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

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

Find the mode of the following frequency distribution :
Class: 10–20, 20–30, 30–40, 40–50, 50–60
Frequency: 15, 10, 12, 17, 4

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Answer:

\frac{385}{9} \approx 42.78

Highest frequency is 17, so the modal class is 40–50.
$l = 40$ , $h = 10$ , $f_{1} = 17$ , $f_{0} = 12$ , $f_{2} = 4$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{17 - 12}{34 - 12 - 4} \times 10$
$= 40 + \frac{50}{18} = 40 + 2.78 = 42.78$ (approx.)

Also asked in: 2022 Standard 30/1/1, 2022 Standard 30/1/3

Q22 marksVery Short AnswerCircles

Two concentric circles are of radii 4 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

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Answer:

27

Let the chord AB of the larger circle touch the smaller circle at M; O is the common centre.
OM $⊥$ AB (radius is perpendicular to tangent), and OM = 3 cm, OA = 4 cm.
AM $= O A^{2} - O M^{2} = 16 - 9 = 7$ cm.
The perpendicular from the centre bisects the chord, so AB $= 27$ cm.

Also asked in: 2022 Standard 30/1/1, 2022 Standard 30/1/3

Q32 marksVery Short AnswerArithmetic Progressions

Find the sum of first 30 terms of AP : $- 30, - 24, - 18$ , ..... .

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Answer: 1710

Here $a = - 30$ , $d = - 24 - (- 30) = 6$ , $n = 30$ .
$S_{30} = \frac{30}{2} [2 (- 30) + (30 - 1) (6)]$
$= 15 [- 60 + 174] = 15 \times 114 = 1710$

Also asked in: 2022 Standard 30/1/1

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

In an AP if $S_{n} = n (4 n + 1)$ , then find the AP.

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Answer: 5, 13, 21, 29, ...

$S_{1} = 1 (4 + 1) = 5$ , so $a_{1} = 5$ .
$S_{2} = 2 (8 + 1) = 18$ , so $a_{2} = S_{2} - S_{1} = 13$ .
$d = 13 - 5 = 8$ .
The AP is 5, 13, 21, 29, ...

Also asked in: 2022 Standard 30/1/1

Q42 marksVery Short AnswerSurface Areas and VolumesNot in current syllabus

A cone of height 28 cm and radius of base 7 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.

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Answer: 7 cm

Volume is unchanged: $\frac{4}{3} π R^{3} = \frac{1}{3} π (7)^{2} (28)$ .
$4 R^{3} = 49 \times 28 \Rightarrow R^{3} = 343$
$R = 7$ cm.

Q52 marksVery Short AnswerQuadratic Equations

Find the value of $m$ for which the quadratic equation
$(m - 1) x^{2} + 2 (m - 1) x + 1 = 0$
has two real and equal roots.

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Answer:

m = 2

For equal roots, $D = b^{2} - 4 a c = 0$ (and $m - 1 \neq = 0$ ).
$[2 (m - 1)]^{2} - 4 (m - 1) (1) = 0$
$4 (m - 1) (m - 1 - 1) = 0 \Rightarrow 4 (m - 1) (m - 2) = 0$
$m = 1$ is rejected since then the equation is not quadratic.
So $m = 2$ .

Also asked in: 2022 Standard 30/1/1, 2022 Standard 30/1/3

Q5 (OR) (OR)2 marksVery Short AnswerQuadratic Equations

Solve the following quadratic equation for x :
$3 x^{2} + 10 x + 73 = 0$

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Answer:

x = - 3

x = - \frac{7}{3} = - \frac{7 3}{3}

Split the middle term: $3 \times 73 = 21 = 3 \times 7$ and $3 + 7 = 10$ .
$3 x^{2} + 3 x + 7 x + 73 = 0$
$3 x (x + 3) + 7 (x + 3) = 0$
$(x + 3) (3 x + 7) = 0$
$x = - 3$ or $x = - \frac{7}{3} = - \frac{7 3}{3}$

Also asked in: 2022 Standard 30/1/1, 2022 Standard 30/1/3

Q62 marksVery Short AnswerQuadratic Equations

The product of Rehan’s age (in years) 5 years ago and his age 7 years from now, is one more than twice his present age. Find his present age.

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Answer: 6 years

Let the present age be $x$ years.
$(x - 5) (x + 7) = 2 x + 1$
$x^{2} + 2 x - 35 = 2 x + 1 \Rightarrow x^{2} = 36$
$x = 6$ (age cannot be negative).
Rehan's present age is 6 years.

Also asked in: 2022 Standard 30/1/1, 2022 Standard 30/1/3

Q73 marksShort AnswerSome Applications of Trigonometry

The angle of elevation of the top of a building from the foot of the tower is $3 0^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $6 0^{\circ}$ . If the tower is 50 m high, then find the height of the building.

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Answer:

\frac{50}{3}

= 16 \frac{2}{3}

Let the tower AB = 50 m and the building CD = $h$ m, with feet B and D a distance $d$ apart.
From D: $tan 6 0^{\circ} = \frac{50}{d} \Rightarrow d = \frac{50}{3}$ m.
From B: $tan 3 0^{\circ} = \frac{h}{d} \Rightarrow h = \frac{d}{3} = \frac{50}{3}$ m.
Height of the building $= 16 \frac{2}{3}$ m (about 16.67 m).

Also asked in: 2022 Standard 30/1/1, 2022 Standard 30/1/3

Q7 (OR) (OR)3 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 $4 5^{\circ}$ respectively. If the bridge is at a height of 3 m from the banks, then find the width of the river.

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Answer:

3 (1 + 3)

\approx 8.19

Let P be the point on the bridge, 3 m above the banks, and A, B the banks on opposite sides; D is the point directly below P.
Angle of depression $3 0^{\circ}$ to A: $tan 3 0^{\circ} = \frac{3}{A D} \Rightarrow A D = 33$ m.
Angle of depression $4 5^{\circ}$ to B: $tan 4 5^{\circ} = \frac{3}{D B} \Rightarrow D B = 3$ m.
Width AB $= 33 + 3 = 3 (1 + 3)$ m $\approx 8.19$ m.

Also asked in: 2022 Standard 30/1/1, 2022 Standard 30/1/3

Q83 marksShort AnswerStatistics

Following is the daily expenditure on lunch by 30 employees of a company :
Daily Expenditure (in Rupees): 100–120, 120–140, 140–160, 160–180, 180–200
Number of Employees: 8, 3, 8, 6, 5
Find the mean daily expenditure of the employees.

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Answer: ₹148

Class marks $x_{i}$ : 110, 130, 150, 170, 190; $Σ f_{i} = 30$ .
$f_{i} x_{i}$ : 880, 390, 1200, 1020, 950; $Σ f_{i} x_{i} = 4440$ .
Mean $= \frac{Σ f _{i} x _{i}}{Σ f _{i}} = \frac{4440}{30} = 148$
Mean daily expenditure is ₹148.

Also asked in: 2022 Standard 30/1/1

Q93 marksShort AnswerStatistics

If the median of the following data is 33 then, find the value of m :
Class: 0–15, 15–30, 30–45, 45–60, 60–75, 75–90
Frequency: 17, 35, 40, 18, m, 2

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Answer:

m = 8

Cumulative frequencies: 17, 52, 92, 110, $110 + m$ , $112 + m$ ; so $N = 112 + m$ .
Median 33 lies in the class 30–45: $l = 30$ , $c f = 52$ , $f = 40$ , $h = 15$ .
$33 = 30 + \frac{\frac{112 + m}{2} - 52}{40} \times 15$
$3 = \frac{15}{40} (4 + \frac{m}{2}) \Rightarrow 8 = 4 + \frac{m}{2}$
$m = 8$ (check: $N = 120$ , $\frac{N}{2} = 60$ lies in 52–92, so 30–45 is the median class).

Q103 marksShort AnswerCirclesNot in current syllabus

Construct a pair of tangents to a circle of radius 5 cm, which are inclined to each other at an angle of $6 0^{\circ}$ .

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Answer: Construction: draw radii OA and OB with

∠ A O B = 12 0^{\circ}

; the perpendiculars to OA at A and to OB at B are the required tangents.

Draw a circle with centre O and radius 5 cm.
The angle between the tangents is $6 0^{\circ}$ , so the angle between the radii to the points of contact is $18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .
Draw radius OA and radius OB with $∠ A O B = 12 0^{\circ}$ .
At A draw AP $⊥$ OA and at B draw BP $⊥$ OB; they meet at P.
PA and PB are the required tangents, since in quadrilateral OAPB, $∠ A P B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ .

Also asked in: 2022 Standard 30/4/3

Q114 marksLong AnswerCircles

In Figure 1, a right triangle ABC in which $∠ B = 9 0^{\circ}$ , AB = 12 cm and BC = 5 cm is shown. Find the radius of the circle inscribed in the triangle ABC.

Diagram for CBSE 2022 Class 10 Maths question 11

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Answer: 2 cm

AC $= 1 2^{2} + 5^{2} = 13$ cm.
Let the incircle (centre O, radius $r$ ) touch BC, CA, AB at D, E, F. OD $⊥$ BC and OF $⊥$ AB, so ODBF is a square: BD = BF = $r$ .
Tangents from a point are equal: CD = CE = $5 - r$ and AF = AE = $12 - r$ .
AC = AE + EC: $13 = (12 - r) + (5 - r) \Rightarrow 2 r = 4$
$r = 2$ cm.

Q124 marksLong AnswerSurface Areas and Volumes

From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and same radius is hollowed out. Find the total surface area of the remaining solid.

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Answer: 2024 cm²

Slant height of cone $l = 2 4^{2} + 7^{2} = 625 = 25$ cm.
TSA = curved surface of cylinder + area of base + curved surface of cone.
Curved surface of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 7 \times 30 = 1320$ cm².
Base $= π r^{2} = \frac{22}{7} \times 7 \times 7 = 154$ cm².
Curved surface of cone $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm².
TSA $= 1320 + 154 + 550 = 2024$ cm².

Also asked in: 2022 Standard 30/1/1

Q12 (OR) (OR)4 marksLong AnswerSurface Areas and VolumesNot in current syllabus

Water in a canal, 8 m wide and 6 m deep, is flowing with a speed of 12 km/hour. How much area will it irrigate in one hour, if 0.05 m of standing water is required ?

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Answer: 11520000 m² (1152 hectares)

In one hour the water travels 12 km = 12000 m.
Volume flowing in one hour $= 8 \times 6 \times 12000 = 576000$ m³.
Area irrigated $\times 0.05 = 576000$
Area $= \frac{576000}{0.05} = 11520000$ m² $= 1152$ hectares.

Also asked in: 2022 Standard 30/1/1

Q134 marksCase StudyArithmetic Progressions

In Mathematics, relations can be expressed in various ways. The matchstick patterns are based on linear relations. Different strategies can be used to calculate the number of matchsticks used in different figures.
One such pattern is shown below. Observe the pattern and answer the following questions using Arithmetic Progression :
(a) Write the AP for the number of triangles used in the figures. Also, write the $n^{t h}$ term of this AP. (2)
(b) Which figure has 61 matchsticks ? (2)

Diagram for CBSE 2022 Class 10 Maths question 13

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Answer: (a) 4, 6, 8, ...;

a_{n} = 2 n + 2

(b) Figure 8

(a) Triangles: Figure 1 has 4, Figure 2 has 6, Figure 3 has 8, so the AP is 4, 6, 8, ...
$a = 4$ , $d = 2$ : $a_{n} = 4 + (n - 1) 2 = 2 n + 2$ .
(b) Matchsticks: Figure 1 uses $4 + 4 \times 2 = 12$ , Figure 2 uses $7 + 6 \times 2 = 19$ , Figure 3 uses $10 + 8 \times 2 = 26$ .
AP 12, 19, 26, ... with $a = 12$ , $d = 7$ : $a_{n} = 12 + (n - 1) 7 = 7 n + 5$ .
$7 n + 5 = 61 \Rightarrow n = 8$ . Figure 8 has 61 matchsticks.

Also asked in: 2022 Standard 30/1/1, 2022 Standard 30/1/3

Q144 marksCase StudySome Applications of Trigonometry

Gadisar Lake is located in the Jaisalmer district of Rajasthan. It was built by the King of Jaisalmer and rebuilt by Gadsi Singh in 14 $^{t h}$ century. The lake has many Chhatris. One of them is shown below :
Observe the picture. From a point A $h$ m above from water level, the angle of elevation of top of Chhatri (point B) is $4 5^{\circ}$ and angle of depression of its reflection in water (point C) is $6 0^{\circ}$ . If the height of Chhatri above water level is (approximately) 10 m, then
(a) draw a well-labelled figure based on the above information; (2)
(b) find the height ( $h$ ) of the point A above water level. (Use $3 = 1.73$ ) (2)

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Answer: (a) Figure: A at height

h

above water; B is 10 m above water and C is 10 m below water on the vertical through the Chhatri; elevation of B

4 5^{\circ}

, depression of C

6 0^{\circ}

. (b)

h = 20 - 103 \approx 2.7

(a) Draw the water line and the vertical line of the Chhatri meeting it at D, with B 10 m above D and its reflection C 10 m below D. A is a point $h$ m above the water; draw the horizontal AM to the vertical BC. Mark $∠ B A M = 4 5^{\circ}$ and $∠ M A C = 6 0^{\circ}$ .
(b) Let AM $= x$ . Then BM $= 10 - h$ and MC $= 10 + h$ .
$tan 4 5^{\circ} = \frac{10 - h}{x} \Rightarrow x = 10 - h$ .
$tan 6 0^{\circ} = \frac{10 + h}{x} \Rightarrow 10 + h = 3 (10 - h)$ .
$h (1 + 3) = 10 (3 - 1) \Rightarrow h = \frac{10 ( 3 - 1 )}{3 + 1} = 5 (3 - 1)^{2} = 20 - 103$
$h = 20 - 17.3 = 2.7$ m.

Also asked in: 2022 Standard 30/1/1, 2022 Standard 30/1/3

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