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CBSE Class 10 Maths Basic 2022 Question Paper 430/4/2 (Term 2) with Solutions

All 18 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/4/2 (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 AnswerArithmetic Progressions

Determine the 36 $^{th}$ term of the A.P. whose first two terms are $- 3$ and 4 respectively.

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Answer:

a_{36} = 242

$a = - 3$ , $d = 4 - (- 3) = 7$ .
$a_{36} = a + 35 d = - 3 + 35 \times 7 = - 3 + 245 = 242$ .

Also asked in: 2022 Basic 430/4/1

Q1 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions

Write the next two terms of the A.P. :
$27, 48, 75, .......$

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Answer:

108

and

147

(i.e.

63

and

73

)

$27 = 33$ , $48 = 43$ , $75 = 53$ .
So $d = 3$ .
Next two terms: $63 = 108$ and $73 = 147$ .

Also asked in: 2022 Basic 430/4/1

Q22 marksVery Short AnswerStatistics

Using the empirical relationship between the three measures of central tendency, find the median of a distribution, whose mean is 169 and mode is 175.

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Answer: Median = 171

Empirical relation: $3 Median = Mode + 2 Mean$ .
$3 Median = 175 + 2 \times 169 = 175 + 338 = 513$ .
Median $= \frac{513}{3} = 171$ .

Also asked in: 2022 Basic 430/4/1

Q32 marksVery Short AnswerStatistics

Find the modal and median classes of the following distribution.
Class : 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100
Frequency : 11, 22, 19, 13, 7

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Answer: Modal class: 20 – 40; median class: 40 – 60

Highest frequency is 22, so the modal class is 20 – 40.
Cumulative frequencies: 11, 33, 52, 65, 72; $N = 72$ , $\frac{N}{2} = 36$ .
The first cumulative frequency greater than 36 is 52, so the median class is 40 – 60.

Q42 marksVery Short AnswerSurface Areas and Volumes

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $π$ .

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Answer:

π cm^{3}

$r = h = 1$ cm.
Volume $= \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π + \frac{2}{3} π$ .
$= π cm^{3}$ .

Q4 (OR) (OR)2 marksVery Short AnswerSurface Areas and Volumes

Find the volume of a solid hemisphere whose total surface area is 462 sq.m.

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Answer:

\frac{2156}{3} m^{3} \approx 718.67 m^{3}

TSA of a solid hemisphere $= 3 π r^{2} = 462$ .
Taking $π = \frac{22}{7}$ : $r^{2} = \frac{462 \times 7}{66} = 49 \Rightarrow r = 7$ m.
Volume $= \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 343 = \frac{2156}{3} \approx 718.67 m^{3}$ .

Q52 marksVery Short AnswerCircles

The distance between two tangents parallel to each other of a circle is 13 cm. Find the radius of the circle.

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Answer: 6.5 cm

The radii to the two points of contact are perpendicular to the parallel tangents, so the two points of contact and the centre lie on one line.
Hence the distance between the parallel tangents is a diameter: $2 r = 13$ cm.
Radius $r = 6.5$ cm.

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

Q62 marksVery Short AnswerQuadratic Equations

Find the value(s) of 'a' for which the quadratic equation $x^{2} - a x + 1 = 0$ has real and equal roots.

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Answer:

a = 2

a = - 2

For real and equal roots, discriminant $= 0$ .
$(- a)^{2} - 4 (1) (1) = 0 \Rightarrow a^{2} = 4$ .
$a = \pm 2$ .

Q73 marksShort AnswerTrianglesNot in current syllabus

Draw a line segment AB of length 6 cm and mark a point X on it such that AX $= \frac{4}{5}$ AB. [Using a scale & compass]

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Answer: Point X on AB with AX : XB = 4 : 1 (AX = 4.8 cm)

AX $= \frac{4}{5}$ AB means AX : XB = 4 : 1.
Draw AB = 6 cm and a ray AY making an acute angle with AB.
Mark 5 equal arcs $A_{1}, \dots, A_{5}$ along AY and join $A_{5} B$ .
Through $A_{4}$ draw a line parallel to $A_{5} B$ meeting AB at X.
By the Basic Proportionality Theorem, AX : XB = 4 : 1, so AX $= \frac{4}{5}$ AB = 4.8 cm.

Q83 marksShort AnswerArithmetic Progressions

In an A.P., the first term is 12 and the common difference is 6. If the last term of the A.P. is 252, then find its middle term.

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Answer: Middle term = 132

$a = 12$ , $d = 6$ , $a_{n} = 252$ .
$12 + (n - 1) 6 = 252 \Rightarrow n - 1 = 40 \Rightarrow n = 41$ .
Number of terms is odd, so the middle term is the $\frac{41 + 1}{2} = 21$ st term.
$a_{21} = 12 + 20 \times 6 = 132$ .

Q93 marksShort AnswerSome Applications of Trigonometry

Two poles of heights 25 m and 35 m stand vertically on the ground. The tops of two poles are connected by a wire, which is inclined to the horizontal at an angle of $3 0^{\circ}$ . Find the length of the wire and the distance between the poles.

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Answer: Length of wire = 20 m; distance between poles

= 103

\approx 17.32

Difference in heights $= 35 - 25 = 10$ m; this is the vertical side of a right triangle whose hypotenuse is the wire $l$ .
$sin 3 0^{\circ} = \frac{10}{l} \Rightarrow \frac{1}{2} = \frac{10}{l} \Rightarrow l = 20$ m.
Horizontal distance $d$ : $tan 3 0^{\circ} = \frac{10}{d} \Rightarrow d = 103$ m.
$d \approx 17.32$ m.

Also asked in: 2022 Basic 430/4/1

Q103 marksShort AnswerQuadratic Equations

If $x = 3$ is one root of the quadratic equation $2 x^{2} + p x + 30 = 0$ , find the value of p and the other root of the quadratic equation.

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Answer:

p = - 16

; other root

= 5

Put $x = 3$ : $2 (9) + 3 p + 30 = 0 \Rightarrow 3 p = - 48 \Rightarrow p = - 16$ .
Equation: $2 x^{2} - 16 x + 30 = 0 \Rightarrow x^{2} - 8 x + 15 = 0$ .
$(x - 3) (x - 5) = 0$ , so $x = 3$ or $x = 5$ .
The other root is 5.

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

Q10 (OR) (OR)3 marksShort AnswerQuadratic Equations

The length of a rectangular park is 5 metres more than twice its breadth. If the area of the park is 250 sq m, find the length and breadth of the park.

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Answer: Length = 25 m, breadth = 10 m

Let the breadth be $x$ m; then length $= (2 x + 5)$ m.
$x (2 x + 5) = 250 \Rightarrow 2 x^{2} + 5 x - 250 = 0$ .
$2 x^{2} + 25 x - 20 x - 250 = 0 \Rightarrow (2 x + 25) (x - 10) = 0$ .
$x = 10$ (breadth cannot be negative).
Breadth = 10 m, length $= 2 (10) + 5 = 25$ m.

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

Q114 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.
In $△ O A P$ and $△ O A S$ : OP = OS (radii), AP = AS (tangents from A), OA common. So $△ O A P ≅ △ O A S$ (SSS), giving $∠ A O P = ∠ A O S$ .
Similarly $∠ B O P = ∠ B O Q$ , $∠ C O Q = ∠ C O R$ , $∠ D O R = ∠ D O S$ .
The eight angles at O add to $36 0^{\circ}$ : $2 (∠ A O P + ∠ B O P + ∠ C O R + ∠ D O R) = 36 0^{\circ}$ .
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 430/3/2, 2022 Basic 430/4/1, 2022 Basic 430/4/3

Q11 (OR) (OR)4 marksLong AnswerCircles

If a circle is touching the side BC of $△ A B C$ at P and is touching AB and AC produced at Q and R respectively (see the figure).
Prove that AQ $= \frac{1}{2}$ (perimeter of $△ A B C$ ).

Diagram for CBSE 2022 Class 10 Maths question 11 (OR)

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Answer: Proved.

Tangents from an external point are equal: AQ = AR, BQ = BP, CP = CR.
Perimeter of $△ A B C$ = AB + BC + CA = AB + BP + PC + CA.
= AB + BQ + CR + CA = AQ + AR.
= 2AQ (since AQ = AR).
Hence AQ $= \frac{1}{2}$ (perimeter of $△ A B C$ ).

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

Q124 marksLong AnswerSome Applications of Trigonometry

As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships are found to be $3 0^{\circ}$ and $4 5^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. [Use $3 = 1.732$ ]

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Answer: 73.2 m

Let AB = 100 m be the lighthouse, with ships at C (angle $4 5^{\circ}$ ) and D (angle $3 0^{\circ}$ ) on the same side, D beyond C.
Angle of depression equals the angle of elevation from the ship (alternate angles).
In $△ A B C$ : $tan 4 5^{\circ} = \frac{100}{B C} \Rightarrow B C = 100$ m.
In $△ A B D$ : $tan 3 0^{\circ} = \frac{100}{B D} \Rightarrow B D = 1003$ m.
$C D = B D - B C = 100 (3 - 1) = 100 \times 0.732 = 73.2$ m.

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

Q134 marksCase StudyStatistics

Electric buses are becoming popular nowadays. These buses have the electricity stored in a battery. Electric buses could have a range of approximately 280 km with just one charge. These buses are superior to diesel buses as they reduce brake wear and also reduce pollution. Transport department of a city wants to buy some electric buses for the city. So, the department wants to know the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 50 existing public transport buses in a day.
Daily distance travelled (in km): 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200
Number of buses: 12, 14, 8, 6, 10
(a) Find the 'median' distance travelled by a bus. (2)
(b) Find the 'mean (average)' distance travelled by a bus. (2)

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Answer: (a) Median

\approx 138.57

km (b) Mean = 145.2 km

(a) Cumulative frequencies: 12, 26, 34, 40, 50; $N = 50$ , $\frac{N}{2} = 25$ .
Median class is 120 – 140: $l = 120$ , $c f = 12$ , $f = 14$ , $h = 20$ .
Median $= 120 + \frac{25 - 12}{14} \times 20 = 120 + \frac{260}{14} \approx 120 + 18.57 = 138.57$ km.
(b) Class marks: 110, 130, 150, 170, 190.
$\sum f_{i} x_{i} = 1320 + 1820 + 1200 + 1020 + 1900 = 7260$ .
Mean $= \frac{7260}{50} = 145.2$ km.

Q144 marksCase StudySurface Areas and Volumes

A company deals in casting and moulding of metal on orders received from its clients.
In one such order, company is supposed to make 50 toys in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of hemisphere. If the radius of the base of the cone is 21 cm and height is 28 cm, then
(a) find the volume of 50 toys; (2)
(b) find the ratio of the volume of hemisphere to the volume of cone. (2)

Diagram for CBSE 2022 Class 10 Maths question 14

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Answer: (a) 1617000 cm

^{3}

(b) 3 : 2

(a) $r = 21$ cm, $h = 28$ cm. Volume of one toy $= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} (2 r + h)$ .
$= \frac{1}{3} \times \frac{22}{7} \times 441 \times 70 = 32340 cm^{3}$ .
Volume of 50 toys $= 50 \times 32340 = 1617000 cm^{3}$ .
(b) $\frac{\frac{2}{3} π r ^{3}}{\frac{1}{3} π r ^{2} h} = \frac{2 r}{h} = \frac{42}{28} = \frac{3}{2}$ , so the ratio is 3 : 2.

Also asked in: 2022 Basic 430/4/3

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