CBSE Class 10 Maths Basic 2025 Question Paper (Set 430/1/1) with Solutions

1. For any two positive integers, $\text{HCF} \times \text{LCM} = a \times b$.
2. With $\text{HCF} = 1$: $1 \times \text{LCM} = ab$, so the LCM is $ab$.

Answer: (D) $ab$

Common mistake: Choosing $a$ or $b$. The LCM equals one of them only when the other divides it, which is not the case for numbers with HCF $1$ (such as $4$ and $9$).

1. $\sqrt{2}$ is irrational.
2. Suppose $3 + \sqrt{2}$ were a rational number $r$. Then $\sqrt{2} = r - 3$ would be rational, which is false.
3. So the sum of a rational number and an irrational number is irrational.

Answer: (B) an irrational number

1. Here $a = 1$, $b = -3$, $c = -2$.
2. Discriminant $D = b^2 - 4ac = (-3)^2 - 4(1)(-2) = 9 + 8 = 17$.

Answer: (B) $17$

Common mistake: Giving $\sqrt{17}$. That is a part of the roots' formula. The discriminant itself is $b^2 - 4ac$.

1. Multiply both sides by $x$: $x^2 + 1 = 3x$.
2. Bring everything to one side: $x^2 - 3x + 1 = 0$. So $a = 1$, $b = -3$, $c = 1$.
3. $a - b + c = 1 - (-3) + 1 = 5$.

Answer: (A) $5$

Common mistake: Forgetting that $b = -3$, which gives $1 - 3 + 1 = -1$.

1. In $(x, y)$ the abscissa is $x$ and the ordinate is $y$. Here abscissa $= 3$ and ordinate $= -5$.
2. Abscissa $-$ ordinate $= 3 - (-5) = 8$.

Answer: (D) $8$

1. The mid-point is the point exactly halfway, so the two parts are equal.
2. The ratio is $1 : 1$.

Answer: (C) $1 : 1$

1. The standard similarity criteria are AAA (or AA), SSS and SAS. These three are valid.
2. RHS is usually taught as a congruence rule, but for right-angled triangles it also works for similarity: if the hypotenuse and one other side of one right triangle are in the same ratio as those of another, the triangles are similar.
3. So every option can be used to show similarity, and none of them is "not a criterion". CBSE’s own marking scheme for this paper says the same: none of the given options is correct.

Answer: None of the given options is correct (as stated in CBSE’s answer key)

Common mistake: Spending time trying to force one letter. If you meet this in the exam, write your reasoning in a line next to your choice.

1. Compare the sides of $\triangle ABC$ with those of $\triangle PQR$: $\dfrac{AB}{QR} = \dfrac{3.8}{7.6} = \dfrac{1}{2}$, $\dfrac{BC}{PQ} = \dfrac{6}{12} = \dfrac{1}{2}$, $\dfrac{AC}{PR} = \dfrac{3\sqrt{3}}{6\sqrt{3}} = \dfrac{1}{2}$.
2. All three ratios are equal, so $\triangle ABC \sim \triangle RQP$ by SSS similarity.
3. Match angles by the sides opposite them. $QR$ (opposite $P$) corresponds to $AB$ (opposite $C$), so $\angle P = \angle C$.
4. $\angle C = 180^\circ - 80^\circ - 60^\circ = 40^\circ$, so $\angle P = 40^\circ$.

Answer: (C) $\angle P = 40^\circ$

Common mistake: Pairing $P$ with $A$ because they are the first letters. In similar triangles the matching vertex is the one opposite the matching side.

1. The radius $OA$ is perpendicular to the tangent $PA$, so $\triangle OAP$ has a right angle at $A$.
2. $\angle OPA = 30^\circ$ and the hypotenuse is $OP = 10$ cm. $AP$ is the side next to the $30^\circ$ angle.
3. $AP = OP \cos 30^\circ = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}$ cm.

Answer: (D) $5\sqrt{3}$ cm

Common mistake: Using $\sin 30^\circ$ instead of $\cos 30^\circ$, which gives $5$ (that is $OA$, not $AP$).

1. (A) $\tan 45^\circ = 1$ and $\cot 45^\circ = 1$. True.
2. (B) $\sin 90^\circ = 1$ and $\tan 45^\circ = 1$. True.
3. (C) $\sin 30^\circ = \dfrac{1}{2}$ but $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$. False.
4. (D) $\sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}}$. True.

Answer: (C) $\sin 30^\circ = \cos 30^\circ$

Common mistake: Not noticing the question asks for the false statement, and picking a true one.

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

Answer: (D) $-1$

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

1. The angle of depression is measured at the observer’s eye, from the horizontal line down to the line of sight.
2. In the figure that is the angle between the dashed horizontal line and the line of sight, which is $z$.
3. ($x$ is equal to $z$ (alternate angles) but it sits at the object, so it is the angle of elevation of the observer from the object.)

Answer: (C) z

1. The shaded region is the triangle together with the segment under the chord, so the chord is inside the region and is not part of its boundary.
2. Together they form a sector. Its boundary is the two radii and the arc: $r + r + l$.
3. Perimeter $= l + 2r$.

Answer: (C) $l + 2r$

Common mistake: Adding the chord $a$. A line inside a region is not part of its perimeter.

1. A quadrant is $\dfrac{1}{4}$ of the circle, so its area is $\dfrac{1}{4}\pi r^2$.
2. Ratio $= \dfrac{1}{4}\pi r^2 : \pi r^2 = 1 : 4$.

Answer: (C) $1 : 4$

1. Curved and total surface area are equal only when the solid has no flat face.
2. A cube and a cuboid have flat faces. A hemisphere has a flat circular top, so its total surface area ($3\pi r^2$) is more than its curved area ($2\pi r^2$).
3. A sphere has one curved surface and nothing else: both are $4\pi r^2$.

Answer: (D) Sphere

1. Total frequency $N = 2 + 3 + 7 + 6 + 6 + 6 = 30$, so $\dfrac{N}{2} = 15$.
2. Cumulative frequencies: $2,\ 5,\ 12,\ 18,\ 24,\ 30$.
3. The first cumulative frequency that is at least $15$ is $18$, so the median class is $55 - 70$.
4. Class mark $= \dfrac{55 + 70}{2} = 62.5$.

Answer: (D) $62.5$

Common mistake: Taking the class with the highest frequency as the median class. That is the modal class, not the median class.

1. The modal class is the class with the highest frequency.
2. The highest frequency is $10$, for the class $4000 - 5000$. Its lower limit is $4000$.

Answer: (B) $4000$

1. A sure event is one that happens every time, with probability 1.
2. A die shows 1, 2, 3, 4, 5 or 6, and each of these is a natural number less than 7.
3. “Between 1 and 6” leaves out 1 and 6, odd numbers leave out 2, 4 and 6, and even numbers leave out 1, 3 and 5. So only the last option always happens.

Answer: (D) Getting a natural number < 7

1. The reason is true: the HCF divides both $a$ and $b$.
2. Because the HCF divides $a$, and $a$ divides the LCM (the LCM is a multiple of $a$), the HCF divides the LCM. So the assertion is true.
3. The reason is exactly the step that proves the assertion, so R explains A.

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

1. For the given system: $\dfrac{a_1}{a_2} = \dfrac{4}{2} = 2$, $\dfrac{b_1}{b_2} = \dfrac{p}{2}$, $\dfrac{c_1}{c_2} = \dfrac{8}{2} = 4$.
2. If $p = 4$: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = 2$ but $\dfrac{c_1}{c_2} = 4$. The lines are parallel, so there is no solution and the system is inconsistent. The assertion is false.
3. If $p \neq 4$, then $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, so there is exactly one solution: consistent. So the system is consistent for every $p$ except $4$.
4. The reason states the standard condition for infinitely many solutions, and it is true.

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

Common mistake: Reading "consistent" as "infinitely many solutions only". A system with a unique solution is consistent too.

1. Clear the fractions. Multiply the first equation by $6$: $3x + 4y = -6$. … (i)
2. Multiply the second by $3$: $3x - y = 9$. … (ii)
3. Subtract (ii) from (i): $5y = -15$, so $y = -3$.
4. Put $y = -3$ in (ii): $3x + 3 = 9$, so $x = 2$.
5. Check in the first equation: $\dfrac{2}{2} + \dfrac{2(-3)}{3} = 1 - 2 = -1$ ✓.

Answer: $x = 2,\ y = -3$

Common mistake: Multiplying only one side when clearing fractions.

1. (a) $PQ \parallel RS$ with transversal $QR$, so $\angle PQO = \angle SRO$ (alternate interior angles).
2. $\angle POQ = \angle SOR$ (vertically opposite angles).
3. Two pairs of equal angles, so $\triangle POQ \sim \triangle SOR$ by AA similarity.
4. (b) $\angle ROQ$ is an exterior angle of $\triangle OSR$ (side $SO$ is extended to $Q$). An exterior angle equals the sum of the two opposite interior angles: $\angle ROQ = \angle OSR + \angle ORS$.
5. $125^\circ = \angle OSR + 70^\circ$, so $\angle OSR = 55^\circ$.
6. $\triangle OSR \sim \triangle OQP$ (with $S \leftrightarrow Q$), so corresponding angles are equal: $\angle OQP = \angle OSR = 55^\circ$.

Answer: (a) $\triangle POQ \sim \triangle SOR$ shown by AA   (b) $\angle OSR = 55^\circ,\ \angle OQP = 55^\circ$

Common mistake: Using $125^\circ$ as an angle inside the triangle. It is outside: the straight line $SOQ$ makes the inside angle $\angle SOR = 55^\circ$.

1. Let the chord $AB$ of the larger circle touch the smaller circle at $C$. The radius $OC = 6$ cm is perpendicular to the tangent $AB$, and it bisects the chord.
2. In right triangle $OCB$: $OB = 10$ (radius of the larger circle), $OC = 6$. So $CB^2 = 10^2 - 6^2 = 100 - 36 = 64$ and $CB = 8$ cm.
3. The chord $AB = 2 \times 8 = 16$ cm.

Answer: $16$ cm

Common mistake: Stopping at $8$ cm, which is only half the chord.

1. (a) $\tan(A + B) = 1 = \tan 45^\circ$, so $A + B = 45^\circ$. … (i)
2. $\tan(A - B) = \dfrac{1}{\sqrt{3}} = \tan 30^\circ$, so $A - B = 30^\circ$. … (ii)
3. Add (i) and (ii): $2A = 75^\circ$, so $A = 37.5^\circ$. Then $B = 45^\circ - 37.5^\circ = 7.5^\circ$.
4. (b) Take a right triangle $ABC$ with the right angle at $B$ and $AB = BC = x$ (an isosceles right triangle).
5. Equal sides have equal angles opposite them, so $\angle A = \angle C$. The angles add to $90^\circ$, so $\angle A = \angle C = 45^\circ$.
6. $\tan 45^\circ = \tan A = \dfrac{BC}{AB} = \dfrac{x}{x} = 1$.

Answer: (a) $A = 37.5^\circ,\ B = 7.5^\circ$   (b) $\tan 45^\circ = 1$

1. Radius $r = 10$ cm. In $\triangle OAB$, $OA = OB = 10$ and $\angle AOB = 60^\circ$, so the other two angles are $60^\circ$ too: the triangle is equilateral.
2. Area of the sector $= \dfrac{60}{360} \times \pi r^2 = \dfrac{1}{6} \times 3.14 \times 100 = 52.33$ cm².
3. Area of the equilateral triangle $= \dfrac{\sqrt{3}}{4} r^2 = \dfrac{1.73}{4} \times 100 = 43.25$ cm².
4. Area of the minor segment $=$ sector $-$ triangle $= 52.33 - 43.25 = 9.08$ cm² (exactly $\dfrac{109}{12}$ cm²).

Answer: $\approx 9.08$ cm²

Common mistake: Giving the sector area. The segment is what is left after removing the triangle.

1. (a) Suppose $\sqrt{3}$ is rational. Then $\sqrt{3} = \dfrac{p}{q}$ with integers $p, q$ ($q \neq 0$) that have no common factor.
2. Squaring: $3q^2 = p^2$. So $3$ divides $p^2$, and since $3$ is prime, $3$ divides $p$. Write $p = 3m$.
3. Then $3q^2 = 9m^2$, so $q^2 = 3m^2$. So $3$ divides $q^2$, and hence $3$ divides $q$.
4. Now $3$ divides both $p$ and $q$, which contradicts "no common factor". So $\sqrt{3}$ is irrational.
5. (b) Work upwards from the bottom. $35 = 5 \times b$, so $b = 7$. Then $70 = 2 \times 35$ ✓.
6. $210 = a \times 70$, so $a = 3$. Then $y = 2 \times 210 = 420$ and $x = 2 \times 420 = 840$.
7. The prime factors of $x$: $840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7$, that is $x = 2^3 \times 3 \times 5 \times 7$.

Answer: (a) $\sqrt{3}$ is irrational   (b) $b = 7,\ a = 3,\ y = 420,\ x = 840 = 2^3 \times 3 \times 5 \times 7$

Common mistake: In the proof, forgetting to say why 3 dividing p squared means 3 divides p (because 3 is prime).

1. A quadratic polynomial with zeroes whose sum is $S$ and product is $P$ is $x^2 - Sx + P$.
2. Here $S = 0$ and $P = -9$, so the polynomial is $x^2 - 0x - 9 = x^2 - 9$.
3. Zeroes: $x^2 - 9 = (x + 3)(x - 3) = 0$, so $x = -3$ or $x = 3$.
4. Check: sum $= -3 + 3 = 0$ ✓, product $= (-3)(3) = -9$ ✓.

Answer: Polynomial $x^2 - 9$; zeroes $-3$ and $3$

Common mistake: Writing $x^2 + Sx + P$. The middle term is $-S$.

1. (a) Line 1, $x + 3y = 6$: points $(0, 2)$, $(3, 1)$, $(6, 0)$.
2. Line 2, $2x - 3y = 12$: points $(0, -4)$, $(3, -2)$, $(6, 0)$.
3. Plot both lines on the same axes. They meet at the single point $(6, 0)$, so $x = 6$ and $y = 0$.
4. Check: $6 + 0 = 6$ ✓ and $12 - 0 = 12$ ✓.
5. (b) Complementary angles add to $90^\circ$: $x + y = 90$. … (i)
6. $x : y = 1 : 2$ means $y = 2x$, i.e. $2x - y = 0$. … (ii)
7. Add (i) and (ii): $3x = 90$, so $x = 30$ and $y = 60$.

Answer: (a) $x = 6,\ y = 0$   (b) $x + y = 90,\ y = 2x$; $x = 30^\circ,\ y = 60^\circ$

Common mistake: Plotting with a wrong scale or too few points. Two correct points define a line, but a third point is a free check.

1. Let rectangle $ABCD$ circumscribe a circle, touching the sides $AB, BC, CD, DA$ at $P, Q, R, S$.
2. Tangents from an outside point are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$.
3. Add the four equalities: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$, that is $AB + CD = AD + BC$.
4. In a rectangle opposite sides are equal: $AB = CD$ and $AD = BC$. So $2AB = 2AD$, giving $AB = AD$.
5. A rectangle whose adjacent sides are equal is a square.

Answer: The circumscribed rectangle is a square

Common mistake: Skipping the labelled figure with the four points of contact. It is worth marks.

1. Left side: $1 + \cot^2 A = \mathrm{cosec}^2 A$ and $1 + \tan^2 A = \sec^2 A$, so $\text{LHS} = \dfrac{\mathrm{cosec}^2 A}{\sec^2 A} = \dfrac{\cos^2 A}{\sin^2 A} = \cot^2 A$.
2. Right side: $1 - \cot A = \dfrac{\sin A - \cos A}{\sin A}$ and $1 - \tan A = \dfrac{\cos A - \sin A}{\cos A}$.
3. $\dfrac{1 - \cot A}{1 - \tan A} = \dfrac{\sin A - \cos A}{\sin A} \times \dfrac{\cos A}{\cos A - \sin A} = -\dfrac{\cos A}{\sin A} = -\cot A$.
4. Squaring: $\left(-\cot A\right)^2 = \cot^2 A$.
5. So $\text{LHS} = \cot^2 A = \text{RHS}$.

Answer: Proved

Common mistake: Working on both sides at once. Simplify the left side and the right side separately, and show they meet at the same expression.

1. Defective pens in the first lot: $200 - 180 = 20$.
2. The customer does not buy a defective pen. $P(\text{not buy}) = \dfrac{20}{200} = \dfrac{1}{10}$.
3. After mixing: total pens $= 200 + 100 = 300$ and good pens $= 180 + 80 = 260$.
4. The customer buys a good pen. $P(\text{buy}) = \dfrac{260}{300} = \dfrac{13}{15}$.

Answer: $\dfrac{1}{10}$ and $\dfrac{13}{15}$

Common mistake: Answering the first part with the probability of a good pen. The question asks for the chance the customer will NOT buy.

1. (a) Let the greater number be $x$ and the smaller be $y$. Then $x^2 - y^2 = 180$ and $y^2 = 8x$.
2. Put $y^2 = 8x$ in the first: $x^2 - 8x = 180$, so $x^2 - 8x - 180 = 0$.
3. Factorise: $(x - 18)(x + 10) = 0$, so $x = 18$ or $x = -10$.
4. The numbers are positive, so $x = 18$. Then $y^2 = 8 \times 18 = 144$ and $y = 12$.
5. Check: $18^2 - 12^2 = 324 - 144 = 180$ ✓.
6. (b) Equal roots need the discriminant to be zero: $k^2 - 4(2)(3) = 0$, so $k^2 = 24$ and $k = \pm 2\sqrt{6}$.
7. For $k = 2\sqrt{6}$: $2x^2 + 2\sqrt{6}x + 3 = 0$. The equal roots are $x = \dfrac{-b}{2a} = \dfrac{-2\sqrt{6}}{4} = -\dfrac{\sqrt{6}}{2}$.
8. For $k = -2\sqrt{6}$: $2x^2 - 2\sqrt{6}x + 3 = 0$. The equal roots are $x = \dfrac{2\sqrt{6}}{4} = \dfrac{\sqrt{6}}{2}$.

Answer: (a) $18$ and $12$   (b) $k = \pm 2\sqrt{6}$; roots $-\dfrac{\sqrt{6}}{2}$ (twice) for $k = 2\sqrt{6}$ and $\dfrac{\sqrt{6}}{2}$ (twice) for $k = -2\sqrt{6}$

Common mistake: Keeping $x = -10$. The question says the numbers are positive, so say clearly why it is rejected.

1. Basic Proportionality Theorem: if a line is drawn parallel to one side of a triangle to meet the other two sides at distinct points, it divides those two sides in the same ratio.
2. Given: $\dfrac{AO}{BO} = \dfrac{CO}{DO}$. To prove: $AB \parallel CD$.
3. Construction: through $O$ draw $OE \parallel AB$, meeting $AD$ at $E$.
4. In $\triangle DAB$, $OE \parallel AB$, so by the theorem $\dfrac{DE}{EA} = \dfrac{DO}{OB}$. … (i)
5. From the given condition, cross-multiplying: $AO \times DO = CO \times BO$, so $\dfrac{DO}{BO} = \dfrac{CO}{AO}$. … (ii)
6. From (i) and (ii): $\dfrac{DE}{EA} = \dfrac{CO}{AO}$.
7. In $\triangle ADC$, this says $E$ and $O$ divide $AD$ and $AC$ in the same ratio, so by the converse of the theorem $OE \parallel DC$.
8. Now $OE \parallel AB$ and $OE \parallel DC$, so $AB \parallel DC$. A quadrilateral with one pair of parallel sides is a trapezium. Hence $ABCD$ is a trapezium.

Answer: ABCD is a trapezium

Common mistake: Writing the theorem but not naming the triangle each time you use it, or using the converse without saying so.

1. (a) Radius $r = 5$ cm. The height of the cone is the diameter: $h = 2r = 10$ cm. Use $\pi = \dfrac{22}{7}$.
2. Volume of the hemisphere $= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3} \times \dfrac{22}{7} \times 125 = \dfrac{5500}{21}$ cm³.
3. Volume of the cone $= \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 25 \times 10 = \dfrac{5500}{21}$ cm³.
4. Volume of the toy $= \dfrac{5500}{21} + \dfrac{5500}{21} = \dfrac{11000}{21} \approx 523.81$ cm³.
5. (b) For the hemisphere to lie fully on the cube, the cube's face must hold the circular base. The smallest edge equals the diameter: $a = 2 \times 3.5 = 7$ cm.
6. The solid's surface is the five exposed faces plus the top face with the circle removed, plus the curved surface of the hemisphere. So the total surface area $= 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$.
7. $= 6 \times 49 + \dfrac{22}{7} \times 3.5 \times 3.5 = 294 + 38.5 = 332.5$ cm².

Answer: (a) $\dfrac{11000}{21} \approx 523.81$ cm³   (b) edge $7$ cm, total surface area $332.5$ cm²

Common mistake: In (b), using $6a^2 + 2\pi r^2$ and forgetting that the circle covered by the hemisphere is no longer exposed.

1. Class marks $x_i$: $10, 30, 50, 70, 90, 110$. Frequencies $f_i$: $10, 35, 50, 60, 30, 15$, with $\sum f_i = 200$.
2. Products $f_i x_i$: $100,\ 1050,\ 2500,\ 4200,\ 2700,\ 1650$. Their sum is $\sum f_i x_i = 12200$.
3. Mean $= \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{12200}{200} = 61$.
4. (Step-deviation check with $a = 50$, $h = 20$: $u_i = -2, -1, 0, 1, 2, 3$ and $\sum f_i u_i = -20 - 35 + 0 + 60 + 60 + 45 = 110$, so mean $= 50 + \dfrac{110}{200} \times 20 = 61$ ✓.)

Answer: Mean lifetime $= 61$ hours

Common mistake: Using the class limits instead of the class marks (midpoints) in the products.

1. (i) The building is $15$ m high and the ladder makes $60^\circ$ with the ground. The ladder $L$ is the hypotenuse: $\sin 60^\circ = \dfrac{15}{L}$, so $L = \dfrac{15}{\sqrt{3}/2} = \dfrac{30}{\sqrt{3}} = 10\sqrt{3}$ m.
2. (ii) Distance $x$ from the building: $\tan 60^\circ = \dfrac{15}{x}$, so $x = \dfrac{15}{\sqrt{3}} = 5\sqrt{3}$ m.
3. (iii)(a) Diagram: a vertical line for the building (height $15$ m) on the left, the ground as a horizontal line of width $y$ (the road) to the base of the opposite wall on the right, and the ladder from that point up to the roof of the building, making $30^\circ$ with the ground. This forms a right triangle with height $15$, base $y$ and the ladder as hypotenuse.
4. $\tan 30^\circ = \dfrac{15}{y}$, so $y = 15 \times \sqrt{3} = 15\sqrt{3}$ m.
5. (iii)(b) In the same triangle: $\sin 30^\circ = \dfrac{15}{l}$, so $\dfrac{1}{2} = \dfrac{15}{l}$ and $l = 30$ m.

Answer: (i) $10\sqrt{3}$ m   (ii) $5\sqrt{3}$ m   (iii)(a) $15\sqrt{3}$ m   (iii)(b) $30$ m

Common mistake: Using $\tan$ for the ladder length. The ladder is the hypotenuse, so use $\sin$ with the height.