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

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

In Fig. 1, perimeter of $△ P QR$ is 20 cm. Find the length of tangent PA.

Diagram for CBSE 2022 Class 10 Maths question 1

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Answer: PA = 10 cm

Tangents from an external point are equal: QA = QC, RB = RC, PA = PB.
Perimeter of $△ P QR$ = PQ + QR + PR = PQ + QC + CR + PR
= PQ + QA + RB + PR = PA + PB = 2PA.
So 2PA = 20, hence PA = 10 cm.

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

Q1 (OR) (OR)2 marksVery Short AnswerCircles

In Fig. 2, BC is tangent to the circle at point B of circle centred at O. BD is a chord of the circle so that $∠ B A D = 5 5^{\circ}$ . Find m $∠ D B C$ .

Diagram for CBSE 2022 Class 10 Maths question 1 (OR)

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Answer:

∠ D B C = 5 5^{\circ}

AB passes through O, so AB is a diameter.
$∠ A D B = 9 0^{\circ}$ (angle in a semicircle).
In $△ A B D$ : $∠ A B D = 18 0^{\circ} - 9 0^{\circ} - 5 5^{\circ} = 3 5^{\circ}$ .
$∠ A B C = 9 0^{\circ}$ (radius OB $⊥$ tangent BC).
$∠ D B C = 9 0^{\circ} - 3 5^{\circ} = 5 5^{\circ}$ .

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

Q22 marksVery Short AnswerStatistics

Find the mode of the following frequency distribution :
Class: 20–30, 30–40, 40–50, 50–60, 60–70
Frequency: 25, 30, 45, 42, 35

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Answer: Mode ≈ 48.33

Maximum frequency 45, so modal class is 40–50.
$l = 40$ , $h = 10$ , $f_{1} = 45$ , $f_{0} = 30$ , $f_{2} = 42$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{15}{90 - 30 - 42} \times 10$
$= 40 + \frac{150}{18} = 40 + 8.33 = 48.33$ .

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

Q32 marksVery Short AnswerArithmetic Progressions

Find the sum of the first fifteen multiples of 8.

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Answer: 960

Multiples: 8, 16, 24, …, 120 with $a = 8$ , $d = 8$ , $n = 15$ .
$S_{15} = \frac{15}{2} [2 (8) + 14 (8)] = \frac{15}{2} \times 128 = 960$ .

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

Q42 marksVery Short AnswerArithmetic Progressions

Which term of the A.P. 3, 8, 13, 18, … is 78 ?

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Answer: 16th term

$a = 3$ , $d = 8 - 3 = 5$ .
$a_{n} = a + (n - 1) d = 78$
$3 + 5 (n - 1) = 78$
$5 (n - 1) = 75 \Rightarrow n - 1 = 15 \Rightarrow n = 16$ .
So 78 is the 16th term.

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

Q4 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions

Find the common difference of an A.P. whose $n^{t h}$ term is given by $a_{n} = 6 n - 5$ .

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Answer: Common difference = 6

$a_{1} = 6 (1) - 5 = 1$ , $a_{2} = 6 (2) - 5 = 7$ .
$d = a_{2} - a_{1} = 7 - 1 = 6$ .
(In general $a_{n + 1} - a_{n} = 6$ .)

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

Q52 marksVery Short AnswerQuadratic Equations

Solve the equation : $3 x^{2} - 8 x - 1 = 0$ for $x$ .

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Answer:

x = \frac{4 + 19}{3}

x = \frac{4 - 19}{3}

$a = 3$ , $b = - 8$ , $c = - 1$ .
$D = 64 - 4 (3) (- 1) = 64 + 12 = 76$ .
$x = \frac{8 \pm 76}{6} = \frac{8 \pm 2 19}{6} = \frac{4 \pm 19}{3}$ .

Q62 marksVery Short AnswerSurface Areas and Volumes

3 cubes each of 8 cm edge are joined end to end. Find the total surface area of the cuboid.

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Answer: 896 cm²

Cuboid formed: $l = 3 \times 8 = 24$ cm, $b = 8$ cm, $h = 8$ cm.
TSA $= 2 (l b + bh + h l) = 2 (24 \times 8 + 8 \times 8 + 8 \times 24)$
$= 2 (192 + 64 + 192) = 2 \times 448 = 896$ cm².

Q73 marksShort AnswerStatistics

Find the mean of the following frequency distribution :
Class: 10–15, 15–20, 20–25, 25–30, 30–35
Frequency: 4, 10, 5, 6, 5

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Answer: Mean ≈ 22.17

Class marks $x_{i}$ : 12.5, 17.5, 22.5, 27.5, 32.5.
$f_{i} x_{i}$ : 50, 175, 112.5, 165, 162.5.
$\sum f_{i} = 30$ , $\sum f_{i} x_{i} = 665$ .
Mean $= \frac{665}{30} = 22.17$ (approx.).

Also asked in: 2022 Basic 430/2/1

Q83 marksShort AnswerTrianglesNot in current syllabus

Draw a line segment of length 7.5 cm and divide it in the ratio 1 : 3.

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Answer: Construction (the dividing point is 1.875 cm from one end).

Draw AB = 7.5 cm and a ray AX making an acute angle with AB.
Mark 4 (= 1 + 3) equal arcs $A_{1}, A_{2}, A_{3}, A_{4}$ along AX.
Join $A_{4} B$ .
Through $A_{1}$ draw a line parallel to $A_{4} B$ meeting AB at P.
Then AP : PB = 1 : 3 (by the Basic Proportionality Theorem); AP = 1.875 cm.

Q93 marksShort AnswerSome Applications of Trigonometry

As observed from the top of a light house 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from $3 0^{\circ}$ to $4 5^{\circ}$ . Determine the distance travelled by the ship during this time.
(Use $3$ = 1.73)

Diagram for CBSE 2022 Class 10 Maths question 9

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Answer: 73 m

Let AB = 100 m be the light house, D and C the initial and final positions of the ship.
$∠ A D B = 3 0^{\circ}$ and $∠ A C B = 4 5^{\circ}$ (alternate angles).
In $△ A B C$ : $tan 4 5^{\circ} = \frac{A B}{B C} \Rightarrow B C = 100$ m.
In $△ A B D$ : $tan 3 0^{\circ} = \frac{A B}{B D} \Rightarrow B D = 1003 = 173$ m.
Distance travelled CD = BD − BC = 173 − 100 = 73 m.

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

Q9 (OR) (OR)3 marksShort AnswerSome Applications of TrigonometryNot in current syllabus

At a point on level ground, the angle of elevation of a vertical tower is, found to be $α$ such that $tan α = \frac{1}{3}$ . After walking 100 m towards the tower, the angle of elevation $β$ becomes such that $tan β = \frac{3}{4}$ . Find the height of the tower.

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Answer: 60 m

Let the height be $h$ and the distance of the second point from the foot be $x$ .
$tan β = \frac{h}{x} = \frac{3}{4} \Rightarrow x = \frac{4 h}{3}$ .
$tan α = \frac{h}{x + 100} = \frac{1}{3} \Rightarrow 3 h = x + 100$ .
$3 h = \frac{4 h}{3} + 100 \Rightarrow \frac{5 h}{3} = 100 \Rightarrow h = 60$ .
Height of the tower = 60 m.

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

Q103 marksShort AnswerStatistics

The median of the following frequency distribution is 35. Find the value of $x$ .
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 6, 3, $x$ , 12, 19

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Answer:

x = 10

$N = 6 + 3 + x + 12 + 19 = 40 + x$ .
Median 35 lies in class 30–40: $l = 30$ , $c f = 9 + x$ , $f = 12$ , $h = 10$ .
$35 = 30 + \frac{\frac{40 + x}{2} - ( 9 + x )}{12} \times 10$
$6 = 20 + \frac{x}{2} - 9 - x = 11 - \frac{x}{2} \Rightarrow x = 10$ .
Check: N = 50, N/2 = 25; cf before 30–40 is 19, so 30–40 is the median class.

Q114 marksLong AnswerQuadratic Equations

The sum of the ages of a boy and his sister (in years) is 25 and product of their ages is 150. Find their present ages.

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Answer: 15 years and 10 years

Let one age be $x$ years; the other is $25 - x$ .
$x (25 - x) = 150 \Rightarrow x^{2} - 25 x + 150 = 0$ .
$(x - 10) (x - 15) = 0 \Rightarrow x = 10$ or $x = 15$ .
So their ages are 15 years and 10 years.

Also asked in: 2022 Basic 430/2/1

Q124 marksLong AnswerCircles

$△ A B C$ circumscribes a circle of radius r such that $∠ B = 9 0^{\circ}$ . If AB = 3 cm and BC = 4 cm, then find the value of r.

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Answer: r = 1 cm

$A C = 3^{2} + 4^{2} = 5$ cm.
Let the circle (centre O) touch AB, BC, CA at P, Q, R.
OPBQ is a square (three right angles, OP = OQ = r), so BP = BQ = r.
AP = AR = 3 − r and CQ = CR = 4 − r.
AC = AR + CR: $(3 - r) + (4 - r) = 5 \Rightarrow 2 r = 2 \Rightarrow r = 1$ cm.
(Check: area $= \frac{1}{2} \times 3 \times 4 = 6 = \frac{1}{2} r (3 + 4 + 5) = 6 r$ .)

Q12 (OR) (OR)4 marksLong AnswerCircles

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

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Answer: Proved.

Let quadrilateral 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.
In $△ O A P$ and $△ O A S$ : OP = OS (radii), OA common, $∠ O P A = ∠ O S A = 9 0^{\circ}$ , so they are congruent (RHS) and $∠ A O P = ∠ A O S$ .
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 up to $36 0^{\circ}$ , so $2 (∠ A O P + ∠ B O P + ∠ C O R + ∠ D O R) = 36 0^{\circ}$ .
Thus $∠ A O B + ∠ C O D = 18 0^{\circ}$ , and similarly $∠ B O C + ∠ A O D = 18 0^{\circ}$ .

Also asked in: 2026 Basic 430/4/3, 2025 Standard 30/1/1, 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q134 marksCase StudySome Applications of Trigonometry

Qutub Minar, located in South Delhi, India was built in the year 1193. It is 72 m high tower. Working on a school project, Charu and Daljeet visited the monument. They used trigonometry to find their distance from the tower. Observe the picture given below. Points C and D represent their positions on the ground in line with the base of tower, the angles of elevation of top of the tower (Point A) are $6 0^{\circ}$ and $4 5^{\circ}$ from points C and D respectively.
(1) Based on above information, draw a well-labelled diagram. (1)
(2) Find the distances CD, BC and BD. (use $3$ = 1.73) (3)

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Answer: (1) Vertical tower AB = 72 m with C and D on the ground in line with B;

∠ A C B = 6 0^{\circ}

∠ A D B = 4 5^{\circ}

(2) BC = 41.52 m, BD = 72 m, CD = 30.48 m

(1) Draw AB vertical (72 m), B on the ground line BCD with C nearer to B; mark $∠ A C B = 6 0^{\circ}$ and $∠ A D B = 4 5^{\circ}$ .
(2) In $△ A B C$ : $tan 6 0^{\circ} = \frac{A B}{B C} \Rightarrow B C = \frac{72}{3} = 243 = 41.52$ m.
In $△ A B D$ : $tan 4 5^{\circ} = \frac{A B}{B D} \Rightarrow B D = 72$ m.
CD = BD − BC = 72 − 41.52 = 30.48 m.

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

Q144 marksCase StudySurface Areas and Volumes

A solid cuboidal toy is made of wood. It has five cone shaped cavities to hold toy carrots.
The dimensions of the toy are cuboid – 10 cm × 10 cm × 8 cm.
Each cone carved out – Radius = 2.1 cm and Height = 6 cm.
(1) Find the volume of wood carved out to make five conical cavities. (2)
(2) Find the volume of the wood in the final product. (2)

Diagram for CBSE 2022 Class 10 Maths question 14

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Answer: (1) 138.6 cm³ (2) 661.4 cm³

(1) Volume of one cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 6 = 27.72$ cm³.
Volume of five cones $= 5 \times 27.72 = 138.6$ cm³.
(2) Volume of cuboid $= 10 \times 10 \times 8 = 800$ cm³.
Volume of wood left $= 800 - 138.6 = 661.4$ cm³.

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

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