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CBSE Class 10 Maths Standard 2022 Question Paper 30/2/1 (Term 2) with Solutions

All 18 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/2/1 (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 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 30/2/3

Q22 marksVery Short AnswerArithmetic Progressions

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

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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 30/2/2

Q2 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions

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

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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 30/2/2

Q32 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 30/2/2, 2022 Standard 30/2/3

Q42 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 4

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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 30/2/2, 2022 Standard 30/2/3

Q4 (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 4 (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 30/2/2, 2022 Standard 30/2/3

Q52 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 30/2/2, 2022 Standard 30/2/3

Q62 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 30/2/2, 2022 Standard 30/2/3

Q73 marksShort AnswerCirclesNot in current syllabus

Draw two concentric circles of radii 2 cm and 5 cm. From a point on the outer circle, construct a pair of tangents to the inner circle.

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Answer: Construction (each tangent is

21 \approx 4.6

cm long).

Draw circles of radii 2 cm and 5 cm with common centre O. Take a point P on the outer circle and join OP.
Bisect OP; let M be its midpoint.
With M as centre and MO as radius draw a circle cutting the inner 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 and PQ is a tangent; similarly PR.
Length check: $P Q = 5^{2} - 2^{2} = 21 \approx 4.6$ cm.

Q83 marksShort AnswerSome Applications of Trigonometry

In Fig. 3, AB is tower of height 50 m. A man standing on its top, observes two cars on the opposite sides of the tower with angles of depression $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the distance between the two cars.

Diagram for CBSE 2022 Class 10 Maths question 8

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Answer:

50 (3 + 1)

\approx 136.6

Angles of depression equal the angles of elevation from the cars: $∠ A C B = 3 0^{\circ}$ , $∠ A D B = 4 5^{\circ}$ .
In right $△ A B C$ : $tan 3 0^{\circ} = \frac{A B}{B C} \Rightarrow B C = 503$ m.
In right $△ A B D$ : $tan 4 5^{\circ} = \frac{A B}{B D} \Rightarrow B D = 50$ m.
Distance $C D = B C + B D = 503 + 50 = 50 (3 + 1)$ m $\approx 136.6$ m.

Q93 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 30/2/2, 2022 Standard 30/2/3

Q9 (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 30/2/2, 2022 Standard 30/2/3

Q103 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 30/2/2, 2022 Standard 30/2/3

Q114 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 11

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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 30/2/2

Q124 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 30/2/3

Q12 (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 30/2/3

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, 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

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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.

Also asked in: 2022 Standard 30/2/2

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 30/2/2, 2022 Standard 30/2/3

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