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

1. $PA = PB$ (tangents from $P$), so $\triangle PAB$ is isosceles and $\angle PAB = \dfrac{180^\circ - 80^\circ}{2} = 50^\circ$.

Answer: (C) $50^\circ$

1. Equal roots need $D = (9k)^2 - 4(5)(5) = 81k^2 - 100 = 0$, so $k^2 = \dfrac{100}{81}$ and $k = \pm\dfrac{10}{9}$.

Answer: (D) $\pm\dfrac{10}{9}$

1. $d = \sqrt{(6 + 1)^2 + (-2 - 5)^2} = \sqrt{49 + 49} = 7\sqrt{2}$.

Answer: (B) $7\sqrt{2}$

1. Sum $= 3 + (-2) = 1$ and product $= -6$. The polynomial is $x^2 - x - 6$. (Option (C) has the same zeroes only when divided by $2$: $2x^2 - x - 12$ has zeroes $\dfrac{1 \pm \sqrt{97}}{4}$, not $3$ and $-2$.)

Answer: (A) $x^2 - x - 6$

1. $x = a$ is a vertical line and $y = b$ is a horizontal line. They meet at the single point with $x = a$ and $y = b$, that is $(a, b)$.

Answer: (A) intersecting at $(a, b)$

1. $DE \parallel BC$, so $\dfrac{AD}{DB} = \dfrac{AE}{EC}$: $\dfrac{2.8}{2.1} = \dfrac{x}{4.8}$, so $x = \dfrac{2.8 \times 4.8}{2.1} = 6.4$ cm.

Answer: (C) 6.4 cm

1. $AB = AC$ makes the triangle isosceles right-angled, so $\angle C = 45^\circ$ and $\sin C = \dfrac{1}{\sqrt{2}}$.

Answer: (D) $\dfrac{1}{\sqrt{2}}$

1. The numbers $1$ to $5$ are favourable: $P = \dfrac{5}{6}$.

Answer: (B) $\dfrac{5}{6}$

1. A tangent through $P$ parallel to $PA$ would have to be a different line through $P$ with the same direction as $PA$. There is no such line, since the only line through $P$ with that direction is $PA$ itself. So there are $0$.

Answer: (D) 0

1. $\dfrac{4}{3} \times \dfrac{22}{7} \times r^3 = \dfrac{88}{21}r^3 = \dfrac{11}{21}$, so $r^3 = \dfrac{1}{8}$ and $r = \dfrac{1}{2}$ cm.

Answer: (C) $\dfrac{1}{2}$ cm

1. The zeroes are the points where the graph meets the $x$-axis. It crosses it at $2$ points.

Answer: (D) 2

1. Area $= \dfrac{60}{360}\pi(5^2 - 3^2) = \dfrac{\pi}{6} \times 16 = \dfrac{8\pi}{3}$ cm².

Answer: (B) $\dfrac{8\pi}{3}$ cm²

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean} = 33 - 20 = 13$.

Answer: (D) 13

1. The outcomes are $HH, HT, TH, TT$; only $HH$ has $2$ heads. $P = \dfrac{1}{4}$.

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

1. $\cos A = \dfrac{1}{2}$ gives $A = 60^\circ$ and $\tan 60^\circ = \sqrt{3}$.

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

1. The biggest sphere has a diameter equal to the edge of the cube, $14$ cm, so the radius is $7$ cm.

Answer: (A) 7 cm

1. $a_{n+1} - a_n = (5(n + 1) - 1) - (5n - 1) = 5$.

Answer: (C) 5

1. Mean $= \dfrac{132 + 5p}{20} = 8.1$, so $132 + 5p = 162$ and $p = 6$.

Answer: (B) 6

1. (R) This is the distance formula. True.
2. (A) Using it with the origin $(0, 0)$, the distance is $\sqrt{a^2 + b^2}$, not $a^2 + b^2$. False.

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

1. (A) $\sqrt{2}(5 - \sqrt{2}) = 5\sqrt{2} - 2$, which is irrational (an irrational number minus a rational one). True.
2. (R) $\sqrt{2} \times \sqrt{2} = 2$ is rational, so the product of two irrationals need not be irrational. False.

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

1. (a) $QP = QR$: $(0 - 5)^2 + (2 + 3)^2 = (0 - x)^2 + (2 - 7)^2$, so $50 = x^2 + 25$ and $x^2 = 25$, giving $x = 5$ or $x = -5$ (the key lists $x = 5$).
2. (b) The centre is the mid-point of the diameter: $\left(\dfrac{1 + 7}{2}, \dfrac{1 + 9}{2}\right) = (4, 5)$.

Answer: (a) $x = \pm 5$   (b) $(4, 5)$

1. $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$, $\tan 45^\circ = 1$, $\sec 30^\circ = \dfrac{2}{\sqrt{3}}$, $\tan 60^\circ = \sqrt{3}$.
2. $4 \times \dfrac{3}{4} \times 1 - 2 \times \dfrac{4}{3} \times 3 = 3 - 8 = -5$.

Answer: $-5$

1. (a) Suppose $x = -7 - 2\sqrt{3}$ is rational. Then $\sqrt{3} = \dfrac{-7 - x}{2}$ is rational, which contradicts the given fact. So $-7 - 2\sqrt{3}$ is irrational.
2. (b) $7 \times 11 \times 13 + 2 \times 11 = 11(7 \times 13 + 2) = 11 \times 93$. It has more than two factors, so it is not prime.

Answer: (a) proved   (b) $= 11 \times 93$, not prime

1. A point on the $y$-axis has $x = 0$. Let it divide $AB$ in the ratio $k : 1$: $\dfrac{-k + 5}{k + 1} = 0$, so $k = 5$. The ratio is $5 : 1$.
2. $y = \dfrac{5(-4) + 1(-6)}{6} = -\dfrac{26}{6} = -\dfrac{13}{3}$. The point is $\left(0, -\dfrac{13}{3}\right)$.

Answer: Ratio $5 : 1$; point $\left(0, -\dfrac{13}{3}\right)$

1. Multiples of $8$ up to $80$: $8, 16, \dots, 80$, which is $10$ cards. $P(\text{divisible by } 8) = \dfrac{10}{80} = \dfrac{1}{8}$.
2. $P(\text{not divisible by } 8) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$.

Answer: $\dfrac{7}{8}$

1. $336 = 2^4 \times 3 \times 7$ and $54 = 2 \times 3^3$.
2. HCF $= 2 \times 3 = 6$ and LCM $= 2^4 \times 3^3 \times 7 = 3024$. (Check: $6 \times 3024 = 18144 = 336 \times 54$ ✓.)

Answer: HCF $= 6$, LCM $= 3024$

1. (a) Let the base be $x$ cm, so the altitude is $(x - 7)$ cm. Pythagoras: $17^2 = x^2 + (x - 7)^2$, so $2x^2 - 14x - 240 = 0$, that is $x^2 - 7x - 120 = 0$.
2. (b) $(x - 15)(x + 8) = 0$. A length cannot be negative, so $x = 15$. The sides other than the hypotenuse are $15$ cm and $8$ cm.

Answer: (a) $x^2 - 7x - 120 = 0$   (b) $15$ cm and $8$ cm

1. (a) $r = 7$ cm. The height of the cylinder is $13 - 7 = 6$ cm. Inner surface area $= 2\pi rh + 2\pi r^2 = 2 \times \dfrac{22}{7} \times 7 \times 6 + 2 \times \dfrac{22}{7} \times 7 \times 7 = 264 + 308 = 572$ cm².
2. (b) $r = 2$ cm. Volume $= \dfrac{1}{3}\pi r^2 h + \dfrac{2}{3}\pi r^3 = \dfrac{1}{3} \times \dfrac{22}{7} \times 4 \times 2 + \dfrac{2}{3} \times \dfrac{22}{7} \times 8 = \dfrac{176}{21} + \dfrac{352}{21} = \dfrac{176}{7}$ cm³ $\approx 25.14$ cm³.

Answer: (a) $572$ cm²   (b) $\dfrac{176}{7} \approx 25.14$ cm³

1. $5x^2 + 3x - 2 = (5x - 2)(x + 1)$, so the zeroes are $\dfrac{2}{5}$ and $-1$.
2. Sum $= -1 + \dfrac{2}{5} = -\dfrac{3}{5} = -\dfrac{3}{5}$ ✓. Product $= -1 \times \dfrac{2}{5} = -\dfrac{2}{5} = \dfrac{-2}{5}$ ✓.

Answer: Zeroes $\dfrac{2}{5}$ and $-1$; relations verified

1. $\angle ABC = \angle ACB$, so $AC = AB$. The given $\dfrac{BC}{BE} = \dfrac{BD}{AC}$ becomes $\dfrac{BC}{BE} = \dfrac{BD}{AB}$, that is $\dfrac{BE}{AB} = \dfrac{BC}{BD}$.
2. In $\triangle ABE$ and $\triangle DBC$: $\angle B$ is common and the sides including it are proportional, $\dfrac{AB}{DB} = \dfrac{BE}{BC}$. So $\triangle ABE \sim \triangle DBC$ by SAS.
3. Then $\angle BAE = \angle BDC$, which are corresponding angles for $AE$ and $DC$ with the transversal $BD$. So $AE \parallel DC$.

Answer: Shown

1. (a) LHS $= \sin^2\theta + \operatorname{cosec}^2\theta + 2\sin\theta\operatorname{cosec}\theta + \cos^2\theta + \sec^2\theta + 2\cos\theta\sec\theta$. With $\sin\theta\operatorname{cosec}\theta = \cos\theta\sec\theta = 1$: $= 1 + (1 + \cot^2\theta) + 2 + (1 + \tan^2\theta) + 2 = 7 + \tan^2\theta + \cot^2\theta$.
2. (b) $\cos A = \dfrac{5}{13}$ gives $\sin A = \dfrac{12}{13}$, $\tan A = \dfrac{12}{5}$ and $\cot A = \dfrac{5}{12}$.
3. LHS $= \dfrac{5/13}{1 - 12/5} + \dfrac{12/13}{1 - 5/12} = \dfrac{5/13}{-7/5} + \dfrac{12/13}{7/12} = -\dfrac{25}{91} + \dfrac{144}{91} = \dfrac{119}{91} = \dfrac{17}{13}$. RHS $= \dfrac{5}{13} + \dfrac{12}{13} = \dfrac{17}{13}$ ✓.

Answer: (a) proved   (b) verified, both sides $\dfrac{17}{13}$

1. $N = 50$. Class marks: $110, 130, 150, 170, 190$. Take $a = 150$, $h = 20$: $u = -2, -1, 0, 1, 2$. $\sum f_iu_i = -24 - 14 + 0 + 6 + 20 = -12$.
2. Mean $= 150 + \dfrac{-12}{50} \times 20 = 150 - 4.8 = 145.2$.
3. Cumulative frequencies: $12, 26, 34, 40, 50$. $\dfrac{N}{2} = 25$ lies in $120 - 140$: $l = 120$, $cf = 12$, $f = 14$, $h = 20$. Median $= 120 + \dfrac{25 - 12}{14} \times 20 = 120 + 18.57 \approx 138.57$.

Answer: Mean $= 145.2$; median $\approx 138.57$

1. (a) $OP$ is the perpendicular bisector of $AB$, so $BQ = QA = 3$ cm, and in the right triangle $OQB$: $OQ = \sqrt{25 - 9} = 4$ cm.
2. Let $PQ = x$ and $PB = y$. In $\triangle PQB$: $x^2 + 9 = y^2$. In $\triangle OBP$ (right angle at $B$): $(x + 4)^2 = y^2 + 25$.
3. $(x + 4)^2 = x^2 + 9 + 25$ gives $8x + 16 = 34$, so $x = \dfrac{9}{4}$. $y^2 = \dfrac{81}{16} + 9 = \dfrac{225}{16}$, so $PB = y = \dfrac{15}{4} = 3.75$ cm.
4. (b) Let the parallelogram $ABCD$ touch the circle at $P, Q, R, S$. Tangent lengths are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$. Adding: $AB + CD = AD + BC$.
5. In a parallelogram $AB = CD$ and $AD = BC$, so $2AB = 2BC$ and $AB = BC$. All four sides are equal, so $ABCD$ is a rhombus.
6. Area of the rhombus $= 4 \times \text{ar}(\triangle AOB) = 4 \times \dfrac{1}{2} \times 10 \times 3 = 60$ cm² (each triangle has base the side $10$ and height the radius $3$).

Answer: (a) $PB = \dfrac{15}{4}$ cm   (b) proved; area $= 60$ cm²

1. Let the poles be $AB$ and $CD$ of height $h$, and $P$ the point on $AC$, at distances $y$ from $A$ and $x$ from $C$.
2. $\tan 60^\circ = \dfrac{h}{y}$ gives $h = y\sqrt{3}$. (i) $\tan 30^\circ = \dfrac{h}{x}$ gives $h = \dfrac{x}{\sqrt{3}}$. (ii)
3. From (i) and (ii): $x = 3y$. With $x + y = 100$: $4y = 100$, so $y = 25$ m and $x = 75$ m.
4. $h = 25\sqrt{3} \approx 43.3$ m.

Answer: Height $= 25\sqrt{3} \approx 43.3$ m; the point is $25$ m and $75$ m from the poles

1. (a) $x + 2y = 8$: points $(0, 4)$, $(8, 0)$, $(2, 3)$. $3x - 2y = 12$: points $(4, 0)$, $(0, -6)$, $(6, 3)$. The lines meet at $(5, 1.5)$, so $x = 5$ and $y = 1.5$. (Check: $5 + 3 = 8$ ✓ and $15 - 3 = 12$ ✓.)
2. (b) Let the tens digit be $y$ and the ones digit $x$. $x + y = 9$ and $9(10y + x) = 2(10x + y)$, so $90y + 9x = 20x + 2y$, that is $88y = 11x$ and $x = 8y$.
3. $8y + y = 9$ gives $y = 1$ and $x = 8$. The number is $18$. (Check: $9 \times 18 = 162 = 2 \times 81$ ✓.)

Answer: (a) $x = 5$, $y = 1.5$   (b) $18$

1. (i) The squares in Fig. (1), (2), (3) are $1, 5, 9, \dots$, an A.P. with first term $1$ and common difference $4$.
2. (ii) The sticks are $4, 16, 28, \dots$, an A.P. with first term $4$ and common difference $12$.
3. (iii)(a) Squares in Fig. (10) $= 1 + 9 \times 4 = 37$. Sticks in Fig. (10) $= 4 + 9 \times 12 = 112$.
4. (iii)(b) $88 = 4 + (m - 1) \times 12$ gives $m - 1 = 7$, so $m = 8$. Squares in Fig. (8) $= 1 + 7 \times 4 = 29$.

Answer: (i) $1$ and $4$   (ii) $4$ and $12$   (iii)(a) $37$ squares, $112$ sticks   (iii)(b) $m = 8$, $29$ squares

1. (i) Four diameters equally placed make $8$ equal angles at $O$: $\angle ROS = \dfrac{360^\circ}{8} = 45^\circ$.
2. (ii) Perimeter of sector $OPQ = OP + OQ + \text{arc } PQ = 21 + 21 + \dfrac{45}{360} \times 2 \times \dfrac{22}{7} \times 21 = 42 + 16.5 = 58.5$ cm.
3. (iii)(a) Area $PQRS = \text{sector } OPQ - \text{sector } ORS = \dfrac{45}{360} \times \dfrac{22}{7} \times (21^2 - 10^2) = \dfrac{1}{8} \times \dfrac{22}{7} \times 341 = \dfrac{3751}{28} \approx 133.96$ cm².
4. (iii)(b) The segment $ACB$ is cut off by the chord $AB$, which subtends $90^\circ$ at $O$ (two sectors of $45^\circ$). Area $= \dfrac{90}{360} \times \dfrac{22}{7} \times 100 - \dfrac{1}{2} \times 10 \times 10 = \dfrac{550}{7} - 50 = \dfrac{200}{7} \approx 28.57$ cm².

Answer: (i) $45^\circ$   (ii) $58.5$ cm   (iii)(a) $\approx 133.96$ cm²   (iii)(b) $\dfrac{200}{7} \approx 28.57$ cm²

1. (i) The table top $BD \parallel AC$ (the ground), so $\angle D = \angle C$ and $\angle B = \angle A$ (alternate interior angles). By AA, $\triangle OAC \sim \triangle OBD$.
2. (ii) Corresponding sides of similar triangles are proportional: $\dfrac{OA}{OB} = \dfrac{AC}{BD}$, so $\dfrac{OA}{AC} = \dfrac{OB}{BD}$.
3. (iii)(a) $\dfrac{OA}{OB} = \dfrac{OC}{OD}$: $\dfrac{3x + 4}{x} = \dfrac{3x + 19}{x + 3}$, so $(3x + 4)(x + 3) = x(3x + 19)$, that is $3x^2 + 13x + 12 = 3x^2 + 19x$ and $6x = 12$, $x = 2$. $OC = 3(2) + 19 = 25$.
4. (iii)(b) $\dfrac{BD}{AC} = \dfrac{OB}{OA} = \dfrac{x}{3x + 4} = \dfrac{2}{10} = \dfrac{1}{5}$.

Answer: (i) proved   (ii) proved   (iii)(a) $x = 2$, $OC = 25$   (iii)(b) $\dfrac{1}{5}$