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CBSE Class 10 Maths Standard 2022 Question Paper 30/2/3 (Term 2) with Solutions

All 18 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/2/3 (2022, Term 2), with answers and step-by-step solutions. Total 40 marks. Tap “Show answer & solution” under any question.

Set 30/1/1 Set 30/1/2 Set 30/1/3 Set 30/2/1 Set 30/2/2 Set 30/2/3 Set 30/3/1 Set 30/3/2 Set 30/3/3 Set 30/4/1 Set 30/4/2 Set 30/4/3

Q12 marksVery Short AnswerCircles

In Fig. 1, AB is diameter of a circle centered at O. BC is tangent to the circle at B. If OP bisects the chord AD and $∠ A O P = 6 0^{\circ}$ , then find m $∠ C$ .

Diagram for CBSE 2022 Class 10 Maths question 1

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Answer: m

∠ C = 6 0^{\circ}

OP bisects chord AD, so $O P ⊥ A D$ (line from centre to midpoint of a chord), i.e. $∠ A P O = 9 0^{\circ}$ .
In $△ A P O$ : $∠ O A P = 18 0^{\circ} - 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$ , so $∠ B A C = 3 0^{\circ}$ .
BC is tangent at B and OB is a radius, so $∠ A B C = 9 0^{\circ}$ .
In $△ A B C$ : $∠ C = 18 0^{\circ} - 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ .

Also asked in: 2022 Standard 30/2/1, 2022 Standard 30/2/2

Q1 (OR) (OR)2 marksVery Short AnswerCircles

In Fig. 2, XAY is a tangent to the circle centered at O. If $∠ A B O = 4 0^{\circ}$ , then find m $∠ B A Y$ and m $∠ A O B$ .

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

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Answer: m

∠ B A Y = 5 0^{\circ}

, m

∠ A O B = 10 0^{\circ}

OA = OB (radii), so $∠ O A B = ∠ O B A = 4 0^{\circ}$ .
XAY is a tangent at A, so $∠ O A Y = 9 0^{\circ}$ .
$∠ B A Y = 9 0^{\circ} - 4 0^{\circ} = 5 0^{\circ}$ .
In $△ A O B$ : $∠ A O B = 18 0^{\circ} - 4 0^{\circ} - 4 0^{\circ} = 10 0^{\circ}$ .

Also asked in: 2022 Standard 30/2/1, 2022 Standard 30/2/2

Q22 marksVery Short AnswerStatistics

If mode of the following frequency distribution is 55, then find the value of $x$ .
Class: 0 – 15, 15 – 30, 30 – 45, 45 – 60, 60 – 75, 75 – 90
Frequency: 10, 7, $x$ , 15, 10, 12

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Answer:

x = 5

Mode 55 lies in 45 – 60, so this is the modal class.
$l = 45$ , $h = 15$ , $f_{1} = 15$ , $f_{0} = x$ , $f_{2} = 10$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$
$55 = 45 + \frac{15 - x}{30 - x - 10} \times 15$
$10 (20 - x) = 15 (15 - x)$
$200 - 10 x = 225 - 15 x \Rightarrow 5 x = 25 \Rightarrow x = 5$ .

Also asked in: 2022 Standard 30/2/1, 2022 Standard 30/2/2

Q32 marksVery Short AnswerArithmetic Progressions

In an A.P. if the sum of third and seventh term is zero, find its $5^{t h}$ term.

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Answer:

a_{5} = 0

$a_{3} + a_{7} = 0 \Rightarrow (a + 2 d) + (a + 6 d) = 0$
$2 a + 8 d = 0 \Rightarrow a + 4 d = 0$
$a_{5} = a + 4 d = 0$ .

Q3 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions

Determine the A.P. whose third term is 5 and seventh term is 9.

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Answer: 3, 4, 5, 6, ...

$a + 2 d = 5$ and $a + 6 d = 9$ .
Subtracting: $4 d = 4 \Rightarrow d = 1$ ; then $a = 3$ .
The A.P. is 3, 4, 5, 6, ...

Q42 marksVery Short AnswerQuadratic Equations

Solve the quadratic equation $x^{2} + 22 x - 6 = 0$ for $x$ .

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Answer:

x = 2

x = - 32

Here $a = 1$ , $b = 22$ , $c = - 6$ .
$D = b^{2} - 4 a c = 8 + 24 = 32$ , so $D = 42$ .
$x = \frac{- 2 2 \pm 4 2}{2}$ .
$x = \frac{2 2}{2} = 2$ or $x = \frac{- 6 2}{2} = - 32$ .
(Check by factorising: $(x - 2) (x + 32) = x^{2} + 22 x - 6$ .)

Also asked in: 2022 Standard 30/2/1

Q52 marksVery Short AnswerArithmetic Progressions

Find the sum of first 20 terms of an A.P. whose $n^{t h}$ term is given as $a_{n} = 5 - 2 n$ .

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Answer:

S_{20} = - 320

$a_{1} = 5 - 2 = 3$ , $a_{20} = 5 - 40 = - 35$ .
$S_{20} = \frac{20}{2} (a_{1} + a_{20}) = 10 (3 - 35)$
$S_{20} = - 320$ .

Also asked in: 2022 Standard 30/2/1, 2022 Standard 30/2/2

Q62 marksVery Short AnswerSurface Areas and VolumesNot in current syllabus

A solid piece of metal in the form of a cuboid of dimensions 11 cm $\times$ 7 cm $\times$ 7 cm is melted to form n number of solid spheres of radii $\frac{7}{2}$ cm each. Find the value of n.

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Answer: n = 3

Volume of cuboid $= 11 \times 7 \times 7 = 539$ cm $^{3}$ .
Volume of one sphere $= \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} = \frac{539}{3}$ cm $^{3}$ .
n $\times \frac{539}{3} = 539$
n $= 3$ .

Also asked in: 2022 Standard 30/2/1, 2022 Standard 30/2/2

Q73 marksShort AnswerStatistics

The mean of the following frequency distribution is 25. Find the value of f.
Class: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50
Frequency: 5, 18, 15, f, 6

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Answer: f = 16

Class marks $x_{i}$ : 5, 15, 25, 35, 45.
$f_{i} x_{i}$ : 25, 270, 375, 35f, 270.
$\sum f_{i} = 44 + f$ , $\sum f_{i} x_{i} = 940 + 35 f$ .
Mean $= \frac{940 + 35 f}{44 + f} = 25$
$940 + 35 f = 1100 + 25 f \Rightarrow 10 f = 160 \Rightarrow f = 16$ .

Also asked in: 2022 Standard 30/2/1, 2022 Standard 30/2/2

Q7 (OR) (OR)3 marksShort AnswerStatistics

Find the mean of the following data using assumed mean method :
Class: 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25
Frequency: 8, 7, 10, 13, 12

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Answer: Mean = 13.9

Class marks $x_{i}$ : 2.5, 7.5, 12.5, 17.5, 22.5. Take assumed mean $a = 12.5$ .
$d_{i} = x_{i} - a$ : $- 10, - 5, 0, 5, 10$ .
$f_{i} d_{i}$ : $- 80, - 35, 0, 65, 120$ .
$\sum f_{i} = 50$ , $\sum f_{i} d_{i} = 70$ .
Mean $= a + \frac{\sum f _{i} d _{i}}{\sum f _{i}} = 12.5 + \frac{70}{50} = 12.5 + 1.4 = 13.9$ .

Also asked in: 2022 Standard 30/2/1, 2022 Standard 30/2/2

Q83 marksShort AnswerSome Applications of Trigonometry

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $3 0^{\circ}$ and $4 5^{\circ}$ . If the bridge is at a height of 8 m from the banks, then find the width of the river.

Diagram for CBSE 2022 Class 10 Maths question 8

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Answer:

8 (1 + 3)

\approx 21.86

Let A be the point on the bridge, AC = 8 m, and B, D the banks with $∠ A B C = 4 5^{\circ}$ , $∠ A D C = 3 0^{\circ}$ (alternate angles).
In right $△ A C B$ : $tan 4 5^{\circ} = \frac{A C}{B C} \Rightarrow B C = 8$ m.
In right $△ A C D$ : $tan 3 0^{\circ} = \frac{A C}{C D} \Rightarrow C D = 83$ m.
Width $B D = 8 + 83 = 8 (1 + 3)$ m $\approx 21.86$ m.

Q93 marksShort AnswerStatistics

Heights of 50 students of class X of a school are recorded and following data is obtained :
Height (in cm): 130-135, 135-140, 140-145, 145-150, 150-155, 155-160
Number of Students: 4, 11, 12, 7, 10, 6
Find the median height of the students.

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Answer: Median height

= 144 \frac{1}{6}

\approx 144.17

Cumulative frequencies: 4, 15, 27, 34, 44, 50.
$n = 50$ , $\frac{n}{2} = 25$ , so the median class is 140-145.
$l = 140$ , $c f = 15$ , $f = 12$ , $h = 5$ .
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 140 + \frac{25 - 15}{12} \times 5$
$= 140 + \frac{50}{12} = 140 + 4.17 = 144.17$ cm (approx.).

Also asked in: 2022 Standard 30/2/1, 2022 Standard 30/2/2

Q103 marksShort AnswerCirclesNot in current syllabus

Construct a pair of tangents to a circle of radius 4 cm from a point P lying outside the circle at a distance of 6 cm from the centre.

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Answer: Construction (each tangent is

25 \approx 4.5

cm long).

Draw a circle of radius 4 cm with centre O and mark P with OP = 6 cm.
Bisect OP; let M be its midpoint.
With centre M and radius MO draw a circle cutting the given circle at Q and R.
Join PQ and PR; these are the required tangents.
Justification: $∠ O QP = 9 0^{\circ}$ (angle in a semicircle), so PQ $⊥$ OQ; similarly PR.
Length check: $P Q = 6^{2} - 4^{2} = 20 \approx 4.5$ cm.

Q114 marksLong AnswerQuadratic Equations

A 2-digit number is such that the product of its digits is 24. If 18 is subtracted from the number, the digits interchange their places. Find the number.

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Answer: 64

Let the tens digit be $x$ and the units digit be $y$ ; the number is $10 x + y$ .
$x y = 24$ and $10 x + y - 18 = 10 y + x \Rightarrow 9 x - 9 y = 18 \Rightarrow x = y + 2$ .
$(y + 2) y = 24 \Rightarrow y^{2} + 2 y - 24 = 0 \Rightarrow (y + 6) (y - 4) = 0$ .
$y = 4$ (a digit cannot be $- 6$ ), so $x = 6$ .
The number is 64. Check: $6 \times 4 = 24$ and $64 - 18 = 46$ .

Also asked in: 2022 Standard 30/2/1

Q11 (OR) (OR)4 marksLong AnswerQuadratic Equations

The difference of the squares of two numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers.

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Answer: 18 and 12 (or 18 and

- 12

)

Let the greater number be $x$ ; then (smaller) $^{2} = 8 x$ .
$x^{2} - 8 x = 180 \Rightarrow x^{2} - 8 x - 180 = 0$ .
$(x - 18) (x + 10) = 0 \Rightarrow x = 18$ or $x = - 10$ .
$x = - 10$ gives (smaller) $^{2} = - 80$ , impossible; so $x = 18$ .
(Smaller) $^{2} = 144 \Rightarrow$ smaller $= 12$ or $- 12$ .
The numbers are 18 and 12 (or 18 and $- 12$ ).

Also asked in: 2022 Standard 30/2/1

Q124 marksLong AnswerCircles

Prove that a parallelogram circumscribing a circle is a rhombus.

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Answer: Proved.

Let parallelogram ABCD touch the circle at P, Q, R, S on AB, BC, CD, DA respectively.
Tangents from an external point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS.
Adding: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), i.e. AB + CD = AD + BC.
In a parallelogram AB = CD and AD = BC, so 2AB = 2AD, i.e. AB = AD.
Thus all four sides are equal: AB = BC = CD = DA.
Hence ABCD is a rhombus.

Also asked in: 2026 Basic 430/4/1, 2023 Standard 30/6/3, 2022 Basic 430/2/1, 2022 Basic 430/2/3

Q134 marksCase StudySome Applications of Trigonometry

Kite Festival
Kite festival is celebrated in many countries at different times of the year. In India, every year 14th January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites.
The picture given below, three kites flying together.
In Fig. 5, the angles of elevation of two kites (Points A and B) from the hands of a man (Point C) are found to be $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Taking AD = 50 m and BE = 60 m, find
(1) the lengths of strings used (take them straight) for kites A and B as shown in the figure. (2)
(2) the distance ‘d’ between these two kites (2)

Diagram for CBSE 2022 Class 10 Maths question 13

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Answer: (1) AC = 100 m, BC =

403

\approx 69.28

m (2) d =

2037

\approx 121.66

(1) In right $△ A D C$ : $sin 3 0^{\circ} = \frac{A D}{A C} \Rightarrow A C = \frac{50}{1/2} = 100$ m.
In right $△ B E C$ : $sin 6 0^{\circ} = \frac{B E}{B C} \Rightarrow B C = \frac{60 \times 2}{3} = 403$ m.
(2) $∠ A C B = 18 0^{\circ} - 3 0^{\circ} - 6 0^{\circ} = 9 0^{\circ}$ .
In right $△ A C B$ : $d^{2} = A C^{2} + B C^{2} = 10000 + 4800 = 14800$ .
$d = 14800 = 2037$ m $\approx 121.66$ m.

Q144 marksCase StudySurface Areas and Volumes

A ‘circus’ is a company of performers who put on shows of acrobats, clowns etc. to entertain people started around 250 years back, in open fields, now generally performed in tents.
One such ‘Circus Tent’ is shown below.
The tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of cylindrical part are 9 m and 30 m respectively and height of conical part is 8 m with same diameter as that of the cylindrical part, then find
(1) the area of the canvas used in making the tent; (3)
(2) the cost of the canvas bought for the tent at the rate ₹ 200 per sq m, if 30 sq m canvas was wasted during stitching. (1)

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Answer: (1) 1650 sq m (2) ₹ 3,36,000

(1) Radius $r = 15$ m, cylinder height $h = 9$ m, cone height 8 m.
Slant height $l = 1 5^{2} + 8^{2} = 17$ m.
Canvas = curved surface of cylinder + curved surface of cone $= 2 π r h + π r l = π r (2 h + l)$ .
$= \frac{22}{7} \times 15 \times (18 + 17) = \frac{22}{7} \times 15 \times 35 = 1650$ sq m.
(2) Canvas bought $= 1650 + 30 = 1680$ sq m.
Cost $= 1680 \times 200 = ₹ 3, 36, 000$ .

Also asked in: 2022 Standard 30/2/1, 2022 Standard 30/2/2

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