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

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

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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.

Q22 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 30/3/2

Q32 marksVery Short AnswerArithmetic Progressions

Find the common difference ‘d’ of an AP whose first term is 10 and the sum of the first 14 terms is 1505.

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Answer:

d = 15

$S_{14} = \frac{14}{2} [2 \times 10 + 13 d] = 7 (20 + 13 d)$ .
$7 (20 + 13 d) = 1505 \Rightarrow 20 + 13 d = 215 \Rightarrow 13 d = 195 \Rightarrow d = 15$ .

Q42 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 30/3/2

Q52 marksVery Short AnswerQuadratic Equations

Solve the quadratic equation for x :
$x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0$

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Answer:

x = a + 2 b

x = a - 2 b

Here $A = 1$ , $B = - 2 a$ , $C = - (4 b^{2} - a^{2})$ .
$D = B^{2} - 4 A C = 4 a^{2} + 4 (4 b^{2} - a^{2}) = 16 b^{2}$ .
$x = \frac{2 a \pm 16 b ^{2}}{2} = \frac{2 a \pm 4 b}{2} = a \pm 2 b$ .
So $x = a + 2 b$ or $x = a - 2 b$ .

Also asked in: 2022 Standard 30/3/1

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

If the quadratic equation
$(1 + a^{2}) x^{2} + 2 ab x + (b^{2} - c^{2}) = 0$
has equal and real roots, then prove that :
$b^{2} = c^{2} (1 + a^{2})$

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Answer: Proved.

For equal roots, $D = 0$ : $(2 ab)^{2} - 4 (1 + a^{2}) (b^{2} - c^{2}) = 0$ .
$\Rightarrow a^{2} b^{2} - (b^{2} - c^{2} + a^{2} b^{2} - a^{2} c^{2}) = 0$
$\Rightarrow - b^{2} + c^{2} + a^{2} c^{2} = 0$
$\Rightarrow b^{2} = c^{2} (1 + a^{2})$ . Hence proved.

Also asked in: 2022 Standard 30/3/1

Q62 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 30/3/2

Q6 (OR) (OR)2 marksVery Short AnswerSurface Areas and Volumes

Three cubes of side 6 cm each, are joined as shown in Figure 2. Find the total surface area of the resulting cuboid.

Diagram for CBSE 2022 Class 10 Maths question 6 (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 $^{2}$ .

Q73 marksShort AnswerSome Applications of Trigonometry

Two boats are sailing in the sea 80 m apart from each other towards a cliff AB. The angles of depression of the boats from the top of the cliff are $3 0^{\circ}$ and $4 5^{\circ}$ respectively, as shown in Figure 3. Find the height of the cliff.

Diagram for CBSE 2022 Class 10 Maths question 7

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Answer:

40 (3 + 1)

m ≈ 109.28 m

Let $A B = h$ . The angle of depression equals the angle of elevation from the boat.
In $△ A B C$ : $tan 4 5^{\circ} = \frac{h}{B C} \Rightarrow B C = h$ .
In $△ A B D$ : $tan 3 0^{\circ} = \frac{h}{B D} \Rightarrow B D = h 3$ .
$C D = B D - B C = h (3 - 1) = 80 \Rightarrow h = \frac{80}{3 - 1} = 40 (3 + 1)$ .
$h \approx 40 \times 2.732 = 109.28$ m.

Q83 marksShort AnswerStatistics

For the following frequency distribution, find the median :
Class: 1400 – 1550, 1550 – 1700, 1700 – 1850, 1850 – 2000
Frequency: 6, 13, 25, 10

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Answer: 1748

Cumulative frequencies: 6, 19, 44, 54. $N = 54$ , $\frac{N}{2} = 27$ .
Median class is 1700 – 1850: $l = 1700$ , $c f = 19$ , $f = 25$ , $h = 150$ .
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 1700 + \frac{27 - 19}{25} \times 150 = 1700 + 48 = 1748$ .

Also asked in: 2022 Standard 30/3/1

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 30/3/2

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 30/3/2

Q103 marksShort AnswerStatistics

Find the mean of the following frequency distribution :
Class: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60
Frequency: 12, 18, 27, 20, 17, 6

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Answer: 28

Class marks $x_{i}$ : 5, 15, 25, 35, 45, 55; $\sum f_{i} = 100$ .
$f_{i} x_{i}$ : 60, 270, 675, 700, 765, 330; $\sum f_{i} x_{i} = 2800$ .
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{2800}{100} = 28$ .

Q114 marksLong AnswerSome Applications of Trigonometry

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is $6 0^{\circ}$ . From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower PQ is $4 5^{\circ}$ . Find the height of the tower PQ and the distance PX. [Use $3$ = 1.73]

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Answer: PQ =

20 (3 + 3)

m ≈ 94.6 m; PX =

20 (3 + 1)

m ≈ 54.6 m

Let $P Q = H$ and $P X = x$ .
From X: $tan 6 0^{\circ} = \frac{H}{x} \Rightarrow H = x 3$ .
From Y (40 m above X): $tan 4 5^{\circ} = \frac{H - 40}{x} \Rightarrow H - 40 = x$ .
$x 3 - 40 = x \Rightarrow x = \frac{40}{3 - 1} = 20 (3 + 1) \approx 20 \times 2.73 = 54.6$ m.
$H = x + 40 = 203 + 60 \approx 34.6 + 60 = 94.6$ m.

Q124 marksLong AnswerCircles

In Figure 4, 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.

Q12 (OR) (OR)4 marksLong AnswerCircles

In Figure 5, 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.

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 30/3/2

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 30/3/2

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