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CBSE Class 10 Maths Standard 2022 Question Paper 30/4/2 (Term 2) with Solutions

All 18 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/4/2 (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 AnswerQuadratic Equations

If the sum of the roots of the quadratic equation $k y^{2} - 11 y + (k - 23) = 0$ is $\frac{13}{21}$ more than the product of the roots, then find the value of $k$ .

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Answer:

k = 21

Sum of roots $= \frac{11}{k}$ , product of roots $= \frac{k - 23}{k}$ .
Given: $\frac{11}{k} - \frac{k - 23}{k} = \frac{13}{21}$
$\frac{34 - k}{k} = \frac{13}{21}$
$21 (34 - k) = 13 k \Rightarrow 714 = 34 k$
$k = 21$

Also asked in: 2022 Standard 30/4/1

Q1 (OR) (OR)2 marksVery Short AnswerQuadratic Equations

If $x = - 2$ is the common solution of quadratic equations $a x^{2} + x - 3 a = 0$ and $x^{2} + b x + b = 0$ , then find the value of $a^{2} b$ .

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Answer:

a^{2} b = 16

Put $x = - 2$ in $a x^{2} + x - 3 a = 0$ : $4 a - 2 - 3 a = 0 \Rightarrow a = 2$ .
Put $x = - 2$ in $x^{2} + b x + b = 0$ : $4 - 2 b + b = 0 \Rightarrow b = 4$ .
$a^{2} b = 2^{2} \times 4 = 16$

Also asked in: 2022 Standard 30/4/1

Q22 marksVery Short AnswerCircles

In Fig. 1, there are two concentric circles with centre O. If ARC and AQB are tangents to the smaller circle from the point A lying on the larger circle, find the length of AC, if AQ = 5 cm.

Diagram for CBSE 2022 Class 10 Maths question 2

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Answer: AC = 10 cm

AQ and AR are tangents from A to the smaller circle, so AR = AQ = 5 cm.
AC is a chord of the larger circle touching the smaller circle at R, so $O R ⊥ A C$ .
The perpendicular from the centre bisects the chord, so RC = AR = 5 cm.
AC = AR + RC = 10 cm.

Also asked in: 2022 Standard 30/4/1, 2022 Standard 30/4/3

Q32 marksVery Short AnswerSurface Areas and Volumes

The curved surface area of a right circular cylinder is 176 sq. cm and its volume is 1232 cu. cm. What is the height of the cylinder ?

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Answer: 2 cm

$2 π r h = 176$ and $π r^{2} h = 1232$ .
Dividing: $\frac{π r ^{2} h}{2 π r h} = \frac{1232}{176} \Rightarrow \frac{r}{2} = 7 \Rightarrow r = 14$ cm
$2 \times \frac{22}{7} \times 14 \times h = 176 \Rightarrow 88 h = 176$
$h = 2$ cm

Also asked in: 2022 Standard 30/4/1, 2022 Standard 30/4/3

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

The largest sphere is carved out of a solid cube of side 21 cm. Find the volume of the sphere.

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Answer: 4851

cm^{3}

Diameter of the largest sphere = side of cube = 21 cm, so $r = 10.5$ cm.
Volume $= \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5$
$= 4851$ $cm^{3}$

Also asked in: 2022 Standard 30/4/1, 2022 Standard 30/4/3

Q42 marksVery Short AnswerArithmetic Progressions

If the first term of an A.P. is 5, the last term is 15 and the sum of first $n$ terms is 30, then find the value of $n$ .

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Answer:

n = 3

$S_{n} = \frac{n}{2} (a + l)$
$30 = \frac{n}{2} (5 + 15) = 10 n$
$n = 3$

Q52 marksVery Short AnswerStatistics

For the following frequency distribution, find the mode :
Class: 25–30, 30–35, 35–40, 40–45, 45–50
Frequency: 12, 5, 14, 8, 9

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Answer: 38

Highest frequency is 14, so the modal class is 35–40.
$l = 35$ , $h = 5$ , $f_{1} = 14$ , $f_{0} = 5$ , $f_{2} = 8$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 35 + \frac{9}{28 - 13} \times 5$
$= 35 + \frac{9}{15} \times 5 = 35 + 3 = 38$

Q62 marksVery Short AnswerStatistics

If the mean of the following frequency distribution is 18, then find the missing frequency ‘f’.
Class: 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25
Frequency: 3, 6, 9, 13, f, 5, 4

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Answer:

f = 8

Class marks: 12, 14, 16, 18, 20, 22, 24.
$\sum f_{i} = 40 + f$
$\sum f_{i} x_{i} = 36 + 84 + 144 + 234 + 20 f + 110 + 96 = 704 + 20 f$
$\frac{704 + 20 f}{40 + f} = 18$
$704 + 20 f = 720 + 18 f \Rightarrow 2 f = 16$
$f = 8$

Q73 marksShort AnswerQuadratic Equations

Find the value of ‘p’ for which the quadratic equation $p (x - 4) (x - 2) + (x - 1)^{2} = 0$ has real and equal roots.

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Answer:

p = 0

p = 3

$p (x^{2} - 6 x + 8) + x^{2} - 2 x + 1 = 0$
$(p + 1) x^{2} - (6 p + 2) x + (8 p + 1) = 0$
For real and equal roots, $D = 0$ :
$(6 p + 2)^{2} - 4 (p + 1) (8 p + 1) = 0$
$36 p^{2} + 24 p + 4 - 32 p^{2} - 36 p - 4 = 0$
$4 p^{2} - 12 p = 0 \Rightarrow 4 p (p - 3) = 0$
$p = 0$ or $p = 3$ (for both, $p + 1 \neq = 0$ , so the equation stays quadratic).

Also asked in: 2022 Standard 30/4/1, 2022 Standard 30/4/3

Q7 (OR) (OR)3 marksShort AnswerQuadratic Equations

Had Aarush scored 8 more marks in a Mathematics test, out of 35 marks, 7 times these marks would have been 4 less than square of his actual marks. How many marks did he get in the test ?

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Answer: 12 marks

Let his actual marks be $x$ .
$7 (x + 8) = x^{2} - 4$
$x^{2} - 7 x - 60 = 0$
$(x - 12) (x + 5) = 0$
$x = 12$ (marks cannot be negative)
Aarush got 12 marks.

Also asked in: 2022 Standard 30/4/1, 2022 Standard 30/4/3

Q83 marksShort AnswerCirclesNot in current syllabus

Construct a pair of tangents to a circle of radius 4 cm which are inclined to each other at an angle of $6 0^{\circ}$ .

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Answer: PA and PB drawn from the construction are the required tangents.

Draw a circle with centre O and radius 4 cm.
Draw any radius OA.
Since the angle between the tangents is $6 0^{\circ}$ , the angle between the radii to the points of contact is $18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .
Draw radius OB such that $∠ A O B = 12 0^{\circ}$ .
At A and B draw lines perpendicular to OA and OB respectively; let them meet at P.
PA and PB are the required tangents, inclined to each other at $6 0^{\circ}$ .
Justification: in quadrilateral OAPB, $∠ O A P = ∠ O B P = 9 0^{\circ}$ and $∠ A O B = 12 0^{\circ}$ , so $∠ A P B = 36 0^{\circ} - 30 0^{\circ} = 6 0^{\circ}$ .

Also asked in: 2022 Standard 30/4/1

Q93 marksShort AnswerSome Applications of Trigonometry

There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are points directly opposite to each other on two banks and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively $3 0^{\circ}$ and $4 5^{\circ}$ , find the height of the tree. (Use $3 = 1.732$ )

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Answer: 36.6 m

Let the height of the tree be $h$ m and its foot be at distances $x$ from P and $100 - x$ from Q.
From P: $tan 3 0^{\circ} = \frac{h}{x} \Rightarrow x = h 3$ .
From Q: $tan 4 5^{\circ} = \frac{h}{100 - x} \Rightarrow 100 - x = h$ .
$h 3 + h = 100 \Rightarrow h = \frac{100}{3 + 1} = 50 (3 - 1)$
$h = 50 \times 0.732 = 36.6$ m

Q103 marksShort AnswerArithmetic Progressions

In an A.P., the sum of first $n$ terms is $\frac{n}{2} (3 n + 5)$ . Find the $2 5^{t h}$ term of the A.P.

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Answer: 76

$S_{n} = \frac{n}{2} (3 n + 5)$
$a_{1} = S_{1} = \frac{1}{2} (8) = 4$
$S_{2} = 1 \times 11 = 11$ , so $a_{2} = 11 - 4 = 7$ and $d = 3$ .
$a_{25} = 4 + 24 \times 3 = 76$

Q114 marksLong AnswerSome Applications of Trigonometry

From the top of an 8 m high building, the angle of elevation of the top of a cable tower is $6 0^{\circ}$ and the angle of depression of its foot is $4 5^{\circ}$ . Determine the height of the tower. (Take $3 = 1.732$ ).

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Answer: 21.856 m (approx.)

Let the building be AB = 8 m and the tower CD, with horizontal distance BD = $x$ .
Angle of depression of the foot is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{8}{x} \Rightarrow x = 8$ m.
Let the top of the tower be $h$ m above the top of the building. $tan 6 0^{\circ} = \frac{h}{x} \Rightarrow h = 83$ m.
Height of tower $= 8 + 83 = 8 + 8 \times 1.732 = 21.856$ m.

Also asked in: 2022 Standard 30/4/1, 2022 Standard 30/4/3

Q124 marksLong AnswerCircles

In Fig.-2, if a circle touches the side QR of $△ P QR$ at S and extended sides PQ and PR at M and N, respectively, then
prove that $P M = \frac{1}{2} (P Q + QR + P R)$

Diagram for CBSE 2022 Class 10 Maths question 12

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Answer: Proved.

Tangents drawn from an external point to a circle are equal.
From P: PM = PN. From Q: QM = QS. From R: RN = RS.
PQ + QR + PR = PQ + QS + SR + PR
= PQ + QM + RN + PR
= (PQ + QM) + (PR + RN) = PM + PN
= 2PM (since PN = PM)
Hence $P M = \frac{1}{2} (P Q + QR + P R)$ .

Also asked in: 2022 Standard 30/4/1

Q12 (OR) (OR)4 marksLong AnswerCircles

In Fig. 3, a triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 6 cm and 8 cm respectively. If the area of $△ A B C$ is 84 $cm^{2}$ , find the lengths of sides AB and AC.

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

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Answer: AB = 13 cm, AC = 15 cm

Let the circle touch AB at F and AC at E, and let AF = AE = $x$ cm (equal tangents from A).
BF = BD = 6 cm and CE = CD = 8 cm.
So AB = $x + 6$ , AC = $x + 8$ , BC = 14.
Area of $△ A B C$ = area OBC + area OCA + area OAB $= \frac{1}{2} \times 4 \times (A B + B C + C A)$
$84 = 2 (2 x + 28) \Rightarrow 2 x + 28 = 42 \Rightarrow x = 7$
AB = 13 cm and AC = 15 cm.

Also asked in: 2022 Standard 30/4/1, 2022 Standard 30/4/3

Q134 marksCase StudySurface Areas and VolumesNot in current syllabus

Khurja is a city in the Indian state of Uttar Pradesh famous for the pottery. Khurja pottery is traditional Indian pottery work which has attracted Indians as well as foreigners with a variety of tea-sets, crockery and ceramic tile works. A huge portion of the ceramics used in the country is supplied by Khurja and is also referred as ‘The Ceramic Town’.
One of the private schools of Bulandshahr organised an Educational Tour for class 10 students to Khurja. Students were very excited about the trip. Following are the few pottery objects of Khurja.
Students found the shapes of the objects very interesting and they could easily relate them with mathematical shapes viz sphere, hemisphere, cylinder etc. Maths teacher who was accompanying the students asked following questions :
(a) The internal radius of hemispherical bowl (filled completely with water) in I is 9 cm and radius and height of cylindrical jar in II is 1.5 cm and 4 cm respectively. If the hemispherical bowl is to be emptied in cylindrical jars, then how many cylindrical jars are required ? (2)
(b) If in the cylindrical jar full of water, a conical funnel of same height and same diameter is immersed, then how much water will flow out of the jar ? (2)

Diagram for CBSE 2022 Class 10 Maths question 13

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Answer: (a) 54 (b)

3 π

cm^{3} = \frac{66}{7}

cm^{3} \approx 9.43

cm^{3}

(a) Volume of hemispherical bowl $= \frac{2}{3} π (9)^{3} = 486 π$ $cm^{3}$ .
Volume of one jar $= π (1.5)^{2} \times 4 = 9 π$ $cm^{3}$ .
Number of jars $= \frac{486 π}{9 π} = 54$ .
(b) Water that flows out = volume of the cone with $r = 1.5$ cm, $h = 4$ cm.
$= \frac{1}{3} π (1.5)^{2} \times 4 = 3 π = 3 \times \frac{22}{7} = \frac{66}{7} \approx 9.43$ $cm^{3}$ .

Also asked in: 2022 Standard 30/4/1, 2022 Standard 30/4/3

Q144 marksCase StudyStatistics

Yoga is an ancient practice which is a form of meditation and exercise. By practising yoga, we not even make our body healthy but also achieve inner peace and calmness. The International Yoga Day is celebrated on $2 1^{s t}$ of June every year since 2015.
To promote Yoga, Green park society in Pune organised a 7-day Yoga camp in their society. The number of people of different age groups who enrolled for this camp is given as follows :
Age Group: 15–25, 25–35, 35–45, 45–55, 55–65, 65–75, 75–85
Number of People: 8, 10, 15, 25, 40, 24, 18
Based on the above, find the following :
(a) Find the median age of people enrolled for the camp. (2)
(b) If $x$ more people of age group 65 – 75 had enrolled for the camp, the mean age would have been 58. Find the value of $x$ . (2)

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Answer: (a) 58 years (b)

x = \frac{145}{6} \approx 24.17

(about 24)

(a) Cumulative frequencies: 8, 18, 33, 58, 98, 122, 140; $N = 140$ , $\frac{N}{2} = 70$ .
Median class is 55–65: $l = 55$ , $c f = 58$ , $f = 40$ , $h = 10$ .
Median $= 55 + \frac{70 - 58}{40} \times 10 = 55 + 3 = 58$ years.
(b) Class marks: 20, 30, 40, 50, 60, 70, 80.
$\sum f_{i} x_{i} = 160 + 300 + 600 + 1250 + 2400 + 1680 + 1440 = 7830$ , $\sum f_{i} = 140$ .
With $x$ more people in 65–75 (class mark 70): $\frac{7830 + 70 x}{140 + x} = 58$
$7830 + 70 x = 8120 + 58 x \Rightarrow 12 x = 290$
$x = \frac{145}{6} \approx 24.17$ , i.e. about 24 people.

Also asked in: 2022 Standard 30/4/1, 2022 Standard 30/4/3

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