CBSE Class 10 Maths Basic 2024 Question Paper (Set 430/5/1) with Solutions

1. $\text{HCF} \times \text{LCM}$ equals the product of the two numbers: $40 \times 30 = 1200$.

Answer: (C) 1200

1. $x^2 + 3x - 10 = (x + 5)(x - 2) = 0$, so $x = -5$ or $x = 2$.

Answer: (B) −5, 2

1. A unique solution needs $\dfrac{2k}{6} \neq \dfrac{5}{5} = 1$, so $2k \neq 6$ and $k \neq 3$.

Answer: (A) $k \neq 3$

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$: $16 = 3M - 56$, so $M = 24$.

Answer: (C) 24

1. The odd primes on a die are $3$ and $5$ ($2$ is even, $1$ is not prime). $P = \dfrac{2}{6} = \dfrac{1}{3}$.

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

1. $\dfrac{\theta}{360} = \dfrac{1}{8}$, so $\theta = 45^\circ$.

Answer: (B) $45^\circ$

1. $424 = 2 \times 212 = 2 \times 2 \times 106 = 2 \times 2 \times 2 \times 53 = 2^3 \times 53$.

Answer: (C) $2^3 \times 53$

1. $8^x$ ends in $8, 4, 2, 6$ for $x = 1, 2, 3, 4, \dots$, all even. But $8^0 = 1$, which is odd. So the exception is $x = 0$.

Answer: (D) 0

1. $\tan 45^\circ = \dfrac{\text{height}}{7} = 1$, so the height is $7$ m.

Answer: (C) 7 m

1. $\dfrac{5}{\cot^2\theta} = 5\tan^2\theta$ and $\dfrac{5}{\cos^2\theta} = 5\sec^2\theta$. The value is $5(\tan^2\theta - \sec^2\theta) = 5 \times (-1) = -5$.

Answer: (C) −5

1. $\tan\theta = \sqrt{3}$ (acute angle), so $\theta = 60^\circ$.

Answer: (B) $60^\circ$

1. $10\pi = \dfrac{\theta}{360} \times 2\pi \times 12 = \dfrac{\theta}{360} \times 24\pi$, so $\theta = \dfrac{10 \times 360}{24} = 150^\circ$.

Answer: (D) $150^\circ$

1. The angle between the tangents and the angle between the radii add up to $180^\circ$: $180^\circ - 130^\circ = 50^\circ$.

Answer: (A) $50^\circ$

1. $D = b^2 - 4ac = 16 - 12 = 4$.

Answer: (C) 4

1. $\left(\dfrac{-2 - 6}{2}, \dfrac{8 + 4}{2}\right) = (-4, 6)$.

Answer: (C) (−4, 6)

1. $OP \perp AP$ and $OQ \perp AQ$, and $\angle PAQ = 90^\circ$, so $OPAQ$ has four right angles. $OP = OQ$ (radii) and $AP = AQ = 2$, so it is a square of side $2$ and the radius is $2$ cm.

Answer: (B) 2 cm

1. $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$, so $\angle B = \angle E = 45^\circ$. $\angle C = 180^\circ - 55^\circ - 45^\circ = 80^\circ$.

Answer: (A) $80^\circ$

1. $DE \parallel BC$, so $\dfrac{AD}{DB} = \dfrac{AE}{EC}$: $\dfrac{2x}{4} = \dfrac{x + 2}{3}$. $6x = 4x + 8$, so $x = 4$.

Answer: (D) 4

1. (R) An impossible event has no favourable outcomes, so $P = 0$. True.
2. (A) A die has no face $8$, so getting $8$ is impossible and $P = 0$. True, and (R) is exactly why.

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

1. (R) The common difference is any term minus the previous term. True.
2. (A) By (R), $d = 1 - 5 = -4$, not $4$. False.

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

1. $\left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{\sqrt{2}}\right)^2 - 1 \times 1 = \dfrac{1}{4} + \dfrac{1}{2} - 1 = -\dfrac{1}{4}$.

Answer: $-\dfrac{1}{4}$

1. (a) Let the ratio be $k : 1$. The $x$-coordinate: $\dfrac{6k - 2}{k + 1} = 3$, so $6k - 2 = 3k + 3$ and $k = \dfrac{5}{3}$. The ratio is $5 : 3$.
2. $y = \dfrac{k \times 3 + 1 \times (-5)}{k + 1} = \dfrac{5 - 5}{\frac{8}{3}} = 0$.
3. (b) Let the point be $P(0, y)$. $PA^2 = PB^2$: $(0 - 6)^2 + (y - 5)^2 = (0 + 4)^2 + (y - 3)^2$, so $36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9$ and $36 = 4y$, $y = 9$. $P = (0, 9)$.

Answer: (a) ratio $5 : 3$, $y = 0$   (b) $(0, 9)$

1. Substituting $x = \dfrac{2}{3}$: $k \times \dfrac{4}{9} - \dfrac{2}{3} - 2 = 0$, so $\dfrac{4k}{9} = \dfrac{8}{3}$.
2. $k = \dfrac{8}{3} \times \dfrac{9}{4} = 6$.

Answer: $k = 6$

1. The radius to the point of contact is perpendicular to the tangent, so $r = \sqrt{OA^2 - AT^2} = \sqrt{100 - 36} = 8$ cm.

Answer: $8$ cm

1. (a) $\alpha + \beta = -\dfrac{7}{2}$ and $\alpha\beta = \dfrac{5}{2}$. $\alpha^2 + \beta^2 + \alpha\beta = (\alpha + \beta)^2 - \alpha\beta = \dfrac{49}{4} - \dfrac{5}{2} = \dfrac{39}{4}$.
2. (b) The zeroes are $\alpha$ and $\dfrac{1}{\alpha}$, so their product is $1 = \dfrac{-(p - 2)}{6}$. Then $p - 2 = -6$ and $p = -4$.

Answer: (a) $\dfrac{39}{4}$   (b) $p = -4$

1. Suppose $6 + 3\sqrt{2} = x$ is rational. Then $\sqrt{2} = \dfrac{x - 6}{3}$, which is rational.
2. This contradicts the given fact that $\sqrt{2}$ is irrational. So $6 + 3\sqrt{2}$ is irrational.

Answer: Proved

1. $6x^2 - 7x - 3 = (2x - 3)(3x + 1)$, so the zeroes are $\dfrac{3}{2}$ and $-\dfrac{1}{3}$.
2. Sum $= \dfrac{3}{2} - \dfrac{1}{3} = \dfrac{7}{6} = -\dfrac{-7}{6}$ ✓. Product $= \dfrac{3}{2} \times \left(-\dfrac{1}{3}\right) = -\dfrac{1}{2} = \dfrac{-3}{6}$ ✓.

Answer: Zeroes $\dfrac{3}{2}, -\dfrac{1}{3}$; verified

1. LHS $= \sin^2\theta + \operatorname{cosec}^2\theta + 2\sin\theta\operatorname{cosec}\theta + \cos^2\theta + \sec^2\theta + 2\cos\theta\sec\theta$.
2. $= (\sin^2\theta + \cos^2\theta) + (1 + \cot^2\theta) + (1 + \tan^2\theta) + 2 + 2 = 7 + \tan^2\theta + \cot^2\theta =$ RHS.

Answer: Proved

1. (a) Let the parallelogram $ABCD$ touch the circle at $P, Q, R, S$ on $AB, BC, CD, DA$. Tangents from a point are equal: $AP = AS$, $PB = BQ$, $CR = CQ$, $DR = DS$.
2. Adding: $AB + CD = AD + BC$. In a parallelogram $AB = CD$ and $AD = BC$, so $2AB = 2AD$ and $AB = AD$. All sides are equal: $ABCD$ is a rhombus.
3. (b) Let $PA$ and $PB$ be tangents from $P$ to a circle with centre $O$. Join $OA$, $OB$ and $OP$. In $\triangle AOP$ and $\triangle BOP$: $OA = OB$ (radii), $OP$ is common and $\angle OAP = \angle OBP = 90^\circ$ (radius and tangent).
4. So $\triangle AOP \cong \triangle BOP$ (RHS) and $PA = PB$ by CPCT.

Answer: Proved (both parts)

1. (a) The corner angle is $60^\circ$, so the horse grazes a $60^\circ$ sector of radius $14$ m: $\dfrac{60}{360} \times \dfrac{22}{7} \times 14 \times 14 = \dfrac{308}{3} \approx 102.67$ m².
2. The field has area $\dfrac{\sqrt{3}}{4} \times 20^2 = 100\sqrt{3} \approx 173.2$ m². The part it cannot graze $= 173.2 - 102.67 \approx 70.5$ m². (The key uses $173$ and gets $70.33$ m².)
3. (b) In $60$ minutes the hand turns $360^\circ$, so in $5$ minutes it turns $30^\circ$. Area $= \dfrac{30}{360} \times \dfrac{22}{7} \times 14 \times 14 = \dfrac{154}{3} \approx 51.33$ cm².

Answer: (a) $\approx 70.5$ m²   (b) $\dfrac{154}{3} \approx 51.33$ cm²

1. $P$ with $AP : PB = 1 : 2$: $\left(\dfrac{1(-4) + 2(5)}{3}, \dfrac{1(3) + 2(-3)}{3}\right) = (2, -1)$.
2. $Q$ with $AQ : QB = 2 : 1$: $\left(\dfrac{2(-4) + 1(5)}{3}, \dfrac{2(3) + 1(-3)}{3}\right) = (-1, 1)$.

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

1. (a) Cone height $= 31 - 7 = 24$ cm, slant height $l = \sqrt{7^2 + 24^2} = 25$ cm.
2. Surface area $=$ curved area of the hemisphere $+$ curved area of the cone $= 2\pi r^2 + \pi r l = \pi r(2r + l) = \dfrac{22}{7} \times 7 \times (14 + 25) = 22 \times 39 = 858$ cm².
3. (b) Radius $4.2$ cm, cylinder height $15$ cm. The surface is the curved area of the cylinder and the two hemispherical hollows: $2\pi r h + 2 \times 2\pi r^2 = 2\pi r(h + 2r)$.
4. $= 2 \times \dfrac{22}{7} \times 4.2 \times (15 + 8.4) = 26.4 \times 23.4 = 617.76$ cm².

Answer: (a) $858$ cm²   (b) $617.76$ cm²

1. Let the incomes be $9x$ and $7x$ and the expenditures $4y$ and $3y$. Saving $=$ income $-$ expenditure: $9x - 4y = 2000$ (i) and $7x - 3y = 2000$ (ii).
2. (i) $\times 3$: $27x - 12y = 6000$. (ii) $\times 4$: $28x - 12y = 8000$. Subtracting, $x = 2000$.
3. The incomes are $9 \times 2000 = ₹18{,}000$ and $7 \times 2000 = ₹14{,}000$. (Check: $y = 4000$, expenditures ₹ $16000$ and ₹ $12000$, savings ₹ $2000$ each ✓.)

Answer: ₹ $18{,}000$ and ₹ $14{,}000$

1. (a) Given $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$. $AD$ and $PM$ are medians, so $BC = 2BD$ and $QR = 2QM$: $\dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$.
2. So $\triangle ABD \sim \triangle PQM$ (SSS) and $\angle B = \angle Q$. In $\triangle ABC$ and $\triangle PQR$: $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ and $\angle B = \angle Q$, so $\triangle ABC \sim \triangle PQR$ by SAS.
3. (b) **Given:** $DE \parallel BC$ in $\triangle ABC$. **To prove:** $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. **Construction:** join $BE$ and $DC$; draw $DM \perp AC$ and $EN \perp AB$.
4. $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \dfrac{AE}{EC}$. The triangles $BDE$ and $CDE$ stand on the same base $DE$ between the same parallels, so their areas are equal. Hence $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

Answer: Proved (both parts)

1. Cumulative frequencies: $7, 15, 22, 42, 56, 65, 70$. $N = 70$ and $\dfrac{N}{2} = 35$.
2. The cumulative frequency passes $35$ in $125 - 145$: $l = 125$, $cf = 22$, $f = 20$, $h = 20$.
3. Median $= 125 + \dfrac{35 - 22}{20} \times 20 = 125 + 13 = 138$ units.

Answer: $138$ units

1. Out of $100$ families: $51$ have $0$, $20$ have $1$, $19$ have $2$, $7$ have $3$ and $3$ have $4$ or more children.
2. (i) $P(2 \text{ or } 3) = \dfrac{19 + 7}{100} = \dfrac{26}{100} = \dfrac{13}{50}$.
3. (ii) $P(\text{more than one}) = \dfrac{19 + 7 + 3}{100} = \dfrac{29}{100}$.
4. (iii)(a) $P(\text{less than } 3) = \dfrac{51 + 20 + 19}{100} = \dfrac{90}{100} = \dfrac{9}{10}$. (b) $P(\text{more than } 2) = \dfrac{7 + 3}{100} = \dfrac{1}{10}$.

Answer: (i) $\dfrac{13}{50}$   (ii) $\dfrac{29}{100}$   (iii)(a) $\dfrac{9}{10}$   (iii)(b) $\dfrac{1}{10}$

1. The pairs per row are $3, 5, 7, \dots$, an A.P. with $a = 3$ and $d = 2$.
2. (i) $a_6 = 3 + 5 \times 2 = 13$ pairs.
3. (ii) The difference between the $6^{\text{th}}$ and $1^{\text{st}}$ rows is $13 - 3 = 10$ pairs.
4. (iii)(a) $S_{15} = \dfrac{15}{2}[2(3) + 14 \times 2] = \dfrac{15}{2} \times 34 = 255$ pairs.
5. (iii)(b) $a_4 = 3 + 3 \times 2 = 9$ pairs, earning $9 \times 500 = ₹4{,}500$.

Answer: (i) $13$   (ii) $10$   (iii)(a) $255$ pairs   (iii)(b) ₹ $4{,}500$

1. (i) In the right triangle $ABP$ ($\angle P = 30^\circ$, $AP = 40$): $\tan 30^\circ = \dfrac{AB}{40}$, so $AB = \dfrac{40}{\sqrt{3}} = \dfrac{40\sqrt{3}}{3}$ m.
2. (ii) In the right triangle $APQ$ ($\angle A = 60^\circ$): $\tan 60^\circ = \dfrac{PQ}{40}$, so $PQ = 40\sqrt{3}$ m.
3. (iii)(a) Draw $BM \perp PQ$ with $BM = AP = 40$. In $\triangle BQM$: $\cos 30^\circ = \dfrac{BM}{BQ}$, so $BQ = \dfrac{40}{\sqrt{3}/2} = \dfrac{80}{\sqrt{3}} = \dfrac{80\sqrt{3}}{3}$ m.
4. (iii)(b) Let $CD$ be the perpendicular from $C$ to the ground, $CD = 20$ m. In the right triangle $ACD$ ($\angle A = 60^\circ$): $\sin 60^\circ = \dfrac{CD}{AC}$, so $AC = \dfrac{20}{\sqrt{3}/2} = \dfrac{40}{\sqrt{3}} = \dfrac{40\sqrt{3}}{3}$ m, the distance of $C$ along the wire from $A$.

Answer: (i) $\dfrac{40\sqrt{3}}{3}$ m   (ii) $40\sqrt{3}$ m   (iii)(a) $\dfrac{80\sqrt{3}}{3}$ m   (iii)(b) $\dfrac{40\sqrt{3}}{3}$ m