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CBSE Class 10 Maths Basic 2022 Question Paper 430/3/1 (Term 2) with Solutions

All 18 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/3/1 (2022, Term 2), with answers and step-by-step solutions. Total 40 marks. Tap “Show answer & solution” under any question.

Set 430/1/1 Set 430/1/2 Set 430/1/3 Set 430/2/1 Set 430/2/2 Set 430/2/3 Set 430/3/1 Set 430/3/2 Set 430/3/3 Set 430/4/1 Set 430/4/2 Set 430/4/3

Q12 marksVery Short AnswerQuadratic Equations

For what value of p, does the quadratic equation $p x^{2} + 2 x + p = 0$ have real and equal roots ?

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Answer:

p = 1

p = - 1

For real and equal roots, the discriminant must be zero.
$D = b^{2} - 4 a c = 2^{2} - 4 (p) (p) = 4 - 4 p^{2}$
$4 - 4 p^{2} = 0 \Rightarrow p^{2} = 1$
So $p = 1$ or $p = - 1$ (both non-zero, so the equation stays quadratic).

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

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

Solve the quadratic equation for x : $6 - x - x^{2} = 0$

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Answer:

x = 2

x = - 3

Multiply by $- 1$ : $x^{2} + x - 6 = 0$
Split the middle term: $x^{2} + 3 x - 2 x - 6 = 0$
$x (x + 3) - 2 (x + 3) = 0 \Rightarrow (x + 3) (x - 2) = 0$
So $x = 2$ or $x = - 3$ .

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q22 marksVery Short AnswerArithmetic Progressions

For an AP with common difference 6, the sum of first ten terms is same as four times the sum of first five terms. Determine the first term of the AP.

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Answer: First term

= 3

Let the first term be $a$ ; $d = 6$ .
$S_{10} = \frac{10}{2} [2 a + 9 (6)] = 10 a + 270$
$S_{5} = \frac{5}{2} [2 a + 4 (6)] = 5 a + 60$ , so $4 S_{5} = 20 a + 240$
$10 a + 270 = 20 a + 240 \Rightarrow 10 a = 30 \Rightarrow a = 3$

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q32 marksVery Short AnswerStatistics

Find mode of the following frequency distribution :
Class: 100–110, 110–120, 120–130, 130–140, 140–150
Frequency: 5, 9, 8, 11, 7

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Answer: Mode

= 130 + \frac{30}{7} \approx 134.29

Highest frequency is 11, so the modal class is 130–140.
$l = 130$ , $h = 10$ , $f_{1} = 11$ , $f_{0} = 8$ , $f_{2} = 7$
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 130 + \frac{11 - 8}{22 - 8 - 7} \times 10$
$= 130 + \frac{30}{7} = 134 \frac{2}{7} \approx 134.29$

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q42 marksVery Short AnswerSurface Areas and Volumes

A cubical block of side 7 cm is surmounted by a hemisphere of largest possible diameter as shown in Figure 1. Find the total surface area of the solid.

Diagram for CBSE 2022 Class 10 Maths question 4

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Answer: 332.5 cm

^{2}

Largest hemisphere has diameter 7 cm, so $r = 3.5$ cm.
TSA = surface area of cube − area of base circle of hemisphere + curved surface area of hemisphere
$= 6 (7)^{2} - π r^{2} + 2 π r^{2} = 294 + π r^{2}$
$= 294 + \frac{22}{7} \times 3.5 \times 3.5 = 294 + 38.5 = 332.5$ cm $^{2}$

Also asked in: 2022 Basic 430/3/2

Q4 (OR) (OR)2 marksVery Short AnswerSurface Areas and VolumesNot in current syllabus

How many solid cones of height 3 cm and radius 2 cm can be formed by melting a solid sphere of radius 3 cm ?

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Answer: 9 cones

Volume of sphere $= \frac{4}{3} π (3)^{3} = 36 π$ cm $^{3}$
Volume of one cone $= \frac{1}{3} π (2)^{2} (3) = 4 π$ cm $^{3}$
Number of cones $= \frac{36 π}{4 π} = 9$

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q52 marksVery Short AnswerArithmetic Progressions

Which term of AP : $\frac{11}{4}$ , $\frac{7}{2}$ , $\frac{17}{4}$ , ... is $\frac{53}{4}$ ?

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Answer: 15th term

$a = \frac{11}{4}$ , $d = \frac{7}{2} - \frac{11}{4} = \frac{3}{4}$
$a_{n} = a + (n - 1) d = \frac{53}{4}$
$\frac{11}{4} + (n - 1) \frac{3}{4} = \frac{53}{4} \Rightarrow (n - 1) \frac{3}{4} = \frac{42}{4}$
$n - 1 = 14 \Rightarrow n = 15$

Q62 marksVery Short AnswerCircles

In Figure 2, PA and PB are tangents to the circle with centre at O. If $∠ A P B = 7 0^{\circ}$ , then find m $∠ A QB$ .

Diagram for CBSE 2022 Class 10 Maths question 6

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Answer:

∠ A QB = 5 5^{\circ}

OA $⊥$ PA and OB $⊥$ PB (radius is perpendicular to tangent), so $∠ O A P = ∠ O B P = 9 0^{\circ}$ .
In quadrilateral OAPB: $∠ A O B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$
Angle subtended by an arc at the centre is double the angle at a point on the remaining part of the circle.
$∠ A QB = \frac{1}{2} ∠ A O B = 5 5^{\circ}$

Also asked in: 2022 Basic 430/3/2

Q73 marksShort AnswerCirclesNot in current syllabus

Draw a circle of radius 2.5 cm. From a point P lying outside the circle at a distance of 6 cm from the centre of the circle, construct tangents PA and PB to the circle.

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Answer: Construction (each tangent

= 6^{2} - 2. 5^{2} = 29.75 \approx 5.45

cm)

Draw a circle with centre O and radius 2.5 cm; mark P with OP = 6 cm.
Draw the perpendicular bisector of OP; let it meet OP at M.
With M as centre and MO as radius, draw a circle cutting the given circle at A and B.
Join PA and PB; these are the required tangents.
Justification: $∠ O A P = ∠ O B P = 9 0^{\circ}$ (angles in a semicircle), so PA and PB are tangents. Each measures $36 - 6.25 \approx 5.45$ cm.

Also asked in: 2022 Basic 430/3/3

Q7 (OR) (OR)3 marksShort AnswerTrianglesNot in current syllabus

Draw a line segment of length 9.5 cm and divide it in the ratio 2 : 3.

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Answer: Construction (parts 3.8 cm and 5.7 cm)

Draw AB = 9.5 cm and a ray AX making an acute angle with AB.
Mark 5 equal arcs $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ on AX.
Join $A_{5} B$ ; through $A_{2}$ draw a line parallel to $A_{5} B$ meeting AB at C.
Then AC : CB = 2 : 3 (by BPT), i.e. AC = 3.8 cm and CB = 5.7 cm.

Also asked in: 2022 Basic 430/3/3

Q83 marksShort AnswerSome Applications of Trigonometry

Two poles of equal heights are standing opposite each other on either side of the road of width 60 m. From a point P between them on the road, the angles of elevation of the top of the poles are $6 0^{\circ}$ and $3 0^{\circ}$ respectively, as shown in Figure 3. Find the height of the poles and distances of the point from the poles.

Diagram for CBSE 2022 Class 10 Maths question 8

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Answer: Height of each pole

= 153

\approx 25.98

m; distances of P from the poles are 15 m (from CD) and 45 m (from AB)

In right $△ D C P$ : $tan 6 0^{\circ} = \frac{h}{x} \Rightarrow h = 3 x$
In right $△ B A P$ : $tan 3 0^{\circ} = \frac{h}{y} \Rightarrow y = 3 h = 3 x$
$x + y = 60 \Rightarrow 4 x = 60 \Rightarrow x = 15$ m, $y = 45$ m
$h = 153 \approx 25.98$ m

Q93 marksShort AnswerStatistics

The mileage (km/l) of 50 cars was recorded by a dealer and tabulated as given below :
Mileage (in km/l): 10–12, 12–14, 14–16, 16–18, 18–20
Number of Cars: 13, 18, 10, 7, 2
Find mean of the above distribution.

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Answer: Mean mileage

= 13.68

km/l

Class marks $x_{i}$ : 11, 13, 15, 17, 19; $f_{i}$ : 13, 18, 10, 7, 2; $\sum f_{i} = 50$
$f_{i} x_{i}$ : 143, 234, 150, 119, 38; $\sum f_{i} x_{i} = 684$
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{684}{50} = 13.68$ km/l

Also asked in: 2022 Basic 430/3/2

Q103 marksShort AnswerStatistics

Determine median of the following frequency distribution :
Class: 15–20, 20–25, 25–30, 30–35, 35–40, 40–45
Frequency: 8, 13, 21, 12, 5, 4

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Answer: Median

= 27.5

Cumulative frequencies: 8, 21, 42, 54, 59, 63; $n = 63$ , $\frac{n}{2} = 31.5$
Median class is 25–30: $l = 25$ , $c f = 21$ , $f = 21$ , $h = 5$
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 25 + \frac{31.5 - 21}{21} \times 5 = 25 + 2.5 = 27.5$

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q114 marksLong AnswerCircles

In Figure 4, quadrilateral ABCD circumscribes a circle centred at O. Prove that AD + BC = AB + CD.

Diagram for CBSE 2022 Class 10 Maths question 11

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Answer: Proved.

Let the circle touch AB, BC, CD, DA at P, Q, R, S respectively.
Tangents drawn from an external point to a circle are equal: AP = AS, BP = BQ, CR = CQ, DR = DS.
AB + CD = (AP + BP) + (DR + CR) = AS + BQ + DS + CQ
= (AS + DS) + (BQ + CQ) = AD + BC
Hence AD + BC = AB + CD.

Also asked in: 2022 Basic 430/3/3

Q11 (OR) (OR)4 marksLong AnswerCircles

In Figure 5, two concentric circles are drawn with centre O. PQ and RS are two chords of the larger circle which are tangents to the smaller circle. Prove that PQ = RS.

Diagram for CBSE 2022 Class 10 Maths question 11 (OR)

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Answer: Proved.

Let PQ touch the smaller circle at M and RS touch it at N; let the radii be R (larger) and r (smaller).
OM $⊥$ PQ and ON $⊥$ RS (radius is perpendicular to the tangent at the point of contact), and OM = ON = r.
The perpendicular from the centre to a chord bisects it, so PQ = 2PM and RS = 2RN.
In right $△ O M P$ : $P M = O P^{2} - O M^{2} = R^{2} - r^{2}$ ; similarly in right $△ O N R$ : $R N = R^{2} - r^{2}$
So PM = RN, hence PQ = 2PM = 2RN = RS.

Also asked in: 2022 Basic 430/3/3

Q124 marksLong AnswerSome Applications of Trigonometry

The angles of depression of the top and bottom of a 6 m tall building from the top of a multi-storeyed building are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use $3 = 1.73$ )

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Answer: Height of multi-storeyed building

= (9 + 33)

= 14.19

m; distance between the buildings

= 14.19

Let the multi-storeyed building be H m tall and the buildings be d m apart.
Angle of depression of the bottom is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{H}{d} \Rightarrow d = H$
Angle of depression of the top is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{H - 6}{d} \Rightarrow d = 3 (H - 6)$
$H = 3 H - 63 \Rightarrow H = \frac{6 3}{3 - 1} = \frac{6 3 ( 3 + 1 )}{2} = 9 + 33$
$H = 9 + 3 (1.73) = 14.19$ m, and $d = H = 14.19$ m

Also asked in: 2022 Basic 430/3/2

Q134 marksCase StudyQuadratic Equations

The tradition of pottery making in India is very old. In fact, it is older than Indus Valley Civilization. The shaping and baking of clay articles has continued through the ages. The picture of a potter is shown below :
A potter makes a certain number of pottery articles in a day. It was observed on a particular day the cost of production of each article (in ₹) was one more than twice the number of articles produced on that day. The total cost of production on that day was ₹ 210.
(a) Taking number of articles produced on that day as x, form a quadratic equation in x. (2)
(b) Find the number of articles produced and the cost of each article. (2)

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Answer: (a)

2 x^{2} + x - 210 = 0

(b) 10 articles; cost of each article ₹ 21

(a) Cost of each article $= ₹ (2 x + 1)$ , so $x (2 x + 1) = 210$ , i.e. $2 x^{2} + x - 210 = 0$ .
(b) $2 x^{2} + 21 x - 20 x - 210 = 0 \Rightarrow x (2 x + 21) - 10 (2 x + 21) = 0 \Rightarrow (2 x + 21) (x - 10) = 0$
$x = 10$ (rejecting $x = - \frac{21}{2}$ since the number of articles cannot be negative)
Cost of each article $= 2 (10) + 1 = ₹ 21$

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q144 marksCase StudySurface Areas and Volumes

The technique of Rainwater harvesting through Recharge pit is very useful. Rainwater is collected on the roof and then flowing through the Recharge pit it goes to the ground. Observe the picture given below :
B (BREADTH) = 3 m
D (DEPTH = 2 m
L (LENGTH) = 3 m
The surface area of the roof floor is 100 m $^{2}$ . The cuboidal pit measures 3 m $\times$ 3 m $\times$ 2 m.
(a) Water standing on the roof is released into the cuboidal pit. If the cuboidal pit is filled completely by the roof water, then find the height of standing water on the roof. (2)
(b) Instead of a cuboidal pit, if a cylindrical pit with diameter 3 m and height 2 m had been built, then which tank would hold more water ? (2)

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Answer: (a) 0.18 m (18 cm) (b) The cuboidal pit (18 m

^{3}

) holds more water than the cylindrical pit (

\approx 14.14

^{3}

)

(a) Volume of cuboidal pit $= 3 \times 3 \times 2 = 18$ m $^{3}$
Volume of water on roof $= 100 \times h$ ; $100 h = 18 \Rightarrow h = 0.18$ m $= 18$ cm
(b) Cylinder: $r = 1.5$ m, $h = 2$ m; volume $= π r^{2} h = \frac{22}{7} \times 2.25 \times 2 \approx 14.14$ m $^{3}$
Since $18 > 14.14$ , the cuboidal pit would hold more water.

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

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