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

All 18 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/3/2 (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 given frequency distribution :
Class: 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65, 65 – 75
Frequency: 6, 11, 22, 23, 14, 5

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Answer: 46

Highest frequency is 23, so the modal class is 45 – 55.
$l = 45$ , $h = 10$ , $f_{1} = 23$ , $f_{0} = 22$ , $f_{2} = 14$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 45 + \frac{23 - 22}{46 - 22 - 14} \times 10$
$= 45 + \frac{1}{10} \times 10 = 46$ .

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

Q22 marksVery Short AnswerArithmetic Progressions

For what value of ‘n’, are the $n^{t h}$ terms of the APs : 9, 7, 5, ..... and 15, 12, 9, ..... the same ?

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Answer:

n = 7

First AP: $a = 9$ , $d = - 2$ , so $a_{n} = 9 - 2 (n - 1) = 11 - 2 n$ .
Second AP: $a = 15$ , $d = - 3$ , so $a_{n} = 15 - 3 (n - 1) = 18 - 3 n$ .
$11 - 2 n = 18 - 3 n \Rightarrow n = 7$ (both 7th terms equal $- 3$ ).

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

Q32 marksVery Short AnswerSurface Areas and VolumesNot in current syllabus

150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, and are completely immersed in water. Find the rise in the level of water in the cylindrical vessel.

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Answer: 5.6 cm

Radius of each marble $= 0.7$ cm; volume of 150 marbles $= 150 \times \frac{4}{3} π (0.7)^{3} = 200 \times 0.343 π = 68.6 π$ cm $^{3}$ .
Radius of vessel $= 3.5$ cm. Let the water rise by h cm: $π (3.5)^{2} h = 12.25 π h$ .
$12.25 π h = 68.6 π \Rightarrow h = \frac{68.6}{12.25} = 5.6$ cm.

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

Q3 (OR) (OR)2 marksVery Short AnswerSurface Areas and Volumes

Three cubes of side 6 cm each, are joined as shown in Figure 1. Find the total surface area of the resulting cuboid.

Diagram for CBSE 2022 Class 10 Maths question 3 (OR)

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Answer: 504 cm

^{2}

Joining three cubes of side 6 cm end to end gives a cuboid of length $3 \times 6 = 18$ cm, breadth 6 cm and height 6 cm.
TSA $= 2 (l b + bh + h l) = 2 (18 \times 6 + 6 \times 6 + 6 \times 18)$
$= 2 (108 + 36 + 108) = 2 \times 252 = 504$ cm $^{2}$ .

Also asked in: 2022 Standard 30/3/1

Q42 marksVery Short AnswerQuadratic Equations

For what value of m, the quadratic equation
$m x^{2} - 2 (m - 1) x + (m + 2) = 0$
has two real and equal roots ?

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Answer:

m = \frac{1}{4}

For real and equal roots, $D = 0$ : $[- 2 (m - 1)]^{2} - 4 m (m + 2) = 0$ .
$4 (m^{2} - 2 m + 1) - 4 m^{2} - 8 m = 0 \Rightarrow - 16 m + 4 = 0$ .
$m = \frac{1}{4}$ (and $m \neq = 0$ , so the equation is quadratic).

Q4 (OR) (OR)2 marksVery Short AnswerQuadratic Equations

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

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Answer: Shorter side 90 m, longer side 120 m

Let the shorter side be x m. Then longer side $= x + 30$ and diagonal $= x + 60$ .
$x^{2} + (x + 30)^{2} = (x + 60)^{2} \Rightarrow 2 x^{2} + 60 x + 900 = x^{2} + 120 x + 3600$ .
$x^{2} - 60 x - 2700 = 0 \Rightarrow (x - 90) (x + 30) = 0 \Rightarrow x = 90$ (x cannot be negative).
Sides are 90 m and 120 m.

Q52 marksVery Short AnswerCircles

In Figure 2, PQ and PR are tangents to the circle centred at O. If $∠$ OPR = $4 5^{\circ}$ , then prove that ORPQ is a square.

Diagram for CBSE 2022 Class 10 Maths question 5

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Answer: Proved.

$O Q ⊥ P Q$ and $O R ⊥ P R$ (radius is perpendicular to the tangent), so $∠ O QP = ∠ O R P = 9 0^{\circ}$ .
In $△ O R P$ , $∠ O R P = 9 0^{\circ}$ and $∠ O P R = 4 5^{\circ}$ , so $∠ P O R = 4 5^{\circ}$ . Hence $O R = R P$ .
$O P$ bisects $∠ QP R$ (tangents from an external point), so $∠ QP R = 2 \times 4 5^{\circ} = 9 0^{\circ}$ .
In quadrilateral $O R P Q$ : $∠ QO R = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ , so all four angles are right angles.
$O Q = O R$ (radii), $P Q = P R$ (tangents from P) and $O R = R P$ , so all four sides are equal.
A quadrilateral with all sides equal and all angles $9 0^{\circ}$ is a square. Hence $O R P Q$ is a square.

Also asked in: 2022 Standard 30/3/1

Q62 marksVery Short AnswerArithmetic Progressions

Find the sum of first 20 terms of an AP in which d = 5 and $a_{20}$ = 135.

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Answer: 1750

$a_{20} = a + 19 d \Rightarrow 135 = a + 95 \Rightarrow a = 40$ .
$S_{20} = \frac{20}{2} (a + a_{20}) = 10 (40 + 135) = 1750$ .

Also asked in: 2022 Standard 30/3/1

Q73 marksShort AnswerSome Applications of Trigonometry

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of $3 0^{\circ}$ with it. The height of the breaking point from the ground is 2 m. Find the total height of the tree.

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Answer: 6 m

Let the broken part (hypotenuse) be L m; the standing part is 2 m.
$sin 3 0^{\circ} = \frac{2}{L} \Rightarrow \frac{1}{2} = \frac{2}{L} \Rightarrow L = 4$ m.
Total height $= 2 + 4 = 6$ m.

Q83 marksShort AnswerStatistics

The percentage of marks obtained by 100 students in an examination are given below :
Percentage of Marks: 30 – 35, 35 – 40, 40 – 45, 45 – 50, 50 – 55, 55 – 60, 60 – 65
Number of Students: 16, 14, 18, 20, 18, 12, 2
Determine the median percentage of marks.

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Answer: 45.5

Cumulative frequencies: 16, 30, 48, 68, 86, 98, 100. $N = 100$ , $\frac{N}{2} = 50$ .
Median class is 45 – 50: $l = 45$ , $c f = 48$ , $f = 20$ , $h = 5$ .
Median $= 45 + \frac{50 - 48}{20} \times 5 = 45 + 0.5 = 45.5$ .

Q93 marksShort AnswerTrianglesNot in current syllabus

Draw a line segment AB of length 8 cm and locate a point P on AB such that AP : PB = 1 : 5.

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Answer: P is located on AB with AP : PB = 1 : 5 (AP =

\frac{4}{3}

cm ≈ 1.33 cm, PB =

\frac{20}{3}

cm ≈ 6.67 cm).

Draw AB = 8 cm.
Draw a ray AX making an acute angle with AB.
Mark 6 (= 1 + 5) equal arcs on AX at points $A_{1}, A_{2}, \dots, A_{6}$ .
Join $A_{6} B$ .
Through $A_{1}$ draw a line parallel to $A_{6} B$ , meeting AB at P.
By BPT, $A P : P B = A A_{1} : A_{1} A_{6} = 1 : 5$ . So $A P = \frac{8}{6} = \frac{4}{3}$ cm.

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

Q9 (OR) (OR)3 marksShort AnswerCirclesNot in current syllabus

Draw a circle of radius 3 cm. From a point P lying outside the circle at a distance of 6 cm from its centre, construct two tangents PA and PB to the circle.

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Answer: Tangents PA and PB constructed; PA = PB =

33

cm ≈ 5.2 cm.

Draw a circle with centre O and radius 3 cm; mark P with OP = 6 cm.
Draw the perpendicular bisector of OP; let M be the midpoint of OP.
With M as centre and MO (= 3 cm) as radius, draw a circle cutting the given circle at A and B.
Join PA and PB; these are the required tangents.
Justification: $∠ O A P = 9 0^{\circ}$ (angle in a semicircle), so PA is a tangent; similarly PB.
Check: $P A = 6^{2} - 3^{2} = 27 = 33$ cm ≈ 5.2 cm.

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

Q103 marksShort AnswerStatistics

The weights (in kg) of 50 wild animals of a National Park were recorded and the following data was obtained :
Weight (in kg): 100 – 110, 110 – 120, 120 – 130, 130 – 140, 140 – 150
Number of animals: 4, 12, 23, 8, 3
Find the mean weight (in kg) of animals, using assumed mean method.

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Answer: 123.8 kg

Class marks $x_{i}$ : 105, 115, 125, 135, 145. Take assumed mean $a = 125$ .
$d_{i} = x_{i} - 125$ : $- 20, - 10, 0, 10, 20$ .
$f_{i} d_{i}$ : $- 80, - 120, 0, 80, 60$ ; $\sum f_{i} d_{i} = - 60$ , $\sum f_{i} = 50$ .
Mean $= a + \frac{\sum f _{i} d _{i}}{\sum f _{i}} = 125 + \frac{- 60}{50} = 125 - 1.2 = 123.8$ kg.

Also asked in: 2022 Standard 30/3/1

Q114 marksLong AnswerSome Applications of Trigonometry

The angle of elevation of an aeroplane from a point on the ground is $6 0^{\circ}$ . After a flight of 30 seconds, the angle of elevation from the same point becomes $3 0^{\circ}$ . If the aeroplane is flying at a constant height of $30003$ m, find the speed of the aeroplane.

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Answer: 200 m/s (= 720 km/h)

Height $h = 30003$ m.
At $6 0^{\circ}$ : horizontal distance $= \frac{h}{t a n 6 0 ^{\circ}} = \frac{3000 3}{3} = 3000$ m.
At $3 0^{\circ}$ : horizontal distance $= \frac{h}{t a n 3 0 ^{\circ}} = 30003 \times 3 = 9000$ m.
Distance flown in 30 s $= 9000 - 3000 = 6000$ m.
Speed $= \frac{6000}{30} = 200$ m/s $= 720$ km/h.

Q124 marksLong AnswerCircles

In Figure 3, two circles with centres at O and O′ of radii 2r and r respectively, touch each other internally at A. A chord AB of the bigger circle meets the smaller circle at C. Show that C bisects AB.

Diagram for CBSE 2022 Class 10 Maths question 12

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Answer: Proved.

The circles touch internally at A, so O, O′ and A are collinear and $O A = 2 r$ is a radius of the bigger circle.
Since $O^{'} A = r$ and $O A = 2 r$ , O lies on the smaller circle and $O A$ is a diameter of the smaller circle.
C lies on the smaller circle, so $∠ O C A = 9 0^{\circ}$ (angle in a semicircle). Hence $O C ⊥ A B$ .
AB is a chord of the bigger circle with centre O, and the perpendicular from the centre to a chord bisects the chord.
Therefore $A C = C B$ , i.e. C bisects AB.

Also asked in: 2022 Standard 30/3/1

Q12 (OR) (OR)4 marksLong AnswerCircles

In Figure 4, O is centre of a circle of radius 5 cm. PA and BC are tangents to the circle at A and B respectively. If OP = 13 cm, then find the length of tangents PA and BC.

Diagram for CBSE 2022 Class 10 Maths question 12 (OR)

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Answer: PA = 12 cm, BC =

\frac{10}{3}

$O A ⊥ P A$ , so in right $△ O A P$ : $P A = O P^{2} - O A^{2} = 169 - 25 = 12$ cm.
B lies on OP with $O B = 5$ cm, so $P B = 13 - 5 = 8$ cm, and $B C ⊥ O P$ (tangent at B).
Let $B C = x$ . Tangents from C are equal, so $C A = C B = x$ and $P C = 12 - x$ .
In right $△ P B C$ : $(12 - x)^{2} = 8^{2} + x^{2} \Rightarrow 144 - 24 x = 64 \Rightarrow x = \frac{10}{3}$ .
So $P A = 12$ cm and $B C = \frac{10}{3}$ cm.

Also asked in: 2022 Standard 30/3/1

Q134 marksCase StudyQuadratic Equations

In the picture given below, one can see a rectangular in-ground swimming pool installed by a family in their backyard. There is a concrete sidewalk around the pool of width x m. The outside edges of the sidewalk measure 7 m and 12 m. The area of the pool is 36 sq. m.
(a) Based on the information given above, form a quadratic equation in terms of x. (2)
(b) Find the width of the sidewalk around the pool. (2)

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Answer: (a)

(12 - 2 x) (7 - 2 x) = 36

, i.e.

2 x^{2} - 19 x + 24 = 0

(b) 1.5 m

(a) Pool dimensions are $(12 - 2 x)$ m by $(7 - 2 x)$ m.
$(12 - 2 x) (7 - 2 x) = 36 \Rightarrow 84 - 38 x + 4 x^{2} = 36 \Rightarrow 4 x^{2} - 38 x + 48 = 0 \Rightarrow 2 x^{2} - 19 x + 24 = 0$ .
(b) $2 x^{2} - 16 x - 3 x + 24 = 0 \Rightarrow (x - 8) (2 x - 3) = 0 \Rightarrow x = 8$ or $x = \frac{3}{2}$ .
$x = 8$ is impossible since $7 - 2 x$ would be negative. So the sidewalk is 1.5 m wide.

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

Q144 marksCase StudySurface Areas and Volumes

John planned a birthday party for his younger sister with his friends. They decided to make some birthday caps by themselves and to buy a cake from a bakery shop. For these two items, they decided the following dimensions :
Cake : Cylindrical shape with diameter 24 cm and height 14 cm.
Cap : Conical shape with base circumference 44 cm and height 24 cm.
Based on the above information, answer the following questions :
(a) How many square cm paper would be used to make 4 such caps ? (2)
(b) The bakery shop sells cakes by weight (0.5 kg, 1 kg, 1.5 kg, etc.). To have the required dimensions, how much cake should they order, if 650 cm $^{3}$ equals 100 g of cake ? (2)

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Answer: (a) 2200 cm

^{2}

(b) 1 kg (cake needed ≈ 975 g)

(a) Base circumference $2 π r = 44 \Rightarrow r = 7$ cm (taking $π = \frac{22}{7}$ ).
Slant height $l = 7^{2} + 2 4^{2} = 25$ cm.
Paper for one cap $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm $^{2}$ ; for 4 caps $= 2200$ cm $^{2}$ .
(b) Cake radius $= 12$ cm; volume $= π r^{2} h = \frac{22}{7} \times 144 \times 14 = 6336$ cm $^{3}$ .
Weight $= \frac{6336}{650} \times 100 \approx 974.8$ g ≈ 0.97 kg.
Since cakes are sold in steps of 0.5 kg, they should order 1 kg of cake.

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

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