CBSE Class 10 Maths Basic 2024 Question Paper (Set 430/1/3) with Solutions

1. Product of zeroes $= \dfrac{c}{a} = \dfrac{-7}{k} = 2$, so $k = -\dfrac{7}{2}$.

Answer: (B) $-\dfrac{7}{2}$

1. $a_{10} = a + 9d$: $-19 = 8 + 9d$, so $9d = -27$ and $d = -3$.

Answer: (D) −3

1. $\left(\dfrac{-1 + 8}{2}, \dfrac{3 + \frac{3}{2}}{2}\right) = \left(\dfrac{7}{2}, \dfrac{9}{4}\right)$.

Answer: (D) $\left(\dfrac{7}{2}, \dfrac{9}{4}\right)$

1. $\cos\theta = \sqrt{1 - \dfrac{1}{9}} = \dfrac{2\sqrt{2}}{3}$, so $\sec\theta = \dfrac{3}{2\sqrt{2}}$.

Answer: (B) $\dfrac{3}{2\sqrt{2}}$

1. $132 = 2^2 \times 3 \times 11$ and $77 = 7 \times 11$. The common prime is $11$: HCF $= 11$.

Answer: (A) 11

1. Equal roots need $D = 25 - 16k = 0$, so $k = \dfrac{25}{16}$.

Answer: (B) $\dfrac{25}{16}$

1. Losing is the complement of winning: $P = 1 - p$.

Answer: (D) $1 - p$

1. $d = \sqrt{(2 + 2)^2 + (-3 - 3)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$.

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

1. $\sin^2\theta + \cos^2\theta = 1$, so the expression is $1 + \sin\theta = 2$, giving $\sin\theta = 1$ and $\theta = 90^\circ$.

Answer: (C) $90^\circ$

1. There are $2$ red queens (hearts and diamonds). $P = \dfrac{2}{52} = \dfrac{1}{26}$.

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

1. The value that divides ordered data into two equal halves is the median.

Answer: (B) median

1. $V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times \dfrac{343}{8} = \dfrac{539}{3}$ cm³.

Answer: (C) $\dfrac{539}{3}$ cu cm

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean} = 69 - 42 = 27$.

Answer: (A) 27

1. $l = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$ cm.

Answer: (D) 25 cm

1. $(\alpha - 1)(9) - 3\alpha + 1 = 0$ gives $9\alpha - 9 - 3\alpha + 1 = 0$, so $6\alpha = 8$ and $\alpha = \dfrac{4}{3}$.

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

1. The other end lies on the $x$-axis, at distance $6$ from $(-4, 0)$: $(-4 + 6, 0) = (2, 0)$ (or $(-10, 0)$, which is not in the options).

Answer: (C) (2, 0)

1. No solution needs $\dfrac{5}{2} = \dfrac{2}{k} \neq \dfrac{-7}{1}$. $\dfrac{5}{2} = \dfrac{2}{k}$ gives $k = \dfrac{4}{5}$, and the third ratio $-7$ is different ✓.

Answer: (B) $\dfrac{4}{5}$

1. A doublet means both dice show the same number: $(1,1), (2,2), \dots, (6,6)$, which is $6$ of $36$. $P = \dfrac{1}{6}$.

Answer: (C) $\dfrac{1}{6}$

1. (A) $\angle OAP = \angle OBP = 90^\circ$ (radius and tangent), so $\angle A + \angle B = 180^\circ$ for the opposite angles at $A$ and $B$ of $OAPB$. A quadrilateral with a pair of supplementary opposite angles is cyclic. True.
2. (R) In a cyclic quadrilateral the opposite angles are supplementary (they add to $180^\circ$), not equal. False.

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

1. (A) $x^2 - 2x - 3 = (x + 1)(x - 3)$, so the zeroes are $-1$ and $3$. True.
2. (R) The zeroes are exactly the $x$-values where the graph meets the $x$-axis, so it meets it at $(-1, 0)$ and $(3, 0)$. True, and this is the reason for (A).

Answer: (A) Both A and R are true and R explains A completely

1. In $\triangle ADC$ and $\triangle BAC$: $\angle ADC = \angle BAC$ (given) and $\angle C$ is common.
2. So $\triangle ADC \sim \triangle BAC$ by AA, and $\dfrac{AC}{BC} = \dfrac{DC}{AC}$. Hence $AC^2 = BC \times DC$.

Answer: Shown

1. (A) From the second equation $y = 2x + 2$. Substituting, $x + 4x + 4 = 9$, so $x = 1$ and $y = 4$. (Check: $1 + 8 = 9$ ✓.)
2. (B) $x + y + 1$ at $(-4, 3)$: $-4 + 3 + 1 = 0$ ✓, so it lies on the first line. $x - y = -4 - 3 = -7 \neq 1$, so it does not lie on the second.

Answer: (A) $x = 1$, $y = 4$   (B) on the first line only

1. (A) Suppose $6 - 4\sqrt{5} = x$ is rational. Then $\sqrt{5} = \dfrac{6 - x}{4}$, a rational number, which contradicts the given fact. So $6 - 4\sqrt{5}$ is irrational.
2. (B) $11 \times 19 \times 23 + 3 \times 11 = 11(19 \times 23 + 3) = 11 \times 440$. It has more than two factors, so it is not prime.

Answer: (A) proved   (B) $= 11 \times 440$, not prime

1. $\sin 30^\circ = \dfrac{1}{2}$, $\cos 45^\circ = \dfrac{1}{\sqrt{2}}$, $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$, $\sin 45^\circ = \dfrac{1}{\sqrt{2}}$.
2. $\dfrac{1}{2} \times \dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2} \times \dfrac{1}{\sqrt{2}} = \dfrac{1 + \sqrt{3}}{2\sqrt{2}}$. (This equals $\sin 75^\circ$.)

Answer: $\dfrac{\sqrt{3} + 1}{2\sqrt{2}}$

1. Let there be $n$ yellow balls: total $= 9 + n$. $P(\text{red}) = \dfrac{4}{9 + n} = \dfrac{1}{5}$, so $9 + n = 20$ and $n = 11$.
2. $P(\text{yellow}) = \dfrac{11}{20}$.

Answer: $\dfrac{11}{20}$

1. They beep together again after the LCM of $20$ and $25$ minutes. $20 = 2^2 \times 5$ and $25 = 5^2$, so LCM $= 2^2 \times 5^2 = 100$ minutes.
2. $100$ minutes after $12{:}00$ noon is $1{:}40$ p.m.

Answer: At $1{:}40$ p.m.

1. Let the angles be $x > y$. $x + y = 180^\circ$ and $x - y = 18^\circ$.
2. Adding, $2x = 198^\circ$, so $x = 99^\circ$ and $y = 81^\circ$.

Answer: $99^\circ$ and $81^\circ$

1. Let $A(-2, 2)$ and $B(7, -4)$. The point $P$ with $AP : PB = 1 : 2$: $\left(\dfrac{1(7) + 2(-2)}{3}, \dfrac{1(-4) + 2(2)}{3}\right) = (1, 0)$.
2. The point $Q$ with $AQ : QB = 2 : 1$: $\left(\dfrac{2(7) + 1(-2)}{3}, \dfrac{2(-4) + 1(2)}{3}\right) = (4, -2)$.

Answer: $(1, 0)$ and $(4, -2)$

1. (A) $OA \perp PA$, so $OA^2 = OP^2 - PA^2 = 400 - 256 = 144$ and $OA = r = 12$ cm.
2. $OQ \perp CD$ (radius and tangent), so in the right triangle $OQD$: $QD = \sqrt{OD^2 - OQ^2} = \sqrt{144 - 36} = \sqrt{108} = 6\sqrt{3}$ cm.
3. The perpendicular from the centre bisects the chord, so $CD = 2 \times 6\sqrt{3} = 12\sqrt{3}$ cm.
4. (B) Let $\angle PTQ = \theta$. $TP = TQ$, so $\angle TPQ = \angle TQP = 90^\circ - \dfrac{\theta}{2}$.
5. $OP \perp PT$, so $\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \dfrac{\theta}{2}\right) = \dfrac{\theta}{2}$. Hence $\angle PTQ = 2\angle OPQ$.

Answer: (A) $CD = 12\sqrt{3}$ cm   (B) proved

1. (A) $r = 7$ cm. The hemispheres take $2 \times 7 = 14$ cm of the height, so the cylinder is $20 - 14 = 6$ cm tall.
2. Surface area $= 2\pi r h + 2 \times 2\pi r^2 = 2\pi r(h + 2r) = 2 \times \dfrac{22}{7} \times 7 \times 20 = 880$ cm².
3. (B) $r = 5$ cm and $h = 14$ cm. Capacity $=$ cylinder $-$ raised hemisphere $= \pi r^2 h - \dfrac{2}{3}\pi r^3 = 3.14 \times 25 \times 14 - \dfrac{2}{3} \times 3.14 \times 125$.
4. $= 1099 - 261.67 \approx 837.33$ cm³ (exactly $\dfrac{2512}{3}$ cm³).

Answer: (A) $880$ cm²   (B) $\dfrac{2512}{3} \approx 837.33$ cm³

1. LHS $= \left(\dfrac{\cos\theta}{\sin\theta} - \dfrac{1}{\sin\theta}\right)^2 = \dfrac{(\cos\theta - 1)^2}{\sin^2\theta} = \dfrac{(1 - \cos\theta)^2}{1 - \cos^2\theta}$.
2. $1 - \cos^2\theta = (1 - \cos\theta)(1 + \cos\theta)$, so LHS $= \dfrac{(1 - \cos\theta)^2}{(1 - \cos\theta)(1 + \cos\theta)} = \dfrac{1 - \cos\theta}{1 + \cos\theta} =$ RHS.

Answer: Proved

1. (A) **Given:** $DE \parallel BC$ in $\triangle ABC$. **To prove:** $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. **Construction:** join $BE$ and $CD$; draw $DM \perp AC$ and $EN \perp AB$.
2. $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \dfrac{\frac{1}{2} \cdot AD \cdot EN}{\frac{1}{2} \cdot DB \cdot EN} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \dfrac{\frac{1}{2} \cdot AE \cdot DM}{\frac{1}{2} \cdot EC \cdot DM} = \dfrac{AE}{EC}$.
3. $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels, so their areas are equal. Hence $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
4. (B) In the figure the equal-ratio sides are the two sides from the apex, $AB$ with $PQ$ and $AC$ with $PR$ (the printed “$BC$” is a typo for $AC$). Given $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$.
5. Extend $AD$ to $E$ with $DE = AD$ and $PM$ to $N$ with $MN = PM$; join $CE$ and $RN$. $\triangle ADB \cong \triangle EDC$ (SAS), so $CE = AB$; likewise $RN = PQ$.
6. $\dfrac{CE}{RN} = \dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM} = \dfrac{AE}{PN}$ (as $AE = 2AD$, $PN = 2PM$), so $\triangle AEC \sim \triangle PNR$ (SSS) and $\angle 1 = \angle 2$. In the same way $\angle 3 = \angle 4$, so $\angle BAC = \angle QPR$.
7. With $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$, we get $\triangle ABC \sim \triangle PQR$ by SAS.

Answer: Proved (both parts)

1. $a = 27$ and $d = -3$. $S_n = \dfrac{n}{2}[54 + (n - 1)(-3)] = \dfrac{n}{2}(57 - 3n) = 105$, so $3n^2 - 57n + 210 = 0$, that is $n^2 - 19n + 70 = 0$ and $(n - 5)(n - 14) = 0$.
2. So $n = 5$ or $n = 14$. Both work: the first $5$ terms $27 + 24 + 21 + 18 + 15 = 105$, and the terms $6$ to $14$ ($12, 9, 6, 3, 0, -3, -6, -9, -12$) add to $0$, so $14$ terms also give $105$.
3. The zero term: $27 + (n - 1)(-3) = 0$ gives $n - 1 = 9$ and $n = 10$: the $10^{\text{th}}$ term is zero.

Answer: $5$ or $14$ terms; the $10^{\text{th}}$ term is zero

1. (A) Let the tower be $h$. The shadow at $60^\circ$ is $\dfrac{h}{\tan 60^\circ} = \dfrac{h}{\sqrt{3}}$ and at $30^\circ$ it is $h\sqrt{3}$. $h\sqrt{3} - \dfrac{h}{\sqrt{3}} = 40$.
2. $\dfrac{2h}{\sqrt{3}} = 40$, so $h = 20\sqrt{3} = 34.6$ m. The original shadow (at $60^\circ$) $= \dfrac{20\sqrt{3}}{\sqrt{3}} = 20$ m.
3. (B) Let the tall building be $CD = h$ and the distance between the buildings $x$. The bottom of the small building is at $45^\circ$: $\tan 45^\circ = \dfrac{h}{x} = 1$, so $x = h$.
4. Its top is $h - 8$ below $D$ at $30^\circ$: $\tan 30^\circ = \dfrac{h - 8}{x}$, so $h - 8 = \dfrac{x}{\sqrt{3}} = \dfrac{h}{\sqrt{3}}$.
5. $h\left(1 - \dfrac{1}{\sqrt{3}}\right) = 8$ gives $h = \dfrac{8\sqrt{3}}{\sqrt{3} - 1} = 4\sqrt{3}(\sqrt{3} + 1) = 12 + 4\sqrt{3} = 12 + 6.92 = 18.92$ m. The distance is also $x = h = 18.92$ m.

Answer: (A) height $= 20\sqrt{3} \approx 34.6$ m; original shadow $= 20$ m   (B) height and distance both $\approx 18.92$ m

1. Sector $OAB = \dfrac{90}{360} \times \dfrac{22}{7} \times 14 \times 14 = 154$ cm².
2. Triangle $OAB$ (right-angled at $O$ with both legs $14$): $\dfrac{1}{2} \times 14 \times 14 = 98$ cm².
3. Minor segment $= 154 - 98 = 56$ cm². Area of the circle $= \dfrac{22}{7} \times 196 = 616$ cm², so the major segment $= 616 - 56 = 560$ cm².

Answer: Minor segment $= 56$ cm²; major segment $= 560$ cm²

1. (i) The sum of the areas is $\pi R^2 + \pi r^2 = 130\pi$, so $R^2 + r^2 = 130$.
2. (ii) The circles touch externally, so $R + r = 14$ and $R = 14 - r$. $(14 - r)^2 + r^2 = 130$ gives $2r^2 - 28r + 66 = 0$, that is $r^2 - 14r + 33 = 0$.
3. (iii)(a) $(r - 3)(r - 11) = 0$, so $r = 3$ or $11$. As $r < R$, $r = 3$ m. Area irrigated $= \pi r^2 = 9\pi$ m² ($\approx 28.3$ m²).
4. (iii)(b) $R = 14 - 3 = 11$ m (the quadratic $R^2 - 14R + 33 = 0$ gives $R = 11$ or $3$, and $R > r$). Area irrigated $= \pi R^2 = 121\pi$ m² ($\approx 380.1$ m²).

Answer: (i) $R^2 + r^2 = 130$   (ii) $r^2 - 14r + 33 = 0$   (iii)(a) $r = 3$ m, area $9\pi$ m²   (iii)(b) $R = 11$ m, area $121\pi$ m²

1. $N = 3 + 5 + 9 + 12 + 5 + 4 + 2 = 40$. Cumulative frequencies: $3, 8, 17, 29, 34, 38, 40$. $\dfrac{N}{2} = 20$.
2. (i) The cumulative frequency passes $20$ in the class $100 - 110$, so that is the median class.
3. (ii) $10$ cm $= 100$ mm. Leaves of length $100$ mm or more: $12 + 5 + 4 + 2 = 23$.
4. (iii)(a) $l = 100$, $cf = 17$, $f = 12$, $h = 10$. Median $= 100 + \dfrac{20 - 17}{12} \times 10 = 102.5$ mm.
5. (iii)(b) The highest frequency is $12$, so the modal class is $100 - 110$: $f_1 = 12$, $f_0 = 9$, $f_2 = 5$. Mode $= 100 + \dfrac{12 - 9}{24 - 9 - 5} \times 10 = 103$ mm.

Answer: (i) $100 - 110$   (ii) $23$   (iii)(a) $102.5$ mm   (iii)(b) modal class $100 - 110$, mode $= 103$ mm

1. (i) $AP = AQ$ (tangents from $A$) and $\angle PAQ = 60^\circ$, so $\triangle APQ$ is isosceles with an apex angle of $60^\circ$, which makes it equilateral. $PQ = AP = 30$ cm.
2. (ii) $OP \perp AP$ and $OQ \perp AQ$, so $\angle POQ = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ$.
3. (iii)(a) $AO$ bisects $\angle PAQ$, so $\angle PAO = 30^\circ$. In the right triangle $APO$: $\cos 30^\circ = \dfrac{AP}{AO} = \dfrac{30}{AO}$, so $AO = \dfrac{30 \times 2}{\sqrt{3}} = 20\sqrt{3}$ cm.
4. (iii)(b) $\tan 30^\circ = \dfrac{OP}{AP}$, so $OP = 30 \times \dfrac{1}{\sqrt{3}} = 10\sqrt{3}$ cm, the radius of the mirror.

Answer: (i) $30$ cm   (ii) $120^\circ$   (iii)(a) $20\sqrt{3}$ cm   (iii)(b) $10\sqrt{3}$ cm