CBSE Class 10 Maths Basic 2023 Question Paper (Set 430/6/1) with Solutions

1. Athletes who took less than $17$ seconds are those in the classes up to $16 - 17$: $2 + 4 + 5 + 71 = 82$. (This is the cumulative frequency at $17$.)

Answer: (C) 82

1. $d = \sqrt{5^2 + 0^2} = 5$.

Answer: (B) 5

1. $\tan A = \dfrac{BC}{AC} = \dfrac{8}{7}$. At $B$ the side next to the angle is $BC$ and the opposite side is $AC$, so $\cot B = \dfrac{BC}{AC} = \dfrac{8}{7}$. (In general $\cot B = \tan(90^\circ - B) = \tan A$.)

Answer: (B) $\dfrac{8}{7}$

1. $\dfrac{1}{4}\pi r^2 = \dfrac{1}{4} \times \dfrac{22}{7} \times 49 = \dfrac{154}{4} = \dfrac{77}{2}$ cm².

Answer: (C) $\dfrac{77}{2}$ cm²

1. LCM $= \dfrac{72 \times 120}{24} = 3 \times 120 = 360$.

Answer: (C) 360

1. There are $2$ black kings (spades and clubs): $P = \dfrac{2}{52} = \dfrac{1}{26}$.

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

1. The zeroes are the points where the graph meets the $x$-axis. This curve lies completely above the axis and never touches it, so there are $0$ zeroes.

Answer: (A) 0

1. Put $x = 6$, $y = k$: $6 - 3k + 6 = 0$, so $3k = 12$ and $k = 4$.

Answer: (D) 4

1. $2304 = 2 \times 1152 = 2^2 \times 576 = 2^3 \times 288 = 2^4 \times 144 = 2^8 \times 9 = 2^8 \times 3^2$. (Check: $256 \times 9 = 2304$ ✓.)

Answer: (A) $2^8 \times 3^2$

1. $8^n = 2^{3n}$ has only the prime factor $2$. To end in $0$ a number needs both $2$ and $5$ as factors, so $8^n$ never ends in $0$. (Its last digits go $8, 4, 2, 6$, repeating.)

Answer: (A) 0

1. The first seven primes are $2, 3, 5, 7, 11, 13, 17$. The middle (fourth) value is $7$.

Answer: (B) 7

1. $\dfrac{6 + a}{2} = 2$ gives $6 + a = 4$, so $a = -2$. (The $y$-coordinate: $\dfrac{3 + 5}{2} = 4$ ✓.)

Answer: (D) −2

1. No solution means parallel lines: $\dfrac{k}{3} = \dfrac{2}{4} \neq \dfrac{5}{1}$. So $k = \dfrac{3}{2}$, and the constants ratio $5$ differs from $\dfrac{1}{2}$ as needed.

Answer: (A) $k = \dfrac{3}{2}$

1. $OQ \perp PQ$ and $OR \perp PR$. In quadrilateral $OQPR$: $\angle QOR = 360^\circ - 90^\circ - 90^\circ - 65^\circ = 115^\circ$.

Answer: (C) $115^\circ$

1. $16x^2 - 9 = (4x - 3)(4x + 3) = 0$, so $x = \dfrac{3}{4}$ or $x = -\dfrac{3}{4}$.

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

1. The middle term is the average of its neighbours: $x = \dfrac{-5 + 3}{2} = -1$.

Answer: (D) −1

1. The odd primes on a die are $3$ and $5$ (the number $2$ is an even prime): $P = \dfrac{2}{6} = \dfrac{1}{3}$.

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

1. The sum of the six numbers is $6 \times 9 = 54$. So $6 + 7 + x + 8 + y + 14 = 54$, that is $35 + x + y = 54$ and $x + y = 19$.

Answer: (B) $x + y = 19$

1. (A) A leap year has $366 = 52 \times 7 + 2$ days, so there are two extra days. $53$ Sundays needs one of them to be a Sunday, which happens in $2$ of the $7$ equally likely pairs: $P = \dfrac{2}{7}$. True.
2. (R) A non-leap year has $365 = 52 \times 7 + 1$ days, so one extra day is a Sunday with $P = \dfrac{1}{7}$. True. But the non-leap year result does not explain the leap year one.

Answer: (B) Both A and R are true, but R is not the correct explanation of A

1. (A) $(\operatorname{cosec}\theta - \cot\theta)(\operatorname{cosec}\theta + \cot\theta) = \operatorname{cosec}^2\theta - \cot^2\theta = 1$, so they are reciprocals. True.
2. (R) The correct identity is $\operatorname{cosec}^2\theta - \cot^2\theta = 1$, so $\cot^2\theta - \operatorname{cosec}^2\theta = -1$. False.

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

1. $5(\sqrt{2})^2 - 3(1)^2 + 5(1) = 10 - 3 + 5 = 12$.

Answer: $12$

1. (a) Sum of zeroes $= 6 + (-3) = 3$ and product $= 6 \times (-3) = -18$. The polynomial is $x^2 - 3x - 18$ (or any non-zero multiple of it).
2. (b) $x^2 + 4x - 12 = (x + 6)(x - 2)$, so the zeroes are $x = -6$ and $x = 2$.

Answer: (a) $x^2 - 3x - 18$   (b) $-6$ and $2$

1. (a) $a = 5$, $b = -10$, $c = k$. Equal roots need $D = 0$: $100 - 20k = 0$, so $k = 5$.
2. (b) Let the roots be $\alpha$ and $7\alpha$. Sum: $8\alpha = \dfrac{8}{3}$, so $\alpha = \dfrac{1}{3}$. Product: $7\alpha^2 = -\dfrac{2k + 1}{3}$, so $\dfrac{7}{9} = -\dfrac{2k + 1}{3}$.
3. $2k + 1 = -\dfrac{7}{3}$, so $2k = -\dfrac{10}{3}$ and $k = -\dfrac{5}{3}$.

Answer: (a) $k = 5$   (b) $k = -\dfrac{5}{3}$

1. There are $20$ equally likely discs.
2. (i) The $2$-digit numbers are $10$ to $20$, that is $11$ numbers: $P = \dfrac{11}{20}$.
3. (ii) Numbers less than $10$ are $1$ to $9$, that is $9$ numbers: $P = \dfrac{9}{20}$.

Answer: (i) $\dfrac{11}{20}$   (ii) $\dfrac{9}{20}$

1. The radius to the point of contact is perpendicular to the tangent, so $OP^2 = OT^2 + PT^2$.
2. $r^2 = 25^2 - 24^2 = 625 - 576 = 49$, so $r = 7$ cm.

Answer: $7$ cm

1. Let Varun's present age be $x$ years. $\dfrac{1}{x - 3} + \dfrac{1}{x + 5} = \dfrac{1}{3}$ gives $\dfrac{2x + 2}{(x - 3)(x + 5)} = \dfrac{1}{3}$.
2. Cross-multiplying: $6x + 6 = x^2 + 2x - 15$, so $x^2 - 4x - 21 = 0$ and $(x - 7)(x + 3) = 0$.
3. $x = 7$ or $x = -3$. An age cannot be negative, so Varun is $7$ years old. (Check: $\dfrac{1}{4} + \dfrac{1}{12} = \dfrac{4}{12} = \dfrac{1}{3}$ ✓.)

Answer: $7$ years

1. Cumulative frequencies: $7, 15, 17, 19, 20$. $N = 20$ and $\dfrac{N}{2} = 10$, which lies in the class $3 - 5$.
2. $l = 3$, $cf = 7$, $f = 8$, $h = 2$. Median $= 3 + \dfrac{10 - 7}{8} \times 2 = 3 + \dfrac{3}{4} = 3.75$.

Answer: Median $= 3.75$

1. (a) In $\triangle ABE$ and $\triangle CFB$: $\angle A = \angle C$ (opposite angles of a parallelogram) and $\angle AEB = \angle CBF$ (alternate interior angles, since $AE \parallel BC$ with transversal $BE$).
2. So $\triangle ABE \sim \triangle CFB$ by AA similarity.
3. (b) $\triangle ABC \sim \triangle PQR$ gives $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$. Since $CM$ and $RN$ are medians, $AB = 2AM$ and $PQ = 2PN$, so $\dfrac{2AM}{2PN} = \dfrac{AC}{PR}$, that is $\dfrac{AM}{PN} = \dfrac{AC}{PR}$.
4. Also $\angle A = \angle P$ (similar triangles). By SAS similarity, $\triangle AMC \sim \triangle PNR$.

Answer: Proved (both parts)

1. Let $A(5, 3)$, $B(4, 5)$, and let $C$ and $D$ trisect $AB$. $C$ divides $AB$ in the ratio $1 : 2$: $C = \left(\dfrac{1 \times 4 + 2 \times 5}{3}, \dfrac{1 \times 5 + 2 \times 3}{3}\right) = \left(\dfrac{14}{3}, \dfrac{11}{3}\right)$.
2. $D$ divides $AB$ in the ratio $2 : 1$: $D = \left(\dfrac{2 \times 4 + 1 \times 5}{3}, \dfrac{2 \times 5 + 1 \times 3}{3}\right) = \left(\dfrac{13}{3}, \dfrac{13}{3}\right)$.

Answer: $\left(\dfrac{14}{3}, \dfrac{11}{3}\right)$ and $\left(\dfrac{13}{3}, \dfrac{13}{3}\right)$

1. Suppose $3 - 2\sqrt{5} = \dfrac{p}{q}$ with integers $p, q$ and $q \neq 0$. Then $\sqrt{5} = \dfrac{3q - p}{2q}$.
2. The right side is rational, but $\sqrt{5}$ is irrational. This contradiction shows $3 - 2\sqrt{5}$ is irrational.

Answer: Proved

1. (a) LHS $= \dfrac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A} = \dfrac{\cos A\,(1 - \sin A)}{\cos A\,(1 + \sin A)} = \dfrac{1 - \sin A}{1 + \sin A}$.
2. Multiply top and bottom by $(1 + \sin A)$: $\dfrac{(1 - \sin A)(1 + \sin A)}{(1 + \sin A)^2} = \dfrac{1 - \sin^2 A}{(1 + \sin A)^2} = \dfrac{\cos^2 A}{(1 + \sin A)^2} =$ RHS.
3. (b) LHS $= \left(\dfrac{1}{\cos\theta} + \dfrac{\sin\theta}{\cos\theta}\right)(1 - \sin\theta) = \dfrac{(1 + \sin\theta)(1 - \sin\theta)}{\cos\theta} = \dfrac{1 - \sin^2\theta}{\cos\theta} = \dfrac{\cos^2\theta}{\cos\theta} = \cos\theta =$ RHS.

Answer: Proved (both parts)

1. (a) Let the point on the bridge be $P$, directly above $Q$ on the river with $PQ = 3$ m, and let the banks be $A$ and $B$. In $\triangle APQ$, $\tan 30^\circ = \dfrac{3}{AQ}$, so $AQ = 3\sqrt{3}$ m.
2. In $\triangle PBQ$, $\tan 45^\circ = \dfrac{3}{BQ}$, so $BQ = 3$ m. The width $AB = 3\sqrt{3} + 3 = 3(1.73 + 1) = 8.19$ m.
3. (b) Let the building be $BD = 20$ m, the tower $BC = h$ and the distance $AD = x$. In $\triangle ABD$, $\tan 45^\circ = \dfrac{20}{x}$, so $x = 20$ m.
4. In $\triangle ACD$, $\tan 60^\circ = \dfrac{20 + h}{x}$, so $\sqrt{3} \times 20 = 20 + h$ and $h = 20(\sqrt{3} - 1) = 20 \times 0.73 = 14.6$ m.

Answer: (a) $\approx 8.19$ m   (b) $20(\sqrt{3} - 1) \approx 14.6$ m

1. $S_n = \dfrac{n}{2}(a + l) = 64$: $\dfrac{n}{2}(22 - 6) = 64$, so $8n = 64$ and $n = 8$.
2. $a_8 = a + 7d = -6$: $22 + 7d = -6$, so $7d = -28$ and $d = -4$. (Check: $22, 18, 14, 10, 6, 2, -2, -6$ sum to $64$ ✓.)

Answer: $n = 8$, $d = -4$

1. The ice-cream is a hemisphere of radius $5$ cm on top of a cone. The conical part has height $10 - 5 = 5$ cm.
2. Volume of one ice-cream $= \dfrac{1}{3}\pi r^2 h + \dfrac{2}{3}\pi r^3 = \dfrac{1}{3}\pi r^2(h + 2r) = \dfrac{1}{3} \times \dfrac{22}{7} \times 25 \times (5 + 10) = \dfrac{22 \times 25 \times 15}{21}$ cm³.
3. Seven cones: $7 \times \dfrac{22 \times 25 \times 15}{21} = \dfrac{22 \times 25 \times 15}{3} = 2750$ cm³.

Answer: $2750$ cm³

1. (a) **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) and $\dfrac{\text{ar}(ADE)}{\text{ar}(CDE)} = \dfrac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \dfrac{AE}{EC}$ (ii).
3. $\triangle BDE$ and $\triangle CDE$ have the same base $DE$ and lie between the parallels $DE$ and $BC$, so $\text{ar}(BDE) = \text{ar}(CDE)$ (iii). From (i), (ii), (iii): $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
4. (b) In $\triangle PQR$, $\angle 1 = \angle 2$ means $PQ = PR$ (sides opposite equal angles). So $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ becomes $\dfrac{QR}{QS} = \dfrac{QT}{PQ}$, that is $\dfrac{QS}{QR} = \dfrac{PQ}{QT}$.
5. In $\triangle PQS$ and $\triangle TQR$: $\angle Q$ is common and $\dfrac{QS}{QR} = \dfrac{PQ}{QT}$. By SAS similarity, $\triangle PQS \sim \triangle TQR$.

Answer: Proved (both parts)

1. (i) Area of the square $= 8^2 = 64$ cm².
2. (ii) $AC = \sqrt{8^2 + 8^2} = \sqrt{128} = 8\sqrt{2}$ cm.
3. (iii) The circle has radius $4$ cm and $\angle POQ = 90^\circ$. Area of sector $OPRQ = \dfrac{90}{360} \times \dfrac{22}{7} \times 4 \times 4 = \dfrac{88}{7}$ cm² $\approx 12.57$ cm².
4. (iii) OR: area of the circle $= \dfrac{22}{7} \times 4 \times 4 = \dfrac{352}{7}$ cm². Remaining part $= 64 - \dfrac{352}{7} = \dfrac{96}{7}$ cm² $\approx 13.71$ cm².

Answer: (i) $64$ cm²   (ii) $8\sqrt{2}$ cm   (iii) $\dfrac{88}{7}$ cm² (OR) $\dfrac{96}{7}$ cm²

1. Let the fixed charge be ₹ $x$ and the charge per km ₹ $y$. $x + 10y = 105$ and $x + 15y = 155$.
2. Subtracting: $5y = 50$, so $y = 10$, and $x = 105 - 100 = 5$.
3. (i) Fixed charges $= ₹\,5$. (ii) Charges per km $= ₹\,10$.
4. (iii) $20 + 10 \times 10 = ₹\,120$. (OR) Home to office $= 5 + 10 \times 10 = 105$ and office to home $= 5 + 25 \times 10 = 255$, so the total is $2x + 35y = 10 + 350 = ₹\,360$.

Answer: (i) ₹ $5$   (ii) ₹ $10$   (iii) ₹ $120$ (OR) ₹ $360$

1. (i) $B$ is the mid-point of $AC$, so $AC = 2 \times AB = 40$ m.
2. (ii) The shortest distance of the road from the centre is the perpendicular $OB$ (radius to the tangent). $OB = \sqrt{25^2 - 20^2} = \sqrt{225} = 15$ m.
3. (iii) The radius is $15$ m, so the circumference $= 2\pi r = 2 \times \dfrac{22}{7} \times 15 = \dfrac{660}{7} \approx 94.3$ m.
4. (iii) OR: area $= \pi r^2 = \dfrac{22}{7} \times 15 \times 15 = \dfrac{4950}{7} \approx 707.1$ m².

Answer: (i) $40$ m   (ii) $15$ m   (iii) $\dfrac{660}{7}$ m (OR) $\dfrac{4950}{7}$ m²