CBSE Class 10 Maths Standard 2024 Question Paper (Set 30/4/1) with Solutions

1. Add the two equations: $(a + b)x + (a + b)y = a^2 - b^2$, that is $(a + b)(x + y) = (a - b)(a + b)$.
2. So $x + y = a - b$.

Answer: (C) $a - b$

1. $\text{LCM} = \dfrac{65 \times 104}{\text{HCF}} = \dfrac{65 \times 104}{13} = 5 \times 104 = 520$.
2. $40x = 520$, so $x = 13$.

Answer: (B) 13

1. $p(1) = 1 - 5 + 6 = 2$ and $p(4) = 16 - 20 + 6 = 2$.
2. $p(1) + p(4) = 2 + 2 = 4$.

Answer: (B) 4

1. $D = b^2 - 4ac = (-2)^2 - 4 \times 3 \times c = 4 - 12c$.
2. $4 - 12c = 16$, so $-12c = 12$ and $c = -1$.

Answer: (C) −1

1. Arc length $= \dfrac{\theta}{360} \times$ circumference $= \dfrac{90}{360} \times$ circumference $= \dfrac{1}{4}$ of the circumference.
2. The ratio is $1 : 4$.

Answer: (B) 1 : 4

1. $\dfrac{\theta}{360} \times \pi \times 12^2 = 60\pi$, so $\dfrac{\theta}{360} \times 144 = 60$.
2. $\theta = \dfrac{60 \times 360}{144} = 150^\circ$.

Answer: (D) $150^\circ$

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$, so $\text{Mode} - \text{Median} = 2\,\text{Median} - 2\,\text{Mean} = 2(\text{Median} - \text{Mean})$.
2. $24 = 2(\text{Median} - \text{Mean})$, so $\text{Median} - \text{Mean} = 12$.

Answer: (A) 12

1. Odd numbers on one die: $1, 3, 5$ — $3$ choices. Both dice: $3 \times 3 = 9$ favourable outcomes out of $36$.
2. $P = \dfrac{9}{36} = \dfrac{1}{4}$.

Answer: (D) $\dfrac{9}{36}$

1. Total surface area of a solid hemisphere $= 2\pi r^2 + \pi r^2 = 3\pi r^2$.
2. The ratio to $r^2$ is $3\pi r^2 : r^2 = 3\pi : 1$.

Answer: (C) $3\pi : 1$

1. $\sin\theta = 1$ gives $\theta = 90^\circ$, so $\dfrac{\theta}{2} = 45^\circ$.
2. $\dfrac{1}{2}\sin 45^\circ = \dfrac{1}{2} \times \dfrac{1}{\sqrt{2}} = \dfrac{1}{2\sqrt{2}}$.

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

1. Parallel lines have equal ratios of the $x$ and $y$ coefficients: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$ with different constants.
2. Compare $5x - 3y - 2 = 0$ with $-15x + 9y - 5 = 0$: $\dfrac{a_1}{a_2} = \dfrac{5}{-15} = -\dfrac{1}{3}$, $\dfrac{b_1}{b_2} = \dfrac{-3}{9} = -\dfrac{1}{3}$ and $\dfrac{c_1}{c_2} = \dfrac{-2}{-5} = \dfrac{2}{5}$.
3. The first two ratios are equal and the third is different, so the lines are parallel. For the other options the $x$ and $y$ ratios differ (for example $\dfrac{5}{15} \neq \dfrac{-3}{9}$), so they are not parallel.

Answer: (D) $-15x + 9y = 5$

1. Let the three numbers be $a - d$, $a$, $a + d$. Their sum is $3a = 30$, so $a = 10$.
2. The middle term is $10$.

Answer: (B) 10

1. $DE \parallel BC$, so $\dfrac{AD}{AB} = \dfrac{AE}{AC}$: $\dfrac{4}{9} = \dfrac{AE}{13.5}$, so $AE = 6$ cm.
2. $EC = AC - AE = 13.5 - 6 = 7.5$ cm.

Answer: (B) 7.5 cm

Common mistake: Giving $AE = 6$ cm (option A). The question asks for $EC$.

1. $\tan\theta = \dfrac{\text{height}}{\text{shadow}} = 1$.
2. So the Sun’s altitude is $\theta = 45^\circ$.

Answer: (B) $45^\circ$

1. Tangents from an external point are equal: $AB = AC$.
2. In the isosceles $\triangle ABC$: $\angle ACB = \angle ABC = 42^\circ$.
3. $\angle BAC = 180^\circ - 42^\circ - 42^\circ = 96^\circ$.

Answer: (A) $96^\circ$

1. The diagonals $AC$ and $BD$ bisect each other, so $A + C = B + D$ coordinate by coordinate.
2. $D = A + C - B = (-2 + 8 - 6,\ 3 + 3 - 7) = (0, -1)$.

Answer: (B) $(0, -1)$

1. $P(E) + P(\overline{E}) = 1$ always, so $q = 1$.
2. $q^2 - 4 = 1 - 4 = -3$.

Answer: (A) −3

1. Tangents drawn from $P$ to the left circle: $PQ = PA$. Tangents from $P$ to the right circle: $PR = PA$.
2. So $PQ = PR = PA = 4.2$ cm and $QR = PQ + PR = 8.4$ cm.

Answer: (C) 8.4 cm

1. (A) The mid-point is equidistant from the ends, so it divides the segment $1 : 1$. True.
2. (R) Let the point divide the segment in the ratio $m : n$: $\dfrac{-2m - 5n}{m + n} = -3$, so $-2m - 5n = -3m - 3n$ and $m = 2n$. The ratio is $2 : 1$, not $1 : 2$. So R is false.

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

1. (R) The circumference of a circle is $2\pi r$. True.
2. (A) $2\pi r = 176$: $2 \times \dfrac{22}{7} \times r = 176$, so $r = \dfrac{176 \times 7}{44} = 28$ cm. True, and it uses the formula in R.

Answer: (A) Both true, and R explains A

1. They toll together again after the LCM of $9, 12, 15$.
2. $9 = 3^2$, $12 = 2^2 \times 3$, $15 = 3 \times 5$. LCM $= 2^2 \times 3^2 \times 5 = 180$.
3. They next toll together after $180$ minutes ($3$ hours).

Answer: 180 minutes (3 hours)

1. (a) The minute hand makes $360^\circ$ in $60$ minutes, so in $5$ minutes it moves $\dfrac{360}{60} \times 5 = 30^\circ$.
2. Area $= \dfrac{30}{360} \times \dfrac{22}{7} \times 14 \times 14 = \dfrac{1}{12} \times 616 = \dfrac{154}{3}$ cm² $\approx 51.33$ cm².
3. (b) Arc length $= \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 42 = \dfrac{1}{6} \times 264 = 44$ cm.

Answer: (a) $\dfrac{154}{3} \approx 51.33$ cm²   (b) $44$ cm

1. (a) Numerator: $5 \times \left(\dfrac{1}{2}\right)^2 + 4 \times \left(\dfrac{2}{\sqrt{3}}\right)^2 - 1^2 = \dfrac{5}{4} + \dfrac{16}{3} - 1 = \dfrac{15 + 64 - 12}{12} = \dfrac{67}{12}$.
2. Denominator: $\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{1}{4} + \dfrac{3}{4} = 1$. So the value is $\dfrac{67}{12}$.
3. (b) $\sin(A - B) = \dfrac{1}{2}$ gives $A - B = 30^\circ$. $\cos(A + B) = \dfrac{1}{2}$ gives $A + B = 60^\circ$.
4. Add: $2A = 90^\circ$, so $A = 45^\circ$ and $B = 15^\circ$.

Answer: (a) $\dfrac{67}{12}$   (b) $A = 45^\circ,\ B = 15^\circ$

1. Take a point $P$ on the major arc and join $AP$ and $BP$. The angle at the centre is double the angle at the circumference on the same arc: $\angle APB = \dfrac{1}{2} \times 145^\circ = 72.5^\circ$.
2. $APBC$ is a cyclic quadrilateral, so opposite angles add to $180^\circ$: $\angle APB + \angle ACB = 180^\circ$.
3. $x = \angle ACB = 180^\circ - 72.5^\circ = 107.5^\circ$.

Answer: $x = 107.5^\circ$

1. $\triangle AHK \sim \triangle ABC$, so corresponding sides are proportional: $\dfrac{HK}{BC} = \dfrac{AK}{AC}$.
2. $\dfrac{6.4}{3.2} = \dfrac{8}{AC}$, so $2 = \dfrac{8}{AC}$ and $AC = 4$ cm.

Answer: $AC = 4$ cm

1. Divide the numerator and the denominator of the LHS by $\cos\theta$: $\dfrac{\tan\theta - 1 + \sec\theta}{\tan\theta + 1 - \sec\theta}$.
2. In the denominator write $1 = \sec^2\theta - \tan^2\theta$: $\tan\theta - \sec\theta + (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = (\sec\theta - \tan\theta)(\sec\theta + \tan\theta - 1)$.
3. So LHS $= \dfrac{\tan\theta - 1 + \sec\theta}{(\sec\theta - \tan\theta)(\sec\theta + \tan\theta - 1)} = \dfrac{1}{\sec\theta - \tan\theta} =$ RHS.

Answer: Proved

1. (a) $8$ equally likely outcomes. (i) At least one head: all except $TTT$: $\dfrac{7}{8}$. (ii) Exactly two tails: $HTT, THT, TTH$: $\dfrac{3}{8}$. (iii) At most one tail: $HHH, HHT, HTH, THH$: $\dfrac{4}{8} = \dfrac{1}{2}$.
2. (b) Total $= 90$. (i) $2$-digit numbers less than $40$: $10$ to $39$, that is $30$ numbers: $\dfrac{30}{90} = \dfrac{1}{3}$.
3. (ii) Divisible by $5$ and greater than $50$: $55, 60, 65, 70, 75, 80, 85, 90$ — $8$ numbers: $\dfrac{8}{90} = \dfrac{4}{45}$.
4. (iii) Perfect squares from $1$ to $90$: $1, 4, 9, 16, 25, 36, 49, 64, 81$ — $9$ numbers: $\dfrac{9}{90} = \dfrac{1}{10}$.

Answer: (a) $\dfrac{7}{8},\ \dfrac{3}{8},\ \dfrac{1}{2}$   (b) $\dfrac{1}{3},\ \dfrac{4}{45},\ \dfrac{1}{10}$

1. Let the number of ₹ $50$ notes be $x$ and of ₹ $100$ notes be $y$. $x + y = 25$. … (i)
2. $50x + 100y = 2000$, that is $x + 2y = 40$. … (ii)
3. Subtract (i) from (ii): $y = 15$. Then $x = 10$.
4. Check: $10 \times 50 + 15 \times 100 = 500 + 1500 = 2000$ ✓.

Answer: $10$ notes of ₹ $50$ and $15$ notes of ₹ $100$

1. (a) $4x^2 + 4x - 3 = 4x^2 + 6x - 2x - 3 = 2x(2x + 3) - (2x + 3) = (2x + 3)(2x - 1)$. The zeroes are $-\dfrac{3}{2}$ and $\dfrac{1}{2}$.
2. Sum $= -\dfrac{3}{2} + \dfrac{1}{2} = -1 = -\dfrac{4}{4}$ ✓. Product $= -\dfrac{3}{2} \times \dfrac{1}{2} = -\dfrac{3}{4} = \dfrac{c}{a}$ ✓.
3. (b) $\alpha + \beta = -1$ and $\alpha\beta = -2$. $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = \dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$.
4. $= \dfrac{1 + 4}{-2} = -\dfrac{5}{2}$.

Answer: (a) zeroes $-\dfrac{3}{2}, \dfrac{1}{2}$, verified   (b) $-\dfrac{5}{2}$

1. Suppose $\dfrac{2 - \sqrt{3}}{5}$ is rational, equal to $\dfrac{p}{q}$ with integers $p, q$ and $q \neq 0$.
2. $2 - \sqrt{3} = \dfrac{5p}{q}$, so $\sqrt{3} = 2 - \dfrac{5p}{q} = \dfrac{2q - 5p}{q}$, which is rational.
3. But $\sqrt{3}$ is irrational (given), a contradiction. So $\dfrac{2 - \sqrt{3}}{5}$ is irrational.

Answer: Proved

1. Let parallelogram $ABCD$ circumscribe a circle touching $AB, BC, CD, DA$ at $P, Q, R, S$. Tangents from a point are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$.
2. Add the four: $AP + BP + CR + DR = AS + BQ + CQ + DS$, so $AB + CD = AD + BC$.
3. In a parallelogram $AB = CD$ and $AD = BC$. So $2AB = 2AD$ and $AB = AD$. All four sides are equal, so $ABCD$ is a rhombus.

Answer: Proved

1. Let the pillars be $AB$ and $CD$, each of height $h$, and let $P$ be on the road at distance $x$ from $CD$ (so $100 - x$ from $AB$).
2. In $\triangle CDP$ (angle $60^\circ$): $\tan 60^\circ = \sqrt{3} = \dfrac{h}{x}$, so $h = \sqrt{3}\,x$. … (i)
3. In $\triangle ABP$ (angle $30^\circ$): $\tan 30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{h}{100 - x}$, so $h = \dfrac{100 - x}{\sqrt{3}}$. … (ii)
4. Equate: $\sqrt{3}\,x = \dfrac{100 - x}{\sqrt{3}}$, so $3x = 100 - x$ and $x = 25$ m.
5. $h = 25\sqrt{3} = 25 \times 1.732 = 43.3$ m. The pillar is $43.3$ m high, and $P$ is $25$ m from one pillar and $75$ m from the other.

Answer: Each pillar $= 43.3$ m; the point is $25$ m and $75$ m from the two pillars

Common mistake: Saying the point is $25$ m from the pillar with $30^\circ$. The $60^\circ$ pillar is the nearer one (the steeper angle means closer).

1. (a) In $\triangle ABE$ and $\triangle CFB$: $\angle EAB = \angle BCF$ (opposite angles of a parallelogram).
2. $AE \parallel BC$ and $BE$ is a transversal: $\angle AEB = \angle CBF$ (alternate angles). So $\triangle ABE \sim \triangle CFB$ by AA.
3. (b) $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$. $AD$ and $PM$ are medians, so $BC = 2BD$ and $QR = 2QM$. Then $\dfrac{AB}{PQ} = \dfrac{2BD}{2QM} = \dfrac{AD}{PM}$, so $\dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$.
4. So $\triangle ABD \sim \triangle PQM$ (SSS), and $\angle B = \angle Q$.
5. In $\triangle ABC$ and $\triangle PQR$: $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ and $\angle B = \angle Q$. So $\triangle ABC \sim \triangle PQR$ (SAS).

Answer: (a) $\triangle ABE \sim \triangle CFB$   (b) $\triangle ABC \sim \triangle PQR$

1. (a) Let the original speed be $x$ km/h. New speed $= (x + 15)$ km/h. The new time is $30$ min $= \dfrac{1}{2}$ hour less.
2. $\dfrac{90}{x} - \dfrac{90}{x + 15} = \dfrac{1}{2}$, so $\dfrac{90 \times 15}{x(x + 15)} = \dfrac{1}{2}$ and $x^2 + 15x - 2700 = 0$.
3. $(x + 60)(x - 45) = 0$. Speed is positive, so $x = 45$ km/h.
4. (b) Real and equal roots need $D = 0$: $[-6(c + 1)]^2 - 4(c + 1) \times 3(c + 9) = 0$.
5. $12(c + 1)[3(c + 1) - (c + 9)] = 0$, that is $12(c + 1)(2c - 6) = 0$. Since $c \neq -1$, $2c - 6 = 0$ and $c = 3$.

Answer: (a) $45$ km/h   (b) $c = 3$

1. Class marks $x_i$: $10, 20, 30, 40, 50, 60$. Frequencies $f_i$: $6, 11, 21, 23, 14, 5$ with $\sum f_i = 80$.
2. $f_i x_i$: $60,\ 220,\ 630,\ 920,\ 700,\ 300$. $\sum f_i x_i = 2830$. Mean $= \dfrac{2830}{80} = 35.375$ years.
3. The highest frequency is $23$, so the modal class is $35 - 45$: $l = 35$, $f_1 = 23$, $f_0 = 21$, $f_2 = 14$, $h = 10$.
4. Mode $= 35 + \dfrac{23 - 21}{2 \times 23 - 21 - 14} \times 10 = 35 + \dfrac{20}{11} \approx 36.81$ years.

Answer: Mean $= 35.375$ years and Mode $\approx 36.81$ years

1. Read from the graph: $F = (-3, 0)$, $G = (1, 4)$, $A = (3, 4)$, $B = (3, 2)$, $C = (-1, -2)$, $D = (-2, -5)$.
2. (i) Mid-point of $FG = \left(\dfrac{-3 + 1}{2},\ \dfrac{0 + 4}{2}\right) = (-1, 2)$.
3. (ii)(a) $AC = \sqrt{(-1 - 3)^2 + (-2 - 4)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$ units.
4. (ii)(b) Section formula with $m : n = 1 : 3$ from $A(3, 4)$ to $B(3, 2)$: $x = \dfrac{1 \times 3 + 3 \times 3}{4} = 3$ and $y = \dfrac{1 \times 2 + 3 \times 4}{4} = \dfrac{14}{4} = \dfrac{7}{2}$. The point is $\left(3, \dfrac{7}{2}\right)$.
5. (iii) From the graph, $D = (-2, -5)$.

Answer: (i) $(-1, 2)$   (ii)(a) $2\sqrt{13}$   (ii)(b) $\left(3, \dfrac{7}{2}\right)$   (iii) $(-2, -5)$