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CBSE Class 10 Maths Basic 2022 Question Paper 430/4/3 (Term 2) with Solutions

All 18 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/4/3 (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

Find the 15 $^{th}$ term from the end (towards first term) of the A.P. 3, 8, 13, ........, 253.

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Answer: 183

$d = 8 - 3 = 5$ , last term $l = 253$ .
Reading from the end, the A.P. has first term 253 and common difference $- 5$ .
15th term from the end $= 253 + 14 (- 5) = 253 - 70 = 183$ .

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

Write the next two terms of the A.P. : $5, 20, 45, ........$

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Answer:

80

and

125

(i.e.

45

and

55

)

$5$ , $20 = 25$ , $45 = 35$ .
So $d = 5$ .
Next two terms: $45 = 80$ and $55 = 125$ .

Q22 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 430/4/2

Q32 marksVery Short AnswerStatistics

Find the sum of the lower limit of the modal class and upper limit of the median class for the following distribution.
Class : 50 – 55, 55 – 60, 60 – 65, 65 – 70, 70 – 75, 75 – 80
Frequency : 10, 15, 8, 13, 9, 5

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Answer: 120

Highest frequency is 15, so the modal class is 55 – 60; lower limit = 55.
Cumulative frequencies: 10, 25, 33, 46, 55, 60; $N = 60$ , $\frac{N}{2} = 30$ .
Median class is 60 – 65; upper limit = 65.
Sum $= 55 + 65 = 120$ .

Q42 marksVery Short AnswerSurface Areas and Volumes

Find the curved surface area of a right circular cone whose height is 15 cm and base radius is 8 cm. [Use $π = \frac{22}{7}$ ]

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Answer:

\frac{2992}{7} cm^{2} \approx 427.43 cm^{2}

Slant height $l = h^{2} + r^{2} = 225 + 64 = 289 = 17$ cm.
CSA $= π r l = \frac{22}{7} \times 8 \times 17 = \frac{2992}{7}$ .
CSA $= 427 \frac{3}{7} \approx 427.43 cm^{2}$ .

Also asked in: 2022 Basic 430/4/1

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

The surface area of a sphere is 616 sq cm. Find its radius. [Use $π = \frac{22}{7}$ ]

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Answer: 7 cm

$4 π r^{2} = 616$ .
$4 \times \frac{22}{7} \times r^{2} = 616 \Rightarrow r^{2} = \frac{616 \times 7}{88} = 49$ .
$r = 7$ cm.

Also asked in: 2022 Basic 430/4/1

Q52 marksVery Short AnswerQuadratic Equations

Solve for $x$ : $2 x^{2} + \frac{7}{2} x + \frac{3}{4} = 0$ .

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Answer:

x = - \frac{1}{4}

x = - \frac{3}{2}

Multiply by 4: $8 x^{2} + 14 x + 3 = 0$ .
$8 x^{2} + 12 x + 2 x + 3 = 0 \Rightarrow 4 x (2 x + 3) + 1 (2 x + 3) = 0$ .
$(4 x + 1) (2 x + 3) = 0$ .
$x = - \frac{1}{4}$ or $x = - \frac{3}{2}$ .

Q62 marksVery Short AnswerStatistics

The frequency distribution table of agriculture holding in a village is given below :
Area of Land (in hectares): 1 – 3, 3 – 5, 5 – 7, 7 – 9, 9 – 11, 11 – 13
Number of families: 20, 45, 80, 55, 40, 12
Find the modal agriculture holding per family.

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Answer: Modal holding

= \frac{37}{6} \approx 6.17

hectares

Highest frequency is 80, so the modal class is 5 – 7.
$l = 5$ , $f_{1} = 80$ , $f_{0} = 45$ , $f_{2} = 55$ , $h = 2$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 5 + \frac{35}{160 - 45 - 55} \times 2$ .
$= 5 + \frac{35}{60} \times 2 = 5 + \frac{7}{6} = \frac{37}{6} \approx 6.17$ hectares.

Also asked in: 2022 Basic 430/4/1

Q73 marksShort AnswerTrianglesNot in current syllabus

Draw a line segment of length 7 cm and divide it in the ratio 5 : 3.

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Answer: Point P on AB with AP : PB = 5 : 3 (AP = 4.375 cm, PB = 2.625 cm)

Draw AB = 7 cm.
Draw a ray AX making an acute angle with AB.
Mark 8 (= 5 + 3) equal arcs $A_{1}, A_{2}, \dots, A_{8}$ along AX.
Join $A_{8} B$ .
Through $A_{5}$ draw a line parallel to $A_{8} B$ meeting AB at P.
By the Basic Proportionality Theorem, AP : PB = 5 : 3.

Also asked in: 2022 Basic 430/4/1

Q83 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 430/4/2

Q8 (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 430/4/2

Q93 marksShort AnswerArithmetic Progressions

The sum of first n terms of an AP is given by $S_{n} = 3 n^{2} + 2 n$ . Find the AP.

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Answer: 5, 11, 17, 23, ...

$a_{1} = S_{1} = 3 + 2 = 5$ .
$S_{2} = 12 + 4 = 16$ , so $a_{2} = S_{2} - S_{1} = 11$ .
$d = 11 - 5 = 6$ .
The AP is 5, 11, 17, 23, ...

Q103 marksShort AnswerSome Applications of Trigonometry

The string of a kite is 100 metres long and it makes an angle of $6 0^{\circ}$ with the horizontal. Find the height of the kite, assuming that there is no slack in the string. [Use $3 = 1.73$ ]

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Answer:

503

m = 86.5 m

Let the height be $h$ m; the string (100 m) is the hypotenuse.
$sin 6 0^{\circ} = \frac{h}{100} \Rightarrow h = 100 \times \frac{3}{2} = 503$ .
$h = 50 \times 1.73 = 86.5$ m.

Q114 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 430/4/2

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.
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 430/4/2

Q12 (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 12 (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 430/4/2

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. Electric 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 430/4/2

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