CBSE Maths › Class 10 PYQs › 2023

CBSE Class 10 Maths Standard 2023 Question Paper 30/1/1 with Solutions

All 44 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/1/1 (2023), with answers and step-by-step solutions. Total 80 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 30/4/1 Set 30/4/2 Set 30/4/3 Set 30/5/1 Set 30/5/2 Set 30/5/3 Set 30/6/1 Set 30/6/2 Set 30/6/3

Q11 markMCQPolynomials

The graph of y = $p (x)$ is given, for a polynomial $p (x)$ . The number of zeroes of $p (x)$ from the graph is

Diagram for CBSE 2023 Class 10 Maths question 1

(A)3
(B)1
(C)2
(D)0

Show answer & solution

Answer: (B) 1

The zeroes of $p (x)$ are the x-coordinates of the points where the graph meets the x-axis.
The graph touches the x-axis at exactly one point.
So $p (x)$ has 1 zero.

Also asked in: 2023 Standard 30/1/2

Q21 markMCQPair of Linear Equations in Two Variables

The value of k for which the pair of equations $k x = y + 2$ and $6 x = 2 y + 3$ has infinitely many solutions,

(A)is k = 3
(B)does not exist
(C)is k = –3
(D)is k = 4

Show answer & solution

Answer: (B) does not exist

Write the equations as $k x - y - 2 = 0$ and $6 x - 2 y - 3 = 0$ .
For infinitely many solutions: $\frac{k}{6} = \frac{- 1}{- 2} = \frac{- 2}{- 3}$ .
But $\frac{- 1}{- 2} = \frac{1}{2}$ and $\frac{- 2}{- 3} = \frac{2}{3}$ are not equal.
So no value of k works.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q31 markMCQArithmetic Progressions

If $p - 1$ , $p + 1$ and $2 p + 3$ are in A.P., then the value of p is

(A)–2
(B)4
(C)0
(D)2

Show answer & solution

Answer: (C) 0

For three terms in A.P., $2 (p + 1) = (p - 1) + (2 p + 3)$ .
$2 p + 2 = 3 p + 2$
$p = 0$

Also asked in: 2023 Standard 30/1/3

Q41 markMCQCoordinate Geometry

In what ratio, does x-axis divide the line segment joining the points A(3, 6) and B(–12, –3) ?

(A)1 : 2
(B)1 : 4
(C)4 : 1
(D)2 : 1

Show answer & solution

Answer: (D) 2 : 1

Let the x-axis divide AB in the ratio k : 1.
y-coordinate of the point $= \frac{k ( - 3 ) + 1 ( 6 )}{k + 1} = 0$ .
$- 3 k + 6 = 0$ , so $k = 2$ .
Ratio = 2 : 1

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q51 markMCQCircles

In the given figure, PQ is tangent to the circle centred at O. If $∠ A O B = 9 5^{\circ}$ , then the measure of $∠ A B Q$ will be

Diagram for CBSE 2023 Class 10 Maths question 5

(A) $47. 5^{\circ}$
(B) $42. 5^{\circ}$
(C) $8 5^{\circ}$
(D) $9 5^{\circ}$

Show answer & solution

Answer: (A)

47. 5^{\circ}

OA = OB (radii), so $∠ O B A = ∠ O A B = \frac{18 0 ^{\circ} - 9 5 ^{\circ}}{2} = 42. 5^{\circ}$ .
The radius OB is perpendicular to the tangent PQ, so $∠ O B Q = 9 0^{\circ}$ .
$∠ A B Q = 9 0^{\circ} - 42. 5^{\circ} = 47. 5^{\circ}$

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q61 markMCQIntroduction to Trigonometry

If $2 tan A = 3$ , then the value of $\frac{4 s i n A + 3 c o s A}{4 s i n A - 3 c o s A}$ is

(A) $\frac{7}{13}$
(B) $\frac{1}{13}$
(C)3
(D)does not exist

Show answer & solution

Answer: (C) 3

$tan A = \frac{3}{2}$ .
Divide numerator and denominator by $cos A$ : $\frac{4 t a n A + 3}{4 t a n A - 3}$ .
$= \frac{6 + 3}{6 - 3} = 3$

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q71 markMCQPolynomials

If $α$ , $β$ are the zeroes of a polynomial $p (x) = x^{2} + x - 1$ , then $\frac{1}{α} + \frac{1}{β}$ equals to

(A)1
(B)2
(C)–1
(D) $\frac{- 1}{2}$

Show answer & solution

Answer: (A) 1

$α + β = - 1$ and $α β = - 1$ .
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- 1}{- 1} = 1$

Also asked in: 2023 Standard 30/1/2

Q81 markMCQQuadratic Equations

The least positive value of k, for which the quadratic equation $2 x^{2} + k x - 4 = 0$ has rational roots, is

(A) $\pm 22$
(B)2
(C) $\pm 2$
(D) $2$

Show answer & solution

Answer: (B) 2

Roots are rational when the discriminant is a perfect square.
$D = k^{2} - 4 (2) (- 4) = k^{2} + 32$ .
k = 1 gives D = 33, not a perfect square.
k = 2 gives $D = 36 = 6^{2}$ , a perfect square (roots 1 and –2).
Least positive value of k is 2.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q91 markMCQIntroduction to Trigonometry

$[\frac{3}{4} tan^{2} 3 0^{\circ} - sec^{2} 4 5^{\circ} + sin^{2} 6 0^{\circ}]$ is equal to

(A)–1
(B) $\frac{5}{6}$
(C) $\frac{- 3}{2}$
(D) $\frac{1}{6}$

Show answer & solution

Answer: (A) –1

$tan^{2} 3 0^{\circ} = \frac{1}{3}$ , $sec^{2} 4 5^{\circ} = 2$ , $sin^{2} 6 0^{\circ} = \frac{3}{4}$ .
Value $= \frac{3}{4} \times \frac{1}{3} - 2 + \frac{3}{4} = \frac{1}{4} - 2 + \frac{3}{4} = - 1$

Also asked in: 2023 Standard 30/1/3

Q101 markMCQSurface Areas and Volumes

Curved surface area of a cylinder of height 5 cm is 94.2 cm $^{2}$ . Radius of the cylinder is (Take $π = 3.14$ )

(A)2 cm
(B)3 cm
(C)2.9 cm
(D)6 cm

Show answer & solution

Answer: (B) 3 cm

CSA $= 2 π r h$ .
$2 \times 3.14 \times r \times 5 = 94.2$
$31.4 r = 94.2$ , so $r = 3$ cm.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q111 markMCQStatistics

The distribution below gives the marks obtained by 80 students on a test :
Marks: Less than 10, Less than 20, Less than 30, Less than 40, Less than 50, Less than 60
Number of Students: 3, 12, 27, 57, 75, 80
The modal class of this distribution is :

(A)10 – 20
(B)20 – 30
(C)30 – 40
(D)50 – 60

Show answer & solution

Answer: (C) 30 – 40

Convert the cumulative frequencies to class frequencies:
0–10: 3, 10–20: 9, 20–30: 15, 30–40: 30, 40–50: 18, 50–60: 5.
The highest frequency is 30, for the class 30–40.
Modal class = 30 – 40

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q121 markMCQSurface Areas and Volumes

The curved surface area of a cone having height 24 cm and radius 7 cm, is

(A)528 cm $^{2}$
(B)1056 cm $^{2}$
(C)550 cm $^{2}$
(D)500 cm $^{2}$

Show answer & solution

Answer: (C) 550 cm

^{2}

Slant height $l = 2 4^{2} + 7^{2} = 625 = 25$ cm.
CSA $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm $^{2}$

Also asked in: 2023 Standard 30/1/2

Q131 markMCQCoordinate Geometry

The distance between the points $(0, 25)$ and $(- 25, 0)$ is

(A) $210$ units
(B) $410$ units
(C) $220$ units
(D)0

Show answer & solution

Answer: (A)

210

units

Distance $= (- 25 - 0)^{2} + (0 - 25)^{2}$
$= 20 + 20 = 40 = 210$ units

Also asked in: 2023 Standard 30/1/3

Q141 markMCQPolynomials

Which of the following is a quadratic polynomial having zeroes $\frac{- 2}{3}$ and $\frac{2}{3}$ ?

(A) $4 x^{2} - 9$
(B) $\frac{4}{9} (9 x^{2} + 4)$
(C) $x^{2} + \frac{9}{4}$
(D) $5 (9 x^{2} - 4)$

Show answer & solution

Answer: (D)

5 (9 x^{2} - 4)

Sum of zeroes = 0, product $= - \frac{4}{9}$ .
Polynomial: $k (x^{2} - \frac{4}{9})$ , i.e. a multiple of $9 x^{2} - 4$ .
$5 (9 x^{2} - 4)$ has zeroes $\pm \frac{2}{3}$ .

Also asked in: 2023 Standard 30/1/3

Q151 markMCQStatistics

If the value of each observation of a statistical data is increased by 3, then the mean of the data

(A)remains unchanged
(B)increases by 3
(C)increases by 6
(D)increases by 3n

Show answer & solution

Answer: (B) increases by 3

New mean $= \frac{\sum ( x _{i} + 3 )}{n} = \frac{\sum x _{i}}{n} + 3$ .
So the mean increases by 3.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q161 markMCQProbability

Probability of happening of an event is denoted by p and probability of non-happening of the event is denoted by q. Relation between p and q is

(A)p + q = 1
(B)p = 1, q = 1
(C)p = q – 1
(D)p + q + 1 = 0

Show answer & solution

Answer: (A) p + q = 1

P(E) + P(not E) = 1.
So p + q = 1.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q171 markMCQProbability

A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought ?

(A)40
(B)240
(C)480
(D)750

Show answer & solution

Answer: (C) 480

P(winning) = number of her tickets ÷ total tickets.
Number of her tickets $= 0.08 \times 6000 = 480$

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q181 markMCQProbability

In a group of 20 people, 5 can't swim. If one person is selected at random, then the probability that he/she can swim, is

(A) $\frac{3}{4}$
(B) $\frac{1}{3}$
(C)1
(D) $\frac{1}{4}$

Show answer & solution

Answer: (A)

\frac{3}{4}

People who can swim = 20 – 5 = 15.
P(can swim) $= \frac{15}{20} = \frac{3}{4}$

Also asked in: 2023 Standard 30/1/2

Q191 markAssertion–ReasonCoordinate Geometry

Assertion (A) : Point P(0, 2) is the point of intersection of y-axis with the line $3 x + 2 y = 4$ .
Reason (R) : The distance of point P(0, 2) from x-axis is 2 units.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.

Show answer & solution

Answer: (B) Both A and R are true but R is not the correct explanation of A.

On the y-axis x = 0, so $2 y = 4$ , y = 2. The line meets the y-axis at (0, 2). A is true.
Distance of (0, 2) from the x-axis = |y| = 2 units. R is true.
R does not explain why (0, 2) lies on the line, so R is not the correct explanation of A.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q201 markAssertion–ReasonReal Numbers

Assertion (A) : The perimeter of $△ A B C$ is a rational number.
Reason (R) : The sum of the squares of two rational numbers is always rational.

Diagram for CBSE 2023 Class 10 Maths question 20

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.

Show answer & solution

Answer: (D) A is false but R is true.

$∠ B = 9 0^{\circ}$ , so $A C = 2^{2} + 3^{2} = 13$ cm.
Perimeter $= 2 + 3 + 13 = 5 + 13$ cm, which is irrational. A is false.
If a, b are rational, $a^{2} + b^{2}$ is rational. R is true.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q212 marksVery Short AnswerPair of Linear Equations in Two Variables

Solve the pair of equations $x = 3$ and $y = - 4$ graphically.

Show answer & solution

Answer: x = 3, y = –4, i.e. the lines meet at (3, –4).

$x = 3$ is a line parallel to the y-axis through (3, 0).
$y = - 4$ is a line parallel to the x-axis through (0, –4).
Draw both lines; they intersect at (3, –4).
Solution: $x = 3$ , $y = - 4$ .

Also asked in: 2023 Standard 30/1/3

Q21 (OR) (OR)2 marksVery Short AnswerPair of Linear Equations in Two Variables

Using graphical method, find whether following system of linear equations is consistent or not :
$x = 0$ and $y = - 7$

Show answer & solution

Answer: Consistent; the lines meet at (0, –7).

$x = 0$ is the y-axis.
$y = - 7$ is a line parallel to the x-axis through (0, –7).
The two lines intersect at exactly one point (0, –7).
So the system has a unique solution and is consistent.

Also asked in: 2023 Standard 30/1/3

Q222 marksVery Short AnswerTriangles

In the given figure, XZ is parallel to BC. AZ = 3 cm, ZC = 2 cm, BM = 3 cm and MC = 5 cm. Find the length of XY.

Diagram for CBSE 2023 Class 10 Maths question 22

Show answer & solution

Answer: XY = 1.8 cm

In $△ A B C$ , $X Z ∥ B C$ , so $\frac{A X}{A B} = \frac{A Z}{A C} = \frac{3}{5}$ .
In $△ A B M$ , $X Y ∥ B M$ , so $△ A X Y \sim △ A B M$ (AA).
$\frac{X Y}{B M} = \frac{A X}{A B} = \frac{3}{5}$
$X Y = \frac{3}{5} \times 3 = \frac{9}{5} = 1.8$ cm

Also asked in: 2023 Standard 30/1/2

Q232 marksVery Short AnswerIntroduction to Trigonometry

If $sin θ + cos θ = 3$ , then find the value of $sin θ \cdot cos θ$ .

Show answer & solution

Answer:

sin θ cos θ = 1

Square both sides: $sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 3$ .
$1 + 2 sin θ cos θ = 3$
$sin θ cos θ = 1$

Also asked in: 2023 Standard 30/1/2

Q23 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry

If $sin α = \frac{1}{2}$ and $cot β = 3$ , then find the value of $cosec α + cosec β$ .

Show answer & solution

Answer:

2 + 2

$sin α = \frac{1}{2}$ gives $α = 4 5^{\circ}$ , so $cosec α = 2$ .
$cot β = 3$ gives $β = 3 0^{\circ}$ , so $cosec β = 2$ .
$cosec α + cosec β = 2 + 2$

Also asked in: 2023 Standard 30/1/2

Q242 marksVery Short AnswerReal Numbers

Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively.

Show answer & solution

Answer: 14

The number divides 85 – 1 = 84 and 72 – 2 = 70 exactly.
$84 = 2^{2} \times 3 \times 7$ , $70 = 2 \times 5 \times 7$ .
HCF(84, 70) $= 2 \times 7 = 14$ .
Required number = 14.

Also asked in: 2023 Standard 30/1/3

Q252 marksVery Short AnswerProbability

A bag contains 4 red, 3 blue and 2 yellow balls. One ball is drawn at random from the bag. Find the probability that drawn ball is (i) red (ii) yellow.

Show answer & solution

Answer: (i)

\frac{4}{9}

(ii)

\frac{2}{9}

Total balls = 4 + 3 + 2 = 9.
(i) P(red) $= \frac{4}{9}$
(ii) P(yellow) $= \frac{2}{9}$

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q263 marksShort AnswerPair of Linear Equations in Two Variables

Half of the difference between two numbers is 2. The sum of the greater number and twice the smaller number is 13. Find the numbers.

Show answer & solution

Answer: 7 and 3

Let the greater number be x and the smaller y.
$\frac{x - y}{2} = 2$ , so $x - y = 4$ ... (1)
$x + 2 y = 13$ ... (2)
Subtract (1) from (2): $3 y = 9$ , $y = 3$ .
$x = 4 + 3 = 7$ .
The numbers are 7 and 3.

Also asked in: 2023 Standard 30/1/2

Q273 marksShort AnswerReal Numbers

Prove that $5$ is an irrational number.

Show answer & solution

Answer: Proved.

Suppose $5$ is rational. Then $5 = \frac{a}{b}$ with integers a, b, $b \neq = 0$ , having no common factor other than 1.
Squaring, $a^{2} = 5 b^{2}$ , so 5 divides $a^{2}$ , hence 5 divides a (5 is prime).
Let $a = 5 c$ . Then $25 c^{2} = 5 b^{2}$ , so $b^{2} = 5 c^{2}$ ; 5 divides $b^{2}$ , hence 5 divides b.
So 5 is a common factor of a and b, a contradiction.
Hence $5$ is irrational.

Q283 marksShort AnswerCoordinate Geometry

If (–5, 3) and (5, 3) are two vertices of an equilateral triangle, then find co-ordinates of the third vertex, given that origin lies inside the triangle. (Take $3 = 1.7$ )

Show answer & solution

Answer: (0, –5.5)

Let A(–5, 3), B(5, 3); AB = 10.
The third vertex C is equidistant from A and B, so it lies on x = 0: C(0, y).
$A C^{2} = 25 + (y - 3)^{2} = 100$ , so $(y - 3)^{2} = 75$ , $y = 3 \pm 53$ .
Origin is inside the triangle, so C is below AB: $y = 3 - 53 = 3 - 8.5 = - 5.5$ .
Third vertex = (0, –5.5)

Also asked in: 2023 Standard 30/1/3

Q293 marksShort AnswerCircles

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $∠ P T Q = 2∠ O P Q$ .

Diagram for CBSE 2023 Class 10 Maths question 29

Show answer & solution

Answer: Proved.

Let $∠ P T Q = θ$ .
TP = TQ (tangents from an external point), so $∠ T P Q = ∠ T QP = \frac{18 0 ^{\circ} - θ}{2} = 9 0^{\circ} - \frac{θ}{2}$ .
OP $⊥$ TP, so $∠ O P T = 9 0^{\circ}$ .
$∠ O P Q = 9 0^{\circ} - ∠ T P Q = 9 0^{\circ} - (9 0^{\circ} - \frac{θ}{2}) = \frac{θ}{2}$ .
Hence $∠ P T Q = 2∠ O P Q$ .

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

Q29 (OR) (OR)3 marksShort AnswerCircles

In the given figure, a circle is inscribed in a quadrilateral ABCD in which $∠ B = 9 0^{\circ}$ . If AD = 17 cm, AB = 20 cm and DS = 3 cm, then find the radius of the circle.

Diagram for CBSE 2023 Class 10 Maths question 29 (OR)

Show answer & solution

Answer: 6 cm

Tangents from an external point are equal: DR = DS = 3 cm.
AR = AD – DR = 17 – 3 = 14 cm, so AQ = AR = 14 cm.
BQ = AB – AQ = 20 – 14 = 6 cm.
In OQBP, $∠ O QB = ∠ O P B = ∠ B = 9 0^{\circ}$ and BQ = BP, so OQBP is a square.
r = OQ = BQ = 6 cm.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q303 marksShort AnswerIntroduction to Trigonometry

Prove that : $\frac{t a n θ + s e c θ - 1}{t a n θ - s e c θ + 1} = \frac{1 + s i n θ}{c o s θ}$

Show answer & solution

Answer: Proved.

Use $sec^{2} θ - tan^{2} θ = 1$ , i.e. $1 = (sec θ - tan θ) (sec θ + tan θ)$ .
Numerator: $tan θ + sec θ - (sec^{2} θ - tan^{2} θ) = (sec θ + tan θ) [1 - sec θ + tan θ]$ .
So LHS $= \frac{( s e c θ + t a n θ ) ( t a n θ - s e c θ + 1 )}{t a n θ - s e c θ + 1} = sec θ + tan θ$ .
$= \frac{1}{c o s θ} + \frac{s i n θ}{c o s θ} = \frac{1 + s i n θ}{c o s θ}$ = RHS.

Q313 marksShort AnswerSurface Areas and Volumes

A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of cylindrical part. Find total height of the room if it contains $(\frac{1408}{21})$ m $^{3}$ of air. $(Take π = \frac{22}{7})$

Show answer & solution

Answer: 6 m

Let the radius be r; height of cylinder h = 2r.
Volume $= π r^{2} (2 r) + \frac{2}{3} π r^{3} = \frac{8}{3} π r^{3}$ .
$\frac{8}{3} \times \frac{22}{7} \times r^{3} = \frac{1408}{21}$ , so $\frac{176}{21} r^{3} = \frac{1408}{21}$ , $r^{3} = 8$ , r = 2 m.
Total height = h + r = 4 + 2 = 6 m.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q31 (OR) (OR)3 marksShort AnswerSurface Areas and Volumes

An empty cone is of radius 3 cm and height 12 cm. Ice-cream is filled in it so that lower part of the cone which is $(\frac{1}{6})^{th}$ of the volume of the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. (Take $π = 3.14$ )

Diagram for CBSE 2023 Class 10 Maths question 31 (OR)

Show answer & solution

Answer: 150.72 cm

^{3}

Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times 3.14 \times 9 \times 12 = 113.04$ cm $^{3}$ .
Filled part of cone $= \frac{5}{6} \times 113.04 = 94.2$ cm $^{3}$ .
Volume of hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times 3.14 \times 27 = 56.52$ cm $^{3}$ .
Volume of ice-cream = 94.2 + 56.52 = 150.72 cm $^{3}$ .

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q325 marksLong AnswerTriangles

If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, prove that the other two sides are divided in the same ratio.

Show answer & solution

Answer: Proved.

Given: In $△ A B C$ , $D E ∥ B C$ with D on AB and E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$ .
$\frac{a r ( A D E )}{a r ( B D E )} = \frac{\frac{1}{2} A D \cdot E N}{\frac{1}{2} D B \cdot E N} = \frac{A D}{D B}$ .
$\frac{a r ( A D E )}{a r ( D E C )} = \frac{\frac{1}{2} A E \cdot D M}{\frac{1}{2} E C \cdot D M} = \frac{A E}{E C}$ .
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .

Q335 marksLong AnswerSome Applications of Trigonometry

The angle of elevation of the top of a tower 24 m high from the foot of another tower in the same plane is $6 0^{\circ}$ . The angle of elevation of the top of second tower from the foot of the first tower is $3 0^{\circ}$ . Find the distance between two towers and the height of the other tower. Also, find the length of the wire attached to the tops of both the towers.

Show answer & solution

Answer: Distance

= 83

m, height of other tower = 8 m, wire

= 87

Let AB = 24 m be the first tower, CD = h the second, and BD = d the distance between their feet.
From D: $tan 6 0^{\circ} = \frac{24}{d}$ , so $d = \frac{24}{3} = 83$ m.
From B: $tan 3 0^{\circ} = \frac{h}{d}$ , so $h = \frac{8 3}{3} = 8$ m.
Wire AC: $A C^{2} = d^{2} + (24 - 8)^{2} = 192 + 256 = 448$ .
$A C = 448 = 87$ m.

Q33 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry

A spherical balloon of radius r subtends an angle of $6 0^{\circ}$ at the eye of an observer. If the angle of elevation of its centre is $4 5^{\circ}$ from the same point, then prove that height of the centre of the balloon is $2$ times its radius.

Show answer & solution

Answer: Proved.

Let E be the eye, C the centre of the balloon, and EA a tangent from E touching the balloon at A.
The two tangents from E make $6 0^{\circ}$ , so $∠ A E C = 3 0^{\circ}$ ; also CA $⊥$ EA, CA = r.
$sin 3 0^{\circ} = \frac{r}{E C}$ , so EC = 2r.
Let CD be the perpendicular from C to the horizontal through E. $∠ C E D = 4 5^{\circ}$ .
$C D = E C sin 4 5^{\circ} = 2 r \times \frac{1}{2} = 2 r$ .
Hence the height of the centre is $2$ times the radius.

Q345 marksLong AnswerAreas Related to Circles

A chord of a circle of radius 14 cm subtends an angle of $6 0^{\circ}$ at the centre. Find the area of the corresponding minor segment of the circle. Also find the area of the major segment of the circle.

Show answer & solution

Answer: Minor segment

= (\frac{308}{3} - 493)

^{2}

≈ 17.89 cm

^{2}

; major segment

= (\frac{1540}{3} + 493)

^{2}

≈ 598.11 cm

^{2}

Area of sector $= \frac{60}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{308}{3}$ cm $^{2}$ ≈ 102.67 cm $^{2}$ .
The triangle formed is equilateral (two radii with $6 0^{\circ}$ between them): area $= \frac{3}{4} \times 1 4^{2} = 493$ cm $^{2}$ ≈ 84.87 cm $^{2}$ .
Minor segment $= \frac{308}{3} - 493$ ≈ 17.89 cm $^{2}$ .
Area of circle $= \frac{22}{7} \times 1 4^{2} = 616$ cm $^{2}$ .
Major segment $= 616 - (\frac{308}{3} - 493) = \frac{1540}{3} + 493$ ≈ 598.11 cm $^{2}$ .

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q355 marksLong AnswerArithmetic Progressions

The ratio of the 11 $^{th}$ term to 17 $^{th}$ term of an A.P. is 3 : 4. Find the ratio of 5 $^{th}$ term to 21 $^{st}$ term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms.

Show answer & solution

Answer: 3 : 7 and 25 : 189

$\frac{a + 10 d}{a + 16 d} = \frac{3}{4}$ , so $4 a + 40 d = 3 a + 48 d$ , $a = 8 d$ .
$\frac{a _{5}}{a _{21}} = \frac{a + 4 d}{a + 20 d} = \frac{12 d}{28 d} = \frac{3}{7}$ .
$S_{5} = \frac{5}{2} (2 a + 4 d) = \frac{5}{2} (20 d) = 50 d$ .
$S_{21} = \frac{21}{2} (2 a + 20 d) = \frac{21}{2} (36 d) = 378 d$ .
$S_{5} : S_{21} = 50 : 378 = 25 : 189$ .

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q35 (OR) (OR)5 marksLong AnswerArithmetic Progressions

250 logs are stacked in the following manner :
22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row ?

Show answer & solution

Answer: 20 rows; 3 logs in the top row

Rows form an A.P.: a = 22, d = –1.
$S_{n} = \frac{n}{2} [44 + (n - 1) (- 1)] = 250$ , so $n (45 - n) = 500$ .
$n^{2} - 45 n + 500 = 0$ , $(n - 20) (n - 25) = 0$ , n = 20 or 25.
For n = 25 the last row would have 22 – 24 = –2 logs, not possible. So n = 20.
Top row $a_{20} = 22 + 19 (- 1) = 3$ logs.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q364 marksCase StudyQuadratic Equations

While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by x units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide.
Based on the above information, answer the following questions :
(I) Write an algebraic equation depicting the above information. (1)
(II) Write the corresponding quadratic equation in standard form. (1)
(III) What should be the new dimensions of the enlarged photo ? (2)
OR
Can any rational value of x make the new area equal to 220 cm $^{2}$ ?

Diagram for CBSE 2023 Class 10 Maths question 36

Show answer & solution

Answer: (I)

(18 + x) (12 + x) = 2 \times 18 \times 12

(II)

x^{2} + 30 x - 216 = 0

(III) 24 cm × 18 cm; OR: No, since

x^{2} + 30 x - 4 = 0

has discriminant 916, not a perfect square, so x is irrational.

(I) Original area = 18 × 12 = 216 cm $^{2}$ . New area = $(18 + x) (12 + x) = 2 \times 216 = 432$ .
(II) $x^{2} + 30 x + 216 = 432$ , i.e. $x^{2} + 30 x - 216 = 0$ .
(III) $(x + 36) (x - 6) = 0$ ; x > 0, so x = 6. New dimensions: 18 + 6 = 24 cm by 12 + 6 = 18 cm.
OR: $(18 + x) (12 + x) = 220$ gives $x^{2} + 30 x - 4 = 0$ .
$D = 900 + 16 = 916$ , which is not a perfect square, so x is irrational. No rational x gives area 220 cm $^{2}$ .

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q374 marksCase StudyStatistics

India meteorological department observes seasonal and annual rainfall every year in different sub-divisions of our country.
It helps them to compare and analyse the results. The table given below shows sub-division wise seasonal (monsoon) rainfall (mm) in 2018 :
Rainfall (mm): 200-400, 400-600, 600-800, 800-1000, 1000-1200, 1200-1400, 1400-1600, 1600-1800
Number of Sub-divisions: 2, 4, 7, 4, 2, 3, 1, 1
Based on the above information, answer the following questions :
(I) Write the modal class. (1)
(II) Find the median of the given data. (2)
OR
Find the mean rainfall in this season.
(III) If sub-division having at least 1000 mm rainfall during monsoon season, is considered good rainfall sub-division, then how many sub-divisions had good rainfall ? (1)

Show answer & solution

Answer: (I) 600-800 (II) Median ≈ 771.43 mm; OR: Mean = 850 mm (III) 7

(I) Highest frequency is 7, for 600–800. Modal class = 600–800.
(II) N = 24, $\frac{N}{2} = 12$ . Cumulative frequencies: 2, 6, 13, 17, 19, 22, 23, 24. Median class 600–800.
Median $= 600 + \frac{12 - 6}{7} \times 200 = 600 + 171.43 = 771.43$ mm (approx.).
OR: Class marks 300, 500, 700, 900, 1100, 1300, 1500, 1700. $\sum f_{i} x_{i} = 600 + 2000 + 4900 + 3600 + 2200 + 3900 + 1500 + 1700 = 20400$ .
Mean $= \frac{20400}{24} = 850$ mm.
(III) Sub-divisions with at least 1000 mm = 2 + 3 + 1 + 1 = 7.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q384 marksCase StudyCircles

The discus throw is an event in which an athlete attempts to throw a discus. The athlete spins anti-clockwise around one and a half times through a circle, then releases the throw. When released, the discus travels along tangent to the circular spin orbit.
In the given figure, AB is one such tangent to a circle of radius 75 cm. Point O is centre of the circle and $∠ A B O = 3 0^{\circ}$ . PQ is parallel to OA.
Based on above information :
(a) find the length of AB. (1)
(b) find the length of OB. (1)
(c) find the length of AP. (2)
OR
find the length of PQ.

Diagram for CBSE 2023 Class 10 Maths question 38

Show answer & solution

Answer: (a)

753

cm (b) 150 cm (c)

\frac{75 3}{2}

cm; OR: PQ = 37.5 cm

(a) OA $⊥$ AB. $tan 3 0^{\circ} = \frac{O A}{A B}$ , so $A B = 753$ cm (≈ 129.9 cm).
(b) $sin 3 0^{\circ} = \frac{O A}{O B}$ , so OB = 150 cm.
(c) Q lies on the circle, so OQ = 75 cm and QB = 150 – 75 = 75 cm; Q is the mid-point of OB.
$P Q ∥ O A$ , so by the mid-point theorem (converse) P is the mid-point of AB.
$A P = \frac{1}{2} A B = \frac{75 3}{2}$ cm (≈ 64.95 cm).
OR: $P Q = \frac{1}{2} O A = \frac{75}{2} = 37.5$ cm.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

← 2024 Basic 430/5/3 2023 Standard 30/1/2 →

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