CBSE Class 10 Maths Standard 2025 Question Paper (Set 30/5/1) with Solutions

1. $(\sqrt{3} + 2)^2 = 3 + 4\sqrt{3} + 4 = 7 + 4\sqrt{3}$ and $(\sqrt{3} - 2)^2 = 3 - 4\sqrt{3} + 4 = 7 - 4\sqrt{3}$.
2. Adding, the $4\sqrt{3}$ terms cancel: $7 + 7 = 14$, a positive rational number.

Answer: (A) positive rational number

1. LCM takes the greater power of each prime.
2. For $b$: $\max(3, m) = 4$, so $m = 4$. For $c$: $\max(n, 2) = 3$, so $n = 3$. (For $a$: $\max(2, 3) = 3$ ✓.)
3. $m + n = 4 + 3 = 7$.

Answer: (B) 7

1. This is the key fact used in proofs of irrationality: if a prime $p$ divides $a^2$, then $p$ divides $a$.
2. (Because the prime factorisation of $a^2$ has the same primes as that of $a$.)

Answer: (A) $a$

1. (A) $x^2 + 1 = x^2 - 2x + 1$ gives $2x = 0$: linear.
2. (B) $x^2 + 2x\sqrt{x} + x = 2x\sqrt{x}$ gives $x^2 + x = 0$: quadratic. ✓
3. (C) $x^3 + 3x^2 = x^3 + 3x^2 + 3x + 1$ gives $3x + 1 = 0$: linear.
4. (D) $x^2 - 1 = x^2 + 2x + 1$ gives $2x + 2 = 0$: linear.

Answer: (B) $\left(x + \sqrt{x}\right)^2 = 2x\sqrt{x}$

Common mistake: Judging by the highest power you can see. Expand both sides first: the higher powers may cancel.

1. Real and distinct roots need $D > 0$: $b^2 - 4b > 0$, that is $b(b - 4) > 0$.
2. This holds when $b < 0$ or $b > 4$. Among the options only $-3$ is in that range.
3. (Check the others: $b = 0$ and $b = 4$ give $D = 0$; $b = 3$ gives $D = 9 - 12 < 0$.)

Answer: (D) −3

1. Let each equal part be $1$ unit. Then $AP = PQ = QR = RB = 1$.
2. $PQ = 1$ and $QB = QR + RB = 2$. So $Q$ divides $PB$ in the ratio $PQ : QB = 1 : 2$.

Answer: (C) 1 : 2

1. Red : black $= 3 : 7$, so out of every $3 + 7 = 10$ balls, $7$ are black.
2. $P(\text{black}) = \dfrac{7}{10} = 0.7$.

Answer: (C) 0.7

Common mistake: Using $\dfrac{3}{7}$, the ratio itself. A probability is a part over the whole: $\dfrac{7}{10}$.

1. Two right triangles have one equal angle ($90^\circ$) but the other angles can differ (for example $30^\circ$-$60^\circ$ and $45^\circ$-$45^\circ$), so they need not be similar. (A) is false.
2. Two squares have all angles $90^\circ$ and sides in the same ratio; two equilateral triangles have all angles $60^\circ$; any two circles have the same shape. So (B), (C) and (D) are true.

Answer: (A) Two right triangles are always similar.

1. $AX = \dfrac{2}{3}AD$, so $XD = AD - AX = \dfrac{1}{3}AD$. Hence $DX : XA = 1 : 2$.
2. Three parallel lines cut the two transversals $AD$ and $BC$ in the same ratio: $\dfrac{DX}{XA} = \dfrac{CY}{YB}$.
3. So $CY : YB = 1 : 2$.

Answer: (D) 1 : 2

Common mistake: Answering $2 : 3$ or $3 : 2$ straight from the fraction. $\dfrac{2}{3}$ is $AX : AD$, not $AX : XD$.

1. (A) Every point of the circle has one tangent, so there are infinitely many. True.
2. (B) From a point outside a circle exactly $2$ tangents can be drawn, not infinitely many. False.
3. (C) Infinitely many secants pass through an outside point (rotate the line). True. (D) The tangent is perpendicular to the radius, hence to the diameter through the point of contact. True.

Answer: (B) Infinite number of tangents can be drawn to a circle from a point outside the circle.

1. The marked $210^\circ$ is the reflex angle at $O$. The angle on the other side is $\angle AOB = 360^\circ - 210^\circ = 150^\circ$.
2. $OA \perp PA$ and $OB \perp PB$, so in quadrilateral $OAPB$: $\angle APB = 360^\circ - 90^\circ - 90^\circ - 150^\circ = 30^\circ$.

Answer: (D) $30^\circ$

Common mistake: Using $210^\circ$ directly. The angle between the radii that goes with the tangents is the smaller one, $150^\circ$.

1. $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$, so $\tan^2 30^\circ = \dfrac{1}{3}$.
2. $\dfrac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \dfrac{2/3}{4/3} = \dfrac{1}{2}$.
3. $\cos 60^\circ = \dfrac{1}{2}$. (This is the identity $\cos 2A = \dfrac{1 - \tan^2 A}{1 + \tan^2 A}$ with $A = 30^\circ$.)

Answer: (B) $\cos 60^\circ$

1. Draw a horizontal line from the observer's eyes to the chimney. The part of the chimney above this line is $h$.
2. $\tan 45^\circ = \dfrac{h}{38.2} = 1$, so $h = 38.2$ m.
3. The eyes are $1.8$ m above the ground, so the chimney's height $= 38.2 + 1.8 = 40$ m.

Answer: (C) $40$ m

Common mistake: Forgetting to add the observer’s height of 1.8 m. The 38.2 m is only the part above eye level.

1. Let the radii be $R$ (large) and $r$ (small). $R + r = 16$.
2. $OP \perp AB$ and $P$ is the mid-point of the chord, so $AP = 8$ cm. In right triangle $OPA$: $R^2 = r^2 + 8^2$, so $R^2 - r^2 = 64$.
3. $(R + r)(R - r) = 64$ gives $16(R - r) = 64$, so $R - r = 4$ cm.
4. (So $R = 10$ and $r = 6$.)

Answer: (B) 4 cm

1. Radius of the cone $= \sqrt{l^2 - h^2} = \sqrt{13^2 - 12^2} = \sqrt{25} = 5$ cm.
2. The hemisphere has the same radius, $5$ cm, so it adds a height of $5$ cm.
3. Entire height $= 12 + 5 = 17$ cm.

Answer: (A) 17 cm

1. The empirical relationship is $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$.
2. Rewrite: $3\,\text{Median} + (-2)\,\text{Mean} = 1\cdot\text{Mode}$. So $x = 3$, $y = -2$, $z = 1$.
3. $x + y + z = 3 - 2 + 1 = 2$.

Answer: (C) 2

1. The number of observations is $n = 100$ (even), so the median is the average of the $\dfrac{n}{2} = 50$th and $51$st observations, after arranging in order.
2. Arrange the marks in increasing order with cumulative frequency: $20$ (6), $25$ (26), $28$ (50), $29$ (78), $33$ (93), $38$ (97), $42$ (99), $43$ (100).
3. The $50$th observation is $28$ (cumulative frequency reaches $50$ at $28$) and the $51$st is $29$.

Answer: (C) 28 and 29

Common mistake: Taking the 50th and 51st entries of the table as printed. The marks must be sorted first.

1. Numbers on a die: $1, 2, 3, 4, 5, 6$. Greater than $3$: $4, 5, 6$.
2. Of these, composite numbers are $4$ and $6$ ($5$ is prime). That is $2$ favourable outcomes.
3. $P = \dfrac{2}{6} = \dfrac{1}{3}$.

Answer: (B) $\dfrac{1}{3}$

1. (A) $\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$, so $\cos\theta = \pm\dfrac{4}{5}$. For an acute angle cosine is positive, so $\cos\theta = +\dfrac{4}{5}$. The assertion (with the minus sign) is false.
2. (R) $\sin^2\theta + \cos^2\theta = 1$ is true for every angle. True.

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

1. (R) Going backward from the last term, each step subtracts $d$, so the $n$th term from the end is $l - (n - 1)d$. True.
2. (A) Here $l = 198$ and $d = 3$. The $10$th term from the end is $198 - 9 \times 3 = 198 - 27 = 171$, not $168$. So the assertion is false.

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

Common mistake: Using $198 - 10 \times 3 = 168$. The first term from the end is $l$ itself, so the $10$th uses $(n - 1) = 9$ steps.

1. (a) The equations are $2x + y = 320$ and $4x + 2y = 600$.
2. $\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$, $\dfrac{b_1}{b_2} = \dfrac{1}{2}$ and $\dfrac{c_1}{c_2} = \dfrac{320}{600} = \dfrac{8}{15}$.
3. Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the lines are parallel and the system has no solution: it is not consistent. (The second says $2x + y = 300$, which contradicts $2x + y = 320$.)
4. (b) Multiply the first equation by $\sqrt{3}$ and the second by $\sqrt{2}$: $\sqrt{6}\,x + 3y = 5\sqrt{3}$ and $\sqrt{6}\,x - 4y = -2\sqrt{3}$.
5. Subtract: $7y = 7\sqrt{3}$, so $y = \sqrt{3}$. Then $\sqrt{2}\,x + 3 = 5$, so $x = \dfrac{2}{\sqrt{2}} = \sqrt{2}$.
6. Check in the second: $\sqrt{3}\cdot\sqrt{2} - \sqrt{8}\cdot\sqrt{3} = \sqrt{6} - 2\sqrt{6} = -\sqrt{6}$ ✓.

Answer: (a) $2x + y = 320,\ 4x + 2y = 600$; not consistent   (b) $x = \sqrt{2},\ y = \sqrt{3}$

1. $BP = \dfrac{4}{7}AB$, so $AP = AB - BP = \dfrac{3}{7}AB$. Hence $AP : PB = 3 : 4$.
2. Section formula with $m : n = 3 : 4$: $x = \dfrac{3 \times 2 + 4 \times (-2)}{3 + 4} = \dfrac{-2}{7}$.
3. $y = \dfrac{3 \times (-4) + 4 \times (-2)}{7} = \dfrac{-20}{7}$.

Answer: $P\left(-\dfrac{2}{7},\ -\dfrac{20}{7}\right)$

Common mistake: Using the ratio $4 : 3$. $BP = \dfrac{4}{7}AB$ is measured from $B$, so $AP : PB = 3 : 4$.

1. (a) $15^\circ = 45^\circ - 30^\circ$, so $\sin 15^\circ = \sin 45^\circ\cos 30^\circ - \cos 45^\circ\sin 30^\circ$.
2. $= \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2} = \dfrac{\sqrt{3} - 1}{2\sqrt{2}} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$.
3. (b) $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - y^2}$ (for an acute angle).
4. $\tan A = \dfrac{\sin A}{\cos A} = \dfrac{y}{\sqrt{1 - y^2}}$.

Answer: (a) $\dfrac{\sqrt{6} - \sqrt{2}}{4}$   (b) $\cos A = \sqrt{1 - y^2},\ \tan A = \dfrac{y}{\sqrt{1 - y^2}}$

1. $\triangle ABD \sim \triangle PQS$, so $\dfrac{AB}{PQ} = \dfrac{BD}{QS}$ and $\angle B = \angle Q$.
2. $AD$ and $PS$ are medians, so $BD = \dfrac{BC}{2}$ and $QS = \dfrac{QR}{2}$. Then $\dfrac{BD}{QS} = \dfrac{BC}{QR}$, so $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$.
3. In $\triangle ABC$ and $\triangle PQR$: $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ and the included angles $\angle B = \angle Q$. So $\triangle ABC \sim \triangle PQR$ by SAS.

Answer: Proved

Common mistake: Forgetting to say that a median joins a vertex to the mid-point, which is why BD = BC/2.

1. A pack has $52$ cards, with $12$ face cards (jack, queen and king of each of $4$ suits). So there are $52 - 12 = 40$ cards that are not face cards.
2. (i) $P(\text{not a face card}) = \dfrac{40}{52} = \dfrac{10}{13}$.
3. (ii) There are $2$ black kings (spades and clubs): $P = \dfrac{2}{52} = \dfrac{1}{26}$.

Answer: (i) $\dfrac{10}{13}$   (ii) $\dfrac{1}{26}$

Common mistake: Counting only 4 face cards (the kings). Jack, queen and king are face cards: 12 in all.

1. (a) Suppose $\sqrt{3} = \dfrac{p}{q}$ with $p, q$ co-prime integers and $q \neq 0$. Squaring: $3q^2 = p^2$.
2. So $3$ divides $p^2$, and so $3$ divides $p$. Write $p = 3a$. Then $9a^2 = 3q^2$, so $q^2 = 3a^2$ and $3$ divides $q$.
3. Now $3$ is a common factor of $p$ and $q$, contradicting that they are co-prime. So $\sqrt{3}$ is irrational.
4. (b)(i) True. $2 \times 3 \times 5 \times 7 + 7 = 7(2 \times 3 \times 5 + 1) = 7 \times 31$, which has more than two factors (it is divisible by $7$ and $31$).
5. (b)(ii) False. $2 \times 3 \times 5 \times 7 + 1 = 210 + 1 = 211$, which is a prime number (it has only two factors, $1$ and $211$).

Answer: (a) proved   (b)(i) True   (b)(ii) False

Common mistake: In (b), saying “true” because the number looks big. Actually factorise (i) and test (ii) for factors.

1. $7x^2 + 18x - 9 = 7x^2 + 21x - 3x - 9 = 7x(x + 3) - 3(x + 3) = (7x - 3)(x + 3)$.
2. The zeroes are $x = \dfrac{3}{7}$ and $x = -3$.
3. The new zeroes are twice these: $\dfrac{6}{7}$ and $-6$.
4. Sum $= \dfrac{6}{7} + (-6) = -\dfrac{36}{7}$ and product $= \dfrac{6}{7} \times (-6) = -\dfrac{36}{7}$.
5. Polynomial: $x^2 + \dfrac{36}{7}x - \dfrac{36}{7}$, or multiplying by $7$: $7x^2 + 36x - 36$.

Answer: Zeroes $-3$ and $\dfrac{3}{7}$; new polynomial $7x^2 + 36x - 36$

1. First line: $y = 2x - 2$. Points: $(0, -2)$, $(1, 0)$, $(2, 2)$.
2. Second line: $y = 4x - 4$. Points: $(0, -4)$, $(1, 0)$, $(2, 4)$.
3. Plot both. They cross at $(1, 0)$, so the solution is $x = 1$, $y = 0$.
4. The lines cut the $y$-axis at $(0, -2)$ and $(0, -4)$. The absolute difference of the ordinates is $|-2 - (-4)| = 2$.

Answer: $x = 1,\ y = 0$; absolute difference $= 2$

1. $PA = PB$ gives $PA^2 = PB^2$: $(x - 3)^2 + (y - 5)^2 = (x - 7)^2 + (y - 1)^2$.
2. $x^2 - 6x + 9 + y^2 - 10y + 25 = x^2 - 14x + 49 + y^2 - 2y + 1$, so $8x - 8y = 16$, that is $x - y = 2$.
3. On the $x$-axis $y = 0$: $x = 2$, so the point is $(2, 0)$.
4. On the $y$-axis $x = 0$: $-y = 2$, so $y = -2$ and the point is $(0, -2)$.

Answer: $x - y = 2$; $(2, 0)$ on the $x$-axis and $(0, -2)$ on the $y$-axis

1. (a) LHS $= \dfrac{1 + \frac{1}{\sin A}}{\frac{1}{\sin A}} = \dfrac{\sin A + 1}{1} = \sin A + 1$.
2. Multiply and divide by $(1 - \sin A)$: $\dfrac{(1 + \sin A)(1 - \sin A)}{1 - \sin A} = \dfrac{1 - \sin^2 A}{1 - \sin A} = \dfrac{\cos^2 A}{1 - \sin A} =$ RHS.
3. (b) $\sin(2A + B) = \dfrac{\sqrt{3}}{2}$ with an acute angle gives $2A + B = 60^\circ$. … (i)
4. $\tan(A + 2B) = 1$ gives $A + 2B = 45^\circ$. … (ii)
5. Add: $3A + 3B = 105^\circ$, so $A + B = 35^\circ$. Subtract (ii) from (i): $A - B = 15^\circ$. Hence $A = 25^\circ$ and $B = 10^\circ$.
6. $4A - 7B = 100^\circ - 70^\circ = 30^\circ$, so $\cot 30^\circ = \sqrt{3}$.

Answer: (a) proved   (b) $\sqrt{3}$

1. Join $OC$. $PA$ and $CA$ are tangents from $A$, with $OP = OC$ (radii) and $OA$ common. So $\triangle POA \cong \triangle COA$ (SSS), giving $\angle POA = \angle COA$.
2. In the same way, $\triangle QOB \cong \triangle COB$, so $\angle QOB = \angle COB$.
3. $PQ$ is a diameter (the tangents are parallel, so $P$, $O$, $Q$ lie on one straight line). So $\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ$.
4. This is $2\angle COA + 2\angle COB = 180^\circ$, so $\angle COA + \angle COB = 90^\circ$.
5. $\angle AOB = \angle COA + \angle COB = 90^\circ$.

Answer: $\angle AOB = 90^\circ$

1. (a) Let the ones digit be $x$ and the tens digit be $y$. The number is $10y + x$.
2. Seven times the sum of the digits: $10y + x = 7(x + y)$, so $3y = 6x$ and $y = 2x$. … (i)
3. Two more than $5$ times the product: $10y + x = 5xy + 2$. … (ii)
4. Put (i) in (ii): $20x + x = 10x^2 + 2$, so $10x^2 - 21x + 2 = 0$ and $(x - 2)(10x - 1) = 0$. The digit $x$ must be a whole number, so $x = 2$ and $y = 4$.
5. The number is $42$. (Check: $7 \times 6 = 42$ ✓ and $5 \times 8 + 2 = 42$ ✓.)
6. (b) Real and equal roots need $D = 0$: $[-(p + 1)]^2 - 4(p + 4)(1) = 0$, so $p^2 + 2p + 1 - 4p - 16 = 0$ and $p^2 - 2p - 15 = 0$.
7. $(p - 5)(p + 3) = 0$, so $p = 5$ or $p = -3$.
8. For $p = 5$: $9x^2 - 6x + 1 = 0$, so $(3x - 1)^2 = 0$ and $x = \dfrac{1}{3}, \dfrac{1}{3}$.
9. For $p = -3$: $x^2 + 2x + 1 = 0$, so $(x + 1)^2 = 0$ and $x = -1, -1$.

Answer: (a) $42$   (b) $p = 5$: roots $\dfrac{1}{3}, \dfrac{1}{3}$; $p = -3$: roots $-1, -1$

Common mistake: In (b), writing only $p = 5$. Both values of $p$ make the roots equal, so give both with their roots.