CBSE Class 10 Maths Basic 2024 Question Paper (Set 430/2/1) with Solutions

1. The smallest $2$-digit number is $10 = 2 \times 5$ and the smallest composite number is $4 = 2^2$. HCF $= 2$.

Answer: (A) 2

1. Infinitely many solutions need $\dfrac{1}{2} = \dfrac{1}{k} = \dfrac{-4}{-8} = \dfrac{1}{2}$, so $k = 2$.

Answer: (C) $k = 2$

1. Substitute $x = 2$: (A) $4 - 8 + 5 = 1$; (B) $4 + 6 - 12 = -2$; (C) $8 - 14 + 6 = 0$ ✓; (D) $12 - 12 - 2 = -2$. Only (C) gives $0$.

Answer: (C) $2x^2 - 7x + 6 = 0$

1. $a_7 = a + 6d = a - 24 = 4$, so $a = 28$.

Answer: (D) 28

1. $\sqrt{5^2 + 4^2} = \sqrt{41}$.

Answer: (B) $\sqrt{41}$

1. Opposite $= 3$, hypotenuse $= 5$, adjacent $= 4$. $\cot A = \dfrac{4}{3}$.

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

1. $\dfrac{\sec^2 A}{\operatorname{cosec}^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A$.

Answer: (D) $\tan^2 A$

1. $\dfrac{2/\sqrt{3}}{1 - 1/3} = \dfrac{2/\sqrt{3}}{2/3} = \sqrt{3} = \tan 60^\circ$. (This is the double-angle formula $\tan 2A$ with $A = 30^\circ$.)

Answer: (C) $\tan 60^\circ$

1. $x^2 - (\text{sum})x + \text{product} = x^2 - (-5)x + 6 = x^2 + 5x + 6$.

Answer: (A) $x^2 + 5x + 6$

1. $3x^2 + 11x - 4 = (3x - 1)(x + 4)$, so the zeroes are $\dfrac{1}{3}$ and $-4$.

Answer: (C) $\dfrac{1}{3}, -4$

1. The highest frequency $22$ is in $0 - 10$, so the modal class is $0 - 10$ (upper limit $10$).
2. $N = 66$, $\dfrac{N}{2} = 33$. Cumulative frequencies: $22, 32, 40, \dots$, so the median class is $20 - 30$ (upper limit $30$). The difference is $30 - 10 = 20$.

Answer: (C) 20

1. A probability always lies between $0$ (impossible event) and $1$ (sure event): $0 \leq P(A) \leq 1$.

Answer: (C) $0 \leq P(A) \leq 1$

1. Total surface area of a solid hemisphere $= 3\pi r^2 = 3\pi \times 49 = 147\pi$ cm².

Answer: (B) $147\pi$ cm²

1. Minor sector $= \dfrac{120}{360} \times \dfrac{22}{7} \times 21^2 = 462$ cm². Major sector (angle $240^\circ$) $= 924$ cm². Difference $= 924 - 462 = 462$ cm².

Answer: (B) $462$ cm²

1. Parallel lines never meet: the slopes are equal but the lines are different, so $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$. This is option (D).

Answer: (D) $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

1. $OA \perp PA$ and $OB \perp PB$, so $\angle AOB = 180^\circ - 80^\circ = 100^\circ$.
2. $OA = OB$, so $\triangle OAB$ is isosceles and $\angle ABO = \dfrac{180^\circ - 100^\circ}{2} = 40^\circ$.

Answer: (A) $40^\circ$

1. A line that cuts a circle at two distinct points is a secant. (A chord is the segment between the two points, and a tangent touches at only one point.)

Answer: (A) secant

1. $\tan\theta = \dfrac{6}{2\sqrt{3}} = \sqrt{3}$, so $\theta = 60^\circ$.

Answer: (C) $60^\circ$

1. (A) This is the basic proportionality theorem (when the line meets the other two sides in distinct points). True.
2. (R) A line parallel to a side can obviously be drawn through any point of the triangle. False.

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

1. (R) Every point of the $y$-axis has $x = 0$. True.
2. (A) $(0, 4)$ has $x = 0$, so it is on the $y$-axis. True, and (R) is exactly the reason.

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

1. $84 = 2 \times 2 \times 3 \times 7$ and $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$.
2. HCF $=$ common primes at the smallest powers $= 2 \times 2 \times 3 = 12$.

Answer: $12$

1. (a) $x + y = 70$ and $x - y = 10$. Adding, $2x = 80$, so $x = 40$ and $y = 30$.
2. (b) Subtracting the first from the second: $2x = -2$, so $x = -1$. Then $-1 - 3y = 7$ gives $y = -\dfrac{8}{3}$.

Answer: (a) $40$ and $30$   (b) $x = -1$, $y = -\dfrac{8}{3}$

1. Total pens $= 145 + 15 = 160$. $P(\text{good}) = \dfrac{145}{160} = \dfrac{29}{32}$.

Answer: $\dfrac{29}{32}$

1. (a) $OA \cdot OB = OC \cdot OD$ gives $\dfrac{OA}{OC} = \dfrac{OD}{OB}$. $\angle AOD = \angle COB$ (vertically opposite angles), and the including sides are proportional, so $\triangle AOD \sim \triangle COB$ by SAS.
2. (b) $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ gives $DE \parallel BC$ (converse of the basic proportionality theorem). Then $\angle D = \angle B$ and $\angle E = \angle C$ (corresponding angles).
3. $\angle D = \angle E$ is given, so $\angle B = \angle C$, and the sides opposite equal angles are equal: $AB = AC$. $\triangle ABC$ is isosceles.

Answer: Proved (both parts)

1. Let $AB$ be a diameter of the circle with centre $O$, and let the tangents at $A$ and $B$ be $\ell$ and $m$.
2. A radius is perpendicular to the tangent at the point of contact: $\angle OAl = 90^\circ$ and $\angle OBm = 90^\circ$. So the alternate interior angles made by the transversal $AB$ with $\ell$ and $m$ are equal ($90^\circ$ each), hence $\ell \parallel m$.

Answer: Proved

1. There are $36$ equally likely outcomes.
2. (a) Even on both: each die shows $2, 4$ or $6$, giving $3 \times 3 = 9$ outcomes. $P = \dfrac{9}{36} = \dfrac{1}{4}$.
3. (b) Sum more than $9$: $(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)$, which is $6$. $P = \dfrac{6}{36} = \dfrac{1}{6}$.

Answer: (a) $\dfrac{1}{4}$   (b) $\dfrac{1}{6}$

1. (a) The chord $AB$ touches the smaller circle at $Q$, so $OQ \perp AB$ and $OQ = 5$. The perpendicular from the centre bisects the chord: $QB = 12$.
2. $OB^2 = OQ^2 + QB^2 = 25 + 144 = 169$, so the radius of the larger circle is $OB = 13$ cm.
3. (b) Let $PQ$ and $PR$ be tangents from $P$ to the circle with centre $O$. $OQ \perp PQ$ and $OR \perp PR$, so $\angle OQP = \angle ORP = 90^\circ$.
4. In the quadrilateral $OQPR$: $90^\circ + \angle QPR + 90^\circ + \angle QOR = 360^\circ$, so $\angle QPR + \angle QOR = 180^\circ$.

Answer: (a) $13$ cm   (b) proved

1. Suppose $x = 7 - 3\sqrt{5}$ is rational. Then $\sqrt{5} = \dfrac{7 - x}{3}$, which is rational.
2. This contradicts the given fact that $\sqrt{5}$ is irrational, so $7 - 3\sqrt{5}$ is irrational.

Answer: Proved

1. (a) $\alpha + \beta = 3$ and $\alpha\beta = 2$. New sum $= 2(\alpha + \beta) + 2 = 8$. New product $= (2\alpha + 1)(2\beta + 1) = 4\alpha\beta + 2(\alpha + \beta) + 1 = 8 + 6 + 1 = 15$.
2. The polynomial is $x^2 - 8x + 15$ (or $k(x^2 - 8x + 15)$).
3. (b) $4x^2 - 4x + 1 = (2x - 1)^2$, so both zeroes are $\dfrac{1}{2}$. Sum $= 1 = -\dfrac{-4}{4}$ ✓. Product $= \dfrac{1}{4} = \dfrac{1}{4}$ ✓.

Answer: (a) $x^2 - 8x + 15$   (b) zeroes $\dfrac{1}{2}, \dfrac{1}{2}$; relations verified

1. LHS $= \dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)} + \dfrac{\cos^2\theta}{\sin\theta(\cos\theta - \sin\theta)} = \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta - \cos\theta)}$.
2. $\sin^3\theta - \cos^3\theta = (\sin\theta - \cos\theta)(1 + \sin\theta\cos\theta)$, so LHS $= \dfrac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \sec\theta\operatorname{cosec}\theta + 1 =$ RHS.

Answer: Proved

1. One wiper cleans a sector: $\dfrac{120}{360} \times \dfrac{22}{7} \times 21 \times 21 = 462$ cm².
2. The two wipers do not overlap, so the area cleaned at each sweep is $2 \times 462 = 924$ cm².

Answer: $924$ cm²

1. Let $x$ toys be produced. The cost of each is $(55 - x)$, so the total is $x(55 - x) = 750$.
2. $x^2 - 55x + 750 = 0$, so $(x - 25)(x - 30) = 0$ and $x = 25$ or $x = 30$.
3. Both are valid (each cost, $30$ or $25$, is positive and the totals are $25 \times 30 = 750$). The industry made $25$ or $30$ toys that day.

Answer: $25$ or $30$ toys

1. Let the tower be $AB = h$, the nearer point $C$ with $AC = x$ (the canal width), and the farther point $D$ with $CD = 20$.
2. In $\triangle ABC$: $\tan 60^\circ = \dfrac{h}{x}$, so $h = x\sqrt{3}$. (i)
3. In $\triangle ABD$: $\tan 30^\circ = \dfrac{h}{x + 20} = \dfrac{1}{\sqrt{3}}$, so $\sqrt{3}\,h = x + 20$. (ii)
4. Substituting (i): $3x = x + 20$, so $x = 10$ m. $h = 10\sqrt{3} = 17.32$ m.

Answer: Height $= 17.32$ m; width of the canal $= 10$ m

1. (a)(i) In $\triangle AEP$ and $\triangle CDP$: $\angle APE = \angle CPD$ (vertically opposite) and $\angle AEP = \angle CDP = 90^\circ$. By AA, $\triangle AEP \sim \triangle CDP$.
2. (ii) In $\triangle ABD$ and $\triangle CBE$: $\angle B$ is common and $\angle ADB = \angle CEB = 90^\circ$. By AA, $\triangle ABD \sim \triangle CBE$.
3. (iii) In $\triangle AEP$ and $\triangle ADB$: $\angle A$ is common and $\angle AEP = \angle ADB = 90^\circ$. By AA, $\triangle AEP \sim \triangle ADB$.
4. (b) $\triangle ABC \sim \triangle PQR$ gives $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ and $\angle B = \angle Q$. $AD$ and $PM$ are medians, so $BC = 2BD$ and $QR = 2QM$: $\dfrac{AB}{PQ} = \dfrac{2BD}{2QM} = \dfrac{BD}{QM}$.
5. In $\triangle ABD$ and $\triangle PQM$: $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$ and $\angle B = \angle Q$. By SAS, $\triangle ABD \sim \triangle PQM$, so $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$.

Answer: Proved (both parts)

1. (a) Cuboid: $20 \times 14 \times 7 = 1960$ m³. The half cylinder has radius $\dfrac{14}{2} = 7$ m and length $20$ m: $\dfrac{1}{2}\pi r^2 h = \dfrac{1}{2} \times \dfrac{22}{7} \times 49 \times 20 = 1540$ m³.
2. Air the shed can hold $= 1960 + 1540 = 3500$ m³. Space left after the machinery $= 3500 - 400 = 3100$ m³.
3. (b) Slant height of the cone $l = \sqrt{6^2 + 8^2} = 10$ cm. The remaining solid has the curved surface of the cylinder, the bottom base, and the curved surface of the cone cavity (the top face is removed entirely).
4. Surface area $= 2\pi rh + \pi r^2 + \pi rl = \pi(2 \times 6 \times 8 + 36 + 6 \times 10) = \pi(96 + 36 + 60) = 192\pi = 192 \times 3.14 = 602.88$ cm².

Answer: (a) $3500$ m³ of air; $3100$ m³ left   (b) $602.88$ cm²

1. The number of coins added each day is $1, 2, 3, \dots$, an A.P. with $a = 1$ and $d = 1$.
2. (i) On the $8^{\text{th}}$ day $8$ coins are added.
3. (ii) Coins after $8$ days $= 1 + 2 + \dots + 8 = 36$, worth $36 \times 5 = ₹180$. (The money added each day is $5, 10, 15, \dots$.)
4. (iii)(a) $\dfrac{n}{2}[2(1) + (n - 1)(1)] = 120$ gives $n^2 + n - 240 = 0$, so $(n + 16)(n - 15) = 0$ and $n = 15$ (the root $-16$ is rejected): $15$ days.
5. (iii)(b) The full bank holds $120$ coins worth $120 \times 5 = ₹600$.

Answer: (i) $8$ coins   (ii) ₹ $180$   (iii)(a) $15$ days   (iii)(b) ₹ $600$

1. (i) The range $68 - 77$ covers the classes $68 - 71$, $71 - 74$ and $74 - 77$: $4 + 3 + 8 = 15$ women.
2. (ii) $N = 30$, $\dfrac{N}{2} = 15$. Cumulative frequencies: $2, 6, 9, 17, 24, 28, 30$. The median class is $74 - 77$.
3. (iii)(a) The modal class is $74 - 77$ ($f_1 = 8$, $f_0 = 3$, $f_2 = 7$, $h = 3$): Mode $= 74 + \dfrac{8 - 3}{16 - 3 - 7} \times 3 = 74 + 2.5 = 76.5$.
4. (iii)(b) Median $= 74 + \dfrac{15 - 9}{8} \times 3 = 74 + 2.25 = 76.25$.

Answer: (i) $15$   (ii) $74 - 77$   (iii)(a) $76.5$   (iii)(b) $76.25$

1. (i) Reading the grid: $A = (1, 9)$ and $B = (5, 13)$.
2. (ii) $C = (9, 13)$ and $D = (13, 9)$. The mid-point is $\left(\dfrac{9 + 13}{2}, \dfrac{13 + 9}{2}\right) = (11, 11)$.
3. (iii)(a) $M = (5, 11)$ and $Q = (9, 3)$: $MQ = \sqrt{(9 - 5)^2 + (3 - 11)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$ units.
4. (iii)(b) $M = (5, 11)$ and $N = (9, 11)$. The point dividing $MN$ in the ratio $1 : 3$ is $\left(\dfrac{1 \times 9 + 3 \times 5}{4}, \dfrac{1 \times 11 + 3 \times 11}{4}\right) = (6, 11)$.

Answer: (i) $A(1, 9)$, $B(5, 13)$   (ii) $(11, 11)$   (iii)(a) $4\sqrt{5}$   (iii)(b) $(6, 11)$