Common mistake: Stopping at QC = 12 cm. The question asks for AC.

1. $\alpha + \beta = \dfrac{3}{4}$ and $\alpha\beta = -\dfrac{7}{4}$.
2. $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{3/4}{-7/4} = -\dfrac{3}{7}$.

Answer: (D) $-\dfrac{3}{7}$

1. There are $4$ aces, so $52 - 4 = 48$ cards are not aces. $P = \dfrac{48}{52} = \dfrac{12}{13}$.

Answer: (D) $\dfrac{12}{13}$

1. (A) A leap year has $366 = 52$ weeks $+ 2$ days. The $2$ extra days include a Sunday in $2$ of the $7$ equally likely pairs, so $P = \dfrac{2}{7}$. True.
2. (R) A non-leap year has $52$ weeks $+ 1$ day, and that day is a Sunday with probability $\dfrac{1}{7}$, not $\dfrac{5}{7}$. False.

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

1. (A) $b - a = c - b$ is the same as $2b = a + c$, so this is the definition of three terms in A.P. True.
2. (R) $1 + 3 + 5 + \dots + (2n - 1) = n^2$. True, but it does not explain the condition in A.

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

1. Let the numbers be $2x$ and $3x$, where $x$ is their HCF. Their LCM is $6x$.
2. $6x = 180$, so $x = 30$. The numbers are $60$ and $90$, and their HCF is $30$.

Answer: 30

1. Let the zeroes be $\alpha$ and $\dfrac{1}{\alpha}$. Their product is $1$.
2. Product of zeroes $= \dfrac{-(k - 2)}{6} = 1$, so $-(k - 2) = 6$ and $k = -4$.

Answer: $k = -4$

1. (A) Sum $= -\dfrac{b}{a} = \dfrac{9}{2}$ and product $= \dfrac{c}{a} = \dfrac{4}{2} = 2$.
2. (B) $D = b^2 - 4ac = 0 - 4 \times 4 \times (-5) = 80 > 0$, so the roots are real and distinct.

Answer: (A) sum $\dfrac{9}{2}$, product $2$   (B) $D = 80$: real and distinct roots

1. Outcomes: $HH, HT, TH, TT$. At most one head: $HT, TH, TT$ — $3$ outcomes.
2. $P = \dfrac{3}{4}$.

Answer: $\dfrac{3}{4}$

1. (A) Numerator $= 5 \times \dfrac{1}{4} + 4 \times \dfrac{4}{3} - 1 = \dfrac{5}{4} + \dfrac{16}{3} - 1 = \dfrac{15 + 64 - 12}{12} = \dfrac{67}{12}$. Denominator $= \dfrac{1}{4} + \dfrac{3}{4} = 1$. Value $= \dfrac{67}{12}$.
2. (B) $\sin(A - B) = 0$ gives $A - B = 0^\circ$. $\cos(A + B) = \dfrac{1}{2}$ gives $A + B = 60^\circ$. So $A = B = 30^\circ$.

Answer: (A) $\dfrac{67}{12}$   (B) $A = B = 30^\circ$

1. (A) $a = -14$ and $a_5 = a + 4d = 2$, so $-14 + 4d = 2$ and $d = 4$.
2. $a_n = 62$: $-14 + (n - 1) \times 4 = 62$, so $(n - 1) \times 4 = 76$ and $n = 20$.
3. (B) $a = 65$ and $d = -4$. The $n$th term is negative when $65 - 4(n - 1) < 0$, that is $69 - 4n < 0$ and $n > \dfrac{69}{4} = 17.25$.
4. The least whole number is $n = 18$. So the first negative term is the $18$th term ($a_{18} = 65 - 68 = -3$).

Answer: (A) $20$ terms   (B) the $18$th term

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 $q^2 = 5a^2$ and $5$ divides $q$.
3. $5$ divides both $p$ and $q$, contradicting that they are co-prime. So $\sqrt{5}$ is irrational.

Answer: Proved by contradiction

1. Let $PA$ and $PB$ be tangents from $P$ to a circle with centre $O$. Join $OA$ and $OB$.
2. A radius is perpendicular to the tangent at the point of contact: $\angle OAP = \angle OBP = 90^\circ$.
3. In quadrilateral $OAPB$ the angles add up to $360^\circ$: $90^\circ + \angle APB + 90^\circ + \angle AOB = 360^\circ$.
4. So $\angle APB + \angle AOB = 180^\circ$: the angles are supplementary.

Answer: Proved

1. (A) LHS $= \dfrac{\sin A(1 - 2\sin^2 A)}{\cos A(2\cos^2 A - 1)}$. Using $\sin^2 A = 1 - \cos^2 A$: $1 - 2\sin^2 A = 2\cos^2 A - 1$.
2. So LHS $= \dfrac{\sin A}{\cos A} \times \dfrac{2\cos^2 A - 1}{2\cos^2 A - 1} = \tan A =$ RHS.
3. (B) $\sec A + \tan A = \dfrac{1 + \sin A}{\cos A}$. So LHS $= \dfrac{1}{\cos A}(1 - \sin A)\dfrac{1 + \sin A}{\cos A} = \dfrac{1 - \sin^2 A}{\cos^2 A} = \dfrac{\cos^2 A}{\cos^2 A} = 1$.

Answer: Both proved

1. Let the chord $AB$ of the larger circle touch the smaller circle at $P$. $OP \perp AB$ (radius and tangent), $OP = 3$ cm and $OA = 5$ cm.
2. The perpendicular from the centre bisects the chord: $AP = PB$. In right $\triangle OPA$: $AP^2 = 5^2 - 3^2 = 16$, so $AP = 4$ cm.
3. $AB = 2 \times 4 = 8$ cm.

Answer: $AB = 8$ cm

1. Write in standard form: $px^2 - 2px + 6 = 0$, with $a = p$, $b = -2p$, $c = 6$.
2. Equal roots need $D = b^2 - 4ac = 0$: $4p^2 - 24p = 0$, so $4p(p - 6) = 0$.
3. $p = 0$ would make the equation $6 = 0$ (not quadratic), so it is rejected. $p = 6$.

Answer: $p = 6$

Common mistake: Keeping $p = 0$. Then the $x^2$ term vanishes and it is not a quadratic equation.

1. (A) Let the tower be $AB = 75$ m with the cars at $P$ ($60^\circ$, nearer) and $Q$ ($30^\circ$, farther).
2. $\tan 60^\circ = \dfrac{75}{BP}$, so $BP = 25\sqrt{3}$ m. $\tan 30^\circ = \dfrac{75}{BQ}$, so $BQ = 75\sqrt{3}$ m.
3. $PQ = BQ - BP = 75\sqrt{3} - 25\sqrt{3} = 50\sqrt{3} = 50 \times 1.73 = 86.5$ m.
4. (B) Let the building be $BE = 7$ m with its top $B$, the cable tower $CD$ with foot $D$ and top $T$. The horizontal distance is $x$.
5. The foot $D$ is $7$ m below $B$ at $30^\circ$ depression: $\tan 30^\circ = \dfrac{7}{x}$, so $x = 7\sqrt{3}$ m.
6. The top $T$ is $(h - 7)$ m above $B$ at $60^\circ$ elevation: $\tan 60^\circ = \dfrac{h - 7}{7\sqrt{3}} = \sqrt{3}$, so $h - 7 = 21$ and $h = 28$ m.

Answer: (A) $86.5$ m   (B) $28$ m

1. (A) In $\triangle CBA$ and $\triangle CAD$: $\angle C = \angle C$ (common) and $\angle BAC = \angle ADC$ (given).
2. So $\triangle CBA \sim \triangle CAD$ by AA. Corresponding sides are proportional: $\dfrac{CB}{CA} = \dfrac{CA}{CD}$, so $CA^2 = CB \cdot CD$.
3. (B) $\triangle ABC \sim \triangle PQR$, so $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ and $\angle B = \angle Q$.
4. $D$ and $M$ are mid-points: $BC = 2BD$ and $QR = 2QM$. So $\dfrac{AB}{PQ} = \dfrac{2BD}{2QM} = \dfrac{BD}{QM}$.
5. In $\triangle ABD$ and $\triangle PQM$: $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$ and $\angle B = \angle Q$, so $\triangle ABD \sim \triangle PQM$ (SAS). Hence $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$.

Answer: (A) $CA^2 = CB \cdot CD$   (B) $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$

1. Radius $r = \dfrac{3}{2}$ cm. Each cone has height $H = 2$ cm. The cylinder has height $h = 12 - 2 - 2 = 8$ cm.
2. Volume of air $= \pi r^2 h + 2 \times \dfrac{1}{3}\pi r^2 H = \pi r^2\left(h + \dfrac{2H}{3}\right)$.
3. $= \dfrac{22}{7} \times \dfrac{9}{4} \times \left(8 + \dfrac{4}{3}\right) = \dfrac{22}{7} \times \dfrac{9}{4} \times \dfrac{28}{3} = 66$ cm³.

Answer: $66$ cm³

Common mistake: Taking the cylinder as 12 cm long. The two cones use 2 + 2 = 4 cm, leaving 8 cm for the cylinder.

1. The total is $200$: $24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 172 + x = 200$, so $x = 28$.
2. Cumulative frequencies: $24, 64, 97, 125, 155, 177, 193, 200$. $\dfrac{N}{2} = 100$, so the median class is $2500 - 3000$ ($l = 2500$, $cf = 97$, $f = 28$, $h = 500$).
3. Median $= 2500 + \dfrac{100 - 97}{28} \times 500 = 2500 + \dfrac{1500}{28} \approx 2553.6$.
4. Mean by the step-deviation method: take $a = 2750$, $h = 500$, $u_i = \dfrac{x_i - 2750}{500}$ for $-3, -2, -1, 0, 1, 2, 3, 4$. $\sum f_i u_i = -72 - 80 - 33 + 0 + 30 + 44 + 48 + 28 = -35$.
5. Mean $= 2750 + \dfrac{-35}{200} \times 500 = 2750 - 87.5 = 2662.5$.

Answer: $x = 28$; median $\approx ₹2553.6$; mean $= ₹2662.5$

1. (i) School $P$: $5x + 4y = 9500$. … (1) School $Q$: $4x + 3y = 7370$. … (2)
2. (ii)(a) Multiply (1) by $3$ and (2) by $4$: $15x + 12y = 28500$ and $16x + 12y = 29480$. Subtract: $x = 980$. Hockey prize $= ₹980$.
3. (ii)(b) From (1): $4y = 9500 - 4900 = 4600$, so $y = 1150$. Cricket $(₹1150)$ is more than hockey $(₹980)$ by $₹170$.
4. (iii) $2$ students each: $2(980 + 1150) = 2 \times 2130 = ₹4260$.

Answer: (i) $5x + 4y = 9500,\ 4x + 3y = 7370$   (ii)(a) ₹ $980$   (ii)(b) cricket, by ₹ $170$   (iii) ₹ $4260$

1. (i) $PQ = 400$ and $PQRS$ is a square, so $QR = SP = 400$. $R = (200, 400)$ and $S = (-200, 400)$.
2. (ii)(a) Area $= 400 \times 400 = 160000$ square units.
3. (ii)(b) Diagonal $PR = \sqrt{400^2 + 400^2} = 400\sqrt{2}$ units.
4. (iii) $C$ is on the $x$-axis and on the line through $A(200, 800)$ and $S(-200, 400)$, which has slope $1$: $C = (-600, 0)$. $S$ divides $CA$ in $K : 1$: $x_S = \dfrac{200K - 600}{K + 1} = -200$, so $200K - 600 = -200K - 200$ and $K = 1$.

Answer: (i) $R(200, 400),\ S(-200, 400)$   (ii)(a) $160000$   (ii)(b) $400\sqrt{2}$   (iii) $K = 1$

1. The parking is a semicircle on the $7$-unit side, so its radius is $r = 3.5$ units.
2. (i) Perimeter of the parking area $= \pi r + 2r = \dfrac{22}{7} \times \dfrac{7}{2} + 7 = 11 + 7 = 18$ units.
3. (ii)(a) Parking area $= \dfrac{1}{2}\pi r^2 = \dfrac{1}{2} \times \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} = \dfrac{77}{4}$. Two quadrants of radius $2$: $2 \times \dfrac{1}{4} \times \dfrac{22}{7} \times 4 = \dfrac{44}{7}$. Total $= \dfrac{77}{4} + \dfrac{44}{7} = \dfrac{715}{28} \approx 25.54$ square units.
4. (ii)(b) Playground area $= 14 \times 7 = 98$. Ratio $= 98 : \dfrac{77}{4} = \dfrac{98 \times 4}{77} = \dfrac{56}{11}$, that is $56 : 11$.
5. (iii) CBSE's key takes the length of fencing as $2(14 + 7) + \pi r = 42 + 11 = 53$ units. Cost $= 53 \times 2 = ₹106$.

Answer: (i) $18$ units   (ii)(a) $\dfrac{715}{28} \approx 25.54$ square units   (ii)(b) $56 : 11$   (iii) ₹ $106$

Common mistake: In (i), giving only the curved part (11 units). The perimeter of the parking area includes the diameter edge of 7 units.