CBSE Class 10 Maths Standard 2025 Question Paper (Set 30/2/1) with Solutions

1. Write $7\cos^2\theta + 3\sin^2\theta = 3(\cos^2\theta + \sin^2\theta) + 4\cos^2\theta = 3 + 4\cos^2\theta$.
2. So $3 + 4\cos^2\theta = 4$, which gives $\cos^2\theta = \dfrac{1}{4}$ and $\cos\theta = \dfrac{1}{2}$ (taking the acute angle).
3. $\cos 60^\circ = \dfrac{1}{2}$, so $\theta = 60^\circ$.

Answer: (C) $60^\circ$

Common mistake: Trying the options one by one. Using $\sin^2\theta + \cos^2\theta = 1$ gets the answer in two lines.

1. The only even prime number is $2$, so there is $1$ favourable outcome.
2. There are $30$ equally likely numbers, so the probability is $\dfrac{1}{30}$.

Answer: (A) $\dfrac{1}{30}$

Common mistake: Counting all even numbers (15) or all primes (10). The number has to be both even and prime.

1. Sum of roots $= 7 + \dfrac{1}{7} = \dfrac{50}{7}$ and product of roots $= 7 \times \dfrac{1}{7} = 1$.
2. The equation is $x^2 - (\text{sum})x + \text{product} = 0$, that is $x^2 - \dfrac{50}{7}x + 1 = 0$.
3. Multiply by $7$: $7x^2 - 50x + 7 = 0$.

Answer: (A) $7x^2 - 50x + 7 = 0$

Common mistake: Using a plus sign in front of $50x$. The middle term is minus the sum of the roots.

1. $16 = 2^4$, $20 = 2^2 \times 5$, $50 = 2 \times 5^2$.
2. LCM $= 2^4 \times 5^2 = 400$. Every common multiple is a multiple of $400$, so the least number divisible by all three is $400$.
3. $400 = 20^2$ is already a perfect square, so the answer is $400$.
4. None of the four options is 400, so the question as printed has no correct option. CBSE’s answer key says the same and awards the mark to every candidate.

Answer: $400$ (not among the options; CBSE awarded the mark to all)

Common mistake: Multiplying the LCM by another factor to "make it a square". Check first whether the LCM is already a perfect square.

1. Both points have $x = 5$, so the diameter is vertical. Its length is $2 - (-2) = 4$.
2. Radius $= \dfrac{4}{2} = 2$. A length cannot be negative, so it is not $\pm 2$.

Answer: (D) $2$

1. Let $A(-5, 0)$, $B(5, 0)$, $C(0, 4)$. $AB = 10$.
2. $AC = \sqrt{5^2 + 4^2} = \sqrt{41}$ and $BC = \sqrt{5^2 + 4^2} = \sqrt{41}$.
3. Two sides are equal and the third is different, so it is isosceles. It is not right-angled because $AC^2 + BC^2 = 82 \neq 100 = AB^2$.

Answer: (B) isosceles triangle

Common mistake: Stopping at "the base is on the x-axis" and calling it right-angled. Check with the distance formula.

1. $MN$ is a diameter, so the angle in a semicircle $\angle MLN = 90^\circ$.
2. In $\triangle MLN$: $\angle LNM = 180^\circ - 90^\circ - 30^\circ = 60^\circ$.
3. The angle between a tangent and a chord equals the angle in the alternate segment: $\angle RLM = \angle LNM = 60^\circ$.

Answer: (B) $60^\circ$

Common mistake: Answering $30^\circ$, the angle at $M$. The tangent-chord angle matches the angle on the opposite side of the chord, at $N$.

1. $PQ \parallel BC$, so by the Basic Proportionality Theorem $\dfrac{AQ}{QC} = \dfrac{AP}{PB} = \dfrac{4}{13}$.
2. So $AQ$ is $4$ parts and $QC$ is $13$ parts out of $4 + 13 = 17$ parts of $AC$: $\dfrac{AQ}{AC} = \dfrac{4}{17}$.
3. $AQ = \dfrac{4}{17} \times 20.4 = 4 \times 1.2 = 4.8$ cm.

Answer: (D) $4.8$ cm

Common mistake: Using $\dfrac{4}{13} \times 20.4$. The ratio $4 : 13$ is for $AQ : QC$, not for $AQ : AC$.

1. Congruent figures have the same shape and size, so they are similar. (A) is correct.
2. A square and a rhombus can have the same area, but a rhombus can have angles other than 90°. Similar figures need equal angles, so they are not always similar. (B) is incorrect.
3. All angles of an equilateral triangle are 60°, so any two are similar. (C) is correct.
4. Similar triangles can have different sizes, so they need not be congruent. (D) is correct.

Answer: (B) A square and a rhombus of the same area are always similar.

Common mistake: Thinking equal area means same shape. Similarity is about equal angles and proportional sides.

1. $4004 = 4 \times 1001 = 2^2 \times 1001$ and $1001 = 7 \times 143 = 7 \times 11 \times 13$.
2. So $4004 = 2^2 \times 7^1 \times 11^1 \times 13^1$.
3. The exponents are $2, 1, 1, 1$ and their sum is $5$.

Answer: (A) 5

Common mistake: Counting the number of prime factors (four) instead of adding their exponents.

1. Balls without a boundary: $42 - 7 = 35$.
2. $P(\text{no boundary}) = \dfrac{35}{42} = \dfrac{5}{6}$.
3. Check: $P(\text{boundary}) = \dfrac{7}{42} = \dfrac{1}{6}$, and $\dfrac{1}{6} + \dfrac{5}{6} = 1$.

Answer: (C) $\dfrac{5}{6}$

Common mistake: Giving $\dfrac{1}{6}$, the probability of hitting a boundary. The question asks for NOT hitting one.

1. A full circle is $360^\circ$ at the centre. Five equal sectors share it equally.
2. Each angle $= \dfrac{360^\circ}{5} = 72^\circ$.

Answer: (D) $72^\circ$

1. $\sin 30^\circ = \dfrac{1}{2}$, $\tan 45^\circ = 1$, $\sec 60^\circ = 2$.
2. So $\dfrac{1}{2} \times 1 = \dfrac{2}{k}$, which gives $k = 4$.

Answer: (A) $4$

1. $x - y = 0$ is the same as $y = x$, a line at $45^\circ$ to the axes. It is not parallel to either axis.
2. Put $(0, 0)$: $0 - 0 = 0$ ✓, so it passes through the origin.
3. Put $(3, 2)$: $3 - 2 = 1 \neq 0$, so it does not pass through $(3, 2)$.

Answer: (C) passing through the origin

1. Write all terms over $4$: $\dfrac{20}{4},\ \dfrac{19}{4},\ \dfrac{18}{4},\ \dfrac{17}{4}, \dots$ So $a = 5$ and $d = -\dfrac{1}{4}$.
2. $a_{10} = a + 9d = 5 - \dfrac{9}{4} = \dfrac{20 - 9}{4} = \dfrac{11}{4}$.

Answer: (A) $\dfrac{11}{4}$

Common mistake: Using $10d$ instead of $9d$. The $n$th term is $a + (n - 1)d$.

1. A zero makes the polynomial $0$: $p(-4) = (-4)^2 - (-4) - (2 + 2k) = 0$.
2. $16 + 4 - 2 - 2k = 0$, so $18 - 2k = 0$ and $k = 9$.

Answer: (B) $9$

Common mistake: Losing a sign: $-(-4) = +4$, and the bracket $(2 + 2k)$ is subtracted as a whole.

1. A line parallel to the $x$-axis is horizontal, so every point on it has the same $y$ value: its equation is $y = c$.
2. Below the $x$-axis means $y$ is negative, and the distance is $3$, so $y = -3$.

Answer: (C) $y = -3$

Common mistake: Writing $x = -3$. That is a vertical line, parallel to the $y$-axis.

1. $40 = 2^3 \times 5$, $110 = 2 \times 5 \times 11$, $360 = 2^3 \times 3^2 \times 5$.
2. Common prime factors with the smallest powers: $2^1 \times 5^1 = 10$.

Answer: (D) 10

1. $d = a_2 - a_1 = 1 - 5 = -4$ (and $-3 - 1 = -4$, $-7 - (-3) = -4$). The common difference is $-4$, not $4$. So the assertion is false.
2. The reason gives the correct rule, $d = a_n - a_{n-1}$, so it is true.

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

Common mistake: Dropping the minus sign. The terms are going down, so d is negative.

1. (A) For infinitely many solutions we need $\dfrac{p}{2} = \dfrac{3}{6} = \dfrac{59}{118}$. Here $\dfrac{3}{6} = \dfrac{1}{2}$ and $\dfrac{59}{118} = \dfrac{1}{2}$, so we need $\dfrac{p}{2} = \dfrac{1}{2}$, i.e. $p = 1$. The assertion is true.
2. (R) A unique solution needs $\dfrac{p}{2} \neq \dfrac{3}{6}$, i.e. $p \neq 1$. The reason is true.
3. But the reason is about a different pair of equations (constants $19$ and $157$) and a different case (unique solution). It does not explain why $p = 1$ gives infinitely many solutions in the assertion. So R is not the explanation of A.

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

Common mistake: Choosing (A) because both statements mention p = 1. Check whether the reason actually explains the assertion.

1. For $21y^2 - y - 2$: sum of zeroes $p + q = -\dfrac{-1}{21} = \dfrac{1}{21}$ and product $pq = \dfrac{-2}{21}$.
2. $(1 - p)(1 - q) = 1 - (p + q) + pq = 1 - \dfrac{1}{21} - \dfrac{2}{21}$.
3. $= \dfrac{21 - 1 - 2}{21} = \dfrac{18}{21} = \dfrac{6}{7}$.

Answer: $\dfrac{6}{7}$

Common mistake: Expanding with $-(p + q)$ but then using $+\dfrac{b}{a}$ for the sum. The sum of zeroes is $-\dfrac{b}{a}$.

1. (a) The table top is the circle with the $90^\circ$ wedge $AOB$ cut out, so the top is the major sector with angle $360^\circ - 90^\circ = 270^\circ$ and radius $42$ cm.
2. Length of the curved edge (major arc) $= \dfrac{270}{360} \times 2\pi r = \dfrac{3}{4} \times 2 \times \dfrac{22}{7} \times 42 = \dfrac{3}{4} \times 264 = 198$ cm.
3. The straight edges are the two radii, $42 + 42 = 84$ cm. Perimeter $= 198 + 84 = 282$ cm.
4. (b) The three shaded sectors have the same radius, so their angles add: $95^\circ + 50^\circ + 35^\circ = 180^\circ$.
5. Area $= \dfrac{180}{360} \times \pi r^2 = \dfrac{1}{2} \times \dfrac{22}{7} \times 5 \times 5 = \dfrac{275}{7}$ cm² $\approx 39.29$ cm².

Answer: (a) $282$ cm   (b) $\dfrac{275}{7} \approx 39.29$ cm²

Common mistake: In (a), using the $90^\circ$ angle for the arc. The table top is the big part, $270^\circ$, and the perimeter also needs the two radii.

1. $\tan A = \sqrt{3} = \tan 60^\circ$, so $A = 60^\circ$.
2. $\sin^2 60^\circ = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}$ and $\cos^2 60^\circ = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$.
3. $\dfrac{\sin^2 A}{1 + \cos^2 A} = \dfrac{3/4}{1 + 1/4} = \dfrac{3/4}{5/4} = \dfrac{3}{5}$.

Answer: $\dfrac{3}{5}$

1. (a) In $\triangle CDA$ and $\triangle CAB$: $\angle ADC = \angle BAC$ (given) and $\angle C$ is common.
2. So $\triangle CDA \sim \triangle CAB$ by AA similarity.
3. Corresponding sides are proportional: $\dfrac{CD}{CA} = \dfrac{CA}{CB}$, so $CA^2 = CD \cdot CB$.
4. (b) $OA \cdot OB = OC \cdot OD$ gives $\dfrac{OA}{OC} = \dfrac{OD}{OB}$.
5. In $\triangle AOD$ and $\triangle COB$: $\dfrac{OA}{OC} = \dfrac{OD}{OB}$ and $\angle AOD = \angle COB$ (vertically opposite angles).
6. So $\triangle AOD \sim \triangle COB$ by SAS similarity, with $A \leftrightarrow C$ and $D \leftrightarrow B$. Hence $\angle A = \angle C$ and $\angle D = \angle B$.

Answer: (a) $CA^2 = CD \cdot CB$   (b) $\angle A = \angle C,\ \angle B = \angle D$

Common mistake: In (b), matching the vertices in the wrong order. The ratio $\dfrac{OA}{OC} = \dfrac{OD}{OB}$ pairs $A$ with $C$ and $D$ with $B$.

1. $XY$ is a tangent at $A$, so $AB \perp XY$. The chord $CD \parallel XY$, so $CD \perp AB$ too, and $AB$ bisects $CD$ at $P$.
2. $AP = 16$ cm and $OA = 10$ cm, so $OP = AP - OA = 16 - 10 = 6$ cm (the chord is on the far side of the centre).
3. In right triangle $OPC$: $CP^2 = OC^2 - OP^2 = 10^2 - 6^2 = 64$, so $CP = 8$ cm.
4. $CD = 2 \times CP = 16$ cm.

Answer: $CD = 16$ cm

Common mistake: Using $OP = 16$. The distance $16$ cm is measured from $A$, not from the centre.

1. (a) Let parallelogram $ABCD$ circumscribe a circle touching $AB, BC, CD, DA$ at $P, Q, R, S$.
2. Tangents from an outside point are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$.
3. Add all four: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$, that is $AB + CD = AD + BC$.
4. In a parallelogram $AB = CD$ and $AD = BC$, so $2AB = 2BC$ and $AB = BC$.
5. So all four sides are equal: $AB = BC = CD = DA$. A parallelogram with all sides equal is a rhombus.
6. (b) Let $PA$ and $PB$ be tangents from $P$ to a circle with centre $O$. Join $OA$ and $OB$.
7. A radius is perpendicular to the tangent at the point of contact, so $\angle OAP = \angle OBP = 90^\circ$.
8. In quadrilateral $OAPB$ the angles add up to $360^\circ$: $\angle APB + \angle AOB + 90^\circ + 90^\circ = 360^\circ$, so $\angle APB + \angle AOB = 180^\circ$.

Answer: Both proved

Common mistake: Writing the equal tangent lengths without stating the reason (tangents from an external point are equal).

1. (a) $\dfrac{1}{\tan^2\theta} = \cot^2\theta$ and $\dfrac{1}{\cot^2\theta} = \tan^2\theta$, so $\text{LHS} = (1 + \cot^2\theta)(1 + \tan^2\theta)$.
2. $= \mathrm{cosec}^2\theta \cdot \sec^2\theta = \dfrac{1}{\sin^2\theta} \cdot \dfrac{1}{\cos^2\theta} = \dfrac{1}{\sin^2\theta\,(1 - \sin^2\theta)}$.
3. $= \dfrac{1}{\sin^2\theta - \sin^4\theta} = \text{RHS}$.
4. (b) Put $c = \mathrm{cosec}\,\theta$. Over a common root: $\dfrac{(c - 1) + (c + 1)}{\sqrt{(c + 1)(c - 1)}} = \dfrac{2c}{\sqrt{c^2 - 1}}$.
5. $c^2 - 1 = \mathrm{cosec}^2\theta - 1 = \cot^2\theta$, so this is $\dfrac{2\,\mathrm{cosec}\,\theta}{\cot\theta} = 2 \times \dfrac{1}{\sin\theta} \times \dfrac{\sin\theta}{\cos\theta} = \dfrac{2}{\cos\theta} = 2\sec\theta$.

Answer: Both identities proved

Common mistake: In (a), forgetting $1 - \sin^2\theta = \cos^2\theta$ when turning $\sec^2\theta$ into the form on the right.

1. Mid-point $P = \left(\dfrac{3 + k}{2},\ \dfrac{4 + 6}{2}\right) = \left(\dfrac{3 + k}{2},\ 5\right)$, so $x = \dfrac{3 + k}{2}$ and $y = 5$.
2. Put in $x + y - 10 = 0$: $\dfrac{3 + k}{2} + 5 - 10 = 0$.
3. $\dfrac{3 + k}{2} = 5$, so $3 + k = 10$ and $k = 7$.

Answer: $k = 7$

Common mistake: Forgetting to divide the sum of coordinates by 2 for the mid-point.

1. The hour hand makes a full turn of $360^\circ$ in $12$ hours, so it turns $\dfrac{360^\circ}{12} = 30^\circ$ every hour.
2. From $5$ a.m. to $8$ a.m. is $3$ hours, so the angle is $3 \times 30^\circ = 90^\circ$. Radius $r = 10$ cm. Use $\pi = \dfrac{22}{7}$.
3. Minor sector $= \dfrac{90}{360} \times \pi r^2 = \dfrac{1}{4} \times \dfrac{22}{7} \times 100 = \dfrac{550}{7} \approx 78.57$ cm².
4. Whole circle $= \pi r^2 = \dfrac{22}{7} \times 100 = \dfrac{2200}{7}$ cm².
5. Major sector $= \dfrac{2200}{7} - \dfrac{550}{7} = \dfrac{1650}{7} \approx 235.71$ cm².

Answer: Minor sector $\dfrac{550}{7} \approx 78.57$ cm²; major sector $\dfrac{1650}{7} \approx 235.71$ cm²

Common mistake: Using $60^\circ$ per hour (that is the minute hand: $6^\circ$ per minute). The hour hand turns only $30^\circ$ per hour.

1. Suppose $\sqrt{3}$ is rational. Then $\sqrt{3} = \dfrac{p}{q}$ with integers $p, q$ ($q \neq 0$) having no common factor.
2. Squaring: $3q^2 = p^2$. So $3$ divides $p^2$, and since $3$ is prime, $3$ divides $p$. Write $p = 3a$.
3. Then $3q^2 = 9a^2$, so $q^2 = 3a^2$. So $3$ divides $q^2$, and hence $3$ divides $q$.
4. Now $3$ is a common factor of $p$ and $q$, which contradicts "no common factor". So $\sqrt{3}$ is irrational.

Answer: Proved by contradiction

Common mistake: Skipping the reason that “3 divides p squared” gives “3 divides p”: it works because 3 is prime.

1. Simple interest $= \dfrac{P \times R \times T}{100}$. Each year adds $\dfrac{2000 \times 7}{100} = ₹140$.
2. End of year $1$: ₹$140$. End of year $2$: $\dfrac{2000 \times 7 \times 2}{100} = ₹280$. End of year $3$: $₹420$.
3. The list $140, 280, 420, \dots$ goes up by $140$ each time (difference $280 - 140 = 140 = 420 - 280$), so yes, it is an AP with $a = 140$ and $d = 140$.
4. Interest at the end of the $27$th year: $a_{27} = a + 26d = 140 + 26 \times 140 = 140 \times 27 = ₹3780$.

Answer: ₹$140$, ₹$280$, ₹$420$; yes, an AP; ₹$3780$ at the end of the $27$th year

Common mistake: Using the (n − 1) rule on the wrong number: here the nth term is simply 140 × n, and the 27th is 140 × 27.

1. (a) Let the lighthouse be $AB = h$, with the ships at $P$ (angle of depression $60^\circ$) and $Q$ (angle of depression $45^\circ$) on either side of $B$.
2. The angle of depression from $A$ equals the angle of elevation of $A$ from the ship (alternate angles). In $\triangle ABP$: $\tan 60^\circ = \dfrac{h}{PB}$, so $PB = \dfrac{h}{\sqrt{3}}$.
3. In $\triangle ABQ$: $\tan 45^\circ = \dfrac{h}{BQ} = 1$, so $BQ = h$.
4. $PQ = PB + BQ = \dfrac{h}{\sqrt{3}} + h = h\left(\dfrac{1 + \sqrt{3}}{\sqrt{3}}\right)$.
5. This equals $100\left(\dfrac{1 + \sqrt{3}}{\sqrt{3}}\right)$, so $h = 100$ m.
6. (b) Let the tall building have height $H$ and the buildings be $d$ apart. The bottom $C$ of the small building is seen at $45^\circ$: $\tan 45^\circ = \dfrac{H}{d}$, so $d = H$.
7. The top $D$ of the small building is $8$ m above $C$, so it is $H - 8$ below the observer's level. $\tan 30^\circ = \dfrac{H - 8}{d} = \dfrac{H - 8}{H} = \dfrac{1}{\sqrt{3}}$.
8. $\sqrt{3}(H - 8) = H$, so $H(\sqrt{3} - 1) = 8\sqrt{3}$ and $H = \dfrac{8\sqrt{3}}{\sqrt{3} - 1} = \dfrac{8\sqrt{3}(\sqrt{3} + 1)}{2} = 4\sqrt{3}(\sqrt{3} + 1) = 12 + 4\sqrt{3}$ m.
9. The distance between the buildings is $d = H = (12 + 4\sqrt{3})$ m $\approx 18.93$ m.