CBSE Class 10 Mathematics (Basic) 2024 Solved Paper
Complete Official Board Exam Questions with Step-by-Step Solutions | MathJax & LaTeX Format
SECTION A (1 Mark Each)
Q1. For what value of $k$ the product of zeroes of the polynomial $kx^2 - 4x - 7$ is $2$?
Explanation:
For a quadratic polynomial $ax^2 + bx + c$, the product of zeroes is given by $\frac{c}{a}$.
Here, $a = k$, $b = -4$, and $c = -7$.
$$\text{Product of zeroes} = \frac{-7}{k} = 2 \implies 2k = -7 \implies k = -\frac{7}{2}$$
Q2. If first term $a = 8$ and $a_{10} = -19$, then the value of common difference $d$ is:
Explanation:
The $n^{\text{th}}$ term of an A.P. is given by $a_n = a + (n - 1)d$.
Given $a = 8$ and $a_{10} = 8 + 9d = -19$.
$$9d = -19 - 8 \implies 9d = -27 \implies d = -3$$
Q3. The mid-point of the line segment joining the points $(-1, 3)$ and $\left(8, \frac{3}{2}\right)$ is:
Explanation:
Using the midpoint formula $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$:
$$x = \frac{-1 + 8}{2} = \frac{7}{2}, \quad y = \frac{3 + \frac{3}{2}}{2} = \frac{\frac{9}{2}}{2} = \frac{9}{4}$$ Midpoint is $\left(\frac{7}{2}, \frac{9}{4}\right)$.
Q4. If $\sin \theta = \frac{1}{3}$, then $\sec \theta$ is equal to:
Explanation:
$$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$$ $$\sec \theta = \frac{1}{\cos \theta} = \frac{3}{2\sqrt{2}}$$
Q5. $\text{HCF}(132, 77)$ is:
Explanation:
Prime factorisation:
$132 = 2^2 \times 3 \times 11$
$77 = 7 \times 11$
Common prime factor is $11$, so $\text{HCF}(132, 77) = 11$.
Q6. If the roots of quadratic equation $4x^2 - 5x + k = 0$ are real and equal, then value of $k$ is:
Explanation:
For real and equal roots, discriminant $D = b^2 - 4ac = 0$.
$$(-5)^2 - 4(4)(k) = 0 \implies 25 - 16k = 0 \implies k = \frac{25}{16}$$
Q7. If probability of winning a game is $p$, then probability of losing the game is:
Explanation:
Since $P(E) + P(\overline{E}) = 1$, $P(\text{losing}) = 1 - P(\text{winning}) = 1 - p$.
Q8. The distance between the points $(2, -3)$ and $(-2, 3)$ is:
Explanation:
$$d = \sqrt{(-2 - 2)^2 + (3 - (-3))^2} = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \text{ units}$$
Q9. For what value of $\theta$, $\sin^2\theta + \sin\theta + \cos^2\theta$ is equal to $2$?
Explanation:
$$(\sin^2\theta + \cos^2\theta) + \sin\theta = 2 \implies 1 + \sin\theta = 2 \implies \sin\theta = 1 \implies \theta = 90^\circ$$
Q10. A card is drawn from a well shuffled deck of 52 playing cards. The probability that drawn card is a red queen, is:
Explanation:
Total cards = $52$. Number of red queens = $2$ (Queen of hearts and Queen of diamonds).
$$P(\text{Red Queen}) = \frac{2}{52} = \frac{1}{26}$$
Q11. If a certain variable $x$ divides a statistical data arranged in order into two equal parts, then the value of $x$ is called the:
Explanation: Median is defined as the middle value of an ordered dataset that divides it into two equal halves.
Q12. The radius of a sphere is $\frac{7}{2}\text{ cm}$. The volume of the sphere is:
Explanation:
$$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} = \frac{539}{3}\text{ cm}^3$$
Q13. The mean and median of a statistical data are $21$ and $23$ respectively. The mode of the data is:
Explanation:
Using the empirical relationship: $\text{Mode} = 3(\text{Median}) - 2(\text{Mean})$
$$\text{Mode} = 3(23) - 2(21) = 69 - 42 = 27$$
Q14. The height and radius of a right circular cone are $24\text{ cm}$ and $7\text{ cm}$ respectively. The slant height of the cone is:
Explanation:
$$l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$$
Q15. If one of the zeroes of the quadratic polynomial $(a-1)x^2 + ax + 1$ is $-3$, then the value of $a$ is:
Explanation:
Since $-3$ is a zero of $p(x) = (a-1)x^2 + ax + 1$, $p(-3) = 0$:
$$(a-1)(-3)^2 + a(-3) + 1 = 0 \implies 9(a-1) - 3a + 1 = 0$$ $$9a - 9 - 3a + 1 = 0 \implies 6a - 8 = 0 \implies a = \frac{8}{6} = \frac{2}{3}$$
Q16. The diameter of a circle is of length $6\text{ cm}$. If one end of the diameter is $(-4, 0)$, the other end on $x$-axis is at:
Explanation:
Let the other end on the $x$-axis be $(x, 0)$.
Distance between $(-4, 0)$ and $(x, 0)$ is $|x - (-4)| = 6$.
$$x + 4 = 6 \implies x = 2 \quad \text{or} \quad x + 4 = -6 \implies x = -10$$ Since $(2,0)$ is given in the options, the other end is $(2, 0)$.
Q17. The value of $k$ for which the pair of linear equations $5x + 2y - 7 = 0$ and $2x + ky + 1 = 0$ don't have a solution, is:
Explanation:
For no solution (parallel lines): $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
$$\frac{5}{2} = \frac{2}{k} \neq \frac{-7}{1} \implies 5k = 4 \implies k = \frac{4}{5}$$
Q18. Two dice are rolled together. The probability of getting a doublet is:
Explanation:
Total outcomes $= 36$. Doublet outcomes $= \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} = 6$.
$$P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}$$
Q19. Assertion (A): If $PA$ and $PB$ are tangents drawn to a circle with centre $O$ from an external point $P$, then the quadrilateral $OAPB$ is a cyclic quadrilateral.
Reason (R): In a cyclic quadrilateral, opposite angles are equal.
Explanation:
Since $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$, their sum is $180^\circ$. Opposite angles sum to $180^\circ$, making $OAPB$ cyclic. So Assertion is true.
However, in a cyclic quadrilateral, opposite angles are supplementary ($180^\circ$), not necessarily equal. Thus, Reason is false.
Q20. Assertion (A): Zeroes of a polynomial $p(x) = x^2 - 2x - 3$ are $-1$ and $3$.
Reason (R): The graph of polynomial $p(x) = x^2 - 2x - 3$ intersects $x$-axis at $(-1, 0)$ and $(3, 0)$.
Explanation:
$p(x) = (x-3)(x+1) = 0 \implies x = 3, -1$. The $x$-intercepts of a graph $y = p(x)$ correspond directly to the real zeroes of the polynomial. Both statements are true and R correctly explains A.
SECTION B (2 Marks Each)
Q21. $D$ is a point on the side $BC$ of $\Delta ABC$ such that $\angle ADC = \angle BAC$. Show that $AC^2 = BC \times DC$.
In $\Delta ADC$ and $\Delta BAC$:
$\angle ADC = \angle BAC$ (Given)
$\angle C = \angle C$ (Common angle)
By AA similarity criterion, $\Delta ADC \sim \Delta BAC$.
Therefore, corresponding sides are proportional: $$\frac{AC}{BC} = \frac{DC}{AC} \implies AC^2 = BC \times DC$$ Hence proved.
Q22. (A) Solve the pair of linear equations algebraically: $x + 2y = 9$ and $y - 2x = 2$.
Equation 1: $x + 2y = 9 \implies x = 9 - 2y$
Equation 2: $-2x + y = 2$
Substitute $x$ into Eq 2: $$-2(9 - 2y) + y = 2 \implies -18 + 4y + y = 2 \implies 5y = 20 \implies y = 4$$ Substitute $y = 4$ into Eq 1: $x = 9 - 2(4) = 1$.
Thus, $x = 1, y = 4$.
OR
(B) Check whether the point $(-4, 3)$ lies on both the lines represented by $x + y + 1 = 0$ and $x - y = 1$.
For $x + y + 1 = 0$: Substitute $(-4, 3) \implies -4 + 3 + 1 = 0$ (True).
For $x - y = 1$: Substitute $(-4, 3) \implies -4 - 3 = -7 \neq 1$ (False).
Since $(-4, 3)$ does not satisfy the second equation, it does not lie on both lines.
Q23. (A) Prove that $6 - 4\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is irrational.
Let $6 - 4\sqrt{5}$ be rational, so $6 - 4\sqrt{5} = \frac{a}{b}$ where $a, b \in \mathbb{Z}, b \neq 0$.
$$4\sqrt{5} = 6 - \frac{a}{b} \implies \sqrt{5} = \frac{6b - a}{4b}$$ Since $a$ and $b$ are integers, $\frac{6b - a}{4b}$ is rational. This implies $\sqrt{5}$ is rational, contradicting the given fact that $\sqrt{5}$ is irrational.
Hence, $6 - 4\sqrt{5}$ is irrational.
OR
(B) Show that $11 \times 19 \times 23 + 3 \times 11$ is not a prime number.
$$11 \times 19 \times 23 + 3 \times 11 = 11 \times (19 \times 23 + 3) = 11 \times (437 + 3) = 11 \times 440$$ Since the expression has factors other than $1$ and itself (e.g., $11$ and $440$), it is a composite number and not prime.
Q24. Evaluate: $\sin A \cos B + \cos A \sin B$, if $A = 30^\circ$ and $B = 45^\circ$.
$$\sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ = \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right)$$ $$= \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} = \frac{1 + \sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{2} + \sqrt{6}}{4}$$
Q25. A bag contains $4$ red, $5$ white and some yellow balls. If probability of drawing a red ball is $\frac{1}{5}$, find probability of drawing a yellow ball.
Let yellow balls $= y$. Total balls $= 4 + 5 + y = 9 + y$.
$$P(\text{Red}) = \frac{4}{9 + y} = \frac{1}{5} \implies 9 + y = 20 \implies y = 11$$ Total balls $= 20$. Probability of yellow ball $= \frac{11}{20}$.
SECTION C (3 Marks Each)
Q26. Two alarm clocks ring their alarms at regular intervals of $20$ minutes and $25$ minutes respectively. If they first beep together at $12$ noon, at what time will they beep again together next time?
To find the time when they beep together, calculate $\text{LCM}(20, 25)$:
$20 = 2^2 \times 5$, $25 = 5^2$
$$\text{LCM}(20, 25) = 2^2 \times 5^2 = 100 \text{ minutes} = 1 \text{ hour } 40 \text{ minutes}$$ Adding $1\text{ h } 40\text{ min}$ to $12\text{ noon}$, they will beep together next at 1:40 PM.
Q27. The greater of two supplementary angles exceeds the smaller by $18^\circ$. Find measures of these two angles.
Let smaller angle $= x^\circ$. Greater angle $= (x + 18)^\circ$.
Since angles are supplementary: $$x + (x + 18) = 180 \implies 2x + 18 = 180 \implies 2x = 162 \implies x = 81^\circ$$ Smaller angle $= 81^\circ$, Greater angle $= 81^\circ + 18^\circ = 99^\circ$.
Q28. Find the coordinates of the points of trisection of the line segment joining $(-2, 2)$ and $(7, -4)$.
Let points of trisection be $P$ (divides $1:2$) and $Q$ (divides $2:1$).
Using section formula $\left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}\right)$:
$$P = \left(\frac{1(7) + 2(-2)}{1+2}, \frac{1(-4) + 2(2)}{1+2}\right) = \left(\frac{3}{3}, \frac{0}{3}\right) = (1, 0)$$ $$Q = \left(\frac{2(7) + 1(-2)}{2+1}, \frac{2(-4) + 1(2)}{2+1}\right) = \left(\frac{12}{3}, \frac{-6}{3}\right) = (4, -2)$$ Trisection points are $(1, 0)$ and $(4, -2)$.
Q29. (A) Radii of concentric circles are $OA = r\text{ cm}$ and $OQ = 6\text{ cm}$. Chord $CD$ of larger circle is tangent to smaller circle at $Q$. $PA$ is tangent to larger circle with $PA = 16\text{ cm}, OP = 20\text{ cm}$. Find length $CD$.
In right $\Delta OAP$: $OP^2 = OA^2 + PA^2 \implies 20^2 = r^2 + 16^2 \implies r^2 = 400 - 256 = 144 \implies r = 12\text{ cm}$.
In right $\Delta OQD$: $OD^2 = OQ^2 + QD^2 \implies 12^2 = 6^2 + QD^2 \implies QD^2 = 144 - 36 = 108 \implies QD = 6\sqrt{3}\text{ cm}$.
Length $CD = 2 \times QD = 2 \times 6\sqrt{3} = 12\sqrt{3}\text{ cm}$.
OR
(B) In figure, two tangents $TP$ and $TQ$ are drawn to circle with centre $O$. Prove that $\angle PTQ = 2\angle OPQ$.
Let $\angle PTQ = \theta$. Since $TP = TQ$, $\Delta TPQ$ is isosceles, so $\angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$.
Since radius is perpendicular to tangent, $\angle OPT = 90^\circ$.
$$\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2}\angle PTQ$$ $$\implies \angle PTQ = 2\angle OPQ$$ Hence proved.
Q30. (A) A solid is a cylinder with hemispherical ends. Total height is $20\text{ cm}$ and diameter of cylinder is $14\text{ cm}$. Find total surface area of solid.
Radius $r = 7\text{ cm}$. Height of cylindrical portion $h = 20 - 7 - 7 = 6\text{ cm}$.
$$\text{TSA} = \text{CSA of cylinder} + 2 \times (\text{CSA of hemisphere}) = 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r)$$ $$\text{TSA} = 2 \times \frac{22}{7} \times 7 \times (6 + 14) = 44 \times 20 = 880\text{ cm}^2$$
OR
(B) A juice glass is cylindrical with a hemispherical raised portion at bottom. Inner diameter is $10\text{ cm}$ and height is $14\text{ cm}$. Find capacity of glass ($\pi = 3.14$).
$r = 5\text{ cm}, h = 14\text{ cm}$.
$$\text{Capacity} = \text{Volume of cylinder} - \text{Volume of hemisphere} = \pi r^2 h - \frac{2}{3}\pi r^3 = \pi r^2 \left(h - \frac{2r}{3}\right)$$ $$= 3.14 \times 25 \times \left(14 - \frac{10}{3}\right) = 78.5 \times \frac{32}{3} \approx 837.33\text{ cm}^3$$
Q31. Prove that: $(\cot \theta - \csc \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$.
$$\text{LHS} = \left(\frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)^2 = \frac{(\cos \theta - 1)^2}{\sin^2 \theta} = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$$ $$= \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta} = \text{RHS}$$ Hence proved.
SECTION D (5 Marks Each)
Q32. (A) State and prove Basic Proportionality Theorem (Thales Theorem).
Proof: Given $\Delta ABC$ with line $DE \parallel BC$.
$\text{Area}(\Delta ADE) = \frac{1}{2} \times AD \times EN$, $\text{Area}(\Delta BDE) = \frac{1}{2} \times DB \times EN$.
Ratio: $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{AD}{DB}$.
Similarly, $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta DEC)} = \frac{AE}{EC}$.
Since $\Delta BDE$ and $\Delta DEC$ lie on same base $DE$ between same parallels $DE$ and $BC$, $\text{Area}(\Delta BDE) = \text{Area}(\Delta DEC)$.
Hence, $\frac{AD}{DB} = \frac{AE}{EC}$. Proved.
Q33. How many terms of the A.P. $27, 24, 21, \dots$ must be taken so that their sum is $105$? Which term of the A.P. is zero?
Here $a = 27, d = -3$.
$$S_n = \frac{n}{2}[2(27) + (n-1)(-3)] = 105 \implies n(57 - 3n) = 210 \implies 3n^2 - 57n + 210 = 0$$ $$n^2 - 19n + 70 = 0 \implies (n - 14)(n - 5) = 0 \implies n = 5 \text{ or } 14$$ For zero term: $a_n = 27 + (n-1)(-3) = 0 \implies 3(n-1) = 27 \implies n = 10$.
Number of terms required $= 5$ or $14$. The $10^{\text{th}}$ term is zero.
Q34. (A) Shadow of a tower is $40\text{ m}$ longer when Sun's altitude is $30^\circ$ than when it was $60^\circ$. Find height of tower and original shadow length. ($\sqrt{3} = 1.73$)
Let height $AB = h$, original shadow $BC = x$.
$\tan 60^\circ = \frac{h}{x} \implies h = x\sqrt{3}$.
$\tan 30^\circ = \frac{h}{x + 40} \implies \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 40} \implies x + 40 = 3x \implies 2x = 40 \implies x = 20\text{ m}$.
Height $h = 20\sqrt{3} = 20 \times 1.73 = 34.6\text{ m}$. Original shadow $= 20\text{ m}$.
OR
(B) Angles of depression of top and bottom of an $8\text{ m}$ tall building from top of multi-storeyed building are $30^\circ$ and $45^\circ$. Find height of multi-storeyed building and distance between buildings. ($\sqrt{3} = 1.73$)
Let height of multi-storeyed building $= h$, distance $= x$.
For bottom ($45^\circ$): $\tan 45^\circ = \frac{h}{x} \implies h = x$.
For top ($30^\circ$): $\tan 30^\circ = \frac{h - 8}{x} \implies \frac{1}{\sqrt{3}} = \frac{h - 8}{h} \implies h = h\sqrt{3} - 8\sqrt{3}$.
$$h(\sqrt{3} - 1) = 8\sqrt{3} \implies h = \frac{8\sqrt{3}}{\sqrt{3} - 1} = 4(3 + \sqrt{3}) \approx 4(3 + 1.732) = 18.93\text{ m}$$ Height $= 18.93\text{ m}$, Distance between buildings $= 18.93\text{ m}$.
Q35. A chord of circle of radius $14\text{ cm}$ subtends an angle of $90^\circ$ at centre. Find area of minor and major segments of circle.
Area of minor sector $= \frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154\text{ cm}^2$.
Area of $\Delta AOB = \frac{1}{2} \times 14 \times 14 = 98\text{ cm}^2$.
$$\text{Area of minor segment} = 154 - 98 = 56\text{ cm}^2$$ $$\text{Area of major segment} = \text{Total area} - \text{Minor segment} = 616 - 56 = 560\text{ cm}^2$$
SECTION E (Case Study Based - 4 Marks Each)
Q36. Two circular water sprinklers cover areas with radii $R$ and $r$ ($R > r$). Circles touch externally. Sum of areas is $130\pi\text{ m}^2$ and distance between centres is $14\text{ m}$.
(i) Obtain quadratic equation involving $R$ and $r$.
(ii) Write quadratic equation in $r$ only.
(iii) Find radius $r$ and area irrigated by it. OR Find radius $R$ and area irrigated by it.
(i) $R + r = 14$ and $\pi R^2 + \pi r^2 = 130\pi \implies R^2 + r^2 = 130$.
(ii) Substitute $R = 14 - r$: $(14 - r)^2 + r^2 = 130 \implies 196 - 28r + 2r^2 = 130 \implies 2r^2 - 28r + 66 = 0 \implies r^2 - 14r + 33 = 0$.
(iii) $(r - 11)(r - 3) = 0 \implies r = 3\text{ m}$ (since $r < R$). Area $= \pi (3)^2 = 9\pi\text{ m}^2 \approx 28.28\text{ m}^2$.
OR $R = 14 - 3 = 11\text{ m}$. Area $= \pi (11)^2 = 121\pi\text{ m}^2 \approx 380.13\text{ m}^2$.
Q37. Data of leaf lengths (in mm): $70-80$ ($3$), $80-90$ ($5$), $90-100$ ($9$), $100-110$ ($12$), $110-120$ ($5$), $120-130$ ($4$), $130-140$ ($2$).
(i) Write median class.
(ii) How many leaves length $\ge 10\text{ cm}$ ($100\text{ mm}$)?
(iii) Find median of data. OR Find modal class and mode.
Cumulative frequencies: $3, 8, 17, 29, 34, 38, 40$. Total $N = 40, \frac{N}{2} = 20$.
(i) Median class is 100-110 mm.
(ii) Leaves $\ge 100\text{ mm} = 12 + 5 + 4 + 2 = 23$ leaves.
(iii) $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h = 100 + \left(\frac{20 - 17}{12}\right) \times 10 = 100 + 2.5 = 102.5\text{ mm}$.
OR Modal class is 100-110 ($f_1 = 12$). $\text{Mode} = 100 + \left(\frac{12 - 9}{2(12) - 9 - 5}\right) \times 10 = 100 + 3 = 103\text{ mm}$.
Q38. Circular mirror hanging with cord $AP, AQ$ as tangents. $AP = 30\text{ cm}$ and $\angle PAQ = 60^\circ$.
(i) Find length $PQ$.
(ii) Find $\angle POQ$.
(iii) Find length $OA$. OR Find radius of mirror.
(i) Since $AP = AQ = 30\text{ cm}$ and $\angle PAQ = 60^\circ$, $\Delta APQ$ is equilateral. Hence $PQ = 30\text{ cm}$.
(ii) $\angle POQ = 180^\circ - \angle PAQ = 180^\circ - 60^\circ = 120^\circ$.
(iii) In right $\Delta OAP$, $\angle PAO = 30^\circ$. $\cos 30^\circ = \frac{AP}{OA} \implies \frac{\sqrt{3}}{2} = \frac{30}{OA} \implies OA = \frac{60}{\sqrt{3}} = 20\sqrt{3}\text{ cm}$.
OR $\tan 30^\circ = \frac{OP}{AP} \implies \frac{1}{\sqrt{3}} = \frac{r}{30} \implies r = \frac{30}{\sqrt{3}} = 10\sqrt{3}\text{ cm}$.
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