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

1. Sum of zeroes $= -\dfrac{b}{a} = -\dfrac{-k\sqrt{2}}{2} = \dfrac{k\sqrt{2}}{2}$.
2. $\dfrac{k\sqrt{2}}{2} = \sqrt{2}$, so $k = 2$.

Answer: (B) 2

1. $P(\text{win}) + P(\text{lose}) = 1$, so $P(\text{lose}) = 1 - 0.79 = 0.21$.
2. (Option C is $0.21\%$, which is $0.0021$, a different number.)

Answer: (D) 0.21

1. Real and equal roots need $D = b^2 - 4ac = 0$.
2. So $b^2 = 4ac$, that is $ac = \dfrac{b^2}{4}$.

Answer: (C) $ac = \dfrac{b^2}{4}$

1. $S_n = \dfrac{n}{2}(a + a_n) = \dfrac{n}{2}(7 + 84) = \dfrac{91n}{2}$.
2. $\dfrac{91n}{2} = \dfrac{2093}{2}$, so $n = \dfrac{2093}{91} = 23$.

Answer: (C) 23

1. $p = 2 \times 3^2 \times a^2 \times b^4$ and $q = 2^2 \times 5 \times a^3 \times b^2$.
2. LCM takes the greatest power of each prime: $2^2 \times 3^2 \times 5 \times a^3 \times b^4 = 180a^3b^4$.

Answer: (D) $180a^3b^4$

1. $D$ is the mid-point of $BC$: $D = \left(\dfrac{6 + 0}{2},\ \dfrac{4 + 0}{2}\right) = (3, 2)$.
2. $AD = \sqrt{(5 - 3)^2 + (-6 - 2)^2} = \sqrt{4 + 64} = \sqrt{68}$ units.

Answer: (A) $\sqrt{68}$ units

1. $(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = \sec^2\theta - \tan^2\theta = 1$.
2. So $m(\sec\theta + \tan\theta) = 1$ and $\sec\theta + \tan\theta = \dfrac{1}{m}$.

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

1. Removing the even numbers $4$ and $16$ leaves $1, 7, 9, 21, 25$: $5$ numbers.
2. Of these only $7$ is prime ($1$ is not prime; $9, 21, 25$ are composite). So $P = \dfrac{1}{5}$.

Answer: (B) $\dfrac{1}{5}$

Common mistake: Counting 1 as a prime number. It is neither prime nor composite.

1. $\sum f_i(x_i - \bar{x}) = \sum f_i x_i - \bar{x}\sum f_i$.
2. Since $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$, we have $\bar{x}\sum f_i = \sum f_i x_i$, so the difference is $0$. The deviations from the mean always add to zero.

Answer: (D) 0

1. Sum of zeroes of $4x^2 - 5x - 6$ is $\dfrac{5}{4}$.
2. The new zeroes are twice these, so their sum is $2 \times \dfrac{5}{4} = \dfrac{5}{2}$.
3. For $x^2 + px + q$, the sum of zeroes is $-p$. So $-p = \dfrac{5}{2}$ and $p = -\dfrac{5}{2}$.

Answer: (A) $-\dfrac{5}{2}$

Common mistake: Forgetting the minus sign: the sum of zeroes is $-p$, not $p$.

1. The $y$-coordinates are equal, so the distance is $|x - 3| = 15$.
2. $x - 3 = 15$ gives $x = 18$ and $x - 3 = -15$ gives $x = -12$.

Answer: (B) −12, 18

1. $\cos(\alpha + \beta) = 0$ gives $\alpha + \beta = 90^\circ$.
2. $\cos\left(\dfrac{90^\circ}{2}\right) = \cos 45^\circ = \dfrac{1}{\sqrt{2}}$.

Answer: (A) $\dfrac{1}{\sqrt{2}}$

1. Surface area of the sphere $= 4\pi r^2$.
2. Each hemisphere has curved surface $2\pi r^2$ plus a flat circle $\pi r^2$, so total surface area $3\pi r^2$. Two together: $6\pi r^2$.
3. Ratio $= 4\pi r^2 : 6\pi r^2 = 2 : 3$.

Answer: (C) 2 : 3

Common mistake: Forgetting the two new flat circular faces created when the sphere is cut.

1. The median is the middle value of data arranged in order. The mode is the most frequent value and the mean is the average.

Answer: (B) median

1. The largest cone has its base inscribed in one face of the cube and its height equal to the edge. So base diameter $= 2$ cm, $r = 1$ cm and $h = 2$ cm.
2. $V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \pi \times 1^2 \times 2 = \dfrac{2\pi}{3}$ cm³.

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

Common mistake: Using $r = 2$. The base circle fits inside a face of side $2$ cm, so its diameter is $2$ and its radius is $1$.

1. Total outcomes $= 36$.
2. Sum $2$: $(1, 1)$ → $1$ way. Sum $3$: $(1, 2), (2, 1)$ → $2$ ways. Sum $5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ → $4$ ways.
3. Favourable $= 1 + 2 + 4 = 7$. $P = \dfrac{7}{36}$.

Answer: (A) $\dfrac{7}{36}$

1. The centre is the mid-point of the diameter. Let the other end be $(x, y)$: $\dfrac{x + 6}{2} = 2$ and $\dfrac{y + 0}{2} = 0$.
2. $x = -2$ and $y = 0$. The other end is $(-2, 0)$.

Answer: (C) $(-2, 0)$

1. The two lines in the figure have different slopes, so they are not parallel and not the same line.
2. Lines that are not parallel meet at exactly one point (when extended), so the pair is consistent with a unique solution.

Answer: (A) consistent with unique solution.

1. (A) Each tangent is perpendicular to the diameter at its end, so the two tangents are parallel. True.
2. (R) The diameter is the longest chord. True.
3. But the parallel tangents come from "tangent $\perp$ radius", not from the diameter being the longest chord. So R does not explain A.

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

1. (A) $x^2$ is a quadratic polynomial whose graph touches the $x$-axis at only one point, $(0, 0)$. So the assertion is false.
2. (R) A polynomial of degree $n$ has at most $n$ zeroes. True.

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

1. Multiply the first by $7$ and the second by $2$: $49x - 14y = 35$ and $16x + 14y = 30$.
2. Add: $65x = 65$, so $x = 1$. Then $7 - 2y = 5$ gives $y = 1$.
3. Verify: $7(1) - 2(1) = 5$ ✓ and $8(1) + 7(1) = 15$ ✓.

Answer: $x = 1,\ y = 1$

1. The lost card is black, so all red cards, including the queen of hearts, are still in the pack. The pack now has $52 - 1 = 51$ cards.
2. There is $1$ queen of hearts. $P = \dfrac{1}{51}$.

Answer: $\dfrac{1}{51}$

1. (A) $2\sqrt{2} \times \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2} + 2\sqrt{3} \times \dfrac{\sqrt{3}}{2} = 1 + 3 = 4$.
2. (B) LHS $= \sin(60^\circ + 30^\circ) = \sin 90^\circ = 1$.
3. RHS $= \sin 60^\circ\cos 30^\circ + \cos 60^\circ\sin 30^\circ = \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{3}{4} + \dfrac{1}{4} = 1$.
4. LHS $=$ RHS, so the identity is verified.

Answer: (A) $4$   (B) verified, both sides are $1$

1. (i) In $\triangle ABD$ and $\triangle CBD$: $\angle ABD = \angle CBD$ (BD bisects $\angle B$) and $\angle ADB = \angle CDB$ (BD bisects $\angle D$).
2. So $\triangle ABD \sim \triangle CBD$ by AA similarity.
3. (ii) $BD$ is common to both triangles, so the similar triangles have a pair of equal corresponding sides ($BD = BD$). They are therefore congruent (ASA), and $AB = BC$.

Answer: (i) $\triangle ABD \sim \triangle CBD$   (ii) $AB = BC$

1. (A) Suppose $5 - 2\sqrt{3}$ is rational, equal to $\dfrac{a}{b}$ with integers $a, b$ and $b \neq 0$.
2. Then $2\sqrt{3} = 5 - \dfrac{a}{b} = \dfrac{5b - a}{b}$, so $\sqrt{3} = \dfrac{5b - a}{2b}$, which is rational.
3. But $\sqrt{3}$ is irrational (given), a contradiction. So $5 - 2\sqrt{3}$ is irrational.
4. (B) $5 \times 11 \times 17 + 3 \times 11 = 11(5 \times 17 + 3) = 11 \times 88 = 968$.
5. The number is a product of two factors, $11$ and $88$, both greater than $1$, so it has more than two factors: it is composite.

Answer: (A) proved   (B) $968 = 11 \times 88$ is composite

1. (A) Let the point divide the segment in the ratio $k : 1$. $x = \dfrac{k \times 2 + 1 \times 1}{k + 1} = \dfrac{8}{5}$.
2. $10k + 5 = 8k + 8$, so $k = \dfrac{3}{2}$. The ratio is $3 : 2$.
3. $y = \dfrac{3 \times 3 + 2 \times 2}{3 + 2} = \dfrac{13}{5}$.
4. (B) $P = \left(-1, \dfrac{5}{2}\right)$, $Q = (1, 6)$, $R = \left(3, \dfrac{5}{2}\right)$, $S = (1, -1)$ (mid-points of $AB$, $BC$, $CD$, $DA$).
5. Mid-point of diagonal $PR = \left(\dfrac{-1 + 3}{2},\ \dfrac{5/2 + 5/2}{2}\right) = \left(1, \dfrac{5}{2}\right)$. Mid-point of diagonal $QS = \left(\dfrac{1 + 1}{2},\ \dfrac{6 - 1}{2}\right) = \left(1, \dfrac{5}{2}\right)$.
6. Both diagonals have the same mid-point, so $PR$ and $QS$ bisect each other.

Answer: (A) ratio $3 : 2$, $y = \dfrac{13}{5}$   (B) shown

1. The fewest rooms means the most teachers in each room. The number in each room must divide $48$, $80$ and $144$, so use the HCF.
2. $48 = 2^4 \times 3$, $80 = 2^4 \times 5$, $144 = 2^4 \times 3^2$. $\text{HCF} = 2^4 = 16$.
3. Rooms: $\dfrac{48}{16} + \dfrac{80}{16} + \dfrac{144}{16} = 3 + 5 + 9 = 17$.

Answer: 17 rooms

Common mistake: Using the LCM. The groups must divide each subject’s count exactly, so the HCF is needed.

1. Write $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$. First term: $\dfrac{\sin\theta/\cos\theta}{(\sin\theta - \cos\theta)/\sin\theta} = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)}$.
2. Second term: $\dfrac{\cos\theta/\sin\theta}{(\cos\theta - \sin\theta)/\cos\theta} = \dfrac{\cos^2\theta}{\sin\theta(\cos\theta - \sin\theta)} = -\dfrac{\cos^2\theta}{\sin\theta(\sin\theta - \cos\theta)}$.
3. Sum $= \dfrac{1}{\sin\theta - \cos\theta}\left[\dfrac{\sin^2\theta}{\cos\theta} - \dfrac{\cos^2\theta}{\sin\theta}\right] = \dfrac{1}{\sin\theta - \cos\theta} \cdot \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta}$.
4. Use $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $\sin^3\theta - \cos^3\theta = (\sin\theta - \cos\theta)(1 + \sin\theta\cos\theta)$.
5. Sum $= \dfrac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta} + 1 = 1 + \sec\theta\,\mathrm{cosec}\,\theta$ = RHS.

Answer: Proved

1. Let Rashmi's present age be $x$ years and Nazma's be $y$ years.
2. Three years ago: $x - 3 = 3(y - 3)$, so $x - 3y + 6 = 0$. … (i)
3. Ten years later: $x + 10 = 2(y + 10)$, so $x - 2y - 10 = 0$. … (ii)
4. Subtract (i) from (ii): $y - 16 = 0$, so $y = 16$. Then $x = 3 \times 16 - 6 = 42$.
5. Check: $42 - 3 = 39 = 3 \times 13$ ✓ and $42 + 10 = 52 = 2 \times 26$ ✓.

Answer: Rashmi is 42 years old and Nazma is 16 years old

1. (A) Let $PQ$ touch the circle at $R$. Join $OR$. Tangents from $Q$: $QA = QR$, and $OA = OR$, with $OQ$ common, so $\triangle AOQ \cong \triangle ROQ$ (SSS) and $\angle AOQ = \angle ROQ$. … (i)
2. Likewise $\triangle BOP \cong \triangle ROP$, so $\angle BOP = \angle ROP$. … (ii)
3. $AOB$ is a straight line: $\angle AOR + \angle ROB = 180^\circ$, that is $2\angle QOR + 2\angle ROP = 180^\circ$.
4. So $\angle QOR + \angle ROP = 90^\circ$, that is $\angle POQ = 90^\circ$.
5. (B) Tangents from $B$: $BR = BS = 24$ cm. $BC = 30$, so $CR = 30 - 24 = 6$ cm. Tangents from $C$: $CQ = CR = 6$ cm.
6. $OPDQ$ has right angles at $P$, $Q$ (radius $\perp$ tangent) and $D$ (given), so it is a square: $DQ = OP = 8$ cm.
7. $DC = DQ + QC = 8 + 6 = 14$ cm.

Answer: (A) $\angle POQ = 90^\circ$   (B) $DC = 14$ cm

1. Let the outer radius be $r_2$ and the inner radius $r_1$. $r_2 - r_1 = 1$. … (i)
2. Volume of metal $= \pi h(r_2^2 - r_1^2) = \dfrac{22}{7} \times 14 \times (r_2^2 - r_1^2) = 44(r_2^2 - r_1^2) = 176$, so $r_2^2 - r_1^2 = 4$.
3. $(r_2 - r_1)(r_2 + r_1) = 4$ and $r_2 - r_1 = 1$, so $r_2 + r_1 = 4$. … (ii)
4. Add (i) and (ii): $2r_2 = 5$, so $r_2 = 2.5$ cm. Then $r_1 = 4 - 2.5 = 1.5$ cm.

Answer: Outer radius 2.5 cm and inner radius 1.5 cm

1. (i) Length of the arc $= \dfrac{60}{360} \times 2\pi r = \dfrac{1}{6} \times 2 \times \dfrac{22}{7} \times 21 = 22$ cm.
2. (ii) Area of the sector $= \dfrac{60}{360} \times \pi r^2 = \dfrac{1}{6} \times \dfrac{22}{7} \times 21 \times 21 = 231$ cm².
3. The angle at the centre is $60^\circ$ and $OA = OB$, so $\triangle OAB$ is equilateral with side $21$ cm. Its area $= \dfrac{\sqrt{3}}{4} \times 21^2 = \dfrac{441\sqrt{3}}{4}$ cm² $\approx 190.95$ cm².
4. Area of the minor segment $=$ sector $-$ triangle $= 231 - \dfrac{441\sqrt{3}}{4} \approx 40.05$ cm².

Answer: (i) $22$ cm   (ii) $\left(231 - \dfrac{441\sqrt{3}}{4}\right)$ cm² $\approx 40.05$ cm²

Common mistake: Giving the sector area (231 cm²) as the segment. Subtract the triangle OAB.

1. (A) $a + a_8 = 32$: $2a + 7d = 32$. … (i) And $a \times a_8 = 60$: $a(a + 7d) = 60$. … (ii)
2. From (i), $a + 7d = 32 - a$. Put in (ii): $a(32 - a) = 60$, so $a^2 - 32a + 60 = 0$ and $(a - 2)(a - 30) = 0$. So $a = 2$ or $a = 30$.
3. If $a = 2$: $7d = 28$, $d = 4$. If $a = 30$: $7d = -28$, $d = -4$.
4. $S_{20} = 10[2a + 19d]$. For $a = 2$, $d = 4$: $10(4 + 76) = 800$. For $a = 30$, $d = -4$: $10(60 - 76) = -160$.
5. (B) $S_9 = \dfrac{9}{2}[2a + 8d] = 153$, so $a + 4d = 17$. … (i)
6. Sum of the last $6$ terms $= S_{40} - S_{34} = 687$: this is $a_{35} + \dots + a_{40} = 6a + 219d = 687$, that is $2a + 73d = 229$. … (ii)
7. (ii) $-$ $2 \times$ (i): $65d = 195$, so $d = 3$ and $a = 17 - 12 = 5$.
8. $S_{40} = \dfrac{40}{2}[2 \times 5 + 39 \times 3] = 20 \times 127 = 2540$.

Answer: (A) $a = 2, d = 4$ with $S_{20} = 800$; or $a = 30, d = -4$ with $S_{20} = -160$   (B) $a = 5, d = 3$, $S_{40} = 2540$

Common mistake: In (A), stopping after the first pair. The quadratic gives two values of a, so there are two answers.

1. (A) Given: $\triangle ABC$ with $DE \parallel BC$, $D$ on $AB$ and $E$ on $AC$. To prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. Construction: join $BE$ and $CD$; draw $DM \perp AC$ and $EN \perp AB$.
2. $\text{ar}(ADE) = \dfrac{1}{2} \times AD \times EN$ and $\text{ar}(BDE) = \dfrac{1}{2} \times DB \times EN$, so $\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)} = \dfrac{AD}{DB}$. Likewise $\dfrac{\text{ar}(ADE)}{\text{ar}(CDE)} = \dfrac{AE}{EC}$.
3. $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels, so $\text{ar}(BDE) = \text{ar}(CDE)$. Hence $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
4. (B) $PA \parallel QB \parallel RC$. In $\triangle CBQ$ and $\triangle CAP$: $\angle C$ is common and $\angle CBQ = \angle CAP = 90^\circ$, so $\triangle CBQ \sim \triangle CAP$ and $\dfrac{y}{x} = \dfrac{BC}{AC}$. … (i)
5. In $\triangle ABQ$ and $\triangle ACR$: $\triangle ABQ \sim \triangle ACR$, so $\dfrac{y}{z} = \dfrac{AB}{AC}$. … (ii)
6. Add (i) and (ii): $\dfrac{y}{x} + \dfrac{y}{z} = \dfrac{BC + AB}{AC} = 1$. Divide by $y$: $\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{y}$.

Answer: (A) Basic Proportionality Theorem proved   (B) $\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{y}$

1. Let the tower be $FT = h$ m, the pole on top $TU = 6$ m, and the distance of $P$ from the foot $PF = x$ m (see the diagram).
2. The angle of depression of $P$ from $T$ is $45^\circ$, so the angle of elevation of $T$ from $P$ is $45^\circ$: $\tan 45^\circ = \dfrac{h}{x} = 1$, so $h = x$. … (i)
3. The top of the pole $U$ is at height $h + 6$ and is seen at $60^\circ$: $\tan 60^\circ = \dfrac{h + 6}{x} = \sqrt{3}$, so $h + 6 = \sqrt{3}\,x$. … (ii)
4. Put $h = x$ in (ii): $x + 6 = \sqrt{3}\,x$, so $x(\sqrt{3} - 1) = 6$ and $x = \dfrac{6}{\sqrt{3} - 1} = \dfrac{6(\sqrt{3} + 1)}{2} = 3(\sqrt{3} + 1)$.
5. With $\sqrt{3} = 1.73$: $x = 3 \times 2.73 = 8.19$ m. So the height of the tower is $h = 8.19$ m and the distance of $P$ from the foot is $8.19$ m.