CBSE Maths › Class 10 PYQs › 2022

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

All 18 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/2/2 (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 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.

Show answer & solution

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 30/2/3

Q22 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 2

Show answer & solution

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 30/2/3

Q2 (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 2 (OR)

Show answer & solution

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 30/2/3

Q32 marksVery Short AnswerArithmetic Progressions

Which term of the A.P. $- \frac{11}{2}, - 3, - \frac{1}{2},$ .... is $\frac{49}{2}$ ?

Show answer & solution

Answer: 13th term

$a = - \frac{11}{2}$ , $d = - 3 - (- \frac{11}{2}) = \frac{5}{2}$ .
$a_{n} = a + (n - 1) d = \frac{49}{2}$
$- \frac{11}{2} + (n - 1) \frac{5}{2} = \frac{49}{2}$
$(n - 1) \frac{5}{2} = 30 \Rightarrow n - 1 = 12 \Rightarrow n = 13$ .
So $\frac{49}{2}$ is the 13th term.

Also asked in: 2022 Standard 30/2/1

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

Find a and b so that the numbers a, 7, b, 23 are in A.P.

Show answer & solution

Answer:

a = - 1

b = 15

In an A.P. each middle term is the average of its neighbours.
7 is between a and b, and b is between 7 and 23.
$b = \frac{7 + 23}{2} = 15$ .
Common difference $d = 15 - 7 = 8$ , so $a = 7 - 8 = - 1$ .
Check: $- 1, 7, 15, 23$ has common difference 8.

Also asked in: 2022 Standard 30/2/1

Q42 marksVery Short AnswerArithmetic Progressions

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

Show answer & solution

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 30/2/3

Q52 marksVery Short AnswerQuadratic Equations

Solve the quadratic equation : $x^{2} - 2 a x + (a^{2} - b^{2}) = 0$ for $x$ .

Show answer & solution

Answer:

x = a + b

x = a - b

$x^{2} - 2 a x + a^{2} - b^{2} = 0$
$(x - a)^{2} - b^{2} = 0$
$(x - a - b) (x - a + b) = 0$
$x = a + b$ or $x = a - b$ .

Q62 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

Show answer & solution

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 30/2/3

Q73 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.

Show answer & solution

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 30/2/3

Q83 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

Show answer & solution

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 30/2/3

Q8 (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

Show answer & solution

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 30/2/3

Q93 marksShort AnswerSome Applications of Trigonometry

Two men on either side of a cliff 75 m high observe the angles of elevation of the top of the cliff to be $3 0^{\circ}$ and $6 0^{\circ}$ . Find the distance between the two men.

Diagram for CBSE 2022 Class 10 Maths question 9

Show answer & solution

Answer:

1003

\approx 173.2

In right $△ B A C$ : $tan 3 0^{\circ} = \frac{A B}{A C} \Rightarrow A C = 753$ m.
In right $△ B A D$ : $tan 6 0^{\circ} = \frac{A B}{A D} \Rightarrow A D = \frac{75}{3} = 253$ m.
Distance $C D = A C + A D = 753 + 253 = 1003$ m $\approx 173.2$ m.

Q103 marksShort AnswerCirclesNot in current syllabus

Construct a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle of $6 0^{\circ}$ .

Show answer & solution

Answer: Construction (angle between the radii to the points of contact is

12 0^{\circ}

Draw a circle of radius 3 cm with centre O and draw a radius OA.
The angle between the radii to the points of contact is $18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ ; draw radius OB with $∠ A O B = 12 0^{\circ}$ .
At A and B draw lines perpendicular to OA and OB; let them meet at P.
PA and PB are the required tangents.
Justification: in quadrilateral OAPB, $∠ A = ∠ B = 9 0^{\circ}$ and $∠ O = 12 0^{\circ}$ , so $∠ A P B = 36 0^{\circ} - 30 0^{\circ} = 6 0^{\circ}$ .

Q114 marksLong AnswerQuadratic Equations

The sum of two numbers is 34. If 3 is subtracted from one number and 2 is added to another, the product of these two numbers becomes 260. Find the numbers.

Show answer & solution

Answer: 23 and 11, or 16 and 18

Let one number be $x$ ; the other is $34 - x$ .
$(x - 3) (34 - x + 2) = 260 \Rightarrow (x - 3) (36 - x) = 260$ .
$- x^{2} + 39 x - 108 = 260 \Rightarrow x^{2} - 39 x + 368 = 0$ .
$(x - 23) (x - 16) = 0 \Rightarrow x = 23$ or $x = 16$ .
Numbers: 23 and 11, or 16 and 18.
Check: $(23 - 3) (11 + 2) = 260$ and $(16 - 3) (18 + 2) = 260$ .

Q11 (OR) (OR)4 marksLong AnswerQuadratic Equations

The hypotenuse (in cm) of a right angled triangle is 6 cm more than twice the length of the shortest side. If the length of third side is 6 cm less than thrice the length of shortest side, then find the dimensions of the triangle.

Show answer & solution

Answer: 10 cm, 24 cm and 26 cm

Let the shortest side be $x$ cm. Hypotenuse $= 2 x + 6$ , third side $= 3 x - 6$ .
$x^{2} + (3 x - 6)^{2} = (2 x + 6)^{2}$
$10 x^{2} - 36 x + 36 = 4 x^{2} + 24 x + 36$
$6 x^{2} - 60 x = 0 \Rightarrow 6 x (x - 10) = 0 \Rightarrow x = 10$ ( $x \neq = 0$ ).
Sides: 10 cm, 24 cm, 26 cm. Check: $100 + 576 = 676 = 2 6^{2}$ .

Q124 marksLong AnswerCircles

In Fig. 4, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q meet at a point T. Find the length of TP.

Diagram for CBSE 2022 Class 10 Maths question 12

Show answer & solution

Answer:

T P = \frac{20}{3}

TP = TQ and OT bisects $∠ P T Q$ , so OT is the perpendicular bisector of PQ: $P R = R Q = 4$ cm, $∠ P R O = 9 0^{\circ}$ .
In $△ P R O$ : $O R = 5^{2} - 4^{2} = 3$ cm.
Let $T R = y$ . In $△ P R T$ : $T P^{2} = y^{2} + 16$ .
OP $⊥$ TP, so in $△ O P T$ : $O T^{2} = T P^{2} + O P^{2}$ .
$(y + 3)^{2} = y^{2} + 16 + 25 \Rightarrow 6 y + 9 = 41 \Rightarrow y = \frac{16}{3}$ .
$T P^{2} = \frac{256}{9} + 16 = \frac{400}{9}$ , so $T P = \frac{20}{3}$ cm.

Also asked in: 2022 Standard 30/2/1

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, shows 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

Show answer & solution

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.

Also asked in: 2022 Standard 30/2/1

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)

Show answer & solution

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 30/2/3

← 2022 Standard 30/2/1 2022 Standard 30/2/3 →

Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on MonoMath CBSE →