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CBSE Class 10 Maths Basic 2022 Question Paper 430/3/2 (Term 2) with Solutions

All 18 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/3/2 (2022, Term 2), with answers and step-by-step solutions. Total 40 marks. Tap “Show answer & solution” under any question.

Set 430/1/1 Set 430/1/2 Set 430/1/3 Set 430/2/1 Set 430/2/2 Set 430/2/3 Set 430/3/1 Set 430/3/2 Set 430/3/3 Set 430/4/1 Set 430/4/2 Set 430/4/3

Q12 marksVery Short AnswerSurface Areas and Volumes

A cubical block of side 7 cm is surmounted by a hemisphere of largest possible diameter as shown in Figure 1. Find the total surface area of the solid.

Diagram for CBSE 2022 Class 10 Maths question 1

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Answer: 332.5 cm

^{2}

Largest hemisphere has diameter 7 cm, so $r = 3.5$ cm.
TSA = surface area of cube − area of base circle of hemisphere + curved surface area of hemisphere
$= 6 (7)^{2} - π r^{2} + 2 π r^{2} = 294 + π r^{2}$
$= 294 + \frac{22}{7} \times 3.5 \times 3.5 = 294 + 38.5 = 332.5$ cm $^{2}$

Also asked in: 2022 Basic 430/3/1

Q1 (OR) (OR)2 marksVery Short AnswerSurface Areas and VolumesNot in current syllabus

How many solid cones of height 3 cm and radius 2 cm can be formed by melting a solid sphere of radius 3 cm ?

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Answer: 9 cones

Volume of sphere $= \frac{4}{3} π (3)^{3} = 36 π$ cm $^{3}$
Volume of one cone $= \frac{1}{3} π (2)^{2} (3) = 4 π$ cm $^{3}$
Number of cones $= \frac{36 π}{4 π} = 9$

Also asked in: 2022 Basic 430/3/1, 2022 Basic 430/3/3

Q22 marksVery Short AnswerCircles

In Figure 2, PA and PB are tangents to the circle with centre at O. If $∠ A P B = 7 0^{\circ}$ , then find m $∠ A QB$ .

Diagram for CBSE 2022 Class 10 Maths question 2

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Answer:

∠ A QB = 5 5^{\circ}

OA $⊥$ PA and OB $⊥$ PB (radius is perpendicular to tangent), so $∠ O A P = ∠ O B P = 9 0^{\circ}$ .
In quadrilateral OAPB: $∠ A O B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$
Angle subtended by an arc at the centre is double the angle at a point on the remaining part of the circle.
$∠ A QB = \frac{1}{2} ∠ A O B = 5 5^{\circ}$

Also asked in: 2022 Basic 430/3/1

Q32 marksVery Short AnswerQuadratic Equations

For what value of p, does the quadratic equation $p x^{2} + 2 x + p = 0$ have real and equal roots ?

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Answer:

p = 1

p = - 1

For real and equal roots, the discriminant must be zero.
$D = b^{2} - 4 a c = 2^{2} - 4 (p) (p) = 4 - 4 p^{2}$
$4 - 4 p^{2} = 0 \Rightarrow p^{2} = 1$
So $p = 1$ or $p = - 1$ (both non-zero, so the equation stays quadratic).

Also asked in: 2022 Basic 430/3/1, 2022 Basic 430/3/3

Q3 (OR) (OR)2 marksVery Short AnswerQuadratic Equations

Solve the quadratic equation for x : $6 - x - x^{2} = 0$

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Answer:

x = 2

x = - 3

Multiply by $- 1$ : $x^{2} + x - 6 = 0$
Split the middle term: $x^{2} + 3 x - 2 x - 6 = 0$
$x (x + 3) - 2 (x + 3) = 0 \Rightarrow (x + 3) (x - 2) = 0$
So $x = 2$ or $x = - 3$ .

Also asked in: 2022 Basic 430/3/1, 2022 Basic 430/3/3

Q42 marksVery Short AnswerArithmetic Progressions

For an AP with common difference 6, the sum of first ten terms is same as four times the sum of first five terms. Determine the first term of the AP.

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Answer: First term

= 3

Let the first term be $a$ ; $d = 6$ .
$S_{10} = \frac{10}{2} [2 a + 9 (6)] = 10 a + 270$
$S_{5} = \frac{5}{2} [2 a + 4 (6)] = 5 a + 60$ , so $4 S_{5} = 20 a + 240$
$10 a + 270 = 20 a + 240 \Rightarrow 10 a = 30 \Rightarrow a = 3$

Also asked in: 2022 Basic 430/3/1, 2022 Basic 430/3/3

Q52 marksVery Short AnswerArithmetic Progressions

Find the $1 5^{t h}$ term of an AP whose first term is 17 and fourth term is 44.

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Answer:

a_{15} = 143

$a = 17$ , $a_{4} = a + 3 d = 44 \Rightarrow 3 d = 27 \Rightarrow d = 9$
$a_{15} = a + 14 d = 17 + 14 \times 9 = 17 + 126 = 143$

Q62 marksVery Short AnswerStatistics

Find mode of the following frequency distribution :
Class: 100–110, 110–120, 120–130, 130–140, 140–150
Frequency: 5, 9, 8, 11, 7

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Answer: Mode

= 130 + \frac{30}{7} \approx 134.29

Highest frequency is 11, so the modal class is 130–140.
$l = 130$ , $h = 10$ , $f_{1} = 11$ , $f_{0} = 8$ , $f_{2} = 7$
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 130 + \frac{11 - 8}{22 - 8 - 7} \times 10$
$= 130 + \frac{30}{7} = 134 \frac{2}{7} \approx 134.29$

Also asked in: 2022 Basic 430/3/1, 2022 Basic 430/3/3

Q73 marksShort AnswerSome Applications of Trigonometry

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are $4 5^{\circ}$ and $6 0^{\circ}$ respectively as shown in Figure 3. Find the height of the transmission tower.

Diagram for CBSE 2022 Class 10 Maths question 7

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Answer: Height of tower

= 20 (3 - 1)

\approx 14.64

In right $△ B C D$ : $tan 4 5^{\circ} = \frac{B C}{C D} \Rightarrow C D = B C = 20$ m
In right $△ A C D$ : $tan 6 0^{\circ} = \frac{A C}{C D} \Rightarrow A C = 203$ m
Height of tower AB $= A C - B C = 203 - 20 = 20 (3 - 1) \approx 20 \times 0.732 = 14.64$ m

Q83 marksShort AnswerStatistics

Determine median of the following frequency distribution :
Class: 15–20, 20–25, 25–30, 30–35, 35–40, 40–45
Frequency: 8, 13, 21, 12, 5, 4

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Answer: Median

= 27.5

Cumulative frequencies: 8, 21, 42, 54, 59, 63; $n = 63$ , $\frac{n}{2} = 31.5$
Median class is 25–30: $l = 25$ , $c f = 21$ , $f = 21$ , $h = 5$
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 25 + \frac{31.5 - 21}{21} \times 5 = 25 + 2.5 = 27.5$

Also asked in: 2022 Basic 430/3/1, 2022 Basic 430/3/3

Q93 marksShort AnswerCirclesNot in current syllabus

Draw two concentric circles of radii 5 cm and 2 cm. From a point P on the outer circle, construct a pair of tangents to the inner circle.

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Answer: Construction (each tangent

= 5^{2} - 2^{2} = 21 \approx 4.58

cm)

Draw concentric circles with centre O and radii 2 cm and 5 cm; take P on the outer circle.
Draw the perpendicular bisector of OP; let it meet OP at M.
With M as centre and MO as radius, draw a circle cutting the inner circle at A and B.
Join PA and PB; these are the required tangents ( $∠ O A P = ∠ O B P = 9 0^{\circ}$ , angles in a semicircle).
Each tangent measures $25 - 4 = 21 \approx 4.58$ cm.

Q9 (OR) (OR)3 marksShort AnswerTrianglesNot in current syllabus

Draw a line segment AB = 9 cm. Divide AB in the ratio 2 : 3.

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Answer: Construction (parts 3.6 cm and 5.4 cm)

Draw AB = 9 cm and a ray AX making an acute angle with AB.
Mark 5 equal arcs $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ on AX.
Join $A_{5} B$ ; through $A_{2}$ draw a line parallel to $A_{5} B$ meeting AB at C.
Then AC : CB = 2 : 3 (by BPT), i.e. AC = 3.6 cm and CB = 5.4 cm.

Q103 marksShort AnswerStatistics

The mileage (km/l) of 50 cars was recorded by a dealer and tabulated as given below :
Mileage (in km/l): 10–12, 12–14, 14–16, 16–18, 18–20
Number of Cars: 13, 18, 10, 7, 2
Find mean of the above distribution.

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Answer: Mean mileage

= 13.68

km/l

Class marks $x_{i}$ : 11, 13, 15, 17, 19; $f_{i}$ : 13, 18, 10, 7, 2; $\sum f_{i} = 50$
$f_{i} x_{i}$ : 143, 234, 150, 119, 38; $\sum f_{i} x_{i} = 684$
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{684}{50} = 13.68$ km/l

Also asked in: 2022 Basic 430/3/1

Q114 marksLong AnswerSome Applications of Trigonometry

The angles of depression of the top and bottom of a 6 m tall building from the top of a multi-storeyed building are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use $3 = 1.73$ )

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Answer: Height of multi-storeyed building

= (9 + 33)

= 14.19

m; distance between the buildings

= 14.19

Let the multi-storeyed building be H m tall and the buildings be d m apart.
Angle of depression of the bottom is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{H}{d} \Rightarrow d = H$
Angle of depression of the top is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{H - 6}{d} \Rightarrow d = 3 (H - 6)$
$H = 3 H - 63 \Rightarrow H = \frac{6 3}{3 - 1} = \frac{6 3 ( 3 + 1 )}{2} = 9 + 33$
$H = 9 + 3 (1.73) = 14.19$ m, and $d = H = 14.19$ m

Also asked in: 2022 Basic 430/3/1

Q124 marksLong AnswerCircles

Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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Answer: Proved.

Let ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S.
Join OA, OB, OC, OD, OP, OQ, OR, OS.
$△ O A P ≅ △ O A S$ (OP = OS radii, AP = AS tangents, OA common; SSS), so $∠ A O P = ∠ A O S$ .
Similarly $∠ B O P = ∠ B O Q$ , $∠ C O Q = ∠ C O R$ , $∠ D O R = ∠ D O S$ .
Sum of angles at O: $2 (∠ A O P + ∠ B O P + ∠ C O R + ∠ D O R) = 36 0^{\circ}$
So $(∠ A O P + ∠ B O P) + (∠ C O R + ∠ D O R) = 18 0^{\circ}$ , i.e. $∠ A O B + ∠ C O D = 18 0^{\circ}$ .
Similarly $∠ B O C + ∠ A O D = 18 0^{\circ}$ . Hence opposite sides subtend supplementary angles at the centre.

Also asked in: 2025 Basic 430/4/1, 2025 Basic 430/4/3, 2022 Basic 430/4/1, 2022 Basic 430/4/2, 2022 Basic 430/4/3

Q12 (OR) (OR)4 marksLong AnswerCircles

In Figure 4, CD is a tangent and AB is a diameter of the circle centred at O. If $∠ D C B = 2 0^{\circ}$ , then find m $∠ A D C$ .

Diagram for CBSE 2022 Class 10 Maths question 12 (OR)

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Answer:

∠ A D C = 12 5^{\circ}

OD $⊥$ CD (radius is perpendicular to tangent), so $∠ O D C = 9 0^{\circ}$ .
In $△ O D C$ : $∠ D O C = 18 0^{\circ} - 9 0^{\circ} - 2 0^{\circ} = 7 0^{\circ}$
$∠ A O D = 18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ (linear pair)
OA = OD (radii), so $∠ O A D = ∠ O D A = \frac{18 0 ^{\circ} - 11 0 ^{\circ}}{2} = 3 5^{\circ}$
$∠ A D C = ∠ A D O + ∠ O D C = 3 5^{\circ} + 9 0^{\circ} = 12 5^{\circ}$

Q134 marksCase StudyQuadratic Equations

The tradition of pottery making in India is very old. In fact, it is older than Indus Valley Civilization. The shaping and baking of clay articles has continued through the ages. The picture of a potter is shown below :
A potter makes a certain number of pottery articles in a day. It was observed on a particular day the cost of production of each article (in ₹) was one more than twice the number of articles produced on that day. The total cost of production on that day was ₹ 210.
(a) Taking number of articles produced on that day as x, form a quadratic equation in x. (2)
(b) Find the number of articles produced and the cost of each article. (2)

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Answer: (a)

2 x^{2} + x - 210 = 0

(b) 10 articles; cost of each article ₹ 21

(a) Cost of each article $= ₹ (2 x + 1)$ , so $x (2 x + 1) = 210$ , i.e. $2 x^{2} + x - 210 = 0$ .
(b) $2 x^{2} + 21 x - 20 x - 210 = 0 \Rightarrow x (2 x + 21) - 10 (2 x + 21) = 0 \Rightarrow (2 x + 21) (x - 10) = 0$
$x = 10$ (rejecting $x = - \frac{21}{2}$ since the number of articles cannot be negative)
Cost of each article $= 2 (10) + 1 = ₹ 21$

Also asked in: 2022 Basic 430/3/1, 2022 Basic 430/3/3

Q144 marksCase StudySurface Areas and Volumes

The technique of Rainwater harvesting through Recharge pit is very useful. Rainwater is collected on the roof and then flowing through the Recharge pit it goes to the ground. Observe the picture given below :
B (BREADTH) = 3 m
D (DEPTH = 2 m
L (LENGTH) = 3 m
The surface area of the roof floor is 100 m $^{2}$ . The cuboidal pit measures 3 m $\times$ 3 m $\times$ 2 m.
(a) Water standing on the roof is released into the cuboidal pit. If the cuboidal pit is filled completely by the roof water, then find the height of standing water on the roof. (2)
(b) Instead of a cuboidal pit, if a cylindrical pit with diameter 3 m and height 2 m had been built, then which tank would hold more water ? (2)

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Answer: (a) 0.18 m (18 cm) (b) The cuboidal pit (18 m

^{3}

) holds more water than the cylindrical pit (

\approx 14.14

^{3}

)

(a) Volume of cuboidal pit $= 3 \times 3 \times 2 = 18$ m $^{3}$
Volume of water on roof $= 100 \times h$ ; $100 h = 18 \Rightarrow h = 0.18$ m $= 18$ cm
(b) Cylinder: $r = 1.5$ m, $h = 2$ m; volume $= π r^{2} h = \frac{22}{7} \times 2.25 \times 2 \approx 14.14$ m $^{3}$
Since $18 > 14.14$ , the cuboidal pit would hold more water.

Also asked in: 2022 Basic 430/3/1, 2022 Basic 430/3/3

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