CBSE Class 10 Maths Basic 2023 Question Paper (Set 430/5/1) with Solutions

1. $a_{14} - a_8 = 6d$, so $6d = 29 - 17 = 12$ and $d = 2$.

Answer: (B) 2

1. $OP \perp PT$, so $\angle OPQ = 90^\circ - 50^\circ = 40^\circ$.
2. $OP = OQ$ (radii), so $\angle OQP = 40^\circ$ and $\angle POQ = 180^\circ - 40^\circ - 40^\circ = 100^\circ$.

Answer: (B) $100^\circ$

1. There is exactly one ‘4 of hearts’ in $52$ cards: $P = \dfrac{1}{52}$.

Answer: (A) $\dfrac{1}{52}$

1. $AB = \sqrt{(-6 - 0)^2 + (2 - 6)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$ units.

Answer: (C) $2\sqrt{13}$ units

1. Real roots need $D \geq 0$: $16 - 4k \geq 0$, so $k \leq 4$.

Answer: (B) $k \leq 4$

1. The LCM takes the highest power of every prime: $2^4 \times 3 \times 5 \times 7 = 16 \times 105 = 1680$.

Answer: (C) 1680

1. Put $x = 2$: $4k + 6 + k = 0$, so $5k = -6$ and $k = -\dfrac{6}{5}$.

Answer: (A) $-\dfrac{6}{5}$

1. Parallel lines need $\dfrac{3}{2} = \dfrac{2m}{5}$, so $2m = \dfrac{15}{2}$ and $m = \dfrac{15}{4}$. (The constants $-2$ and $1$ give a ratio different from $\dfrac{3}{2}$, so the lines really are distinct.)

Answer: (D) $\dfrac{15}{4}$

1. The perimeters of similar triangles are in the ratio of corresponding sides: $\dfrac{AB}{DE} = \dfrac{32}{24}$, so $DE = 10 \times \dfrac{24}{32} = 7.5$ cm.

Answer: (B) 7.5 cm

1. $D = (2\sqrt{6})^2 - 4(3)(2) = 24 - 24 = 0$. The roots are real and equal (each $\dfrac{\sqrt{6}}{3}$, which is irrational).

Answer: (C) real and equal

1. Take a right triangle with opposite side $a$ and hypotenuse $b$. The adjacent side is $\sqrt{b^2 - a^2}$, so $\cos\theta = \dfrac{\sqrt{b^2 - a^2}}{b}$ and $\sec\theta = \dfrac{b}{\sqrt{b^2 - a^2}}$.

Answer: (B) $\dfrac{b}{\sqrt{b^2 - a^2}}$

1. $2 + 4 + \dots + 200 = 2(1 + 2 + \dots + 100) = 2 \times 5050 = 10100$. (Or $n(n + 1) = 100 \times 101$.)

Answer: (A) 10100

1. $DE \parallel BC$, so $\triangle ADE \sim \triangle ABC$: $\dfrac{AD}{AB} = \dfrac{DE}{BC}$. With $AB = 2 + 3 = 5$: $\dfrac{2}{5} = \dfrac{2.5}{x}$, so $x = 6.25$ cm.

Answer: (C) 6.25 cm

1. Two parallel tangents touch the circle at the ends of a diameter, so the distance between them is the diameter $= 6$ cm.

Answer: (B) 6 cm

1. $N = 10 + 12 + 14 + 13 + 17 = 66$, so $\dfrac{N}{2} = 33$. Cumulative frequencies: $10, 22, 36, 49, 66$. The first one at least $33$ is $36$, which belongs to $60 - 80$.

Answer: (D) 60 – 80

1. $\sec^2\theta - 1 = \tan^2\theta$, so the expression is $\dfrac{\tan^2\theta\cos^2\theta}{\sin\theta} = \dfrac{\sin^2\theta}{\sin\theta} = \sin\theta = \dfrac{3}{4}$.

Answer: (B) $\dfrac{3}{4}$

1. The proportion matches $AB \leftrightarrow PQ$ and $BC \leftrightarrow PR$. The angle between $AB$ and $BC$ is $\angle B$, and the angle between $PQ$ and $PR$ is $\angle P$. For SAS similarity these included angles must be equal: $\angle B = \angle P$.

Answer: (C) $\angle B = \angle P$

1. Mode $= 3\,\text{Median} - 2\,\text{Mean} = 90 - 64 = 26$.

Answer: (B) 26

1. (A) The outcomes are $HH, HT, TH, TT$. No tail means $HH$ only: $P = \dfrac{1}{4}$. True.
2. (R) A probability always lies from $0$ to $1$. True. But this fact does not produce the value $\dfrac{1}{4}$, so it is not the explanation.

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

1. (A) The largest sphere has diameter $a$, so $r = \dfrac{a}{2}$ and its surface area is $4\pi r^2 = 4\pi\dfrac{a^2}{4} = \pi a^2$. True.
2. (R) $\dfrac{4}{3}\pi r^3$ is the volume of a sphere. The surface area is $4\pi r^2$. False.

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

1. $576 = 2^6 \times 3^2$ and $512 = 2^9$.
2. LCM $= 2^9 \times 3^2 = 512 \times 9 = 4608$.

Answer: LCM $= 4608$

1. (a) $\dfrac{\frac{1}{2} + 1}{\frac{2}{\sqrt{3}} + 1} = \dfrac{\frac{3}{2}}{\frac{2 + \sqrt{3}}{\sqrt{3}}} = \dfrac{3\sqrt{3}}{2(2 + \sqrt{3})}$. Rationalising, this is $\dfrac{3\sqrt{3}(2 - \sqrt{3})}{2} = \dfrac{6\sqrt{3} - 9}{2}$.
2. (b) LHS $= \sin 90^\circ = 1$. RHS $= \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{1}{4} + \dfrac{3}{4} = 1$. LHS $=$ RHS.

Answer: (a) $\dfrac{3\sqrt{3}}{2(2 + \sqrt{3})} = \dfrac{6\sqrt{3} - 9}{2}$   (b) verified

1. $OB \perp AB$, so $\triangle OBA$ is right-angled at $B$. $\sin 60^\circ = \dfrac{OB}{OA} = \dfrac{5}{OA}$, so $OA = \dfrac{10}{\sqrt{3}} = \dfrac{10\sqrt{3}}{3}$ cm.
2. $\tan 60^\circ = \dfrac{OB}{AB} = \dfrac{5}{AB}$, so $AB = \dfrac{5}{\sqrt{3}} = \dfrac{5\sqrt{3}}{3}$ cm. Tangents from $A$ are equal, so $AC = AB = \dfrac{5\sqrt{3}}{3}$ cm.

Answer: $OA = \dfrac{10\sqrt{3}}{3}$ cm, $AC = \dfrac{5\sqrt{3}}{3}$ cm

1. (a) Midpoint of $AC = \left(\dfrac{1 + 3}{2}, \dfrac{2 + 8}{2}\right) = (2, 5)$ and midpoint of $BD = \left(\dfrac{5 - 1}{2}, \dfrac{4 + 6}{2}\right) = (2, 5)$. The diagonals bisect each other, so $ABCD$ is a parallelogram.
2. (b) $AB^2 = 3^2 + 4^2 = 25$, $BC^2 = 7^2 + 1^2 = 50$ and $AC^2 = 4^2 + 3^2 = 25$. Since $BC^2 = AB^2 + AC^2$, the triangle is right-angled at $A$.

Answer: (a) a parallelogram   (b) right-angled at $A$

1. $a = \dfrac{1}{15}$ and $d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5 - 4}{60} = \dfrac{1}{60}$.
2. $S_{15} = \dfrac{15}{2}\left[2 \times \dfrac{1}{15} + 14 \times \dfrac{1}{60}\right] = \dfrac{15}{2}\left[\dfrac{8}{60} + \dfrac{14}{60}\right] = \dfrac{15}{2} \times \dfrac{22}{60} = \dfrac{11}{4}$.

Answer: $\dfrac{11}{4}$

1. (a) Let there be $x$ notes of ₹ $50$ and $y$ notes of ₹ $100$. $x + y = 25$ and $50x + 100y = 2000$, that is $x + 2y = 40$.
2. Subtracting: $y = 15$ and $x = 10$. Sabina received $10$ notes of ₹ $50$ and $15$ notes of ₹ $100$. (Check: $500 + 1500 = 2000$ ✓.)
3. (b) Let Amit be $x$ and Baljeet $y$ years old. $x - 5 = 3(y - 5)$ gives $x - 3y = -10$. $x + 10 = 2(y + 10)$ gives $x - 2y = 10$.
4. Subtracting: $y = 20$ and $x = 50$. Amit is $50$ years old and Baljeet is $20$.

Answer: (a) $10$ notes of ₹ $50$, $15$ of ₹ $100$   (b) Amit $50$ years, Baljeet $20$ years

1. Suppose $7 - 2\sqrt{3} = \dfrac{a}{b}$ with integers $a, b$ and $b \neq 0$. Then $\sqrt{3} = \dfrac{7b - a}{2b}$.
2. The right side is rational but $\sqrt{3}$ is irrational. This contradiction shows $7 - 2\sqrt{3}$ is irrational.

Answer: Proved

1. Class marks $x$: $7.5, 22.5, 37.5, 52.5, 67.5, 82.5$. Take $a = 37.5$ and $h = 15$, so $u = \dfrac{x - 37.5}{15} = -2, -1, 0, 1, 2, 3$.
2. $fu = -24, -15, 0, 20, 32, 18$, so $\sum fu = 31$ and $\sum f = 80$.
3. Mean $= 37.5 + 15 \times \dfrac{31}{80} = 37.5 + 5.8125 \approx 43.3$.

Answer: Mean $\approx 43.3$

1. (a) Let $AP : PB = k : 1$. The $y$-coordinate: $\dfrac{k(-4) + 3}{k + 1} = -2$, so $-4k + 3 = -2k - 2$ and $k = \dfrac{5}{2}$. The ratio is $5 : 2$.
2. The $x$-coordinate: $a = \dfrac{\frac{5}{2}(2) - 4}{\frac{5}{2} + 1} = \dfrac{1}{\frac{7}{2}} = \dfrac{2}{7}$.
3. (b) $D$ is the midpoint of $BC$: $\left(\dfrac{2 - 6}{2}, \dfrac{-2 - 7}{2}\right) = \left(-2, -\dfrac{9}{2}\right)$. $E$ is the midpoint of $AC$: $\left(\dfrac{4 - 6}{2}, \dfrac{-2 - 7}{2}\right) = \left(-1, -\dfrac{9}{2}\right)$.
4. $DE = \sqrt{1^2 + 0^2} = 1$ and $AB = \sqrt{2^2 + 0^2} = 2$, so $DE = \dfrac{1}{2}AB$ ✓.

Answer: (a) ratio $5 : 2$, $a = \dfrac{2}{7}$   (b) verified

1. LHS $= \left(\dfrac{1}{\sin A} - \sin A\right)\left(\dfrac{1}{\cos A} - \cos A\right) = \dfrac{1 - \sin^2 A}{\sin A} \times \dfrac{1 - \cos^2 A}{\cos A} = \dfrac{\cos^2 A}{\sin A} \times \dfrac{\sin^2 A}{\cos A} = \sin A\cos A$.
2. RHS $= \dfrac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \dfrac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A\cos A}} = \dfrac{\sin A\cos A}{1} = \sin A\cos A$. LHS $=$ RHS.

Answer: Proved

1. Tangents from an external point are equal: $AD = AF$, $BD = BE$ and $CE = CF$.
2. $AB = AC$ means $AD + DB = AF + FC$. Since $AD = AF$, this gives $DB = FC$.
3. Then $BE = BD = FC = CE$, so $BE = EC$: $BC$ is bisected at $E$.

Answer: Proved

1. Let the tower be $AB = h$, the person's new position $C$ with $BC = x$, and the old position $D$ with $CD = 48$ so $BD = x + 48$.
2. In $\triangle ABD$: $\tan 45^\circ = \dfrac{h}{x + 48}$, so $h = x + 48$. In $\triangle ABC$: $\tan 60^\circ = \dfrac{h}{x}$, so $h = \sqrt{3}x$.
3. Equate: $\sqrt{3}x = x + 48$, so $x(\sqrt{3} - 1) = 48$ and $x = \dfrac{48}{\sqrt{3} - 1} = 24(\sqrt{3} + 1) = 24 \times 2.732 = 65.57$ m.
4. $h = x + 48 = 113.57$ m.

Answer: Height $\approx 113.57$ m; distance now $\approx 65.57$ m

1. (a) Area of sector $AOB = \dfrac{90}{360} \times \dfrac{22}{7} \times 7 \times 7 = \dfrac{77}{2}$ cm². Area of $\triangle AOB = \dfrac{1}{2} \times 7 \times 7 = \dfrac{49}{2}$ cm².
2. Shaded segment $= \dfrac{77}{2} - \dfrac{49}{2} = 14$ cm². Arc $AB = \dfrac{90}{360} \times 2 \times \dfrac{22}{7} \times 7 = 11$ cm.
3. (b) Shaded area $= \dfrac{60}{360} \times \dfrac{22}{7} \times (10.5^2 - 3.5^2) = \dfrac{1}{6} \times \dfrac{22}{7} \times 98 = \dfrac{154}{3} \approx 51.3$ cm².
4. Arc $CD = \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 10.5 = 11$ cm.

Answer: (a) $14$ cm², arc $11$ cm   (b) $\dfrac{154}{3} \approx 51.3$ cm², arc $11$ cm

1. **Given:** in $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. **To prove:** $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. **Construction:** join $BE$ and $CD$, and draw $DM \perp AC$ and $EN \perp AB$.
2. $\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)} = \dfrac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \dfrac{AD}{DB}$. (i)
3. $\dfrac{\text{ar}(ADE)}{\text{ar}(CDE)} = \dfrac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \dfrac{AE}{EC}$. (ii)
4. $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels, so $\text{ar}(BDE) = \text{ar}(CDE)$. (iii)
5. From (i), (ii) and (iii): $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

Answer: Proved (basic proportionality theorem)

1. (a) Let the numbers be $x$ and $x + 5$. $\dfrac{1}{x} - \dfrac{1}{x + 5} = \dfrac{1}{10}$ gives $\dfrac{5}{x(x + 5)} = \dfrac{1}{10}$, so $x^2 + 5x = 50$ and $x^2 + 5x - 50 = 0$.
2. $(x + 10)(x - 5) = 0$, so $x = -10$ or $x = 5$. The numbers are $5$ and $10$, or $-10$ and $-5$. (Check for $5, 10$: $\dfrac{1}{5} - \dfrac{1}{10} = \dfrac{1}{10}$ ✓.)
3. (b) Equal roots need $D = k^2 - 4(2)(8) = k^2 - 64 = 0$, so $k = \pm 8$.
4. For $k = 8$: $2x^2 + 8x + 8 = 2(x + 2)^2 = 0$, so $x = -2$ (twice). For $k = -8$: $2x^2 - 8x + 8 = 2(x - 2)^2 = 0$, so $x = 2$ (twice).

Answer: (a) $5, 10$ or $-10, -5$   (b) $k = 8$, $x = -2$; $k = -8$, $x = 2$

1. (i) The curve crosses the $x$-axis at two points, so it has $2$ zeroes.
2. (ii) The zeroes are where the graph meets the $x$-axis. If it never meets the $x$-axis, there are $0$ zeroes.
3. (iii)(a) $p(2) = 4 + 2(a + 1) + b = 0$ gives $2a + b = -6$. $p(-3) = 9 - 3(a + 1) + b = 0$ gives $-3a + b = -6$. Subtracting: $5a = 0$, so $a = 0$ and $b = -6$. (Check: the sum of zeroes $-1 = -(a + 1)$ ✓ and the product $-6 = b$ ✓.)
4. (iii)(b) $p(-4) = 16 + 8 - (7p + 3) = 21 - 7p = 0$, so $p = 3$.

Answer: (i) $2$   (ii) $0$   (iii) $a = 0$, $b = -6$ (OR) $p = 3$

1. (i) Mallet volume $= \pi r^2 h = 3.14 \times 2 \times 2 \times 10 = 125.6$ cm³.
2. (ii) Inner surface of the bowl $= 2\pi r^2 = 2 \times 3.14 \times 25 = 157$ cm².
3. (iii)(a) Metal $=$ outer hemisphere $-$ inner hemisphere $= \dfrac{2}{3}\pi(6^3 - 5^3) = \dfrac{2}{3} \times 3.14 \times 91 \approx 190.5$ cm³.
4. (iii)(b) Total surface area of the mallet $= 2\pi r(r + h) = 2 \times 3.14 \times 2 \times 12 = 150.72 \approx 150.7$ cm².

Answer: (i) $125.6$ cm³   (ii) $157$ cm²   (iii) $\approx 190.5$ cm³ (OR) $\approx 150.7$ cm²

1. The probability of a colour is its central angle divided by $360^\circ$.
2. (i) $P(\text{white}) = \dfrac{120}{360} = \dfrac{1}{3}$.
3. (ii) $P(\text{blue or green}) = \dfrac{60 + 60}{360} = \dfrac{1}{3}$.
4. (iii)(a) Yellow is $90^\circ$, which is $\dfrac{1}{4}$ of the circle. If the total is $x$, then $\dfrac{x}{4} = 15$, so $x = 60$ students.
5. (iii)(b) $P(\text{red or blue}) = \dfrac{30 + 60}{360} = \dfrac{1}{4}$.