CBSE Class 10 Maths Standard 2025 Question Paper (Set 30/1/1) with Solutions

1. For $3x^2 + 6x + k$: sum of zeroes $\alpha + \beta = -\dfrac{6}{3} = -2$, and product $\alpha\beta = \dfrac{k}{3}$.
2. Put these in the given condition: $-2 + \dfrac{k}{3} = -\dfrac{2}{3}$.
3. So $\dfrac{k}{3} = 2 - \dfrac{2}{3} = \dfrac{4}{3}$, which gives $k = 4$.

Answer: (D) $4$

Common mistake: Writing $\alpha + \beta = +\dfrac{b}{a}$ (forgetting the minus sign) gives $k = -8$.

1. Put $x = 1,\ y = 2$ in the first equation: $2 - 6 + a = 0$, so $a = 4$.
2. Put $x = 1,\ y = 2$ in the second: $2 + 6 - b = 0$, so $b = 8$.
3. Compare: $b = 8 = 2 \times 4 = 2a$, so $2a = b$.

Answer: (B) $2a = b$

1. Mid-point $= \left(\dfrac{-4 + 4}{2},\ \dfrac{5 + 6}{2}\right) = \left(0,\ \dfrac{11}{2}\right)$.
2. The x-coordinate is 0 and the y-coordinate is not, so the point is on the y-axis (it is not the origin).

Answer: (B) y-axis

Common mistake: Calling it the origin. The origin needs both coordinates to be 0.

1. $4\sin\theta = 9 - 7 = 2$, so $\sin\theta = \dfrac{1}{2}$.
2. $\sin 30^\circ = \dfrac{1}{2}$ and $\theta$ is acute, so $\theta = 30^\circ$.

Answer: (B) $30^\circ$

1. $\dfrac{1}{\cos\theta} = \sec\theta$, so the bracket is $\sec\theta \times \sec\theta = \sec^2\theta$.
2. The expression is $\tan^2\theta - \sec^2\theta$.
3. Since $\sec^2\theta = 1 + \tan^2\theta$, we get $\tan^2\theta - (1 + \tan^2\theta) = -1$.

Answer: (C) $-1$

Common mistake: Remembering the identity as $\sec^2\theta - \tan^2\theta = 1$ and answering $1$, without noticing the order is reversed.

1. $98 = 2 \times 7^2$ and $28 = 2^2 \times 7$. HCF $= 2 \times 7 = 14$, so $m = 14$.
2. LCM $= 2^2 \times 7^2 = 196$, so $n = 196$. (Check: $14 \times 196 = 2744 = 98 \times 28$.)
3. $n - 7m = 196 - 98 = 98$.

Answer: (C) $98$

1. A tangent is perpendicular to the radius at the point of contact.
2. The two ends of a diameter lie on one straight line through the centre, so both tangents are perpendicular to that same line.
3. Two lines perpendicular to the same line are parallel.

Answer: (A) parallel

1. Two pairs of equal angles ($\angle B = \angle E$ and $\angle C = \angle F$) give similarity by the AA criterion.
2. $AB$ and $DE$ are corresponding sides and $AB = 3\,DE$, so the ratio of sides is $3 : 1$, not $1 : 1$.
3. Triangles of equal size would be congruent. These are not the same size.

Answer: (D) similar but not congruent

Common mistake: Treating "similar" as "congruent". Congruent needs equal sides, similar only needs equal angles.

1. $(-1)^8 = 1$ because $8$ is even.
2. So $(-1)^n + 1 = 0$, which means $(-1)^n = -1$.
3. $(-1)$ raised to a power is $-1$ only when the power is odd.

Answer: (C) any odd number

1. A zero of a polynomial is where its graph meets the x-axis.
2. Each parabola meets the x-axis at two points, and both parabolas meet it at the same two points.
3. So the distinct zeroes are only 2, not 4.

Answer: (C) 2

Common mistake: Adding the zeroes of the two graphs (2 + 2 = 4). The question asks for distinct zeroes, and they share both.

1. $S_1 = 2 + 3 = 5$, so $a_1 = 5$.
2. $S_2 = 2(4) + 3(2) = 14$.
3. Second term $= S_2 - S_1 = 14 - 5 = 9$.

Answer: (B) $9$

Common mistake: Answering $14$, which is the sum of two terms, not the second term.

1. Use the empirical relation: Mode $= 3 \times$ Median $- 2 \times$ Mean.
2. $15x = 3\,\text{Median} - 36x$, so $3\,\text{Median} = 51x$.
3. Median $= 17x$.

Answer: (C) $17x$

Common mistake: Swapping the mean and the mode in the relation.

1. Face cards are Jack, Queen and King: 3 in each of the 4 suits.
2. Red suits are hearts and diamonds, so red face cards $= 3 \times 2 = 6$.
3. Probability $= \dfrac{6}{52} = \dfrac{3}{26}$.

Answer: (D) $\dfrac{3}{26}$

Common mistake: Counting all 12 face cards and answering $\dfrac{3}{13}$.

1. $\sqrt{3} \approx 1.732$ and $\sqrt{5} \approx 2.236$, so we need a rational number between $1.732$ and $2.236$.
2. Option A never ends and never repeats, so it is irrational (and it is below 1.732 anyway).
3. Option B is a repeating decimal, so it is rational, but $2.3266\ldots$ is bigger than $2.236$.
4. $\pi$ is irrational. Option D, $1.857142$, is a terminating decimal, so it is rational, and $1.732 < 1.857142 < 2.236$.

Answer: (D) $1.857142$

Common mistake: Picking B because it is rational, without checking that it lies between the two roots.

1. Area of sector $= \dfrac{\theta}{360} \times \pi r^2 = \dfrac{72}{360} \times \pi r^2 = \dfrac{\pi r^2}{5}$.
2. $\dfrac{\pi r^2}{5} = 40\pi$, so $r^2 = 200$.
3. $r = \sqrt{200} = 10\sqrt{2}$ units.

Answer: (D) $10\sqrt{2}$ units

Common mistake: Stopping at $r^2 = 200$ and picking $200$.

1. $A$, $O$, $B$ lie on one straight line, so $\angle POA + \angle POB = 180^\circ$, giving $\angle POA = 180^\circ - 115^\circ = 65^\circ$.
2. The tangent $PA$ is perpendicular to the radius $OA$, so $\angle PAO = 90^\circ$.
3. In $\triangle APO$: $\angle APO = 180^\circ - 90^\circ - 65^\circ = 25^\circ$.

Answer: (A) $25^\circ$

Common mistake: Using $115^\circ$ as $\angle POA$ and getting $\angle APO = 90^\circ - 115^\circ$, which is impossible. Use the straight line $AOB$ first.

1. The height is opposite to the 30° angle, and the string is the hypotenuse.
2. $\sin 30^\circ = \dfrac{\text{height}}{\text{string}}$, so $\dfrac{1}{2} = \dfrac{150}{L}$.
3. $L = 300$ m.

Answer: (B) $300$ m

Common mistake: Using $\cos 30^\circ$ or $\tan 30^\circ$, which would need the horizontal distance.

1. Arc length $= \dfrac{\theta}{360} \times 2\pi r$.
2. $20 = \dfrac{\theta}{360} \times 2\pi \times \dfrac{60}{\pi} = \dfrac{\theta}{360} \times 120$.
3. $\theta = \dfrac{20 \times 360}{120} = 60^\circ$.

Answer: (B) $60^\circ$

1. Picking a number from 1 to 20 means you will certainly get some number from 1 to 20. Every outcome is favourable, so the probability is $\dfrac{20}{20} = 1$. The assertion is true.
2. An event with probability $1$ is called a sure event. The reason is true.
3. The reason is exactly why the assertion holds: the event is certain, so its probability is 1. So R explains A.

Answer: (A) Both true, and R is the correct explanation

Common mistake: Reading the assertion as "the probability of picking one particular number" (which is $\dfrac{1}{20}$). It says selecting a number from 1 to 20, which is certain.

1. Two hemispheres of the same radius, joined flat face to flat face, make a full sphere. The assertion is true.
2. The total surface area of a sphere is $4\pi r^2$, not $3\pi r^2$. ($3\pi r^2$ is the total surface area of a solid hemisphere.) The reason is false.

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

Common mistake: Confusing the sphere's $4\pi r^2$ with the solid hemisphere's $3\pi r^2$.

1. (a) Standard values: $\cos 60^\circ = \dfrac{1}{2}$, $\cos 0^\circ = 1$, $\sin 30^\circ = \dfrac{1}{2}$, $\cot 45^\circ = 1$.
2. So $\dfrac{x}{2} + y + \dfrac{1}{2} - 1 = 5$, which gives $\dfrac{x}{2} + y = \dfrac{11}{2}$.
3. Multiply by 2: $x + 2y = 11$.
4. (b) $\tan 60^\circ = \sqrt{3}$, so $\tan^2 60^\circ = 3$. $\sin 60^\circ = \cos 30^\circ = \dfrac{\sqrt{3}}{2}$, so $\sin^2 60^\circ + \cos^2 30^\circ = \dfrac{3}{4} + \dfrac{3}{4} = \dfrac{3}{2}$.
5. The value is $3 \div \dfrac{3}{2} = 2$.

Answer: (a) $x + 2y = 11$   (b) $2$

1. Multiply by 3 to clear the fractions: $3x^2 + 4x - 4 = 0$.
2. Split the middle term: $3x^2 + 6x - 2x - 4 = 3x(x + 2) - 2(x + 2) = (3x - 2)(x + 2)$.
3. Set each factor to zero: $x = \dfrac{2}{3}$ or $x = -2$.

Answer: Zeroes: $\dfrac{2}{3}$ and $-2$

Common mistake: Giving the factors instead of the zeroes, or flipping the sign of a zero.

1. Radius $= \dfrac{10\sqrt{2}}{2} = 5\sqrt{2}$, so $r^2 = 50$.
2. The distance from the centre $(2a,\ a - 7)$ to $(11,\ -9)$ equals the radius: $(11 - 2a)^2 + (-9 - a + 7)^2 = 50$.
3. That is $(11 - 2a)^2 + (a + 2)^2 = 50$. Expand: $121 - 44a + 4a^2 + a^2 + 4a + 4 = 50$.
4. $5a^2 - 40a + 75 = 0$, so $a^2 - 8a + 15 = 0$, i.e. $(a - 3)(a - 5) = 0$.

Answer: $a = 3$ or $a = 5$

Common mistake: Stopping after one value. The question asks for value(s), and both work.

1. (a) Corresponding sides are in the same ratio, with $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$. $AC$ corresponds to $PR$, so the ratio is $\dfrac{PR}{AC} = \dfrac{6}{8} = \dfrac{3}{4}$.
2. $PQ = AB \times \dfrac{3}{4} = 4.5$ cm and $QR = BC \times \dfrac{3}{4} = 3$ cm.
3. $PQ + QR = 4.5 + 3 = 7.5$ cm.
4. (b) In $\triangle PQR$, $\angle 1 = \angle 2$ means $\angle PQR = \angle PRQ$, so the sides opposite them are equal: $PR = PQ$.
5. Replace $PR$ by $PQ$ in the given ratio: $\dfrac{QR}{QS} = \dfrac{QT}{PQ}$, which can be written $\dfrac{QS}{QR} = \dfrac{PQ}{QT}$.
6. In $\triangle PQS$ and $\triangle TQR$: $\dfrac{PQ}{QT} = \dfrac{QS}{QR}$, and $\angle PQS = \angle TQR$ (the same angle at $Q$).
7. So $\triangle PQS \sim \triangle TQR$ by the SAS similarity criterion.

Answer: (a) $7.5$ cm   (b) shown by SAS similarity

Common mistake: Matching the wrong sides in part (a). Write the similarity in vertex order first (A to P, B to Q, C to R).

1. The radius $OA$ is perpendicular to the tangent $PA$, so $\triangle OAP$ has a right angle at $A$.
2. $OA^2 = OP^2 - PA^2 = 26^2 - 10^2 = 676 - 100 = 576$.
3. $OA = 24$ m.

Answer: Radius $= 24$ m

Common mistake: Adding instead of subtracting: $OP^2 + PA^2$ treats $OP$ as a leg, but $OP$ is the hypotenuse.

1. (a) $BCD$ is a tangent at $C$ and $OC$ is a radius, so $OC \perp BD$. Hence $\angle OCD = 90^\circ$, i.e. $\angle OCA + \angle ACD = 90^\circ$. … (1)
2. In $\triangle OAC$, $OA = OC$ (radii), so $\angle OAC = \angle OCA$ (angles opposite equal sides).
3. $O$ lies on line $BA$, so $\angle BAC$ is the same angle as $\angle OAC$. Put $\angle OCA = \angle BAC$ in (1): $\angle BAC + \angle ACD = 90^\circ$.
4. (b) Let $ABCD$ circumscribe a circle with centre $O$, touching $AB, BC, CD, DA$ at $P, Q, R, S$. Join $O$ to $A, B, C, D$ and to $P, Q, R, S$.
5. Tangents from an outside point are equal: $AP = AS$. Also $OP = OS$ (radii) and $OA$ is common, so $\triangle OAP \cong \triangle OAS$ (SSS). Hence $\angle AOP = \angle AOS$; call it $\angle 1$.
6. In the same way $\angle BOP = \angle BOQ = \angle 2$, $\angle COQ = \angle COR = \angle 3$, $\angle DOR = \angle DOS = \angle 4$.
7. The angles around $O$ add up to $360^\circ$: $2(\angle 1 + \angle 2 + \angle 3 + \angle 4) = 360^\circ$, so $\angle 1 + \angle 2 + \angle 3 + \angle 4 = 180^\circ$.
8. $\angle AOB = \angle 1 + \angle 2$ and $\angle COD = \angle 3 + \angle 4$, so $\angle AOB + \angle COD = 180^\circ$. Likewise $\angle AOD + \angle BOC = \angle 1 + \angle 4 + \angle 2 + \angle 3 = 180^\circ$.

Answer: (a) $\angle BAC + \angle ACD = 90^\circ$   (b) opposite sides subtend supplementary angles

Common mistake: Drawing the radius to the point of contact is the step most answers skip. Without OC, there is nothing to connect the tangent to the angle.

1. (a) Write everything in $\sin$ and $\cos$. $\dfrac{\tan\theta}{1 - \cot\theta} = \dfrac{\sin\theta/\cos\theta}{(\sin\theta - \cos\theta)/\sin\theta} = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)}$.
2. $\dfrac{\cot\theta}{1 - \tan\theta} = \dfrac{\cos\theta/\sin\theta}{(\cos\theta - \sin\theta)/\cos\theta} = \dfrac{\cos^2\theta}{\sin\theta(\cos\theta - \sin\theta)} = -\dfrac{\cos^2\theta}{\sin\theta(\sin\theta - \cos\theta)}$.
3. Add them: $\dfrac{1}{\sin\theta - \cos\theta}\left[\dfrac{\sin^2\theta}{\cos\theta} - \dfrac{\cos^2\theta}{\sin\theta}\right] = \dfrac{1}{\sin\theta - \cos\theta} \cdot \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta}$.
4. Use $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: this becomes $\dfrac{\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta}$.
5. $= \dfrac{1}{\sin\theta\cos\theta} + 1 = 1 + \sec\theta\,\mathrm{cosec}\,\theta = \text{RHS}$.
6. (b) Take the common denominator $(\sin A - \cos A)(\sin A + \cos A) = \sin^2 A - \cos^2 A$.
7. Numerator: $(\sin A + \cos A)^2 + (\sin A - \cos A)^2 = 2(\sin^2 A + \cos^2 A) = 2$.
8. Denominator: $\sin^2 A - \cos^2 A = \sin^2 A - (1 - \sin^2 A) = 2\sin^2 A - 1$. So LHS $= \dfrac{2}{2\sin^2 A - 1} =$ RHS.

Answer: Both identities proved

Common mistake: Starting from both sides at once. Work on the left side only and arrive at the right side.

1. Let the y-axis divide the segment in the ratio $k : 1$ at the point $(0, y)$.
2. Section formula for the x-coordinate: $\dfrac{k \times (-1) + 1 \times 5}{k + 1} = 0$, so $-k + 5 = 0$ and $k = 5$.
3. The ratio is $5 : 1$.
4. $y = \dfrac{5 \times (-4) + 1 \times (-6)}{5 + 1} = \dfrac{-26}{6} = -\dfrac{13}{3}$.

Answer: Ratio $5 : 1$, point of intersection $\left(0, -\dfrac{13}{3}\right)$

Common mistake: Using the x-coordinate for the y-axis condition the wrong way round. On the y-axis, x is 0.

1. Suppose $\dfrac{1}{\sqrt{5}}$ is rational. Then $\dfrac{1}{\sqrt{5}} = \dfrac{p}{q}$ for integers $p, q$ with $q \neq 0$, and $p, q$ have no common factor.
2. Flip both sides: $\sqrt{5} = \dfrac{q}{p}$. So $\sqrt{5}$ would be rational.
3. But $\sqrt{5}$ is irrational. To see this, suppose $\sqrt{5} = \dfrac{a}{b}$ in lowest terms. Then $5b^2 = a^2$, so $5$ divides $a^2$. Since $5$ is prime, $5$ divides $a$. Write $a = 5c$: $5b^2 = 25c^2$, so $b^2 = 5c^2$, and $5$ divides $b$. Now $5$ divides both $a$ and $b$, which contradicts "lowest terms".
4. So our assumption was wrong, and $\dfrac{1}{\sqrt{5}}$ is irrational.

Answer: Proved by contradiction

Common mistake: Writing "so it is irrational" without showing the contradiction with the common factor.

1. Let the common radius be $r$. The height of the cylinder is twice that, $h = 2r$.
2. Volume of air $=$ cylinder $+$ hemisphere $= \pi r^2 h + \dfrac{2}{3}\pi r^3 = \pi r^2(2r) + \dfrac{2}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$.
3. $\dfrac{8}{3} \times \dfrac{22}{7} \times r^3 = \dfrac{1408}{21}$, which is $\dfrac{176}{21}\,r^3 = \dfrac{1408}{21}$.
4. $r^3 = \dfrac{1408}{176} = 8$, so $r = 2$ m.
5. Height of the cylindrical part $= 2r = 4$ m.

Answer: Height $= 4$ m

Common mistake: Answering $r = 2$ m. The question asks for the height of the cylinder, which is $2r$.

1. Total outcomes when two dice are thrown $= 6 \times 6 = 36$.
2. Pairs with difference 2: $(1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4)$. That is $8$ pairs.
3. Probability $= \dfrac{8}{36} = \dfrac{2}{9}$.

Answer: $P = \dfrac{2}{9}$

Common mistake: Counting only 4 pairs, e.g. $(1,3)$ but not $(3,1)$. The two dice are different, so both orders count.

1. Let ₹$x$ be invested in $A$ ($8\%$) and ₹$y$ in $B$ ($9\%$).
2. First case: $\dfrac{8x}{100} + \dfrac{9y}{100} = 1860$, so $8x + 9y = 186000$. … (i)
3. Interchanged: $\dfrac{9x}{100} + \dfrac{8y}{100} = 1860 + 20 = 1880$, so $9x + 8y = 188000$. … (ii)
4. Add (i) and (ii): $17x + 17y = 374000$, so $x + y = 22000$. … (iii)
5. Subtract (i) from (ii): $x - y = 2000$. … (iv)
6. Add (iii) and (iv): $2x = 24000$, so $x = 12000$. Then $y = 22000 - 12000 = 10000$.
7. Check: $8 \times 12000 + 9 \times 10000 = 96000 + 90000 = 186000$ ✓.

Answer: ₹12,000 in scheme A and ₹10,000 in scheme B

Common mistake: Forgetting to divide by 100, or using ₹20 less instead of ₹20 more in the second equation.

1. (a) In a parallelogram $AD \parallel BC$, so $AD \parallel BE$. Consider $\triangle ADF$ and $\triangle EBF$.
2. $\angle AFD = \angle EFB$ (vertically opposite angles).
3. $\angle ADF = \angle FBE$ (alternate angles, since $AD \parallel BE$ with transversal $BD$).
4. So $\triangle ADF \sim \triangle EBF$ by AA similarity. Corresponding sides are proportional: $\dfrac{DF}{FB} = \dfrac{FA}{EF}$.
5. Cross-multiply: $DF \times EF = FB \times FA$.
6. (b) $AD^2 = BD \times DC$ gives $\dfrac{AD}{DC} = \dfrac{BD}{AD}$, i.e. $\dfrac{BD}{AD} = \dfrac{AD}{DC}$.
7. In $\triangle DBA$ and $\triangle DAC$: $\angle BDA = \angle ADC = 90^\circ$, and $\dfrac{BD}{AD} = \dfrac{AD}{DC}$. So $\triangle DBA \sim \triangle DAC$ by SAS similarity.
8. Hence $\angle DBA = \angle DAC$ and $\angle BAD = \angle DCA$.
9. In right triangle $ADB$: $\angle DBA + \angle BAD = 90^\circ$. Replace $\angle DBA$ by $\angle DAC$: $\angle DAC + \angle BAD = 90^\circ$.
10. But $\angle BAD + \angle DAC = \angle BAC$. So $\angle BAC = 90^\circ$.

Answer: (a) $DF \times EF = FB \times FA$   (b) $\angle BAC = 90^\circ$

Common mistake: Skipping the figure. A clear, labelled figure is worth marks in proof questions.

1. (a) Let the other two sides be $x$ cm and $y$ cm. Perimeter: $x + y + 25 = 60$, so $y = 35 - x$.
2. Pythagoras: $x^2 + y^2 = 25^2 = 625$, so $x^2 + (35 - x)^2 = 625$.
3. $x^2 + 1225 - 70x + x^2 = 625$, so $2x^2 - 70x + 600 = 0$, i.e. $x^2 - 35x + 300 = 0$.
4. $(x - 15)(x - 20) = 0$, so $x = 15$ or $20$, and the other side is $20$ or $15$. The sides are $15$ cm and $20$ cm.
5. (b) Let the speed be $x$ km/h. At speed $(x - 8)$ the time is $\dfrac{480}{x - 8}$, which is $3$ hours more than $\dfrac{480}{x}$.
6. $\dfrac{480}{x - 8} - \dfrac{480}{x} = 3$. Combine: $\dfrac{480 \times 8}{x(x - 8)} = 3$, so $x(x - 8) = 1280$.
7. $x^2 - 8x - 1280 = 0$. Factorise: $(x - 40)(x + 32) = 0$, so $x = 40$ or $x = -32$.
8. Speed cannot be negative, so x = 40.