CBSE Class 10 Maths Basic 2023 Question Paper (Set 430/2/1) with Solutions

1. $a = 14$, $d = 5$ and $a_n = 119$: $14 + (n - 1) \times 5 = 119$ gives $(n - 1) \times 5 = 105$, so $n = 22$.

Answer: (C) 22

1. On the $x$-axis $y = 0$. If the ratio is $k : 1$, then $y = \dfrac{6k - 3}{k + 1} = 0$, so $k = \dfrac{1}{2}$.
2. The ratio is $\dfrac{1}{2} : 1 = 1 : 2$. (The point is $(3, 0)$.)

Answer: (D) 1 : 2

1. $9(\sec^2 A - \tan^2 A) = 9 \times 1 = 9$, using $\sec^2 A - \tan^2 A = 1$.

Answer: (A) 9

1. $\sin 60^\circ = \dfrac{h}{50}$, so $h = 50 \times \dfrac{\sqrt{3}}{2} = 25\sqrt{3}$ m.

Answer: (D) $25\sqrt{3}$ m

1. $OQ \perp PQ$ and $OR \perp PR$, so in quadrilateral $OQPR$: $\angle QOR = 360^\circ - 90^\circ - 90^\circ - 50^\circ = 130^\circ$.
2. The angle at the centre is twice the angle at the major arc: $\angle QTR = \dfrac{130^\circ}{2} = 65^\circ$.

Answer: (C) $65^\circ$

1. A sector of angle $\alpha$ is $\dfrac{\alpha}{360}$ of the whole circle, so its area is $\dfrac{\alpha}{360} \times \pi R^2$. (Option B is the arc length, not the area.)

Answer: (D) $\dfrac{\alpha}{360} \times \pi R^2$

1. HCF $\times$ LCM $=$ product of the numbers, so LCM $= \dfrac{360 \times 64}{8} = 360 \times 8 = 2880$.

Answer: (D) 2880

1. $2\pi r h = 88$: $2 \times \dfrac{22}{7} \times r \times 14 = 88$, so $88r = 88$ and $r = 1$ cm. The diameter is $2$ cm.

Answer: (A) 2 cm

1. The composite numbers on a die are $4$ and $6$ ($1$ is neither prime nor composite). $P = \dfrac{2}{6} = \dfrac{1}{3}$.

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

1. Equal roots mean $D = 0$: $b^2 - 4(9)\left(\dfrac{1}{4}\right) = b^2 - 9 = 0$, so $b = \pm 3$.

Answer: (D) $\pm 3$

1. The cone height is $2r$: cone volume $= \dfrac{1}{3}\pi r^2(2r) = \dfrac{2}{3}\pi r^3$. The hemisphere volume is $\dfrac{2}{3}\pi r^3$.
2. Total $= \dfrac{2}{3}\pi r^3 + \dfrac{2}{3}\pi r^3 = \dfrac{4}{3}\pi r^3$.

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

1. The zeroes are the points where the graph meets the $x$-axis. It crosses once on the left and once on the right. The valley in the middle stays above the axis without touching it. So there are $2$ zeroes.

Answer: (A) 2

1. $\dfrac{a_1}{a_2} = \dfrac{1}{2}$ and $\dfrac{b_1}{b_2} = \dfrac{2}{-4} = -\dfrac{1}{2}$. The ratios differ, so the lines intersect at exactly one point: a unique solution.

Answer: (C) is consistent with a unique solution

1. A probability lies between $0$ and $1$. $\dfrac{1}{0.5} = 2 > 1$, so it cannot be a probability. The others are $0.5$, $0.05$ and about $0.036$.

Answer: (C) $\dfrac{1}{0.5}$

1. $2\left(\dfrac{1}{2}\right)^2 + 3(\sqrt{3})^2 - \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{1}{2} + 9 - \dfrac{1}{2} = 9$.

Answer: (D) 9

1. On the line $x = 4$ put $x = 4$ in $y = x$: $y = 4$. So $P = (4, 4)$.

Answer: (B) (4, 4)

1. Mode $= 3\,\text{Median} - 2\,\text{Mean}$: $21 = 75 - 2\,\text{Mean}$, so $2\,\text{Mean} = 54$ and Mean $= 27$.

Answer: (A) 27

1. $\dfrac{1 - \cos^2 A}{1 - \sin^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A = \dfrac{4}{25}$.

Answer: (B) $\dfrac{4}{25}$

1. (A) $x^2 + 4x = x(x + 4) = 0$ gives $x = 0$ and $x = -4$: two real zeroes. True.
2. (R) $x^2 + ax = x(x + a) = 0$ gives $0$ and $-a$, not $a$. False.

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

1. (R) The primes among $1$ to $6$ are $2, 3, 5$. True.
2. (A) That is $3$ favourable outcomes, so $P = \dfrac{3}{6} = \dfrac{1}{2}$, not $\dfrac{2}{3}$. False.

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

1. In $\triangle BAC$, $DE \parallel AC$, so by the basic proportionality theorem $\dfrac{BE}{EC} = \dfrac{BD}{DA}$.
2. It is given that $\dfrac{BE}{EC} = \dfrac{BC}{CP}$, so $\dfrac{BD}{DA} = \dfrac{BC}{CP}$.
3. In $\triangle BAP$ the line $DC$ divides $BA$ and $BP$ in the same ratio, so $DC \parallel AP$ by the converse of the basic proportionality theorem.

Answer: Proved

1. (a) $540 = 2^2 \times 3^3 \times 5$ and $630 = 2 \times 3^2 \times 5 \times 7$. HCF $=$ product of the smallest powers of the common primes $= 2 \times 3^2 \times 5 = 90$.
2. (b) $15^n = (3 \times 5)^n = 3^n \times 5^n$. A number ending in $0$ is divisible by $10$, so its prime factors must include both $2$ and $5$. The only primes in $15^n$ are $3$ and $5$, so it can never end in $0$.

Answer: (a) $90$   (b) proved

1. (a) $PQ^2 = (6 - 1)^2 + (-1 - 3)^2 = 25 + 16 = 41$ and $QR^2 = (x - 1)^2 + (8 - 3)^2 = (x - 1)^2 + 25$.
2. $PQ = QR$ gives $(x - 1)^2 + 25 = 41$, so $(x - 1)^2 = 16$ and $x - 1 = \pm 4$: $x = 5$ or $x = -3$.
3. (b) With $A(-2, 0)$, $B(2, 3)$, $C(1, -3)$: $AB = \sqrt{16 + 9} = 5$, $BC = \sqrt{1 + 36} = \sqrt{37}$ and $CA = \sqrt{9 + 9} = 3\sqrt{2}$. All three sides differ, so the triangle is scalene.

Answer: (a) $x = 5$ or $-3$   (b) scalene

1. Write $p(x) = 3x^2 + (2k + 1)x - (k + 5)$. Sum of zeroes $= -\dfrac{2k + 1}{3}$ and product $= -\dfrac{k + 5}{3}$.
2. Sum $= \dfrac{1}{2} \times$ product: $-\dfrac{2k + 1}{3} = -\dfrac{k + 5}{6}$, so $2(2k + 1) = k + 5$, that is $4k + 2 = k + 5$ and $k = 1$.

Answer: $k = 1$

1. Cards left $= 52 - 13 = 39$. Kings left $= 3$ (the king of diamonds is gone). $P = \dfrac{3}{39} = \dfrac{1}{13}$.

Answer: $\dfrac{1}{13}$

1. The shaded region is the big sector $OAC$ with the small sector $OED$ removed, both with the angle $60^\circ$.
2. Area of sector $OAC = \dfrac{60}{360} \times \pi \times 5^2 = \dfrac{25\pi}{6}$ cm². Area of sector $OED = \dfrac{60}{360} \times \pi \times 2^2 = \dfrac{4\pi}{6}$ cm².
3. Shaded area $= \dfrac{25\pi - 4\pi}{6} = \dfrac{21\pi}{6} = \dfrac{21}{6} \times \dfrac{22}{7} = 11$ cm².

Answer: $11$ cm²

1. Suppose $4 + 2\sqrt{3} = \dfrac{p}{q}$ with integers $p, q$ and $q \neq 0$. Then $\sqrt{3} = \dfrac{p - 4q}{2q}$.
2. The right side is a rational number, but $\sqrt{3}$ is irrational. This contradiction means the assumption is wrong, so $4 + 2\sqrt{3}$ is irrational.

Answer: Proved

1. (a) Let the circle touch $AB, BC, CD, DA$ at $P, Q, R, S$. Tangents from an outside point are equal: $AP = AS$, $BP = BQ$, $CR = CQ$ and $DR = DS$.
2. $AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS) = (AS + DS) + (BQ + CQ) = AD + BC$.
3. (b) The same result holds for a parallelogram $ABCD$ drawn around a circle: $AB + CD = AD + BC$. In a parallelogram $AB = CD$ and $AD = BC$, so $2AB = 2AD$ and $AB = AD$. Adjacent sides are equal, so $ABCD$ is a rhombus.

Answer: Proved (both parts)

1. (a) LHS $= \dfrac{(1 - \cos\theta)^2}{(1 + \cos\theta)(1 - \cos\theta)} = \dfrac{(1 - \cos\theta)^2}{\sin^2\theta} = \left(\dfrac{1 - \cos\theta}{\sin\theta}\right)^2 = (\operatorname{cosec}\theta - \cot\theta)^2 =$ RHS.
2. (b) LHS $= \left(1 + \dfrac{\cos^2 A}{\sin^2 A}\right)\left(1 + \dfrac{\sin^2 A}{\cos^2 A}\right) = \dfrac{\sin^2 A + \cos^2 A}{\sin^2 A} \times \dfrac{\cos^2 A + \sin^2 A}{\cos^2 A} = \dfrac{1}{\sin^2 A \cos^2 A}$.
3. Since $\cos^2 A = 1 - \sin^2 A$: $\dfrac{1}{\sin^2 A(1 - \sin^2 A)} = \dfrac{1}{\sin^2 A - \sin^4 A} =$ RHS.

Answer: Proved (both parts)

1. $2x^2 - 7x - 15 = 2x^2 - 10x + 3x - 15 = 2x(x - 5) + 3(x - 5) = (2x + 3)(x - 5)$. The zeroes are $x = -\dfrac{3}{2}$ and $x = 5$.
2. Sum of zeroes $= -\dfrac{3}{2} + 5 = \dfrac{7}{2} = -\dfrac{(-7)}{2} = -\dfrac{b}{a}$ ✓.
3. Product of zeroes $= -\dfrac{3}{2} \times 5 = -\dfrac{15}{2} = \dfrac{c}{a}$ ✓.

Answer: Zeroes $-\dfrac{3}{2}$ and $5$; the relations are verified

1. Midpoint of $AC = \left(\dfrac{-1 + 2}{2}, \dfrac{0 + 2}{2}\right) = \left(\dfrac{1}{2}, 1\right)$. Midpoint of $BD = \left(\dfrac{3 - 2}{2}, \dfrac{1 + 1}{2}\right) = \left(\dfrac{1}{2}, 1\right)$.
2. The diagonals bisect each other, so $ABCD$ is a parallelogram.
3. $AC = \sqrt{3^2 + 2^2} = \sqrt{13}$ and $BD = \sqrt{5^2 + 0^2} = 5$. The diagonals are not equal ($\sqrt{13} \neq 5$), so it is not a rectangle.

Answer: A parallelogram, but not a rectangle

1. (a) Let the building be $AB = 60$, the tower $CD = h$ and the distance $AC = x$. From the top $B$, the angle of depression of the bottom $C$ is $60^\circ$, so in $\triangle BAC$, $\tan 60^\circ = \dfrac{60}{x}$ and $x = \dfrac{60}{\sqrt{3}} = 20\sqrt{3} \approx 34.64$ m.
2. The top $D$ is $30^\circ$ below $B$, and the level drop is $60 - h$: $\tan 30^\circ = \dfrac{60 - h}{x}$, so $60 - h = \dfrac{20\sqrt{3}}{\sqrt{3}} = 20$ and $h = 40$ m.
3. (b) Let the building be $CD = h$, $BC = x$ and $AB = 30$. At $B$: $\tan 45^\circ = \dfrac{h}{x}$ gives $h = x$. At $A$: $\tan 30^\circ = \dfrac{h}{x + 30}$ gives $\sqrt{3}h = x + 30$.
4. Put $x = h$: $h(\sqrt{3} - 1) = 30$, so $h = \dfrac{30}{\sqrt{3} - 1} = 15(\sqrt{3} + 1) = 15 \times 2.732 = 40.98$ m. The distance of the base from $A$ is $x + 30 = 70.98$ m.

Answer: (a) tower $40$ m, distance $20\sqrt{3} \approx 34.64$ m   (b) height $\approx 40.98$ m, distance $\approx 70.98$ m

1. Total $N = 3 + 5 + 16 + 12 + 13 + 20 + 6 + 5 = 80$. Take the assumed mean $a = 35$ and $h = 10$, so $u = \dfrac{x - 35}{10}$ for the class marks $5, 15, \dots, 75$.
2. $u = -3, -2, -1, 0, 1, 2, 3, 4$ and $fu = -9, -10, -16, 0, 13, 40, 18, 20$, so $\sum fu = 56$.
3. Mean $= 35 + 10 \times \dfrac{56}{80} = 35 + 7 = 42$.
4. Cumulative frequencies: $3, 8, 24, 36, 49, 69, 75, 80$. $\dfrac{N}{2} = 40$ lies in $40 - 50$, so the median class is $40 - 50$ with $l = 40$, $cf = 36$, $f = 13$, $h = 10$.
5. Median $= 40 + \dfrac{40 - 36}{13} \times 10 = 40 + \dfrac{40}{13} \approx 43.1$.

Answer: Mean $= 42$; median $\approx 43.1$

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 $DE$ and $BC$, 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) $S_7 = \dfrac{7}{2}(2a + 6d) = -14$ gives $a + 3d = -2$. $S_{11} = \dfrac{11}{2}(2a + 10d) = -55$ gives $a + 5d = -5$.
2. Subtracting: $2d = -3$, so $d = -\dfrac{3}{2}$ and $a = -2 - 3d = \dfrac{5}{2}$.
3. $S_n = \dfrac{n}{2}\left[2\left(\dfrac{5}{2}\right) + (n - 1)\left(-\dfrac{3}{2}\right)\right] = \dfrac{n}{4}(13 - 3n)$.
4. (b) $a_1 = S_1 = 3 + 1 = 4$. $S_2 = 12 + 2 = 14$, so $a_2 = S_2 - S_1 = 10$ and $d = 10 - 4 = 6$.
5. $a_{15} = a + 14d = 4 + 14 \times 6 = 88$.

Answer: (a) $S_n = \dfrac{n(13 - 3n)}{4}$   (b) $a = 4$, $d = 6$, $a_{15} = 88$

1. (i) Red segment $= \dfrac{90}{360}\pi r^2 - \dfrac{1}{2}r^2$ (sector minus the right triangle with legs $r$). Yellow segment $= \dfrac{60}{360}\pi r^2 - \dfrac{\sqrt{3}}{4}r^2$ (sector minus the equilateral triangle of side $r$).
2. So $\left(\dfrac{\pi}{4} - \dfrac{1}{2}\right)r^2 + \left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right)r^2 = \dfrac{770}{3}$.
3. (ii) With $\pi = \dfrac{22}{7}$ and $\sqrt{3} = 1.732$ the bracket is $0.7857 - 0.5 + 0.5238 - 0.4330 = 0.3765$, so $r^2 = \dfrac{256.67}{0.3765} \approx 681.7$ and $r \approx 26.1$ m.
4. (iii)(a) Red segment $= \left(\dfrac{\pi}{4} - \dfrac{1}{2}\right)r^2 \approx 0.2857 \times 681.7 \approx 194.7$ sq m (CBSE gives about $194.63$).
5. (iii)(b) Yellow segment $= \left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}\right)r^2 \approx 0.0908 \times 681.7 \approx 61.9$ sq m (CBSE gives about $62.03$). The two add to $256\dfrac{2}{3}$ ✓. The small differences are only rounding.

Answer: (i) equation above   (ii) $r \approx 26.1$ m   (iii) red $\approx 194.7$ sq m (OR) yellow $\approx 61.9$ sq m

1. (i) $OA \perp NA$ (radius and tangent), so in the right triangle $OAN$, $\tan 30^\circ = \dfrac{OA}{AN} = \dfrac{5}{AN}$. Hence $AN = 5\sqrt{3}$ cm.
2. (ii) $\angle ANB = 2 \times 30^\circ = 60^\circ$. In quadrilateral $OANB$, $\angle AOB = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ$.
3. (iii)(a) $NA = NB = 5\sqrt{3}$ (equal tangents) and $\angle ANB = 60^\circ$, so $\triangle NAB$ is equilateral and $AB = 5\sqrt{3}$ cm. Total length $= 3 \times 5\sqrt{3} = 15\sqrt{3}$ cm.
4. (iii)(b) If $\angle ANO = 45^\circ$ then $\angle ANB = 90^\circ$, and $\angle AOB = 360^\circ - 90^\circ - 90^\circ - 90^\circ = 90^\circ$. All four angles of $OANB$ are $90^\circ$, so it is a rectangle, and $OA = OB$ (radii) makes it a square.

Answer: (i) $5\sqrt{3}$ cm   (ii) $120^\circ$   (iii) $15\sqrt{3}$ cm (OR) a square

1. (i) One hollow: $\pi r^2 h = \dfrac{22}{7} \times 2 \times 2 \times 3 = \dfrac{264}{7} \approx 37.7$ cm³.
2. (ii) Lateral surface area $= 2(l + b)h = 2(14 + 17) \times 4 = 248$ cm².
3. (iii)(a) Seven hollows: $7 \times \dfrac{264}{7} = 264$ cm³. The block is $14 \times 17 \times 4 = 952$ cm³. Remaining $= 952 - 264 = 688$ cm³.
4. (iii)(b) Top surface to paint $= 14 \times 17 - 7 \times \pi r^2 = 238 - 7 \times \dfrac{22}{7} \times 4 = 238 - 88 = 150$ cm².