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

1. $288 = 2 \times 144 = 2 \times 2^4 \times 3^2 = 2^5 \times 3^2$. (Check: $32 \times 9 = 288$ ✓.)

Answer: (C) $2^5 \times 3^2$

1. $\cos\theta = \dfrac{1}{2}$, so $\theta = 60^\circ$.

Answer: (B) $60^\circ$

1. There are $26$ red cards out of $52$: $P = \dfrac{26}{52} = \dfrac{1}{2}$.

Answer: (D) $\dfrac{1}{2}$

1. $D = (-5)^2 - 4(2)(-3) = 25 + 24 = 49$.

Answer: (B) 49

1. $d = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$ units.

Answer: (D) $3\sqrt{2}$ units

1. $a_7 = 28 + 6(-4) = 4$.

Answer: (B) 4

1. The zeroes are where the graph meets the $x$-axis. This curve lies above the $x$-axis and never touches it, so there are $0$ zeroes.

Answer: (A) 0

1. The perimeters of similar triangles are in the same ratio as the sides: $4 : 7$. (The areas would be in the ratio $16 : 49$.)

Answer: (A) 4 : 7

1. $AB \parallel CD$, so $\triangle OAB \sim \triangle ODC$ (vertically opposite angles and alternate angles).
2. $\dfrac{OB}{OC} = \dfrac{AB}{CD}$: $\dfrac{3}{OC} = \dfrac{5}{2}$, so $OC = \dfrac{6}{5}$ cm.

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

1. Sum $= -\dfrac{b}{a} = -5$ and product $= \dfrac{c}{a} = 6$. (The zeroes are $-2$ and $-3$.)

Answer: (B) $-5, 6$

1. Every number on a die ($1$ to $6$) is less than $7$, so it is a sure event: $P = 1$.

Answer: (B) 1

1. $\tan\theta = \dfrac{100}{100\sqrt{3}} = \dfrac{1}{\sqrt{3}}$, so $\theta = 30^\circ$.

Answer: (D) $30^\circ$

1. $V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi r^2 (3r) = \pi r^3$.

Answer: (D) $\pi r^3$

1. Two parallel tangents touch the circle at the two ends of a diameter, so the distance between them is the diameter, $7$ cm.

Answer: (A) 7 cm

1. $\dfrac{AB}{PQ} = \dfrac{2.2}{4.4} = \dfrac{1}{2}$ and $\dfrac{BC}{QR} = \dfrac{3.5}{7} = \dfrac{1}{2}$: two sides in proportion, with the included angles $\angle B = \angle Q = 50^\circ$ equal. That is SAS similarity.

Answer: (C) SAS (Side – Angle – Side) Similarity

1. $x + y = 180$ and $x - y = 18$ give $x = 99$ and $y = 81$. The larger angle is $99^\circ$.

Answer: (B) $99^\circ$

1. Tangents from a point to a circle are equal: $BP = BR = 6$, $AR = AQ = 5$ and $CP = CQ = 4$.
2. The perimeter is $2(6 + 5 + 4) = 30$ cm.

Answer: (A) 30 cm

1. $OC \perp BC$, so $BC = \sqrt{15^2 - 9^2} = \sqrt{144} = 12$ cm. $BD = BC = 12$ cm (equal tangents), so $BC + BD = 24$ cm.

Answer: (C) 24 cm

1. (A) This is the standard tangent–radius theorem. True. (R) Equal tangent lengths from an external point. True.
2. But (R) is a different theorem and does not explain (A); the perpendicularity comes from the radius being the shortest distance from the centre to the tangent line.

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

1. (R) Parallel lines (no solution) need equal first two ratios and a different third ratio. True.
2. (A) $\dfrac{3}{15} = \dfrac{1}{5}$, $\dfrac{5}{25} = \dfrac{1}{5}$ and $\dfrac{-4}{-25} = \dfrac{4}{25} \neq \dfrac{1}{5}$. So the lines are parallel and the system is inconsistent, exactly by the rule in (R). True.

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

1. (a) $x = \dfrac{2(-3) + 3(7)}{5} = \dfrac{15}{5} = 3$ and $y = \dfrac{2(4) + 3(-1)}{5} = \dfrac{5}{5} = 1$. The point is $(3, 1)$.
2. (b) $AB^2 = 100$: $(11 - 3)^2 + (y + 1)^2 = 100$, so $(y + 1)^2 = 36$ and $y + 1 = \pm 6$, giving $y = 5$ or $y = -7$.

Answer: (a) $(3, 1)$   (b) $y = 5$ or $-7$

1. $(\sqrt{3})^2 - 2(2)^2 - 2\left(\dfrac{1}{\sqrt{3}}\right)^2 = 3 - 8 - \dfrac{2}{3} = -\dfrac{17}{3}$.

Answer: $-\dfrac{17}{3}$

1. $92 = 2^2 \times 23$ and $510 = 2 \times 3 \times 5 \times 17$.
2. HCF $= 2$ and LCM $= 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$. (Check: $2 \times 23460 = 46920 = 92 \times 510$ ✓.)

Answer: HCF $= 2$, LCM $= 23460$

1. (a) From the first, $y = 6 - x$. Then $2x - 18 + 3x = 4$ gives $5x = 22$, so $x = \dfrac{22}{5}$ and $y = \dfrac{8}{5}$.
2. (b) $\dfrac{a_1}{a_2} = \dfrac{5}{-10} = -\dfrac{1}{2}$, $\dfrac{b_1}{b_2} = \dfrac{-3}{6} = -\dfrac{1}{2}$ and $\dfrac{c_1}{c_2} = \dfrac{-11}{-22}$ (writing both as $ax + by + c = 0$) $= \dfrac{1}{2}$. The first two are equal and the third differs: the lines are parallel and the pair is inconsistent.

Answer: (a) $x = \dfrac{22}{5}$, $y = \dfrac{8}{5}$   (b) inconsistent

1. In $\triangle ABC$ and $\triangle AMP$: $\angle ABC = \angle AMP = 90^\circ$ (given) and $\angle BAC = \angle MAP$ (common angle).
2. By AA similarity, $\triangle ABC \sim \triangle AMP$.

Answer: Proved

1. (a) LHS $= \dfrac{1}{\cos\theta}(1 - \sin\theta)\left(\dfrac{1}{\cos\theta} + \dfrac{\sin\theta}{\cos\theta}\right) = \dfrac{(1 - \sin\theta)(1 + \sin\theta)}{\cos^2\theta} = \dfrac{1 - \sin^2\theta}{\cos^2\theta} = 1$.
2. (b) LHS $= \dfrac{1 + \frac{1}{\cos\theta}}{\frac{1}{\cos\theta}} = \cos\theta + 1$. Multiply and divide by $(1 - \cos\theta)$: $\dfrac{(1 + \cos\theta)(1 - \cos\theta)}{1 - \cos\theta} = \dfrac{1 - \cos^2\theta}{1 - \cos\theta} = \dfrac{\sin^2\theta}{1 - \cos\theta} =$ RHS.

Answer: Proved (both parts)

1. $AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{34}$, $BC = \sqrt{(4 + 1)^2 + (2 + 1)^2} = \sqrt{34}$, $CD = \sqrt{(-4 + 1)^2 + (4 + 1)^2} = \sqrt{34}$ and $DA = \sqrt{(-4 - 1)^2 + (4 - 7)^2} = \sqrt{34}$. All four sides are equal.
2. Diagonals: $AC = \sqrt{(1 + 1)^2 + (7 + 1)^2} = \sqrt{68}$ and $BD = \sqrt{(4 + 4)^2 + (2 - 4)^2} = \sqrt{68}$. They are equal.
3. A quadrilateral with four equal sides and equal diagonals is a square.

Answer: Shown

1. **Given:** a circle with centre $O$ and tangents $PQ$ and $PR$ 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).
3. So $\triangle OPQ \cong \triangle OPR$ (RHS) and $PQ = PR$ by CPCT.

Answer: Proved

1. $\alpha + \beta = -3$ and $\alpha\beta = 2$.
2. New sum $= (\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = -1$. New product $= (\alpha + 1)(\beta + 1) = \alpha\beta + (\alpha + \beta) + 1 = 2 - 3 + 1 = 0$.
3. The polynomial is $x^2 + x$ (or $k(x^2 + x)$). (Check: the zeroes of $x^2 + 3x + 2$ are $-1$ and $-2$, so the new zeroes are $0$ and $-1$ ✓.)

Answer: $x^2 + x$

1. Suppose $3 + 7\sqrt{2} = \dfrac{p}{q}$ with integers $p, q$ and $q \neq 0$. Then $\sqrt{2} = \dfrac{p - 3q}{7q}$, a rational number.
2. This contradicts the given fact that $\sqrt{2}$ is irrational. So $3 + 7\sqrt{2}$ is irrational.

Answer: Proved

1. (a) In $\triangle ABE$, $DF \parallel AE$, so by the basic proportionality theorem $\dfrac{BD}{DA} = \dfrac{BF}{FE}$. (i)
2. In $\triangle ABC$, $DE \parallel AC$, so $\dfrac{BD}{DA} = \dfrac{BE}{EC}$. (ii) From (i) and (ii): $\dfrac{BF}{FE} = \dfrac{BE}{EC}$.
3. (b) $\dfrac{AO}{BO} = \dfrac{CO}{OD}$ gives $\dfrac{OA}{OC} = \dfrac{OB}{OD}$, and $\angle AOB = \angle COD$ (vertically opposite angles). By SAS, $\triangle AOB \sim \triangle COD$.
4. So $\angle CAB = \angle ACD$, which are alternate interior angles for $AB$ and $CD$ with the transversal $AC$. Hence $AB \parallel CD$, and $ABCD$ is a trapezium.

Answer: Proved (both parts)

1. (a) **This question is flawed.** Let the shorter side be $x$. The longer side is $x + 80$ and the diagonal is $x + 60$. But a diagonal is always longer than either side, so the diagonal $x + 60$ cannot be shorter than the side $x + 80$. No such rectangle exists. CBSE’s marking scheme notes the error and awards full marks to everyone who attempted it.
2. (b) Let the father be $x$ years old, so the son is $(45 - x)$ years old. Five years ago: $(x - 5)(40 - x) = 124$.
3. $-x^2 + 45x - 200 = 124$ gives $x^2 - 45x + 324 = 0$, so $(x - 36)(x - 9) = 0$. $x = 9$ is rejected (the father cannot be $9$), so $x = 36$.
4. The father is $36$ years old and the son is $9$ years old. (Check: five years ago $31 \times 4 = 124$ ✓.)

Answer: (a) no such rectangle (flawed question)   (b) father $36$ years, son $9$ years

1. Radius $r = 7$ cm. The height of the cylinder is $13 - 7 = 6$ cm.
2. Inner surface area $= 2\pi r h + 2\pi r^2 = 2\pi r(h + r) = 2 \times \dfrac{22}{7} \times 7 \times 13 = 44 \times 13 = 572$ cm².
3. Volume $= \pi r^2 h + \dfrac{2}{3}\pi r^3 = \pi r^2\left(h + \dfrac{2r}{3}\right) = \dfrac{22}{7} \times 49 \times \left(6 + \dfrac{14}{3}\right) = 154 \times \dfrac{32}{3} = \dfrac{4928}{3} \approx 1642.67$ cm³.

Answer: Inner surface area $= 572$ cm²; volume $= \dfrac{4928}{3} \approx 1642.67$ cm³

1. Class marks: $125, 175, 225, 275, 325$. $\sum f_ix_i = 4(125) + 5(175) + 12(225) + 2(275) + 2(325) = 500 + 875 + 2700 + 550 + 650 = 5275$.
2. Mean $= \dfrac{5275}{25} = 211$.
3. The modal class is $200 - 250$ (frequency $12$): $l = 200$, $f_1 = 12$, $f_0 = 5$, $f_2 = 2$, $h = 50$. Mode $= 200 + \dfrac{12 - 5}{24 - 5 - 2} \times 50 = 200 + \dfrac{350}{17} = \dfrac{3750}{17} \approx 220.59$.

Answer: Mean $= ₹211$; mode $\approx ₹220.59$

1. (a) Let the tower be $AB = h$, the nearer point $C$ with $AC = x$, and the farther point $D$ with $CD = 20$. $\tan 60^\circ = \dfrac{h}{x}$ gives $h = x\sqrt{3}$. $\tan 30^\circ = \dfrac{h}{x + 20}$ gives $\sqrt{3}h = x + 20$.
2. Substituting: $3x = x + 20$, so $x = 10$ and $h = 10\sqrt{3} \approx 17.3$ m.
3. (b) Let the higher plane be at $4000$ m and the lower at height $h$, with the observation point at horizontal distance $x$ from the common vertical. $\tan 60^\circ = \dfrac{4000}{x}$ gives $x = \dfrac{4000}{\sqrt{3}}$. $\tan 45^\circ = \dfrac{h}{x} = 1$ gives $h = x = \dfrac{4000}{\sqrt{3}}$.
4. The vertical distance $= 4000 - \dfrac{4000}{\sqrt{3}} = 4000\left(1 - \dfrac{\sqrt{3}}{3}\right) = 4000(1 - 0.577) \approx 1693.33$ m with $\sqrt{3} = 1.73$ in the form $\dfrac{\sqrt{3}}{3}$ (the exact value is about $1690.6$ m).

Answer: (a) $10\sqrt{3} \approx 17.3$ m   (b) about $1693.33$ m

1. The pots per row are $2, 5, 8, \dots$, an A.P. with $a = 2$ and $d = 3$.
2. (i) $a_{10} = 2 + 9 \times 3 = 29$.
3. (ii) $a_5 - a_2 = 3d = 9$.
4. (iii) $S_n = \dfrac{n}{2}[2(2) + (n - 1)3] = 100$ gives $3n^2 + n - 200 = 0$, that is $(3n + 25)(n - 8) = 0$ and $n = 8$ (the root $-\dfrac{25}{3}$ is rejected): $8$ rows.
5. (iii) OR: $S_{12} = \dfrac{12}{2}[2(2) + 11 \times 3] = 6 \times 37 = 222$ pots.

Answer: (i) $29$   (ii) $9$   (iii) $8$ rows   (OR) $222$ pots

1. (i) Area of the square $= 40^2 = 1600$ cm².
2. (ii) Area of the circle (radius $10$) $= \dfrac{22}{7} \times 10 \times 10 = \dfrac{2200}{7} \approx 314.29$ cm².
3. (iii) The four quadrants of radius $10$ together make one full circle: $4 \times \dfrac{1}{4}\pi r^2 = \dfrac{2200}{7}$ cm². Removed area $= \dfrac{2200}{7} + \dfrac{2200}{7} = \dfrac{4400}{7}$ cm². Remaining $= 1600 - \dfrac{4400}{7} = \dfrac{6800}{7} \approx 971.43$ cm².
4. (iii) OR: the combined area removed is $\dfrac{4400}{7} \approx 628.57$ cm².

Answer: (i) $1600$ cm²   (ii) $\dfrac{2200}{7}$ cm²   (iii) remaining $\dfrac{6800}{7} \approx 971.43$ cm²   (OR) removed $\dfrac{4400}{7} \approx 628.57$ cm²

1. The people with type AB $= 50 - (21 + 22 + 5) = 2$.
2. (i) $P(\text{type O}) = \dfrac{21}{50}$.
3. (ii) $P(\text{type AB}) = \dfrac{2}{50} = \dfrac{1}{25}$.
4. (iii) Neither A nor B means type O or AB: $P = \dfrac{21 + 2}{50} = \dfrac{23}{50}$. (OR) A or B or O: $P = \dfrac{21 + 22 + 5}{50} = \dfrac{48}{50} = \dfrac{24}{25}$.

Answer: (i) $\dfrac{21}{50}$   (ii) $\dfrac{1}{25}$   (iii) $\dfrac{23}{50}$   (OR) $\dfrac{24}{25}$