CBSE Maths › Class 10 PYQs › 2022

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

All 18 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/2/3 (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 AnswerArithmetic Progressions

Which term of the A.P. 3, 8, 13, 18, … is 78 ?

Show answer & solution

Answer: 16th term

$a = 3$ , $d = 8 - 3 = 5$ .
$a_{n} = a + (n - 1) d = 78$
$3 + 5 (n - 1) = 78$
$5 (n - 1) = 75 \Rightarrow n - 1 = 15 \Rightarrow n = 16$ .
So 78 is the 16th term.

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

Q1 (OR) (OR)2 marksVery Short AnswerArithmetic Progressions

Find the common difference of an A.P. whose $n^{t h}$ term is given by $a_{n} = 6 n - 5$ .

Show answer & solution

Answer: Common difference = 6

$a_{1} = 6 (1) - 5 = 1$ , $a_{2} = 6 (2) - 5 = 7$ .
$d = a_{2} - a_{1} = 7 - 1 = 6$ .
(In general $a_{n + 1} - a_{n} = 6$ .)

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

Q22 marksVery Short AnswerCircles

In Fig. 1, perimeter of $△ P QR$ is 20 cm. Find the length of tangent PA.

Diagram for CBSE 2022 Class 10 Maths question 2

Show answer & solution

Answer: PA = 10 cm

Tangents from an external point are equal: QA = QC, RB = RC, PA = PB.
Perimeter of $△ P QR$ = PQ + QR + PR = PQ + QC + CR + PR
= PQ + QA + RB + PR = PA + PB = 2PA.
So 2PA = 20, hence PA = 10 cm.

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

Q2 (OR) (OR)2 marksVery Short AnswerCircles

In Fig. 2, BC is tangent to the circle at point B of circle centred at O. BD is a chord of the circle so that $∠ B A D = 5 5^{\circ}$ . Find m $∠ D B C$ .

Diagram for CBSE 2022 Class 10 Maths question 2 (OR)

Show answer & solution

Answer:

∠ D B C = 5 5^{\circ}

AB passes through O, so AB is a diameter.
$∠ A D B = 9 0^{\circ}$ (angle in a semicircle).
In $△ A B D$ : $∠ A B D = 18 0^{\circ} - 9 0^{\circ} - 5 5^{\circ} = 3 5^{\circ}$ .
$∠ A B C = 9 0^{\circ}$ (radius OB $⊥$ tangent BC).
$∠ D B C = 9 0^{\circ} - 3 5^{\circ} = 5 5^{\circ}$ .

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

Q32 marksVery Short AnswerSurface Areas and Volumes

A metallic hollow cylindrical pipe has outer and inner radii as 6 cm and 4 cm respectively. Find the volume of the metal used in the pipe of length of 14 cm.

Show answer & solution

Answer: 880 cm³

Volume of metal $= π (R^{2} - r^{2}) h = \frac{22}{7} \times (36 - 16) \times 14$
$= 22 \times 20 \times 2 = 880$ cm³.

Q42 marksVery Short AnswerQuadratic Equations

Find the nature of the roots of the quadratic equation :
$4 x^{2} - 5 x - 1 = 0$

Show answer & solution

Answer: The roots are real and distinct (unequal), since

D = 41 > 0

Here $a = 4$ , $b = - 5$ , $c = - 1$ .
$D = b^{2} - 4 a c = 25 - 4 (4) (- 1) = 25 + 16 = 41$ .
Since $D > 0$ , the equation has two distinct real roots.
(As 41 is not a perfect square, the roots are irrational.)

Also asked in: 2022 Basic 430/2/1

Q52 marksVery Short AnswerArithmetic Progressions

Find the sum of the first fifteen multiples of 8.

Show answer & solution

Answer: 960

Multiples: 8, 16, 24, …, 120 with $a = 8$ , $d = 8$ , $n = 15$ .
$S_{15} = \frac{15}{2} [2 (8) + 14 (8)] = \frac{15}{2} \times 128 = 960$ .

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

Q62 marksVery Short AnswerStatistics

Find the mode of the following frequency distribution :
Class: 20–30, 30–40, 40–50, 50–60, 60–70
Frequency: 25, 30, 45, 42, 35

Show answer & solution

Answer: Mode ≈ 48.33

Maximum frequency 45, so modal class is 40–50.
$l = 40$ , $h = 10$ , $f_{1} = 45$ , $f_{0} = 30$ , $f_{2} = 42$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{15}{90 - 30 - 42} \times 10$
$= 40 + \frac{150}{18} = 40 + 8.33 = 48.33$ .

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

Q73 marksShort AnswerStatistics

The median of following frequency distribution is 25. Find the value of $x$ .
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 6, 9, 10, 8, $x$

Show answer & solution

Answer:

x = 7

$N = 6 + 9 + 10 + 8 + x = 33 + x$ .
Median 25 lies in class 20–30: $l = 20$ , $c f = 15$ , $f = 10$ , $h = 10$ .
$25 = 20 + \frac{\frac{33 + x}{2} - 15}{10} \times 10$
$5 = \frac{33 + x}{2} - 15 \Rightarrow \frac{33 + x}{2} = 20 \Rightarrow x = 7$ .
Check: N = 40, N/2 = 20 lies in 20–30.

Also asked in: 2022 Basic 430/2/1

Q83 marksShort AnswerStatistics

Find mean of the following frequency distribution :
Class: 0–20, 20–40, 40–60, 60–80, 80–100
Frequency: 6, 8, 5, 9, 7

Show answer & solution

Answer: Mean

= \frac{362}{7} \approx 51.71

Class marks $x_{i}$ : 10, 30, 50, 70, 90.
$f_{i} x_{i}$ : 60, 240, 250, 630, 630.
$\sum f_{i} = 35$ , $\sum f_{i} x_{i} = 1810$ .
Mean $= \frac{1810}{35} = \frac{362}{7} \approx 51.71$ .

Q93 marksShort AnswerSome Applications of Trigonometry

As observed from the top of a light house 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from $3 0^{\circ}$ to $4 5^{\circ}$ . Determine the distance travelled by the ship during this time.
(Use $3$ = 1.73)

Diagram for CBSE 2022 Class 10 Maths question 9

Show answer & solution

Answer: 73 m

Let AB = 100 m be the light house, D and C the initial and final positions of the ship.
$∠ A D B = 3 0^{\circ}$ and $∠ A C B = 4 5^{\circ}$ (alternate angles).
In $△ A B C$ : $tan 4 5^{\circ} = \frac{A B}{B C} \Rightarrow B C = 100$ m.
In $△ A B D$ : $tan 3 0^{\circ} = \frac{A B}{B D} \Rightarrow B D = 1003 = 173$ m.
Distance travelled CD = BD − BC = 173 − 100 = 73 m.

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

Q9 (OR) (OR)3 marksShort AnswerSome Applications of TrigonometryNot in current syllabus

At a point on level ground, the angle of elevation of a vertical tower is, found to be $α$ such that $tan α = \frac{1}{3}$ . After walking 100 m towards the tower, the angle of elevation $β$ becomes such that $tan β = \frac{3}{4}$ . Find the height of the tower.

Show answer & solution

Answer: 60 m

Let the height be $h$ and the distance of the second point from the foot be $x$ .
$tan β = \frac{h}{x} = \frac{3}{4} \Rightarrow x = \frac{4 h}{3}$ .
$tan α = \frac{h}{x + 100} = \frac{1}{3} \Rightarrow 3 h = x + 100$ .
$3 h = \frac{4 h}{3} + 100 \Rightarrow \frac{5 h}{3} = 100 \Rightarrow h = 60$ .
Height of the tower = 60 m.

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

Q103 marksShort AnswerCirclesNot in current syllabus

Draw two concentric circles of radii 2 cm and 5 cm. From a point P on outer circle, construct a pair of tangents to the inner circle.

Show answer & solution

Answer: Construction (pair of tangents drawn; each tangent

= 21 \approx 4.58

cm).

Draw concentric circles with centre O and radii 2 cm and 5 cm; take P on the outer circle.
Bisect OP; let M be its midpoint.
With centre M and radius MO draw a circle cutting the inner circle at Q and R.
Join PQ and PR; these are the required tangents.
Justification: $∠ O QP = 9 0^{\circ}$ (angle in a semicircle), so PQ $⊥$ OQ; similarly PR.
Length of each tangent $= 5^{2} - 2^{2} = 21 \approx 4.58$ cm.

Q114 marksLong AnswerQuadratic Equations

The sum of two numbers is 45. If 5 is subtracted from each of them, the product of these numbers becomes 124. Find the numbers.

Show answer & solution

Answer: 9 and 36

Let the numbers be $x$ and $45 - x$ .
$(x - 5) (45 - x - 5) = 124 \Rightarrow (x - 5) (40 - x) = 124$ .
$- x^{2} + 45 x - 200 = 124 \Rightarrow x^{2} - 45 x + 324 = 0$ .
$(x - 9) (x - 36) = 0 \Rightarrow x = 9$ or $36$ .
The numbers are 9 and 36 (check: $4 \times 31 = 124$ ).

Q124 marksLong AnswerCircles

Prove that a parallelogram circumscribing a circle is a rhombus.

Show answer & solution

Answer: Proved.

Let parallelogram ABCD circumscribe a circle, touching AB, BC, CD, DA at P, Q, R, S.
Tangents from an external point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS.
Adding: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), i.e. AB + CD = AD + BC.
In a parallelogram AB = CD and AD = BC, so 2AB = 2BC, i.e. AB = BC.
Hence AB = BC = CD = DA, so ABCD is a rhombus.

Also asked in: 2026 Basic 430/4/1, 2023 Standard 30/6/3, 2022 Standard 30/2/3, 2022 Basic 430/2/1

Q12 (OR) (OR)4 marksLong AnswerCircles

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

Show answer & solution

Answer: Proved.

Let a circle with centre O touch tangent XY at P, and let PQ be perpendicular to XY at P.
Suppose PQ does not pass through O. Join OP.
The radius through the point of contact is perpendicular to the tangent, so $∠ O P Y = 9 0^{\circ}$ .
Also $∠ QP Y = 9 0^{\circ}$ , so $∠ O P Y = ∠ QP Y$ .
This is possible only if O lies on PQ, contradicting our assumption.
Hence the perpendicular at the point of contact passes through the centre.

Also asked in: 2022 Basic 430/2/1

Q134 marksCase StudySome Applications of Trigonometry

Qutub Minar, located in South Delhi, India, was built in the year 1193. It is 72 m high tower. Working on a school project, Charu and Daljeet visited the monument. They used trigonometry to find their distance from the tower. Observe the picture given below. Points C and D represent their positions on the ground in line with the base of tower, the angles of elevation of top of the tower (Point A) are $6 0^{\circ}$ and $4 5^{\circ}$ from points C and D respectively.
(1) Based on above information, draw a well-labelled diagram. (1)
(2) Find the distances CD, BC and BD. (use $3$ = 1.73) (3)

Show answer & solution

Answer: (1) Vertical tower AB = 72 m with C and D on the ground in line with B;

∠ A C B = 6 0^{\circ}

∠ A D B = 4 5^{\circ}

(2) BC = 41.52 m, BD = 72 m, CD = 30.48 m

(1) Draw AB vertical (72 m), B on the ground line BCD with C nearer to B; mark $∠ A C B = 6 0^{\circ}$ and $∠ A D B = 4 5^{\circ}$ .
(2) In $△ A B C$ : $tan 6 0^{\circ} = \frac{A B}{B C} \Rightarrow B C = \frac{72}{3} = 243 = 41.52$ m.
In $△ A B D$ : $tan 4 5^{\circ} = \frac{A B}{B D} \Rightarrow B D = 72$ m.
CD = BD − BC = 72 − 41.52 = 30.48 m.

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

Q144 marksCase StudySurface Areas and Volumes

A solid cuboidal toy is made of wood. It has five cone shaped cavities to hold toy carrots.
The dimensions of the toy are cuboid – 10 cm × 10 cm × 8 cm.
Each cone carved out – Radius = 2.1 cm and Height = 6 cm.
(1) Find the volume of wood carved out to make five conical cavities. (2)
(2) Find the volume of the wood in the final product. (2)

Diagram for CBSE 2022 Class 10 Maths question 14

Show answer & solution

Answer: (1) 138.6 cm³ (2) 661.4 cm³

(1) Volume of one cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 6 = 27.72$ cm³.
Volume of five cones $= 5 \times 27.72 = 138.6$ cm³.
(2) Volume of cuboid $= 10 \times 10 \times 8 = 800$ cm³.
Volume of wood left $= 800 - 138.6 = 661.4$ cm³.

Also asked in: 2022 Basic 430/2/1, 2022 Basic 430/2/2

← 2022 Basic 430/2/2 2022 Basic 430/3/1 →

Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on MonoMath CBSE →