CBSE Class 10 Maths Basic 2023 Question Paper (Set 430/4/1) with Solutions

1. A polynomial with zeroes of sum $S$ and product $P$ is $x^2 - Sx + P = x^2 - (-3)x + 2 = x^2 + 3x + 2$. (Check: the zeroes are $-1$ and $-2$.)

Answer: (A) $x^2 + 3x + 2$

1. HCF $\times$ LCM $=$ the product of the two numbers $= 70 \times 40 = 2800$.

Answer: (C) 2800

1. Perimeter $= \pi r + 2r = \dfrac{22}{7} \times 7 + 14 = 22 + 14 = 36$ cm. (Do not forget the straight edge, the diameter.)

Answer: (D) 36 cm

1. $5 - 3\sqrt{5} + \sqrt{5} = 5 - 2\sqrt{5}$. A rational number minus a non-zero multiple of the irrational $\sqrt{5}$ is irrational.

Answer: (C) an irrational number

1. $p(-2) = 4 - 10 + 6 = 0$. So $-2$ is a zero of $p(x)$.

Answer: (B) 0

1. A probability lies between $0$ and $1$. $\dfrac{5}{3} > 1$, so it cannot be a probability. The others are $0.1$, $0.03$ and $\dfrac{1}{3}$.

Answer: (B) $\dfrac{5}{3}$

1. $\dfrac{a_1}{a_2} = \dfrac{1}{-3}$, $\dfrac{b_1}{b_2} = \dfrac{2}{-6} = \dfrac{1}{-3}$ and $\dfrac{c_1}{c_2} = \dfrac{5}{1}$. The first two ratios are equal and the third is different, so the lines are parallel and there is no solution.

Answer: (D) no solution

1. Similar triangles have equal corresponding angles, so $\angle B = \angle E = 83^\circ$. Then $\angle C = 180^\circ - 47^\circ - 83^\circ = 50^\circ$.

Answer: (B) $50^\circ$

1. The solution satisfies both equations. Put $x = 2$, $y = 1$ in $x + ky = 5$: $2 + k = 5$, so $k = 3$.

Answer: (C) 3

1. $\sin 90^\circ = 1$ and $\cos 0^\circ = 1$, so the value is $5(1) - 2(1) = 3$.

Answer: (C) 3

1. Arc length $= \dfrac{60}{360} \times 2\pi r = \dfrac{1}{6} \times 2 \times \dfrac{22}{7} \times 14 = \dfrac{1}{6} \times 88 = \dfrac{44}{3}$ cm.

Answer: (A) $\dfrac{44}{3}$ cm

1. $\tan\theta = \dfrac{30}{30} = 1$, so $\theta = 45^\circ$.

Answer: (B) $45^\circ$

1. Counting: $2$ appears $2$ times, $3$ appears $3$ times, $4$ appears $4$ times, $5$ appears $2$ times, $6$ and $7$ once each. The most frequent value is $4$.

Answer: (C) 4

1. There are $2$ red queens (hearts and diamonds) among $52$ cards: $P = \dfrac{2}{52} = \dfrac{1}{26}$. (A queen of any colour would be $\dfrac{1}{13}$.)

Answer: (B) $\dfrac{1}{26}$

1. The roots add to $0$, so the other root is $-2$. The roots are $2$ and $-2$, and the equation is $(x - 2)(x + 2) = x^2 - 4 = 0$.

Answer: (D) $x^2 - 4 = 0$

1. (A) $2x^2 - 4x + 2 = 4x^2 - 2x + 1$ gives $2x^2 + 2x - 1 = 0$: quadratic. (B) $2x - x^2 = x^2 + 5$ gives $2x^2 - 2x + 5 = 0$: quadratic.
2. (C) $2x^2 + 2\sqrt{6}x + 3 + x^2 = 3x^2 - 5x$ gives $(2\sqrt{6} + 5)x + 3 = 0$. The $x^2$ terms cancel, so it is linear, not quadratic.
3. (D) $x^4 + 4x^3 + 4x^2 = x^4 + 3 + 4x^3$ gives $4x^2 - 3 = 0$: quadratic.

Answer: (C) $(\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x$

1. From a point on the circle only one tangent can be drawn, the line perpendicular to the radius at that point. (From an outside point there are two; from an inside point none.)

Answer: (A) One

1. The radius to the point of contact is perpendicular to the tangent, so $OA^2 = 3^2 + 4^2 = 25$ and $OA = 5$ cm.

Answer: (B) 5 cm

1. Both statements are standard true theorems. But equal tangent lengths do not explain why a tangent is perpendicular to the radius, so (R) is not the correct explanation of (A).

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

1. (A) Roots that are reciprocals have product $1$: $\dfrac{k - 4}{4} = 1$, so $k = 8$. True.
2. (R) $D = (-1)^2 - 4(1)(1) = -3 < 0$, so the roots are not real. False.

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

1. $\sin\alpha = \dfrac{1}{2}$ gives $\alpha = 30^\circ$ and $\cos\alpha = \dfrac{\sqrt{3}}{2}$.
2. $3\cos\alpha - 4\cos^3\alpha = \dfrac{3\sqrt{3}}{2} - 4 \times \dfrac{3\sqrt{3}}{8} = \dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2} = 0$. (This is $\cos 3\alpha = \cos 90^\circ = 0$.)

Answer: $0$

1. (A) $x = \dfrac{2(4) + 3(-1)}{5} = \dfrac{5}{5} = 1$ and $y = \dfrac{2(-3) + 3(7)}{5} = \dfrac{15}{5} = 3$. The point is $(1, 3)$.
2. (B) The diagonals of a parallelogram bisect each other, so the midpoint of $AC$ equals the midpoint of $BD$: $\left(\dfrac{2 + 6}{2}, \dfrac{3 + 7}{2}\right) = \left(\dfrac{-5 + p}{2}, \dfrac{6 + 4}{2}\right)$, that is $(4, 5) = \left(\dfrac{p - 5}{2}, 5\right)$.
3. $\dfrac{p - 5}{2} = 4$ gives $p = 13$.

Answer: (A) $(1, 3)$   (B) $p = 13$

1. (A) $a = 3$, $b = -2$, $c = \dfrac{1}{3}$: $D = b^2 - 4ac = 4 - 4(3)\left(\dfrac{1}{3}\right) = 4 - 4 = 0$. Since $D = 0$, the roots are real and equal.
2. (B) $x^2 - x - 2 = (x - 2)(x + 1) = 0$, so $x = 2$ or $x = -1$.

Answer: (A) $D = 0$, real and equal roots   (B) $x = 2, -1$

1. $OT \perp PT$ (the radius is perpendicular to the tangent), so $\angle OTP = 90^\circ$.
2. In $\triangle OTP$: $\angle TOP = 180^\circ - 90^\circ - 30^\circ = 60^\circ$.
3. The line through $T$ and $O$ is straight, so $x = 180^\circ - 60^\circ = 120^\circ$.

Answer: $x = 120^\circ$

1. In $\triangle OPQ$, $AB \parallel PQ$, so by the basic proportionality theorem $\dfrac{OA}{AP} = \dfrac{OB}{BQ}$. (i)
2. In $\triangle OPR$, $AC \parallel PR$, so $\dfrac{OA}{AP} = \dfrac{OC}{CR}$. (ii)
3. From (i) and (ii): $\dfrac{OB}{BQ} = \dfrac{OC}{CR}$. In $\triangle OQR$ the line $BC$ divides $OQ$ and $OR$ in the same ratio, so $BC \parallel QR$ by the converse of the basic proportionality theorem.

Answer: Shown

1. $x^2 + 6x + 8 = (x + 4)(x + 2)$. The zeroes are $-4$ and $-2$.
2. Sum of zeroes $= -4 + (-2) = -6 = -\dfrac{6}{1} = -\dfrac{b}{a}$ ✓.
3. Product of zeroes $= (-4)(-2) = 8 = \dfrac{8}{1} = \dfrac{c}{a}$ ✓.

Answer: Zeroes $-4$ and $-2$; the relations are verified

1. LHS $= \dfrac{\sec^2 A}{\operatorname{cosec}^2 A} = \dfrac{1/\cos^2 A}{1/\sin^2 A} = \dfrac{\sin^2 A}{\cos^2 A}$, using $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \operatorname{cosec}^2 A$.
2. $\dfrac{\sin^2 A}{\cos^2 A} = \dfrac{1 - \cos^2 A}{\cos^2 A} = \dfrac{1}{\cos^2 A} - 1 = \sec^2 A - 1 =$ RHS.

Answer: Proved

1. (A) Let the fixed charge be ₹ $x$ and the charge per extra day ₹ $y$. $7$ days has $4$ extra days and $5$ days has $2$ extra days: $x + 4y = 27$ and $x + 2y = 21$.
2. Subtracting: $2y = 6$, so $y = 3$ and $x = 21 - 6 = 15$. The fixed charge is ₹ $15$ and each extra day costs ₹ $3$.
3. (B) Infinitely many solutions need $\dfrac{3}{a + b} = \dfrac{4}{2(a - b)} = \dfrac{12}{24} = \dfrac{1}{2}$.
4. So $a + b = 6$ and $2(a - b) = 8$, that is $a - b = 4$. Adding: $2a = 10$, so $a = 5$ and $b = 1$.

Answer: (A) fixed ₹ $15$, extra day ₹ $3$   (B) $a = 5$, $b = 1$

1. The sample space is $\{1, 2, 3, 4, 5, 6\}$, with $6$ equally likely outcomes.
2. (i) The only even prime is $2$: $P = \dfrac{1}{6}$.
3. (ii) Numbers greater than $4$ are $5, 6$: $P = \dfrac{2}{6} = \dfrac{1}{3}$.
4. (iii) Odd numbers are $1, 3, 5$: $P = \dfrac{3}{6} = \dfrac{1}{2}$.

Answer: (i) $\dfrac{1}{6}$   (ii) $\dfrac{1}{3}$   (iii) $\dfrac{1}{2}$

1. Area of the sector $= \dfrac{90}{360} \times \dfrac{22}{7} \times 7 \times 7 = \dfrac{1}{4} \times 154 = \dfrac{77}{2} = 38.5$ cm².
2. Area of the circle $= \dfrac{22}{7} \times 7 \times 7 = 154$ cm². The major sector is the rest: $154 - 38.5 = 115.5$ cm².

Answer: Sector $38.5$ cm²; major sector $115.5$ cm²

1. (A) **Given:** a circle with centre $O$ and tangents $PQ$, $PR$ from the external point $P$. **To prove:** $PQ = PR$. **Construction:** join $OP$, $OQ$, $OR$.
2. In $\triangle OPQ$ and $\triangle OPR$: $OP$ is common, $OQ = OR$ (radii) and $\angle OQP = \angle ORP = 90^\circ$ (radius and tangent). So $\triangle OPQ \cong \triangle OPR$ (RHS) and $PQ = PR$ by CPCT.
3. (B) Join $OA$ and $OP$. $AB$ touches the small circle at $P$, so $OP \perp AB$. The perpendicular from the centre bisects the chord: $AP = PB$.
4. In the right triangle $OPA$: $AP^2 = OA^2 - OP^2 = 25 - 9 = 16$, so $AP = 4$ cm and $AB = 8$ cm.

Answer: (A) proved   (B) $AB = 8$ cm

1. (A) Let the tower be $AB = h$, the shadow at $60^\circ$ be $BC = x$ and the longer shadow be $BD = x + 40$. In $\triangle ABC$, $\tan 60^\circ = \dfrac{h}{x}$, so $h = \sqrt{3}x$.
2. In $\triangle ABD$, $\tan 30^\circ = \dfrac{h}{x + 40}$, so $x + 40 = \sqrt{3}h = 3x$. Hence $2x = 40$, $x = 20$ m and $h = 20\sqrt{3} \approx 34.6$ m.
3. (B) Let the building be $PB = 7$ m, the distance to the tower foot $PQ = x$ and the tower height above the level of $B$ be $h$. In the right triangle at the foot, $\tan 45^\circ = \dfrac{7}{x}$, so $x = 7$ m.
4. $\tan 60^\circ = \dfrac{h}{x}$ gives $h = 7\sqrt{3}$ m. The tower stands $7$ m (up to the level of $B$) plus $h$ above the ground: $7 + 7\sqrt{3} = 7(1 + \sqrt{3}) \approx 19.1$ m.

Answer: (A) $20\sqrt{3} \approx 34.6$ m   (B) $7(1 + \sqrt{3}) \approx 19.1$ m

1. (A) $a_1 = 5 + 6 = 11$ and $a_2 = 5 + 12 = 17$, so $d = 6$.
2. $S_{25} = \dfrac{25}{2}[2(11) + 24 \times 6] = \dfrac{25}{2}(22 + 144) = \dfrac{25}{2} \times 166 = 2075$.
3. $\dfrac{a_{20}}{a_{45}} = \dfrac{5 + 120}{5 + 270} = \dfrac{125}{275} = \dfrac{5}{11}$, a ratio of $5 : 11$.
4. (B) $a_1 = S_1 = 3 + 5 = 8$ and $S_2 = 12 + 10 = 22$, so $a_2 = 22 - 8 = 14$ and $d = 14 - 8 = 6$.
5. $a_k = 8 + (k - 1) \times 6 = 164$ gives $(k - 1) \times 6 = 156$, so $k - 1 = 26$ and $k = 27$.

Answer: (A) $S_{25} = 2075$, ratio $5 : 11$   (B) $k = 27$

1. Cumulative frequencies: $5, 14, 31, 59, 83, 93, 100$. $N = 100$ and $\dfrac{N}{2} = 50$.
2. The first cumulative frequency above $50$ is $59$, so the median class is $160 - 170$ with $l = 160$, $cf = 31$, $f = 28$, $h = 10$.
3. Median $= 160 + \dfrac{50 - 31}{28} \times 10 = 160 + \dfrac{190}{28} \approx 160 + 6.8 = 166.8$ units.

Answer: Median $\approx 166.8$ units

1. $h = 7$ m and $r = \dfrac{7}{2}$ m. The two hemispheres together make a sphere.
2. Total surface area $= 2\pi r h + 4\pi r^2 = 2\pi r(h + 2r) = 2 \times \dfrac{22}{7} \times \dfrac{7}{2} \times (7 + 7) = 22 \times 14 = 308$ m².
3. Volume $= \pi r^2 h + \dfrac{4}{3}\pi r^3 = \dfrac{22}{7} \times \dfrac{49}{4} \times 7 + \dfrac{4}{3} \times \dfrac{22}{7} \times \dfrac{343}{8} = \dfrac{539}{2} + \dfrac{539}{3} = \dfrac{2695}{6} \approx 449.17$ m³.
4. Ratio of the cylinder to one hemisphere $= \dfrac{\pi r^2 h}{\frac{2}{3}\pi r^3} = \dfrac{3h}{2r} = \dfrac{21}{7} = 3 : 1$.

Answer: Surface area $308$ m²; volume $\dfrac{2695}{6} \approx 449.17$ m³; ratio $3 : 1$

1. (i) The diagonals of a rectangle bisect each other, so the point is the midpoint of $AC$: $\left(\dfrac{1 + 7}{2}, \dfrac{1 + 5}{2}\right) = (4, 3)$.
2. (ii) $AC = \sqrt{(7 - 1)^2 + (5 - 1)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$.
3. (iii)(a) $AB = \sqrt{(7 - 1)^2 + 0} = 6$ and $BC = \sqrt{0 + (5 - 1)^2} = 4$. Area $= 6 \times 4 = 24$ square units.
4. (iii)(b) $\dfrac{AB}{AC} = \dfrac{6}{2\sqrt{13}} = \dfrac{3}{\sqrt{13}}$, a ratio of $3 : \sqrt{13}$.

Answer: (i) $(4, 3)$   (ii) $2\sqrt{13}$   (iii) $24$ square units (OR) $3 : \sqrt{13}$

1. (i) The fruits are shared equally, so the number of guests divides both $36$ and $60$. $36 = 2^2 \times 3^2$ and $60 = 2^2 \times 3 \times 5$, so HCF $= 12$. At most $12$ guests.
2. (ii) Each guest gets $36 \div 12 = 3$ apples and $60 \div 12 = 5$ bananas.
3. (iii)(a) $42 = 2 \times 3 \times 7$, so HCF$(36, 60, 42) = 2 \times 3 = 6$. At most $6$ guests.
4. (iii)(b) A dozen bananas cost ₹ $60$, so $60$ bananas $= 5$ dozen $= 5 \times 60 = 300$. Apples: $36 \times 15 = 540$. Mangoes: $42 \times 20 = 840$. Total $= 300 + 540 + 840 = ₹\,1680$.

Answer: (i) $12$   (ii) $3$ apples, $5$ bananas   (iii) $6$ guests (OR) ₹ $1680$

1. (i) In Figure A the angles match ($85^\circ, 70^\circ, 105^\circ, 100^\circ$) and every side of A(ii) is twice the matching side of A(i) ($1.2 \to 2.4$, $2 \to 4$, $2.4 \to 4.8$, $2.5 \to 5$), so the figures are similar. In Figure C the triangles are identical, so they are similar as well. The CBSE key gives **Figure A and Figure C**. (Strictly, any two squares are similar, so B(iii) and B(iv) also qualify, but the key does not list B.)
2. (ii) Congruent figures are identical in size and shape. Only Figure C, with equal sides $2, 3, 4.5$ in both triangles, is congruent (SSS).
3. (iii)(a) If two triangles are congruent, their corresponding angles are equal, so by AA they are similar. The converse fails: similar triangles have sides in the same ratio, and that ratio need not be $1$. For example, a triangle with sides $3, 4, 5$ and one with sides $6, 8, 10$ are similar but not congruent.
4. (iii)(b) One pair of corresponding sides must be equal (then the ratio of similarity is $1$ and the triangles are congruent).

Answer: (i) Figures A and C (as per the key)   (ii) Figure C   (iii) proved (OR) one pair of corresponding sides equal