CBSE Class 10 Mathematics (Basic) 2025 Solved Paper

CBSE Class 10 Mathematics (Basic) 2025 Solved Paper

Complete Official Board Exam Questions with Step-by-Step Solutions | MathJax & LaTeX Format

SECTION A (1 Mark Each)

Q1. If $\text{HCF}(x,20)=2$ and $\text{LCM}(x,20)=60$, then value of $x$ is:

(A) 3     (B) 6     (C) 20     (D) 10
Answer: (B) 6
Explanation: Product of two numbers $= \text{HCF} \times \text{LCM}$: $$x \times 20 = 2 \times 60 \implies x = \frac{120}{20} = 6$$

Q2. The distance between the points $(-6, 9)$ and $(2, 7)$ is:

(A) $2\sqrt{17}$     (B) $4\sqrt{17}$     (C) $2\sqrt{5}$     (D) $2\sqrt{15}$
Answer: (A) $2\sqrt{17}$
Explanation: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(2-(-6))^2 + (7-9)^2} = \sqrt{8^2 + (-2)^2} = \sqrt{68} = 2\sqrt{17}$ units.

Q3. If $n^{\text{th}}$ term of an A.P. is $5n-6$ then its common difference is:

(A) $-6$     (B) 51     (C) 5     (D) 6
Answer: (C) 5
Explanation: $T_1 = 5(1)-6 = -1$, $T_2 = 5(2)-6 = 4$. Common difference $d = T_2 - T_1 = 4 - (-1) = 5$.

Q4. One of the zeroes of the polynomial $p(x)=kx^2-9x+3$ is $-\frac{3}{2}$. The value of $k$ is:

(A) $-\frac{3}{2}$     (B) $\frac{22}{3}$     (C) $\frac{14}{3}$     (D) $-\frac{22}{3}$
Answer: (D) $-\frac{22}{3}$
Explanation: $p\left(-\frac{3}{2}\right) = k\left(-\frac{3}{2}\right)^2 - 9\left(-\frac{3}{2}\right) + 3 = 0 \implies \frac{9}{4}k + \frac{27}{2} + 3 = 0 \implies 9k + 66 = 0 \implies k = -\frac{22}{3}$.

Q5. Three coins are tossed together. The probability that only one coin shows tail, is:

(A) $\frac{1}{2}$     (B) $\frac{3}{8}$     (C) $\frac{7}{8}$     (D) 1
Answer: (B) $\frac{3}{8}$
Explanation: Total outcomes $= 8$. Favourable outcomes $= \{HHT, HTH, THH\} = 3$. $P(E) = \frac{3}{8}$.

Q6. Two right circular cylinders of equal volumes have their heights in the ratio 1: 2. The ratio of their radii is:

(A) $\sqrt{2}:1$     (B) 1:2     (C) $1:4$     (D) $1:\sqrt{2}$
Answer: (A) $\sqrt{2}:1$
Explanation: $\pi r_1^2 h_1 = \pi r_2^2 h_2 \implies \frac{r_1^2}{r_2^2} \times \frac{1}{2} = 1 \implies \frac{r_1}{r_2} = \frac{\sqrt{2}}{1}$.

Q7. If $\sqrt{2}\sin\theta=1$, then $\cot\theta \times \csc\theta$ is equal to:

(A) $\frac{1}{\sqrt{2}}$     (B) $\frac{1}{2\sqrt{2}}$     (C) $\sqrt{2}$     (D) $\frac{1}{2}$
Answer: (C) $\sqrt{2}$
Explanation: $\sin\theta = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ$. $\cot 45^\circ \times \csc 45^\circ = 1 \times \sqrt{2} = \sqrt{2}$.

Q8. $PQ$ and $PR$ are tangents to the circle of radius $3\text{ cm}$ and centre $O$. If length of each tangent is $4\text{ cm}$, then perimeter of $\Delta OQP$ is:

(A) $5\text{ cm}$     (B) $12\text{ cm}$     (C) $9\text{ cm}$     (D) $8\text{ cm}$
Answer: (B) $12\text{ cm}$
Explanation: $OP = \sqrt{3^2 + 4^2} = 5\text{ cm}$. Perimeter of $\Delta OQP = 5 + 4 + 3 = 12\text{ cm}$.

Q9. $\alpha, \beta$ are zeroes of the polynomial $2x^2+5x+1$. The value of $\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$ is:

(A) $-\frac{5}{4}$     (B) 5     (C) $\frac{5}{4}$     (D) $-5$
Answer: (D) $-5$
Explanation: $\alpha+\beta = -\frac{5}{2}$, $\alpha\beta = \frac{1}{2}$. $\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-5/2}{1/2} = -5$.

Q10. The $20^{\text{th}}$ term of the A.P.: $10\sqrt{2}, 6\sqrt{2}, 2\sqrt{2}, \dots$ is:

(A) $-76+10\sqrt{2}$     (B) $-62\sqrt{2}$     (C) $-66\sqrt{2}$     (D) $86\sqrt{2}$
Answer: (C) $-66\sqrt{2}$
Explanation: $a=10\sqrt{2}, d=-4\sqrt{2}$. $T_{20} = 10\sqrt{2} + 19(-4\sqrt{2}) = -66\sqrt{2}$.

Q11. If $\sec\theta-\tan\theta=2$ then $\sec\theta+\tan\theta$ is equal to:

(A) $\frac{1}{2}$     (B) $\sqrt{2}$     (C) $\frac{1}{\sqrt{2}}$     (D) 2
Answer: (A) $\frac{1}{2}$
Explanation: $(\sec\theta-\tan\theta)(\sec\theta+\tan\theta) = \sec^2\theta-\tan^2\theta = 1 \implies \sec\theta+\tan\theta = \frac{1}{2}$.

Q12. A card is drawn from a deck of 52 playing cards. The probability that drawn card shows number '9' is:

(A) $\frac{1}{26}$     (B) $\frac{4}{13}$     (C) $\frac{1}{52}$     (D) $\frac{1}{13}$
Answer: (D) $\frac{1}{13}$
Explanation: Favourable cards with number '9' $= 4$. $P(E) = \frac{4}{52} = \frac{1}{13}$.

Q13. Length of arc subtending an angle of $210^\circ$ at centre is $\frac{44}{3}\text{ cm}$. The radius of circle is:

(A) $2\sqrt{2}\text{ cm}$     (B) $4\text{ cm}$     (C) $8\text{ cm}$     (D) $\frac{1}{4}\text{ cm}$
Answer: (B) $4\text{ cm}$
Explanation: $\text{Arc length} = \frac{\theta}{360^\circ} \times 2\pi r \implies \frac{44}{3} = \frac{210}{360} \times 2 \times \frac{22}{7} \times r \implies r = 4\text{ cm}$.

Q14. The value of $m$ for which lines $14x+my=20$ and $-3x+2y=16$ are parallel, is:

(A) $-\frac{3}{14}$     (B) $-\frac{7}{3}$     (C) $-\frac{28}{3}$     (D) $-\frac{3}{28}$
Answer: (C) $-\frac{28}{3}$
Explanation: For parallel lines, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \implies \frac{14}{-3} = \frac{m}{2} \implies m = -\frac{28}{3}$.

Q15. Curved surface area of a cone with base radius $7\text{ cm}$ is $550\text{ cm}^2$. Its slant height is:

(A) $25\text{ cm}$     (B) $14\text{ cm}$     (C) $20\text{ cm}$     (D) $24\text{ cm}$
Answer: (A) $25\text{ cm}$
Explanation: $\text{CSA} = \pi r l \implies 550 = \frac{22}{7} \times 7 \times l \implies l = 25\text{ cm}$.

Q16. If $\sin A = \frac{2}{3}$, then $\cos A$ is equal to:

(A) $\frac{3}{2}$     (B) $\frac{\sqrt{5}}{3}$     (C) $\frac{1}{3}$     (D) $\frac{1}{\sqrt{3}}$
Answer: (B) $\frac{\sqrt{5}}{3}$
Explanation: $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$.

Q17. $PA, PB$ are tangents to circle with centre $O$. If $\angle AOB = 105^\circ$, then $\angle OAP + \angle APB$ is:

(A) $75^\circ$     (B) $175^\circ$     (C) $180^\circ$     (D) $165^\circ$
Answer: (D) $165^\circ$
Explanation: $\angle APB = 180^\circ - 105^\circ = 75^\circ$. $\angle OAP = 90^\circ$. Sum $= 90^\circ + 75^\circ = 165^\circ$.

Q18. In $\Delta ABC$, $PQ \parallel BC$. Given $AP=2.4\text{ cm}, PB=3.6\text{ cm}, BC=5.4\text{ cm}$. $PQ$ is equal to:

(A) $2.7\text{ cm}$     (B) $1.8\text{ cm}$     (C) $3.6\text{ cm}$     (D) $2.16\text{ cm}$
Answer: (D) $2.16\text{ cm}$
Explanation: $\frac{AP}{AB} = \frac{PQ}{BC} \implies \frac{2.4}{2.4+3.6} = \frac{PQ}{5.4} \implies \frac{2.4}{6.0} = \frac{PQ}{5.4} \implies PQ = 2.16\text{ cm}$.

Q19. Assertion (A): Median marks of students in a class test is 16. It means half of the class got marks less than 16.
Reason (R): Median divides the distribution in two equal parts.

Answer: (D) Assertion (A) is false, but Reason (R) is true.
Explanation: Median means half the class got marks less than or equal to 16. Reason is a correct definition.

Q20. Assertion (A): If $P(E)=\frac{1}{999}$ then $P(\overline{E})=0.001$.
Reason (R): $P(E)+P(\overline{E})=1$.

Answer: (D) Assertion (A) is false, but Reason (R) is true.
Explanation: $P(\overline{E}) = 1 - \frac{1}{999} = \frac{998}{999} \approx 0.999 \neq 0.001$.

SECTION B (2 Marks Each)

Q21. Show that $45^n$ cannot end with the digit 0, $n$ being a natural number. Write the prime number '$a$' which on multiplying with $45^n$ makes the product end with the digit 0.

Solution:

$45^n = (3^2 \times 5)^n = 3^{2n} \times 5^n$. For a number to end with digit 0, it must have prime factors 2 and 5. Since 2 is absent in $45^n$, it cannot end with 0. Multiplying by prime number $a = 2$ provides factor 2, making the product end with digit 0.

Q22. Point $P(x,0)$ divides the line segment joining $(2,8)$ and $(-3,-5)$ in a ratio. Find ratio and value of $x$.

Solution:

Let ratio be $k:1$. Using section formula for $y$-coordinate:

$$0 = \frac{k(-5) + 1(8)}{k+1} \implies -5k + 8 = 0 \implies k = \frac{8}{5} \quad (\text{Ratio } 8:5)$$ $$x = \frac{\left(\frac{8}{5}\right)(-3) + 1(2)}{\frac{8}{5}+1} = \frac{-\frac{24}{5} + 2}{\frac{13}{5}} = \frac{-14}{13}$$

Q23. (a) A coin is dropped on a rectangular region $3\text{ m} \times 2\text{ m}$. What is the probability that it lands inside a circle of radius $0.7\text{ m}$?

Solution:

$\text{Area of rectangle} = 3 \times 2 = 6\text{ m}^2$. $\text{Area of circle} = \frac{22}{7} \times 0.7 \times 0.7 = 1.54\text{ m}^2$.

$$P(E) = \frac{1.54}{6} \approx 0.2567$$

OR

(b) A die is thrown twice. What is probability that (i) difference is 3? (ii) sum is 8?

Solution: Total outcomes $= 36$.
(i) Pairs with diff 3 $= \{(1,4),(4,1),(2,5),(5,2),(3,6),(6,3)\} \implies P = \frac{6}{36} = \frac{1}{6}$.
(ii) Pairs with sum 8 $= \{(2,6),(3,5),(4,4),(5,3),(6,2)\} \implies P = \frac{5}{36}$.

Q24. Prove that the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining points of contact at centre.

Proof: In quadrilateral $OAPB$, radius is perpendicular to tangent $\implies \angle OAP = 90^\circ, \angle OBP = 90^\circ$. Sum of angles $= 360^\circ \implies 90^\circ + 90^\circ + \angle APB + \angle AOB = 360^\circ \implies \angle APB + \angle AOB = 180^\circ$. Hence proved.

Q25. Given $\angle ADE = \angle ACB$ and $\frac{AD}{DB} = \frac{AE}{EC}$. Prove that $\Delta ABC$ is an isosceles triangle.

Proof: Since $\frac{AD}{DB} = \frac{AE}{EC} \implies DE \parallel BC$ (Converse of BPT). Thus $\angle ADE = \angle ABC$ (Corresponding angles). Given $\angle ADE = \angle ACB \implies \angle ABC = \angle ACB \implies AB = AC$. Hence $\Delta ABC$ is isosceles.

SECTION C (3 Marks Each)

Q26. Find zeroes of $p(x) = 6x^2 + 13x - 5$ and verify relation with coefficients.

Solution: $6x^2 + 15x - 2x - 5 = (3x-1)(2x+5) = 0 \implies \text{Zeroes } \alpha = -\frac{5}{2}, \beta = \frac{1}{3}$.
Sum $= -\frac{5}{2} + \frac{1}{3} = -\frac{13}{6} = -\frac{b}{a}$. Product $= -\frac{5}{2} \times \frac{1}{3} = -\frac{5}{6} = \frac{c}{a}$.

Q27. (a) Find sum of A.P. $7, 10\frac{1}{2}, 14, \dots, 84$.

Solution: $a=7, d=\frac{7}{2}, l=84 \implies 84 = 7 + (n-1)\frac{7}{2} \implies n=23$.
$S_{23} = \frac{23}{2}(7+84) = \frac{23 \times 91}{2} = 1046\frac{1}{2}$.

Q28. Prove that $\sqrt{3}$ is an irrational number.

Proof: Let $\sqrt{3} = \frac{a}{b}$ (coprime integers, $b \neq 0$). Squaring: $3b^2 = a^2 \implies 3$ divides $a \implies a = 3c$. Substituting gives $3b^2 = 9c^2 \implies b^2 = 3c^2 \implies 3$ divides $b$. Contradiction to coprime assumption. Hence $\sqrt{3}$ is irrational.

Q29. If $A(-5,y), B(2,-2), C(8,4), D(x,5)$ form a parallelogram $ABCD$, find $x, y$ and side lengths.

Solution: Midpoints of $AC$ and $BD$ coincide: $\left(\frac{-5+8}{2}, \frac{y+4}{2}\right) = \left(\frac{2+x}{2}, \frac{-2+5}{2}\right) \implies x=1, y=-1$.
Sides: $AB = \sqrt{(2+5)^2 + (-2+1)^2} = \sqrt{50}$ units, $BC = \sqrt{(8-2)^2 + (4+2)^2} = \sqrt{72} = 6\sqrt{2}$ units.

Q30. A chord of radius $10\text{ cm}$ subtends $60^\circ$ at centre $O$. Find area of shaded segment ($\pi=3.14, \sqrt{3}=1.73$).

Solution: $\text{Area of sector} = \frac{60}{360} \times 3.14 \times 10^2 = 52.33\text{ cm}^2$. $\text{Area of equilateral } \Delta = \frac{\sqrt{3}}{4} \times 10^2 = 43.25\text{ cm}^2$.
$\text{Shaded area} = 52.33 - 43.25 = 9.08\text{ cm}^2$.

Q31. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at centre.

Proof: Tangents from external points subtend equal angles at centre. $2(\angle 2 + \angle 3 + \angle 6 + \angle 7) = 360^\circ \implies \angle AOB + \angle COD = 180^\circ$.

SECTION D (5 Marks Each)

Q32. (a) For $p^2x^2 + (p^2-q^2)x - q^2 = 0$, (i) Show $D$ is perfect square (ii) Find roots.

Solution:
(i) $D = (p^2-q^2)^2 - 4(p^2)(-q^2) = p^4 + q^4 + 2p^2q^2 = (p^2+q^2)^2$ (Perfect Square).
(ii) Roots: $\alpha = \frac{-(p^2-q^2) + (p^2+q^2)}{2p^2} = \frac{q^2}{p^2}$, $\beta = \frac{-(p^2-q^2) - (p^2+q^2)}{2p^2} = -1$.

Q33. Find Mean and Mode of class intervals 15-20 (6), 20-25 (16), 25-30 (17), 30-35 (4), 35-40 (5), 40-45 (2).

Solution:
$\sum f_i = 50, \sum f_i x_i = 1335 \implies \text{Mean} = \frac{1335}{50} = 26.7$.
Modal class $= 25-30 \implies \text{Mode} = 25 + \frac{17-16}{2(17)-16-4} \times 5 = 25 + \frac{5}{14} = 25.36$.

Q34. State and prove Basic Proportionality Theorem (BPT).

Proof: Given line $DE \parallel BC$. $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{AD}{DB}$ and $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta DEC)} = \frac{AE}{EC}$. Since $\text{Area}(\Delta BDE) = \text{Area}(\Delta DEC)$, we get $\frac{AD}{DB} = \frac{AE}{EC}$.

Q35. Elevation of top of pedestal is $30^\circ$, top of flagstaff ($5\text{ m}$) is $60^\circ$. Find height of pedestal $h$ and distance $x$.

Solution: $\tan 30^\circ = \frac{h}{x} \implies x = h\sqrt{3}$. $\tan 60^\circ = \frac{5+h}{x} \implies \sqrt{3} = \frac{5+h}{h\sqrt{3}} \implies 3h = 5+h \implies h = 2.5\text{ m}$.
Distance $x = 2.5 \times 1.73 = 4.325\text{ m}$.

SECTION E (Case Study Based - 4 Marks Each)

Q36. Plan A: $3C=20t$, Plan B: $3C=10t+300$.
(i) Initial amount of Plan B? (ii) Minutes bought in Plan A for ₹250? (iii) Duration when both charge same amount?

Solution:
(i) $t=0 \implies 3C=300 \implies C = \text{₹}100$.
(ii) $C=250 \implies 3(250) = 20t \implies t = 37.5\text{ mins}$.
(iii) $20t = 10t + 300 \implies 10t = 300 \implies t = 30\text{ mins}$ (Charge ₹200).

Q37. Box $42\times 91 \times 50\text{ cm}$ with 600 balls of diameter $7\text{ cm}$.
(i) Volume of one ball? (ii) Surface area of 10 balls? (iii) Empty space in box?

Solution:
(i) $V = \frac{4}{3} \pi (3.5)^3 = 179.67\text{ cm}^3$.
(ii) Surface area of 10 balls $= 10 \times 4\pi (3.5)^2 = 1540\text{ cm}^2$.
(iii) Empty space $= (42 \times 91 \times 50) - (600 \times 179.67) = 191100 - 107802 = 83298\text{ cm}^3$.

Q38. Triangular plot diagram: $BH=40\text{ m}, HD=10\text{ m}, DG=40\text{ m}, GF=38\text{ m}, FE=53\text{ m}, AF=53.58\text{ m}$.
(i) Find length $BC$. (ii) Find length $AG$. (iii) Calculate perimeter of $\Delta ABC$.

Solution:
(i) In $\Delta BDC$, $\cos 45^\circ = \frac{50}{BC} \implies BC = 50\sqrt{2} = 70.50\text{ m}$.
(ii) In $\Delta AGF$, $\tan 45^\circ = \frac{38}{AG} \implies AG = 38\text{ m}$.
(iii) $AC = \frac{38+40}{\sin 60^\circ} = 89.96\text{ m}$. Perimeter $= 38+40+10+40+70.50+89.96 = 288.46\text{ m}$.