Here are the answers to the questions on polynomials from Chapter 2:1. (b) -3, -22. (a) 6 3. (a) x2 - 6x - 4 = 04. (a) m, m + 35. x2 - 5x + 66. The zeroes of the polynomial f(x) = x2 - 5x + 6 are:(a) 3, 2 (b) 2, 3 (c) -3, -2 (d) -2, -36. (c) -3, -2

STUDY MATERIAL SESSION (2022-23)
CLASS X MATHEMATICS (BASIC & STANDARD)
CHIEF PATRON
SH. B.L. MORODIA
DEPUTY COMMISSIONER
PATRON PATRON
SH. D R MEENA SH. M R RAWAL
ASSISTANT COMMISSIONER ASSISTANT COMMISSIONER
COORDINATOR
SH. R C BHURIA
PRINCIPAL, KV NO-5, JAIPUR

CONTENT PREPRATION TEAM MEMBERS
S. No. NAME OF TGTs NAME OF KV
1 SH. S C JAIN KV, No-3, JAIPUR
2 SH. M K VERMA KV, No-1, JAIPUR
3 SMT. SUNITA CHHIPA KV, No-2, JAIPUR
4 SMT. PREM SUDHA YADAV KV, No-5,JAIPUR
5 SH. DEEPAK KUMAR CHAUDHARY KV, No-4, JAIPUR
COURSE STRUCTURE
Units Unit Name Marks
I NUMBER SYSTEMS 06
II ALGEBRA 20
III COORDINATE GEOMETRY 06
IV GEOMETRY 15
V TRIGONOMETRY 12
VI MENSURATION 10
VII STATISTICS & PROBABILTY 11
TOTAL MARKS 80

INDEX
S.No CONTENT PAGE NO.
1 Syllabus (2022 – 23) 5 – 7
2 Chapter 1 - Real Numbers 8 – 9
3 Chapter 2 - Polynomials 10 - 11
4 Chapter 3 - Pair of Linear Equations in Two Variables 12 – 14
5 Chapter 4 - Quadratic Equations 15 – 16
6 Chapter 5 - Arithmetic Progressions 17 – 18
7 Chapter 6 - Triangles 19 – 20
8 Chapter 7 - Coordinate Geometry 21 – 22
9 Chapter 8 - Introduction to Trigonometry 23 – 25
10 Chapter 9 - Some Applications of Trigonometry 26 – 28
11 Chapter 10 - Circles 29 – 32
12 Chapter 12 - Areas related to Circles 33 – 34
13 Chapter 13 - Surface Areas and Volumes 35 – 38
14 Chapter 14 - Statistics 39 – 44
15 Chapter 15 - Probability 45 – 46
16 Sample Paper & Marking Scheme – 1 (Standard Mathematics) 47 – 58
21 Sample Paper & Marking Scheme – 1 (Basic Mathematics) 112 – 120
25 Sample Paper & Marking Scheme – 5 (Basic Mathematics) 156 - 166

SYLLABUS
UNIT I: NUMBER SYSTEMS
1. REAL NUMBER (15) Periods
Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and
after illustrating and motivating through examples, Proofs of irrationality of
UNIT II: ALGEBRA
1. POLYNOMIALS (8) Periods
Zeros of a polynomial. Relationship between zeros and coefficients of quadratic
Polynomials.
2.
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES (15) Periods
Pair of linear equations in two variables and graphical method of their solution,
consistency/inconsistency.
Algebraic conditions for number of solutions. Solution of a pair of linear equations in two
variables algebraically - by substitution, by elimination. Simple situational problems.
3. QUADRATIC EQUATIONS (15) Periods
Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of quadratic
equations (only real roots) by factorization, and by using quadratic formula. Relationship
between discriminant and nature of roots.
Situational problems based on quadratic equations related to day to day activities to be
incorporated.
4. ARITHMETIC PROGRESSIONS (10) Periods
Motivation for studying Arithmetic Progression. Derivation of the nth term and sum of the
first n terms of A.P. and their application in solving daily life problems.
UNIT III: COORDINATE GEOMETRY
Coordinate Geometry (15) Periods
Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula.
Section formula (internal division).
UNIT IV: GEOMETRY
1. TRIANGLES
Definitions, examples, counter examples of similar triangles.

1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other
two sides in distinct points, the other two sides are divided in the same ratio.
2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is
parallel to the third side.
3. (Motivate) If in two triangles, the corresponding angles are equal, their
corresponding sides are proportional and the triangles are similar.
4. (Motivate) If the corresponding sides of two triangles are proportional, their
corresponding angles are equal and the two triangles are similar.
5. (Motivate) If one angle of a triangle is equal to one angle of another triangle
and the sides including these angles are proportional, the two triangles are
similar.
2. CIRCLES (10) Periods
Tangent to a circle at, point of contact
1. (Prove) The tangent at any point of a circle is perpendicular to the radius
through the point of contact.
2. (Prove) The lengths of tangents drawn from an external point to a circle are
equal.
UNIT V: TRIGONOMETRY
1. INTRODUCTION TO TRIGONOMETRY
(10)
Periods
Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their
existence (well defined); motivate the ratios whichever are defined at 0o and 90o.
Values of the trigonometric ratios of 300, 450 and 600. Relationships between the
ratios.
2. TRIGONOMETRIC IDENTITIES (15) Periods
Proof and applications of the identity sin2A + cos2A = 1. Only simple identities to be
given.
3. HEIGHTS AND DISTANCES: Angle of elevation, Angle of Depression. (10)Periods
Simple problems on heights and distances. Problems should not involve more than
two right triangles. Angles of elevation / depression should be only 30°, 45°, and
60°.

7
UNIT VI: MENSURATION
1. AREAS RELATED TO CIRCLES (12) Periods
Area of sectors and segments of a circle. Problems based on areas and perimeter /
circumference of the above said plane figures. (In calculating area of segment of a circle,
problems should be restricted to central angle of 60°, 90° and 120° only.
2. SURFACE AREAS AND VOLUMES (12) Periods
Surface areas and volumes of combinations of any two of the following: cubes, cuboids,
spheres, hemispheres and right circular cylinders/cones.
UNIT VII: STATISTICS AND PROBABILITY
1. STATISTICS (18) Periods
Mean, median and mode of grouped data (bimodal situation to be avoided).
2. PROBABILITY (10) Periods
Classical definition of probability. Simple problems on finding the probability of an
event.

8
CHAPTER 1
The Fundamental Theorem of Arithmetic
Every composite number can be expressed (factorised) as a product of primes, and this
factorisation is unique apart from the order in which the prime factors occur.
The prime factorisation of a natural number is unique except for the order of its factors.
Property of HCF and LCM of two positive integers ‘a’ and ‘b’:
 HCF(a,b) X LCM(a,b) = a X b
LCM(a,b) =
a×b
HCF (a,b)
 HCF(a,b) =
a×b
LCM (a,b)
PRIME FACTORISATION METHOD TO FIND HCF AND LCM
HCF(a,b)=Product of the smallest power of each common prime factor in the numbers.
LCM(a,b)=Product of the greatest power of each prime factor involved in the numbers.
S.NO QUESTION
1 Complete the missing entries in the following factor
tree:
(a) 42 and 21 (b) 24 and 12 (c) 7 and 3 (d) 84 and 42
2 The H.C.F. and the L.C.M. of 12, 21, 15 respectively are:
(a) 3, 140 (b) 12, 420 (c) 3, 420 (d) 420, 3 Ans. (c) 3, 420
3 The H.C.F. of smallest prime number and the smallest composite number is .......... .
(a) 1 (b) 2 (c) 4 (d) none of these
4 225 can be expressed as
(a) 52× 32 (b) 52= 5 × 32 (c) 52× 3 (d)52× 325
5 The largest number which divides 70 and 125, leaving remainders 5 and 8
respectively, is
(a) 13 (b) 65 (c) 875 (d) 1750
6 Assertion : √2 is an irrational number
Reason; 2 is the smallest prime number
(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
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true

9
7 Assertion : 𝐹𝑜𝑟 𝑎𝑛𝑦 𝑡𝑤𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎 𝑎𝑛𝑑 𝑏 ,HCF (a,b)×LCM (a,b) =a× 𝑏
Reason; HCF of two number is 4 and their product is 192 then LCM is 28
8 Assertion : √3 +√5is an irrational number
Reason; sum of any two irrational number is always irrational
SHORT ANSWER TYPE QUESTION
9 Check whether 5x 3x 11 + 11 and 5 x 7 + 7 x 3 are composite numbers and
justify.
10 Check whether 6n can end with the digit 0, where n is any natural number
11 Given that LCM(26,169) = 338, write HCF(26,169).
12 Find the HCF and LCM of 6,72 and 120 using the prime factorization method
13 A class of 20 boys and 15 girls is divided into n groups so that each group has x boys
and y girls. Find x, y and n.?
14 Prove that √2 is an irrational number.
15 Prove that √3 is an irrational number.
16 Prove that 5√2 is an irrational number.
17 Prove that 7 + √2 is an irrational number.
18 Prove that √2 + √3 is an irrational number
ANSWERS
1)A 2) C 3)B 5)A 6)B 7)C 8)C
9) composite number 10) 6n cannot end with digit zero 11)13
12) HCF=6 LCM=36 13) x=4, y=3 and n=7
Question No (14-18) correct proof

10
CHAPTER 2
S. No Questions
1 Write the zero of the polynomial f(x) = x 2 – x – 6
(a) – 3, 2 (b) – 3, – 2 (c) 3, 2 (d) 3, – 2
2 For what value of k is – 4 a zero of the polynomial f(x) = x 2 – x – (2k + 2)?
(a) 6 (b) – 6 (c) 9 (d) – 9
3 If a and b are the zeroes of a polynomial such that a + b = – 6 and ab = – 4, then
write the polynomial.
(a) x2 – 6x – 4 = 0 (b) x2 + 6x – 4 = 0 (c) x2 + 6x + 4 = 0 (d) x2 – 6x + 4 = 0
4 The zeroes of the polynomial x 2 – 3x – m(m + 3) are:
(a) m, m + 3 (b) – m, m + 3 (c) m, – (m + 3) (d) – m, – (m + 3)
5 The quadratic polynomial, the sum of whose zeroes is – 5 and their product is 6, is:
(a) x2 + 5x + 6 (b) x2 – 5x + 6 (c) x2 – 5x – 6 (d) – x2 + 5x + 6
6 ASSERTION;-2 and -3 are the zeroes of x2+5x+6
Reason; a real number K is set to be zero of P(x) if P(k) =0

11
7 A cubic polynomial has exactly 3 zeroes
Reason; number of zero of a polynomial P(x) is equal to its degree
8 There exist only one real zero of P(x) =(x-2)(x2+3)
Reason; sum of zeroes of quadratic polynomials ax2+bx+c is
−𝑏
𝑎
9 If the product of zeros of ax2–6x–6 is 4, find the value of a.
Hence find the sum of its zeros.
10 If zeros of x2–kx+6 are in the ratio3:2, find k.
11 If one zero of the quadratic polynomial (k2+k)x2+68x+6k is reciprocal of the
other, find k.
12 If α and β are the zeros of the polynomial x2–5x+m such that α–β=1, find m.
13 If the sum of squares of zeros of the polynomial x2–8x+k is 40, find the value of k.
14 Find the value of k such that 3x2+2kx+x–k–5 has the sum of zeros as half of their
product.
15 If α and β are zeros of y2+5y+m , find the value of m such that (α+β)2 – αβ = 24
16
If α and β are zeros of x2–x–2, find a polynomial whose zeros are (2α+1) and (2β+1)
17 If α and β are zeros of the polynomial x2+4x+3, find the polynomial whose zeros are
1+
𝛽
𝛼
and 1+
𝛼
𝛽
18 Obtain the zeroes of 4√3x2+5x-2√3 and verify relation between zeroes and coefficient
ANSWERS
1)D 2)C 3)B 4)B 5)A 6)A 7)B 8)B 9)a=-3/2, sum of zeroes =-4
10)-5,5 11)5 12)6 13)12 14)1 15)1 16)x2-4x-5
17)
1
3
(3x2-16x+16) 18)
−2
√3
,
√3
4

12
CHAPTER 3
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
ALGEBRAIC INTERPRETATION OF PAIR OF LINEARE QUATIONS IN TWO VARIABLES
The pair of linear equations represented by these lines a1x+b1y+c1=0 and a2x+b2y+c2=0
S.No. Pair oflines Compare
the ratios
Graphical
representation
Algebraic
Interpretation
1
a1x+b1y+c1=0
a2x+b2y+c2=0
𝑎1
𝑎2
≠
𝑏2
𝑏2
Intersecting lines Unique
solution(Exactly one
solution)
2
a1x+b1y+c1=0
a2x+b2y+c2=0
a1
=
b1
=
c1
a2 b2 c2
Coincident
lines
Infinitely many
solutions
3
a1x+b1y+c1=0
a2x+b2y+c2=0
a1
=
b1
≠
c1
a2 b2 c2
Parallel lines No solution
S.NO QUESTIONS
1 The values of x and y in 2x + 3y = 2 and x – 2y = 8 are:
(a) – 4, 2 (b) – 4, – 2 (c) 4, – 2 (d) 4, 2 Ans. (c) 4, – 2
2 The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is
inconsistent is:
(a) 3 (b) – 4 (c) 8 (d) 4 (c) 10
3 The pair of linear equations 2x + 3y = 4 and 3x + 4y = 9 has:
(a) infinitely many solutions (b) no solution (c) one unique solution (d) two
solutions
4 The pair of equation x + 2y + 5 = 0 and – 3x – 6y + 1 = 0 have:
(a) a unique solution (b) exactly two solutions (c) infinitely many solutions (d) no
solution
5 If a pair of linear equations is consistent, then the lines will be:
(a) parallel (b) always coincident (c) intersecting or coincident
(d) always intersecting
6 System of linear equation given by x-7y+16=0 and 7x-49y-112=0 ade dependent
Reason; Dependent system of equations can be obtained from each other by
multiplying with a suitable constant

13
7 Assertion; there is no value of k for which the system of equations x+2y=3 ,
5x+ky+7=0 has infinite many solution
Reason; system of linear equations have an infinite solutions if
𝑎1
𝑎2
=
𝑏1
𝑏2
=
𝑐1
𝑐2
8 Assertion; If 2x+3y=12 and 3x-2y=5 then x=3 , y=2
Reason; method of elimination involves writing y in terms of x from any one of two
equations and then putting this value of y in other equation to get value of x. Finally
substituting value of x in any one equation gives value of y
9 Form a pair of linear equations for: The sum of the numerator and denominator of the
fraction is 3 less than twice the denominator. If the numerator and denominator
both are decreased by 1, the numerator becomes half the denominator.
10 For what value of p the pair of linear equations (p+2)x – (2p+1)y = 3(2p–1) and
2x – 3y = 7 has a unique solution.
11 ABCDE is a pentagon with BE||CD and BC||DE, BC is perpendicular to CD. If the
perimeter of ABCDE is 21cm, find x and y.
12 Solve for x and y : 3x+2y = 11 and 2x + 3y = 4. Also find p if p = 8x+5y

14
13 For what value of k, the following system of equations will be inconsistent
kx+3y=k–3 and 12x+ky=k
14 For what values of a and b the following pair of linear equations have infinite number of
solutions?
2x + 3y = 7 , a(x + y) – b(x – y) = 3a + b – 2
15 Aman travels 600 km to his home partly by train and partly by bus. He takes 8
hours, if he travels 120 km by train and rest by bus. Further, it takes 20 minutes
longer, if he travels 200 km by train and rest by bus. Find the speeds of the train and
the bus.
16 A and B are two points 150 km apart on a highway. Two cars start with different
speeds from A and B at same time. If they move in same direction, they meet in
15hours. If they move in opposite direction, they meet in one hour. Find their
speeds
17 A boat covers 32 km upstream and 36km downstream in 7hours. Also it Covers 40km
upstream and 48km downstream in 9hours. Find the speed of boat in still water and
that of the stream.
18 The sum of the numerator and denominator of a fraction is 4 more than twice the
numerator. If the numerator and denominator are increased by 3, they are in the
ratio2:3. Determine the fraction.
19 8 Women and 12 men can complete a work in10days while 6women and 8 men can
complete the same work in 14 days. Find the time taken by one woman alone and that
one man alone to finish the work
20 Determine graphically, the vertices of the triangle formed by the lines
y = x, 3y = x and x + y = 8.
ANSWERS
1)C 2)D 3)C 4)D 5)C 6)D 7)A
8)A 9)x – y = -3, 2x - y=1 10)p not equal to 4 11)x=5,y=0
(12)x=5,y=-2,p=30 13) k=-6 14)a=5,b=1 15)=60km/hr,80km/hr
16) 80km/hr,70km/hr 17)10km/hr,2km/hr 18)5/9
19) 1 women in 14 days,1 man in 28 days 20) vertices of triangle ,(0,0),(4,4),(6,2)

15
Chapter 4
QUADRATIC EQUATIONS
An equation involving single variable with a term having highest degree 2 of variable is called
quadratic equation.
In general form, ax2 + bx + c = 0, a ≠ 0 is a quadratic equation in variable x.
SOLUTION OF A QUADRATIC EQUATION
The zeroes of the quadratic polynomial or the roots of the quadratic equation ax2 + bx + c = 0
are called the solutions of the quadratic equation. Solutions of a Quadratic Equation can be found
by using following methods:
(i) By Factorisation Method: To find the solution of a quadratic equation by factorisation
method, first represent the given equation as a product of two linear factors by splitting
the middle term or by using identities and then equate each of the factor equal to zero to
get the desired roots.
(ii) By Quadratic Formula: For a quadratic equation ax2 + bx + c = 0, we have
x=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
NATURE OF ROOTS OF QUADRATIC EQUATION: By quadratic formula
The roots of the quadratic equation are given by x =
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
; where
(b2 – 4ac) is called discriminant of the quadratic equation and denoted by D.
The following cases arise:
i. If D = b2 – 4ac > 0, then the roots of the equation are real and distinct.
ii. If D = b2 – 4ac = 0, then roots of the equation are equal and real.
iii. If D = b2 – 4ac < 0, then there does not exist any real root.
iv. If D = b2 – 4ac > 0, and perfect square, then the roots are real, rational and unequal.
v. If D = b2– 4ac > 0 and not a perfect square, then the roots are real, irrational and unequal
QUADRATIC EQUATION WHEN THE ROOTS ARE GIVEN
The quadratic equation whose roots are a and b is given as x2 – (a +b)x +ab = 0
S.NO QUESTION
1 The quadratic equation ax2 – 4ax + 2a + 1 = 0 has repeated roots, if a =
(a) 0 (b) 1/2 (c) 2 (d) 4
2 The roots of the equation 2x - -
3
𝑥
= 1are
(a)
1
2
,−1 (b) 3 ,2 (c)-1,
3
2
(d) none of these
3 The two roots of a quadratic equation are 2 and – 1. The equation is
(a) x2 + 2x – 2 = 0 (b) x2 + x + 2 = 0
(c) x2 – 2x + 2 = 0 (d) x2 – x – 2 = 0
4 ax2 + bx + c = 0, a > 0, b = 0, c > 0 has
(a) two equal roots (b) one real roots (c) two distinct real roots
(d) no real roots
5 1
3
is a root of the equation x2+ kx –
5
9
= 0, then find the value of k.
(a)
3
4
(b)
4
3
(c)
2
3
(d) 3
6 Assertion: 2x2 – 4x + 3 = 0 is a quadratic equation.
Reason : All polynomials of degree n, when n is a whole number can be treated as
quadratic equation.

16
Choose the correct answer out of the following choices :
(a) Assertion and Reason both are correct statements and Reason is the correct
explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the
correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement
7 Assertion: 3y2 + 17y – 30 = 0 have distinct roots.
Reason: The quadratic equation ax2 + bx + c = 0 have distinct roots (real) if D > 0.
(d) Assertion is wrong statement but Reason is correct statement.
8 Assertion: Both the roots of the equation x2 – x + 1 = 0 are real.
Reason: The roots of the equation ax2 + bx + c = 0 are real if and only if b2 – 4ac ≥ 0
9 solve for x
1
𝑥+4
−
1
𝑥−7
=
11
30
10 solve for x
𝑥−4
𝑥−5
+
𝑥−6
𝑥−7
=
10
3
11 solve for x
1
𝑎+𝑏+𝑥
=
1
𝑥
+
1
𝑎
+
1
𝑏
12 If x =
2
3
and x = – 3 are roots of the quadratic equation ax2 + 7x + b = 0.
Find the value of a and b.
13 Find value of p for which the product of roots of the quadratic equation
px2 + 6x + 4p = 0 is equal to the sum of the roots.
14 A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of
the slow train is 10 km/hr less than the fast train, find the speeds of the two trains.
15 The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return
downstream to the orignal point in 4 hrs 30 minutes. Find the speed of the stream.
16 Sum of areas of two squares is 400 cm2. If the difference of their perimeter is 16 cm.
Find the side of each square.
17 If the roots of the quadratic equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal,
prove: 2b = a + c.
18 A two digit number is such that the product of its digits is 18. When 63 is subtracted
from the number, the digit interchange their places. Find the number.
ANSWERS
1)B 2)C 3)D 4)D 5)B 6)C 7)A 8)D 9) x=1, -4/3
10) x=8, 11/2 11)x=-a, -b 12)a=3, b=-6 13)p= -3/2 14)5km/hr
15)5km/hr 16)x=16m, y=12m or x=12m, y=16m 17) correct proof 18) number 92

17
Chapter 5
ARITHMETIC PROGRESSION
Standard form of an AP is given as a, a+d, a+2d, a+3d,….
Where a is the first term and d is the common difference.
nth term an of the AP with first term a and common difference d is given by an = a+ (n-1)d
nth Term from the end of an AP is given by ln=l-(n-1)d
Sum of n terms of of an AP Sn =
𝑛
2
{ 2a+(n-1)d}
Sum of n terms of of an AP Sn =
𝑛
2
{ a+l} where l is last term of the series
Also an=Sn – Sn-1
S.NO QUESTIONS
1 If k, 2k – 1 and 2k + 1 are three consecutive terms of an A.P., then the value of k is
(a) 2 (b) 3 (c) –3 (d) 5
2 The next term of the A.P. 18, 50 95 ... is
(a) 146 (b) 128 (c) 162 (d) 200
3 The value of a30 – a20 for the A.P. 2, 7, 12, 17, ... is
(a) 100 (b) 10 (c) 50 (d) 20
4 In an A.P., if a = −10, n = 6 and an = 10, then the value of d is
(a) 0 (b) 4 (c) −4 (d) 10/3
5 If the sum of first m terms of an A.P. is 2m2 + 3m, then what is its second term?
(a) 9 (b) 10 (c) 11 (d) 12
6 Assertion : The nth term of a sequence is 3n – 2. It is an A.P.
Reason : A sequence is not an A.P. if its nth term is not a linear expression in n.
7 Assertion : The 10th term from the end of the A.P.7, 10, 13, ...., 184 is 163.
Reason : In an A.P. with first term a, common difference d and last term l, the nth
term from the end is l – (n – 1)d

18
8 Assertion : The common difference of the A.P. 19, 18, 17, .... is 1.
Reason : Let a1, a2, a3, a4, ... is an A.P. Then, common difference of this A.P. will be
the difference between any two consecutive terms, i.e., common difference (d) = a2
– a1 or a3 – a2 or a4 – a3 and so on
9 Find the 9th term from the end of the A.P. 5, 9, 13, ..., 185.
10 Check whether −150 is a term of the A.P. : 17, 12, 7, 2,… or not.
11 What will be the 21st term of the A.P. whose first two terms are –3 and 4?
12 If the sum of first n terms of an A.P. is given by Sn = 5n2 + 3n, then find its nth term.
13 Which term of the A.P. : 21, 42, 63, 84,... is 210
14 Determine the A.P. whose 3rd term is 5 and the 7th term is 9.
15 The sum of the first n terms of an A.P. is 5n – n2 . Find the nth term of this A.P.
16 If Sn denotes the sum of first n terms of an A.P., prove that S30 = 3[S20 – S10]
17 If the sum of n terms of an A.P. is (pn + qn2 ), where p and q are constants, find the
common difference
18 Jasleen saved Rs 4 during first week of the year and then increased her weekly
savings by Rs 1.75 each week. In which week, will her weekly savings be `Rs19.75?
ANSWERS
1)B 2)C 3)C 4)B 5)A 6)A 7)D 8)D 9) 153 10) not a term
11)a21=137 12)an=10n-2 13)t10=210 14)ap=3,4,5,6,7,….
15)Tn= -2n+6 16)correct proof 17)d=2q 18) 10 week

19
CHAPTER 6
Similar Triangles: Two triangles are said to be similar if their corresponding angles are equal and
their corresponding sides are proportionate.
Criteria for Similarity:
In ΔABC and ΔDEF
(i) AAA Similarity: ΔABC ~ ΔDEF when
∠A=∠D, ∠B=∠E and ∠C=∠F
(ii) SAS Similarity:
ΔABC ~ ΔDEF when
𝐴𝐵
𝐷𝐸
=
𝐵𝐶
𝐸𝐹
and ∠B = ∠E
(iii) SSS Similarity: ΔABC ~ ΔDEF if 𝐴𝐵
𝐷𝐸
=
𝐴𝐶
𝐷𝐹
=
𝐵𝐶
𝐸𝐹
Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle to intersect the other
two sides in distinct points, the other two sides are divided in the same ratio.
S.NO QUESTIONS
1 If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not
true?
(a) EF/PR = DF/PQ (b) DE/PQ = EF/RP (c) DE/QR = DF/PQ (d) EF/RP = DE/QR
2 If ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF, then which of the following is not true?
(a) BC . EF = AC. FD (b) AB . EF = AC. DE (c) BC . DE = AB. EF d) BC . DE = AB. FD
3 D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD =
2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is-
(a) 2.5 (b) 3 (c) 5 (d) 6
4 A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a
tower casts a shadow 28 m long. Find the height of the tower.
(a) 42 (b) 32 (c) 5 (d)16
5
In this figure if DEIIBC. Then find the value of x
(a) 10 (b) 11 (c) 12 (d)13
.6 In ∆ABC, D and E are mid-points of AC and BC respectively such that DE
|| AB. If AD = 2x,
BE = 2x – 1, CD = x + 1 and CE = x – 1, then find the value of x
7 Students of a school decided to participate in ‘Save girl child’ campaign.
They decided to decorate a triangular path as shown. If AB = AC and BC2
= AC × CD, then prove that BD = BC.

20
8 The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the
first triangle is 9 cm, find the corresponding side of the second triangle.
9 In ∆ABC, D and E are points on the sides of AB and AC such that DE || BC. If AD = 2.5 cm, BD
= 3 cm, AE = 3.75 cm, find the length of AC.
10 Prove that a line drawn through the mid-point of one side of a triangle parallel to another
side bisects the third side.
11 ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
Show that
𝐴𝑂
𝐵𝑂
=
𝐶𝑂
𝐷𝑂
12 Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O.
Using a similarity criterion for two triangles, show that OA/OC=OB/OD∙
13 In the given figure, altitudes AD and CE of ∆ABC intersect
each other at the point P. Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
14 In the given figure, ABC and AMP are two right triangles, right angled at B and M
respectively. Prove that:
(i) ∆ABC~ ∆AMP
(ii)
𝐶𝐴
𝑃𝐴
=
𝐵𝐶
𝑀𝑃
15 CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides
AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that
(i) ∆DCB~ ∆HGE (ii)
𝐶𝐷
𝐺𝐻
=
𝐴𝐶
𝐹𝐺
16 Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ
and QR and median PM of ∆PQR. Show that ∆ABC ~ ∆PQR.
17 Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ
and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
18 State and proof Basic proportionality theorem.
ANSWERS
1)B 2)C 3)B 4)A 5)B 6)x=1/3
7)correct proof 8)DE=5.4cm 9)AC=8.25cm
Q(10-18) correct proof

21
CHAPTER 7
COORDINATE GEOMETRY
The system of geometry where the position of points on the plane is described using an ordered
pair of numbers.
Distance Formula: Distance between two given points A(x1,y1) And B(x2,y2) =
√(𝑥2 − 𝑥1)2 − (𝑦2 − 𝑦1)2
Section formula
The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and
B(x2, y2) internally in the ratio m1 : m2 are
𝑥 =
𝑚1𝑥2+𝑚2𝑥1
𝑚1+𝑚2
𝑦 =
𝑚1𝑦2+𝑚2𝑦1
𝑚1+𝑚2
The coordinates of point M(x,y) which is the midpoint of point ( x1, y1) and (x2, y2) are
X =
x1 + x2
2
, y =
y1 + y2
2
S.NO QUESTION
1 The distance of the point (2,3) from the x-axis is (a) 2 (b) 3 (c) 1 (d) 5
2 The distance of the point P(-6,8) from the origin is (a) 8 (b) 2√7 (c) 10 (d) 6
3 The distance between the points A(0,6) and B(0,-2) is (a) 6 (b) 8 (c) 4 (d) 2
4 The distance between the points (0,5) and (-5,0) is (a) 5 (b) 5√2 (c) 2√5 (d) 10
5 If the distance between the points (2,-2) and (-1,y) is 5, then the value of y is
(a) -2 (b) 2 (c) -1 (d) 1
6 If the distance between the points (4,p) and (1,0) is 5, then value of p is
(a) 4 only (b) ±4 (c) -4 only (d) 0
7 The mid-point of the line segment joining the points A(-2,8) and B(- 6,-4) is
(a) (-4,-6) (b) (2,6) (c)(-4,2) (d) (4,2)
8 The values of y, for which the distance between the points P(2,-3) and Q(10,y) is 10
units, are (a) 9,6 (b) 3,-9 (c) -3,9 (d) 9,-6
9 ABCD is a rectangle whose three vertices are A(0,3), B(0,0) and C(5,0). Then the
length of its diagonal is (a) 5 (b) 3 (c) √34 (d) 4
10 Find the value of 𝑥 for which the distance between the points P(4,- 5) and Q(12,𝑥) is
10 units.
11 Find a point on x-axis which is equidistant from the points (7,6) and (-3,4).
12 Find a point on y-axis which is equidistant from the points (-5,2) and (9,-2).
13 Find a relation between x and y such that the point P(x, y) is equidistant from the
points A(2,5) and B(-3,7).
14 Find the ratio in which the point (x,1) divides the line segment joining the points
(-3,5) and (2,-5). Also find the value of x
15 Find the coordinates of the point which divides the line segment joining the points
(4,-3) and (8,5) in the ratio 3:1.

22
16 The coordinates of the mid-point of the line segment joining the points (3p,4) and
(-2,2q) are (5,p).Find the values of p and q.
17 If the mid-points of the line segment joining the points A(3,4) and B(k,6) is P(x, y)
and x+y-10=0, find the value of k
18 The mid-points of the sides of a triangle are (3,4), (4,6) and (5,7). Find the
coordinates of vertices of the triangle.
19 A person is riding his bike on a straight road towards East from his college to city A
and then to city B. At some point in between city A and city B, he suddenly realises
that there is not enough petrol for the journey. Also, there is no petrol pump on the
road between these two cities.
Based on the above information, answer the following questions
(I)The value of y is equal to
(II)The value of x is equal to
(III)If M is any point exactly in between city A and city B, then coordinates of M are
OR
The ratio in which A divides the line segment joining the points O and M is (a) 1:2
(b) 2:1
ANSWERS
1)B 2)C 3)B 4)C 5)B 6)B 7)B 8)B 9)C
10(1,-11) 11)(3,0) 12) P=(0,-7) 13)10x-4y+20=0 14)k=3/2,or x=0
15)7,3 16)t=4,q=2 17)k=7 18) A(6,9) B(4,5) C(2,3)
19)(i) y=2 (ii) x=8 (iii) 5,5 or 2,3

23
Chapter 8
Introduction to Trigonometry
Important points
Sometimes we observe imaginary triangle in nature,
e.g. if we look at the top of a tower a right angle can be imagined. As shown in figure
We need to find height BC or distance AB or AC.
These all can be found by using mathematical techniques which comes under
a branch of mathematics called Trigonometry.
Consider a right angled triangle ABC right angled at B.
Fig 1 Fig2
Observing the above two triangles we see that one side i.e. hypotenuse (longest side of right triangle)
is fixed it is opposite to right angle, but other sides varies in respect of angle under consideration.
Here it is to note that we write:
Side opposite to given angle as PERPENDICULAR (P), Side adjacent to given angle as BASE (B)
And the longest side HYPOTENUSE (H).
TRIGONOMETRIC RATIOS
Let’s define certain ratios involving sides of right triangles and call them TRIGONOMETRIC RATIOS
So we have,

24
sin =P/H cos =B/H tan =P/B
cosec =H/P sec =H/B cot =B/P
TRIGNOMETRIC TABLE
0 30 45 60 90
Sin 0 1/2 1/√2 √3/2 1
Cos 1 √3/2 1/√2 1/2 0
Tan 0 1/√3 1 √3 n.d
Cot n.d √3 1 1/√3 0
Sec 1 2/√3 √2 2 n.d
cosec n.d 2 √2 2/√3 1
S.NO QUESTION
1 If x=2 sin2θ and y=2cos2 θ+1 then x + y is equal to
(a) 3 (b) 2 (c) 1 (d) 0
2 If tan A = 4/3 , then the value of cosC is
(a) 3/4 (b)4/5 (c) 1 (d) none of these
3 In ∆OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm, then the values of sin Q.
(a)7/25 (b) 24/25 (c) 1 (d) none of these
4 Given 15 cot A = 8, then sin C =
(a) 0 (b) 8/17 (c) 1 (d) none of these
5 In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm, then the value of sin P
is (a) 5/13 (b) 12/13 (c) 13/12 (d) 0
6 Q.1. Assertion: The value of sin600 cos300 + sin300 cos600 is 1
Reason: sin900=1 and cos900=0
7 Assertion: In a right ΔABC, right angled at B, if tanA=1, then 2sinA.cosA=1
Reason: cosecA is the abbreviation used for cosecant of angle A.
8 Assertion: sin(A+B)=sinA + sinB

25
Reason: For any value of θ, 1+tan2θ = sec2θ
9 If 3 cot A = 4, find the value of (Cosec2 A + 1)/( Cosec A2 – 1)
10 If tan (3x – 15°) = 1 then find the value of x.
11 If sec θ = x +
1
4𝑥
, prove that sec θ + tan θ = 2x or 1/2x
12 Prove that :.
tan θ – cot θ
sinθ cos θ
= tan2 θ- cot2 θ
13 Prove that
Sec θ + tan θ – 1
tan θ – Sec θ +1
=
𝑐𝑜𝑠𝜃
1−𝑠𝑖𝑛𝜃
14 Prove that: sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ.
15 Prove that: sin6 θ + cos6 θ = 1 – 3 sin2 θ cos2 θ.
16 Evaluate sin 60° cos 30° + sin 30° cos 60°
17 Evaluate 2 tan2 45° + cos2 30° – sin2 60°
18
Evaluate
sin 30° + tan 45° – cosec 60°
sec 30° + cos 60° + cot 45°
ANSWER
1)A 2)B 3)A 4)B 5)B 6)B 7)B
8)D 9)13/7 10)20
Q (11to15) correct proof
16)1 17)2 18)
43−24√3
11

26
Chapter 9
Application of Trigonometry
Important Points: -
Trigonometry can be used to measure the height of a building or trees, mountains etc. It is the study
of relationship between the ratios of the right-angled triangle’s sides and its angles. Trigonometry is
being used for finding the heights and distances of various objects without measuring them. In
solving problems of heights and distances two types of angles are involved:
1. The angle of Elevation
2. The angle of Depression
Before knowing these angles, it is necessary to know about the following terms.
 Horizontal Plane: A plane parallel to the earth is called the Horizontal Plane
 Horizontal Line: A line drawn parallel to horizontal plane is called a horizontal line.
Example of:
Angle of Elevation
Angle of Depression
Use of Right-angled Triangle in Trigonometry: -
∆𝐴𝑀𝑃 is right angled at M, PM is perpendicular, AM is base and AP is hypotenuse and ∠𝑃𝐴𝑀 = 𝜃
Most commonly used trigonometric angles are 𝟑𝟎𝟎
, 𝟒𝟓𝟎
and 𝟔𝟎𝟎

27
S.NO QUESTIONS
1 The length of the shadow of a tower on the plane ground is √3 times the height of the
tower. The angle of elevation of sun is : (a) 45° (b) 30° (c) 60° (d) 90°
2 The tops of the poles of height 16 m and 10 m are connected by a wire of length l
metres. If the wire makes an angle of 30° with the horizontal, the l = (a) 26 m (b) 16
m (c) 12 cm (d) 10 m
3 A pole of height 6 m casts a shadow 2 √3 m long on the ground. the angle of elevation
of the sun is (a) 30° (b) 60° (c) 45° (d) 90
4 A ladder leaning aginast a wall makes an angle of 60° with the horizontal. If the foot
of the ladder is 2.5 m away from the wall, then the length of the ladder is —
(a) 3 m (b) 4 m (c) 5 m (d) 6 m
5 If a tower is 30 m hight, costs a shadow 10 √3 m long on the ground, then the angle of
elevation of the sun is:
(a) 30° (b) 45° (c) 60° (d) 90°
6 A tower is 50 m high. When the sun’s altitude is 45° then what will be the length of its
shadow?
7 The length of shadow of a pole 50 m high is
50
3
m. find the sun’s altitude.
8 Find the angle of elevation of a point which is at a distance of 30 m from the base of a
tower 10 √3 m high.
9 A kite is flying at a height of 50√3m from the horizontal. It is attached with a string
and makes an angle 60° with the horizontal. Find the length of the string.
10 The upper part of a tree broken over by the wind makes an angle of 30° with the
ground and the distance of the root from the point where the top touches the ground
is 25 m. What was the total height of the tree?
11 A man standing on the deck of a ship, 10 m above the water level observes the angle
of elevation of the top of a hill as 60° and angle of depression the bottom of a hill as
30°. Find the distance of the hill from the ship and height of the hill.
12 A bird is sitting on the top of a tree, which is 80 m high. The angle of elevation of the
bird, from a point on the ground is 45°. The bird flies away from the point of
observation horizontally and remains at a constant height. After 2 seconds, the angle
of elevation of the bird from the point of observation becomes 30°. Find the speed of
flying of the bird.
13 The shadow of a tower standing on a level ground is found to be 40 m longer when
the Sun’s altitude is 30° than when it is 60°. Find the height of the tower
14 The angles of depression of the top and the bottom of an 8 m tall building from the
top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the
multi-storeyed building and the distance between the two buildings.
15 From a point on a bridge across a river, the angles of depression of the banks on
opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of
3 m from the banks, find the width of the river.
16 From the top of a 7 m high building, the angle of elevation of the top of a cable tower
is 60° and the angle of depression of its foot is 45°. Determine the height of the tower

28
17 A straight highway leads to the foot of a tower. A man standing at the top of the tower
observes a car at an angle of depression of 30°, which is approaching the foot of the
tower with a uniform speed. Six seconds later, the angle of depression of the car is
found to be 60°. Find the time taken by the car to reach the foot of the tower from this
point
18 A hot air balloon is a type of aircraft. It is lifted by heating the air inside the balloon,
usually with fire. Hot air weighs less than the same volume of cold air (it is less
dense), which means that hot air will rise up or float when there is cold air around it,
just like a bubble of air in a pot of water. The greater the difference between the hot
and the cold, the greater the difference in density, and the stronger the balloon will
pull up.
Lakshman is riding on a hot air balloon. After reaching at height x at point P , he spots
a lorry parked at B on the ground at an angle of depression of 30c. The balloon rises
further by 50 metres at point Q and now he spots the same lorry at an angle of
depression of 45c and a car parked at C at an angle of depression of 30c.
(i) What is the relation between the height x of the balloon at point P and distance d
between point A and B ?
(ii) When balloon rises further 50 metres, then what is the relation between new
height y and d ? (iii) What is the new height of the balloon at point Q ?
Or
What is the distance AB on the ground ?
ANSWER
1)B 2)C 3)B 4)C 5)C 6)15m 7)60O
8)30O 9)100m 10)43.3m 11)40m,17.32m 12)29.28m
13)h=23√3 m 14)4(3+√3) 15)3(√3+1) 16)7(√3+1)
17)3sec 18)case study(i) √3AP (ii) AB=AQ (iii) AQ=25(√3+3) OR 25(√3+3)

29
Chapter 10
Circles
PREREQUISITES TERMONIOLOGY:
1) RADIUS: The distance from the centre to any point on the
surface of a circle is called “Radius”.
2) SECANT: A secant to a circle is a line that cuts the circle at
two distinct points.
3) CHORD: A chord is a line segment whose end points lie on
the circle itself. Diameter is the longest chord in a circle.
4) TANGENT:
A tangent to a circle is a line that touches the circle at exactly one point. For every point on the
circle, there is a unique tangent passing through it.
The point where the tangent touches the circle is called “Point of contact”.
Key points:
 No tangent can be drawn to a circle which passes through a point that lies inside it.
 When a point of tangency lies on the circle, there is exactly one tangent to a circle that passes
through it.
 When the point lies outside of the circle, there are accurately two tangents to a circle through
it.
THEOREMS:
1) The perpendicular from the centre of the circle to a chord bisects the chord.
2) The angle subtended by an arc at the centre of the circle is double the angle
subtended by it at any point on the remaining part of the circle.
3) Angles in the same segment of the circle are always equal.
4) Angle formed in a semicircle is always 90.
5) The sum of all the angles of a quadrilateral is 360.

30
IMPORTANT THEOREMS (WITH PROOF):
Theorem 1: The tangent at any point of a circle is perpendicular to
the radius through the point of contact.
Given: A circle C (O, r) and a tangent XY at point “P”.
To prove: OP is perpendicular to tangent XY.
Construction: A point Q on the tangent line XY, other than P. Join
the points OQ.
Proof:
Point Q should lie outside the circle. Because if point Q lies inside the circle, XY will not be a tangent
to the circle and XY would become a secant of a circle.
Now, OQ should be greater than the radius of the circle OP as it lies outside the circle.
Thus, OQ > OP
As, this condition is obeyed for all points on line XY except P, OP should be the shortest of all
distances from the centre of the circle “O” to the points of line XY.
Therefore, OP is perpendicular to XY.
Hence, proved.
Converse of Theorem 1:
A line drawn through the end of a radius and perpendicular to it is a tangent to the circle.
Theorem 2: The lengths of tangents drawn from an external point to a circle are equal.
Given: A circle C (O, r) and two tangents say PQ and PR from an external point P.
To prove: PQ = PR
Construction: Join OQ, OR and OP.
Proof: In ∆ OQP and ∆ ORP,
OQ = OR = r (radii of the same circle)
OP = OP (common)
∠OQP = ∠ORP = 90
(The angles formed between the tangents and radii are right
angles.)
Hence, using the RHS congruence rule,
∆ OQP ≅∆ ORP
Thus, PQ = PR (Using CPCT)
Hence the theorem is proved.
S.No QUESTIONS
1 Two balls of equal size are touching each other externally at point C and AB is common
tangent to the balls. Then ∠ ACB= (a) 600 (b) 450 (c) 300 (d) 900
2 Radha and Shyama were arguing that how many parallel tangents can a circle have? Can
you help them ?
(a) 1 (b) 2 (c) infinite (d) none of these
3 Three friends Ram, Shyam and Rahim are playing in a
triangular park in which there is a circular rose garden as
shown in the fig. Three friends are standing at points A,B and
C respectively. By the information given in the figure, can you
calculate perimeter of the park ?
(a) 30 cm (b) 60cm (c) 45cm (d) 15cm

31
4 If four sides of the quadrilateral ABCD are tangents to a circle , then
(a) AC+AD=BD+CD (c) AB+CD=BC+AD (b) AB+CD=AC+BC (d)AC+AD=BC+DB
5 AP and AQ are tangents drawn from a point A to a circle with centre O and radius 9 cm.
If OA=15 cm, then AP+AQ=
(a) 12cm (b) 18cm (c) 24cm (d) 36cm
6 If common tangents AB and CD of two wheels with
centre O and O’ intersect at E, then find OEO’=?
(a) a triangle (b) a line
(c) an arc (d) none of these
7 PQ and PR are two tangents from P to a circle with
centreA . If ∠QAR=1300 , find ∠QPR=?
(a) 400 (b) 500 (c)600 (d)20O
8 In two concentric circles, if length of one chord AB touching inner
circle is 12cm then find the length of chord CD ?
(a) 10cm (b) 15cm (c) 12cm (d) 6cm
9 The length of the tangent from a point which is at a distance of 10cm from the centre of
the circle having radius 6cm is ? (a) 8cm (b) 10cm (c) 4cm (d)16cm
10 If AB= 14cm and PE=5cm, then AE=?
(a) 7cm (b) 8cm (c) 19cm (d)9cm
11 Prove that The tangent at any point of a circle is perpendicular to the radius through the
point of contact
12 Prove that The lengths of tangents drawn from an external point to a circle are equal.
13 Two tangents TP and TQ are drawn to a circle with centre O from an external point T.
Prove that ∠ PTQ = 2 ∠ OPQ.
14 PQ is a chord of length 8 cm of a circle of radius 5 cm. The
tangents at P and Q intersect at a point T (see Fig). Find the length
TP
15 Prove that the tangents drawn at the ends of a diameter of a circle are parallel

32
16 A quadrilateral ABCD is drawn to circumscribe a circle (see Fig).
Prove that AB + CD = AD + BC
17 Prove that the parallelogram circumscribing a circle is a rhombus
18 In Fig. , XY and X′Y′ are two parallel tangents to a circle with
centre O and another tangent AB with point of contact C
intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°
19 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 8 cm and 6 cm respectively
(see Fig.). Find the sides AB and AC
20 Prove that opposite sides of a quadrilateral circumscribing a circle subtend
supplementary angles at the centre of the circle
ANSWERS
1)D 2)B 3)A 4)C 5)C 6)B 7)B
8)C 9)A 10)D Q(11-13)correct proof 14)20/3
Q(15-18) correct proof
19)AB=15cm,AC=13cm 20) correct proof

33
Chapter 12
Areas Related to Circles
IMPORTANT FORMULAS & CONCEPTS
Perimeter and Area of a Circle
1. Perimeter/circumference of a circle = 𝜋 × diameter
= 𝜋 × 2𝑟 (where 𝑟 is the radius of the circle)
= 2𝜋𝑟
2. Area of a circle= 𝜋𝑟2
, where 𝜋 =
22
7
Areas of Sector and Segment of a Circle
1. Area of the sector of angle 𝜃 =
𝜃
360∘ × 𝜋𝑟2
, where 𝑟 is the radius of the circle
and 𝜃 the angle of the sector in degrees
2. Length of an arc of a sector of angle 𝜃 =
𝜃
360∘ × 2𝜋𝑟, where 𝑟 is the radius of
the circle and 𝜃 the angle of the sector in degrees
3. Area of the segment APB = Area of the sector OAPB - Area of △ OAB
=
𝜃
360∘
× 𝜋𝑟2
− area of △ OAB =
𝜃
360∘
× 𝜋𝑟2
−
𝑟2
2
𝑠𝑖𝑛𝜃
4. Area of the major sector OAQB = 𝜋𝑟2
- Area of the minor sector OAPB
5. Area of major segment AQB = 𝜋𝑟2
- Area of the minor segment APB
Area of segment of a circle = Area of the corresponding sector - Area of the corresponding triangle
S No QUESTIONS
1 If the difference between the circumference and the radius of a circle is 37 cm, then
using π = 22/7 the radius of the circle (in cm) is:
(a) 154 (b) 44 (c) 14 (d) 7
2 In the given figure AOB is a sector of circle of radius 10.5 cm. The
perimeter of the sector (in cm) is:
(a) 32 (b) 21 (c) 11 (d) 35
3 The area of the sector of a circle of radius 6 cm whose central angle is 30°.
(a) 9.42 cm2 (b) 7.42 cm2 (c) 8.42 cm2 (d) 6.42 cm2
4 If π is taken as 22/7 the distance (in meters) covered by a wheel of diameter 35 cm, in
one revolution is:
(a) 2.2 (b) 1.1 (c) 9.625 (d) 96.25
5 The circumference of a circular field is 528 cm. Then the radius will be:
(a) 84 cm (b) 64 cm (c) 55 cm (d) 45 cm
6 If the circumference and the area of a circle are numerically equal, then diameter of the
circle is:
(a) 5 (b) 2p (c) 2 (d) 4
7 The diameter of the driving wheel of a bus is 140 cm. How many revolutions per
minute must the wheel make in order to keep a speed of 66 kmph?
(a) 200 (b) 240 (c) 250 (d) 260
8 A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the
wheel:
(a) 11 m (b) 14 m (c) 12 m (d) 10 m

34
9 The inner circumference of a circular race track, 14 m wide, is 440 m. Find the radius of
the outer circle:
(a) 85 m (b) 82 m (c) 80 m (d) 84 m
10 If the radius of a circle is doubled, its area is increased by:
(a) 100% (b) 200% (c) 300% (d) 400%
11 Area of a sector of a circle of radius 36 cm is 54π cm2 . Find the length of the
corresponding arc of the sector.
12 The length of the minute hand of a clock is 5 cm. Find the area swept by the minute
hand during the time period 6:05 am to 6:40 am.
13 In a circle with centre O and radius 4 cm, and of angle 30°. Find the area of minor
sector and major sector AOB. (π = 3.14)
14 Find the area of the largest triangle that can be inscribed in a semicircle of radius r unit.
15 The cost of fencing a circular field at the rate of Rs 24 per metre is Rs 5280. The field is
to be ploughed at the rate of Rs 0.50 per m2 . Find the cost of ploughing the field
16 A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm
sweeping through an angle of 115°. Find the total area cleaned at each sweep of the
blades.
17 A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by
means of a 5 m long rope . Find (i) the area of that part of the field in which the horse
can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5
m.
18 An umbrella has 8 ribs which are equally spaced .Assuming umbrella to be a flat circle
of radius 45 cm, find the area between the two consecutive ribs of the umbrella
ANSWERS
1)D 2)A 3)A 4)B 5)A 6)D 7)C
8)B 9)D 10)C 11)3π 12) 45
5
6
cm2
13) area of minior sector=4.19cm2, area of major sector=46.1cm2 14)r2unit
15)1925 16)1254.96cm2 17)area of quadrant=19.625 m2,increase in area=58.875m2
18)
22275
28
cm2

35
Chapter 13
Surface areas and Volumes
Formulas to be used in calculating combined shapes surface area and volume
S.n
o
Shape Combinatio
n of figures
Surface area Surface area
formula
Volume Volume formula
1 Cube +
cube=cuboi
d
Total surface
area
of cuboid
2(lb+bh+lh) Volume of
cuboid
l × b × h
2 Cube +
cuboid=
cuboid
Total surface
area of cuboid
2(lb+bh+lh) Volume of
cuboid
l × b × h
3 Cylinder +
hemisphere
CSA of the
cylindrical
part + CSA
of
(2πrh+2πr2)
=2πr(h + r)
Volume of
the
cylindrical
part +
Πr2h+ (2/3)πr3
Shape Parameters Surface Area (Square
units)
Volume (Cubic
units)
Cuboid
Length = l
Breadth = b
Height = h
TSA = 2(lb + bh + lh)
LSA = 2h(l + b)
V = l × b × h
Cube Length = Breadth =
Height = l
TSA = 6l2
LSA = 4l2
V = l3
Cylinder Radius = r
Height = h
CSA = 2π × r × h
TSA = 2πr(h + r)
V = πr2h
Cone
Radius = r
Height = h
Slant Height = l
CSA = πrl
TSA = πr(l + r)
V = (1/3)πr2h
Sphere Radius = r CSA = TSA = 4πr2 V = (4/3)πr3
Hemisphere Radius = r CSA = 2πr2
TSA = 3πr2
V = (2/3)πr3

36
fixed on
ground
hemispheric
al part
volume of
hemispheric
al part
4 2x
hemisphere
+ cylinder
CSA of the
cylindrical
part + CSA
of
hemispheric
al part
2x2πr2+2πr2
=6πr2
Volume of
the
cylindrical
part +
volume of
two
hemispheric
al part
πr2h+
2x(2/3)πr3
5
fixed on ground
Cone +
cylinder
CSA of the
cylindrical
part + CSA
of
hemispheric
al part
Πrl + 2πrh
=πr(l + 2h)
Volume of
the
cylindrical
part of +
volume
hemispheric
al part
Πr2h+⅓πr2h
6 2x cone +
cylinder
2x CSA of
cone + CSA
of cylinder
2πrl+2πrh
=2πr(l+ h)
Volume of
cylinder +
2x volume of
cone +
Πr2h+[2×(⅓πr
2h )]
7 Cone +
hemisphere
s
Csa of
conical part
+ csa of
hemispheric
al part
Πrl+2πr2 Volume of
the conical
part +
volume of
hemispheric
al part
⅓πr2h +
(2/3)πr3
8 Cone +
cone = 2x
cone
2xcsa of
conical part
2x πrl 2x volume of
conical part
2x ⅓πr2h
9 Cone +
cylinder +
hemisphere
Csa of
conical part
+ csa of
cylinder
+csa of
hemispheric
al part
Πrl+2πrh+2π
r2
Volume of
conical part
+ volume of
cylinder
+volume of
hemispheric
al part
⅓πr2h
+πr2h+(2/3)πr
3
S.NO QUESTIONS
1 The volume of a cube is 2744 cm3 . Its surface area is
(a) 196 cm2 (b) 1176 cm2 (c) 784 cm2 (d) 588 cm2
2 The ratio of the total surface area to the lateral surface area of a cylinder with base radius
80 cm and height 20 cm is (a) 1 : 2 (b) 2 : 1 (c) 3 : 1 (d) 5 : 1
3 The height of a cylinder is 14 cm and its curved surface area is 264 cm2 . The volume of
the cylinder is (a) 296 cm3 (b) 396 cm3 (c) 369 cm3 (d) 503 cm3
4 The ratio of the volumes of two spheres is 8 : 27. The ratio between their surface areas is
(a) 2 : 3 (b) 4 : 27 (c) 8 : 9 (d) 4 : 9

37
5 The radii of the base of a cylinder and a cone are in the ratio 3 : 4 and their heights are in
the ratio 2 : 3, then ratio of their volumes is (a) 9 : 8 (b) 9 : 4 (c) 3 : 1 (d) 27 : 64
6 The areas of three adjacent faces of a rectangular block are in the ratio of 2 : 3 : 4 and its
volume is 9000 cu. cm, then the length of the shortest side is (a) 10 cm (b) 12 cm (c) 15
cm (d) 18 cm
7 The height of a conical tent is 14 m and its floor area is 346.5 m2 . The length of canvas, 1.1
m wide, required for it is (a) 490 m (b) 525 m (c) 665 m (d) 860 m
8 A metal sheet 27 cm long, 8 cm broad and 1 cm thick is melted into a cube. The difference
between surface areas of two solids is (a) 284 cm2 (b) 285 cm2 (c) 286 cm2 (d) 287 cm2
9 Assertion : If the areas of three adjacent faces of a cuboid are x, y, z respectively then the
volume of the cuboid is √𝑥𝑦𝑧.
Reason : Volume of a cuboid whose edges are l, b and h is lbh units.
(b) Assertion and Reason both are correct statements but Reason is not the correct
10 Assertion : From a solid cylinder, whose height is 12 cm and diameter 10 cm a conical
cavity of same height and same diameter is hollowed out. Then, volume of the cone is
2200
7
cm3 .
Reason : If a conical cavity of same height and same diameter is hollowed out from a
cylinder of height h and base radius r, then volume of the cone will be half of the volume
of the cylinder.
11 The decorative block shown in Fig. 13.7 is made of two solids — a
cube and a hemisphere. The base of the block is a cube with edge 5
cm, and the hemisphere fixed on the top has a diameter of 4.2 cm.
Find the total surface area of the block. (Take π = 22/7
12 Mayank made a bird-bath for his garden in the shape of a cylinder
with a hemispherical depression at one end (see Fig). The height of
the cylinder is 1.45 m and its radius is 30 cm. Find the toal surface
area of the bird-bath(Take π = 22/7)
13 A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius.
The total height of the toy is 15.5 cm. Find the total surface area of the toy

38
14 A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter
the hemisphere can have? Find the surface area of the solid.
15 A wooden article was made by scooping out a hemisphere from each end of a
solid cylinder, as shown in Fig. If the height of the cylinder is 10 cm, and its
base is of radius 3.5 cm, find the total surface area of the article
16 A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height
of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy.
If a right circular cylinder circumscribes the toy, find the difference of the volumes of the
cylinder and the toy. (Take π = 3.14)
17 A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is
surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the
pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14)
18 A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a
hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such
that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the
cylinder is 60 cm and its height is 180 cm.
ANSWERS
1)b 2)d 3)b 4)d 5)a 6)c 7)b
8)c 9)a 10)c 11)163.86cm2 12)3.3m2
13)214.5cm2 14)332.5cm2 15)374cm2 16)25.12cm
17)892.26Kg 18)1130400cm3

39
Chapter 14
Statistics
Mean: The arithmetic mean of a given data is the sum of the values of all the observations divided by
the total number of observations. There are two different formulas for calculating the mean for
ungrouped data and the mean for grouped data.
Type 1 (Mean of raw data)
Suppose we have n values in a set of data namely as x1, x2, x3……… x n, then the mean of data is given
by using the formula
Type –2 Mean of grouped data(If Frequency and variable are given)
To calculate the mean of grouped data we have two different methods –
i. Direct method and
ii. Assumed mean method.
The mean of grouped data deals with the
frequencies of different observations or
variables that are grouped together. If
the values of the observations are x1, x2,
x3,............xn and their corresponding
frequencies are f1, f2, f3,...........fn , then the
mean of the data is given by,
Mean, x̄ = (x1f1 + x2f2 + ... + xnfn) / (f1 +
f2 + ... + fn)
x̄ = ∑xi fi / ∑fi, where i = 1, 2, 3, 4,......n
Type- 3 Mean for grouped data. (If Frequency and Class Intervals are given)
Here are the steps that can be followed to find the mean for grouped data using the direct method,
1. Create a table containing four columns such as class interval, class marks, frequency and product of
class marks and frequency.
2. Calculate Mean by the Formula Mean i i
i
f x
x
f



Where fi is the frequency and xi is the mid value of
the class interval.
3. To calculate the mid value xi, we use this formula xi = (upper limit + lower limit)/2.
Example. The table below gives information about the percentage distribution of female employees in
a company of various branches and a number of departments. Find the mean percentage of female
employee by the assumed mean method.
Female Employee 5-15 15-25 25-35 35-45 45-55 55-65 65-75
No of Departments 1 2 4 4 7 11 6

40
Solution:
Female
Employee(C I)
No of
Departments( fi )
Class Marks
X i
di = xi - a fi.di
5-15 1 10 -30 -3
15-25 2 20 -20 -40
25-35 4 30 -10 -40
35-45 4 40 0 0
45-55 7 50 10 70
55-65 11 60 20 220
65-75 6 70 30 180
Total Σfi =35 Σfidi = 360
Assumed mean = a = 40
Mean = a+ (Σfidi /Σfi) = 40+ (360/35) = 40+(72/7) = 40 + 10.28 = 50.28
Hence, the mean percentage of female employees is 50.28.
Median
Median for raw data: The middle most term of the data arranged in ascending or descending order is
called the Median of the data. After arranging the data in ascending or descending order, the
following method is applied:
If number of values or observations in the given data is odd, then the median is given by
 
th
n 1
2

 
 
 
observation.
If the number of values or observations is even, then the median is given by the average of
2
th
n
 
 
 
and 1
2
th
n
 

 
 
observation.
Cumulative Frequency: The cumulative frequency of a class is the frequency obtained by adding the
frequencies of all the classes preceding the give class
Median for grouped data:
The median for grouped data can be calculated by using the formula,
Where
l = lower limit of median class
N = Sum of frequencies
cf = cumulative frequency of the preceding class
f = frequency of Median class
h = class size
Steps to Find Median of Grouped Data
Median of grouped data is in the form of a frequency distribution arranged in ascending order and is
continuous. To find the median of any given data is simple since the median is the middlemost value
of the data. Since the data is grouped, it is divided into class intervals. The steps to find the median of
grouped data are.
Step 1: Construct the frequency distribution table with class intervals and frequencies.

41
Step 2: Calculate the cumulative frequency of the data by adding the preceding value of the frequency
with the current value.
Step 3: Find the value of N by adding the values in frequency.
Step 4: Find the lower limit of the class interval and the cumulative frequency.
Step 5: Apply the formula for median for grouped data: Median = 2
n
cf
l h
f
 

 
 
 
Example
Q: Find the median of the following data is
Marks obtained 0-10 10-20 20-30 30-40 40-50
No of Students 5 7 4 8 6
Solution:
Marks
obtained
No of
students
Cumulative
frequency
0-10 5 5
10-20 7 5+7=12
20-30 4 12+4=16
30-40 8 16+8=24
40-50 6 24+6=30
Total N=30
Now N/2 = 30/2 = 15
15 lies in the in cumulative frequency having class intervals 20-30
20-30 is the median class
Now using formula
We have
Lower limit of median class = l = 20
Cumulative frequency preceding the median class= cf =12
Frequency of the median class =f= 4
Class size= difference of limits = h = 10
Median = 20 + (15 -12)/4 * 10
= 20 + 30/4
Median = 20 + 7.5 = 27.5
Mode
Mode: The mode for grouped data is given by the formula
Where,
l = Lower limit of the modal class
h = Class size
f1 = Frequency of the modal class
f0 = Frequency of the class preceding the modal class
f2 = Frequency of the class succeeding the modal class

42
Example
The following data gives the information on the observed life times (in hours) of 150 electrical
components. Find the mode of the data
Life time( In Hours) 0-20 20-40 40-60 60-80 80-100
Frequency 15 10 35 50 40
Solution:
The class 60-80 have the maximum frequency as 50
S.No QUESTION
1 Find the class marks of classes 10–20 and 35–55. (a) 10, 35 (b) 20, 55 (c) 15, 45 (d)
17.5, 45
2 If di = xi – 13, ∑fi di = 30 and ∑fi =120 , then mean, x is equal to
(a) 13 (b) 12.75 (c) 13.25 (d) 14.25
3 The mean of first ten odd natural numbers is (a) 5 (b) 10 (c) 20 (d) 19
4 If the mean of x, x + 3, x + 6, x + 9 and x + 12 is 10, then x equals
(a) 1 (b) 2 (c) 4 (d) 6
5 For a frequency distribution, mean, median and mode are connected by the relation
(a) Mode = 3 Mean – 2 Median (b) Mode = 2 Median – 3 Mean
(c) Mode = 3 Median – 2 Mean (d) Mode = 3 Median + 2 Mean
6 Assertion : Consider the following frequency distribution:
The modal class is 10-15.
Reason : The class having maximum frequency is called the modal class.
Class interval 10-15 15-20 20-25 25-30 30-35
frequency 5 9 12 6 8

43
7 Find the mean of the following distribution:
Class interval 0-6 6-12 12-18 18-24 24-30
frequency 7 5 10 12 2
8 Find the median of the following data :
C I 0-10 10-20 20-30 30-40 40-50 TOTAL
Frequency 8 16 36 34 6 100
9 The following table gives the literacy rate of 40 cities :
Literacy rate 30-40 40-50 50-60 60-70 70-80 80-90
No of cities 6 7 10 6 8 3
Find the modal literacy rate
10 Find the missing frequencies f1, f2 and f3 in the following frequency distribution, when
it is given that f2 : f3 = 4 : 3, and mean = 50
CI 0-20 20-40 40-60 60-80 80-100 TOTAL
Frequency 17 f1 f2 f3 19 120
11 Distance Analysis of Public Transport Buses
Transport department of a city wants to buy some
Electric buses for the city. For which they want to
analyse the distance travelled by existing public
transport buses in a day
Daily distance
travelled (in km)
200-
209
210-
219
220-
229
230-
239
240-
249
Number of buses 4 14 26 10 6
(i)Find the difference between upper limit of a class and lower limit of its succeeding
class
(ii) Find the median class
(iii) The cumulative frequency of the class preceding the median class is_____
Or Find the median of distance travelled
12 The following distribution shows the daily pocket allowance of children of a locality.
The mean pocket allowance is Rs. 18. Find the missing frequency f
Daily pocket
allowance
11-13 13-15 15-17 17-19 19-21 21-23 23-25
No of children 7 6 9 13 f 5 4

44
13 A class teacher has the following absentee record of 40 students of a class for the
whole term. Find the mean number of days a student was absent
No of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40
No of students 11 10 7 4 4 3 1
14 If the median of the distribution given below is 28.5, find the values of x and y
CI 0-10 10-20 20-30 30-40 40-50 50-60 TOTAL
FREQUENCY 5 X 20 15 Y 5 60
15 A life insurance agent found the following data for distribution of ages of 100 policy
holders. Calculate the median age, if policies are given only to persons having age 18
years onwards but less than 60 year
Age (in years) Number of policy holders
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 10
ANSWERS
1)C 2)C 3)B 4)C 5)C 6)D 7)14.5
8)27.22 9)54.29 10) f1=28,f2=32,f3=24
11)Case Study (i)1 (ii) 219.5-229.5 (iii) 18 or 224.12 12)20 13)12.48 days
14)x=8,y=7 15)35.76

45
Chapter 15
Probability
1. The theoretical (classical) probability of an event E, written as P(E), is defined as
𝐏 (𝐄) =
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 𝐟𝐚𝐯𝐨𝐮𝐫𝐚𝐛𝐥𝐞 𝐭𝐨 𝐄
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐚𝐥𝐥 𝐩𝐨𝐬𝐬𝐢𝐛𝐥𝐞 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐱𝐩𝐞𝐫𝐢𝐦𝐞𝐧𝐭
Where, we assume that the outcomes of the experiment are equally likely.
2. The probability of a sure event (or certain event) is 1.
3. The probability of an impossible event is 0.
4. The probability of an event E is a number P(E) such that 0≤P(E) 1
5. An event having only one outcome is called an elementary event. The sum of the probabilities of all
the elementary events of an experiment is 1.
6. For any event E, ( ) ( ) 1
P E P E
 
where E̅ stands for ‘not E’. E and E̅ are called complementary events.
S No QUESTIONS
1 Which of the following cannot be the probability of an event?
(A) 0.7 (B) 2 3 (C) – 1.5 (D) 15%
2 Which of the following can be the probability of an event?
(A) – 0.04 (B) 1.004 (C) 18 /23 (D) 8/ 7
3 An event is very unlikely to happen, its probability is closest to
(A) 0.0001 (B) 0.001 (C) 0.01 (D) 0.1
4 Out of one digit prime numbers, one number is selected at random. The probability of
selecting an even number is:
(A) 1/2 (B) 1/4 (C) 4/9 (D) 2/5
5 When a die is thrown, the probability of getting an odd number less than3 is:
(A) 1/6 (B) 1/3 (C) 1/2 (D) 0
6 A card is drawn from a well shuffled pack of 52 playing cards. The event E is that the
card drawn is not a face card. The number of outcomes favourable to the event E is
(A) 51 (B) 40 (C) 36 (D) 12
7 In a family of 3 children, the probability of having at least one boy is:
(A) 7/ 8 (B) 1/ 8 (C) 5/ 8 (D) 3/ 4
8 The probability of a number selected at random from the numbers 1, 2, 3, .... 15 is a
multiple of 4 is: (A) 4/15 (B) 2/15 (C) 1/5 (D) 1/3
9 The probability that a non-leap year selected at random will contains 53 Mondays is:
(A) 1/7 (B) 2/7 (C) 3/7 (D) 5/7
10 A bag contains 6 red and 5 blue balls. One ball is drawn at random. The probability that
the ball is blue is: (A) 2/11 (B) 5/6 (C) 5/11 (D) 6/11
11 Two dice, one blue and one grey, are thrown at the same time. Write down all the
possible outcomes. What is the probability that the sum of the two numbers appearing
on the top of the dice is (i) 8? (ii) 13 (iii) less than or equal to 12?
12 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just
look at a pen and tell whether or not it is defective. One pen is taken out at random from
this lot. Determine the probability that the pen taken out is a good one

46
13 A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at
random from the box, find the probability that it bears (i) a two-digit number (ii) a
perfect square number (iii) a number divisible by 5
14 A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri
will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one
pen at random and gives it to her. What is the probability that (i) She will buy it ? (ii)
She will not buy it?
15 A letter of English alphabet is chosen at random. Determine the probability that the
chosen letter is a consonant
16 A card is drawn from a well-shuffled deck of 52 playing cards. Then what is the
probability that the card will not be a diamond?
17 The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18.
What is the number of rotten apples in the heap?
18 Cards bearing numbers 2 to 21 are placed in a bag and mixed thoroughly. A card is
taken out of the bag at random. What is the probability that the number on the card
taken out is an even number?
19 A card is drawn out from a well-shuffled deck of 52 cards. What is the probability of
getting a red queen?
20 Two different dice are tossed together. Find the probability that (i) the number on each
dice is odd, and (ii) the sum on the numbers, appearing on the two dice, is 5.
21 Rahim tosses two different coins simulta-neously. Find the probability of getting at least
one tail.
22 Two different dice are rolled simultaneously. Find the probability that the sum of the
numbers appearing on the two dice is 10.
23 Two dice are thrown simultaneously. Find the probability that the sum of the numbers
appearing on the two dice is more than 9.
24 . A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability
that the card is neither a red card nor a jack.
25 A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at
random from the box, then find the probability that it will be : (i) a blue card, (ii) not a
yellow card, and (iii) neither yellow nor a blue card.
ANSWERS
1)C 2)C 3)A 4)B 5)A 6)B 7)A
8)C 9)A 10)C 11)5/36,0,1 12)11/12 13) 9/10,1/10,1/5
14)31/36 ,5/16 15)21/26 16)3/4 17)162 18)1/2 19)1/26
20)1/4,1/9 21)3/4 22)1/12 23)1/6 24)6/13 25)1/7,3/7,2/7

47
Class - X Session 2022-23
Subject - Mathematics (Standard - 041)
Sample Question Paper 1
Time Allowed: 3 Hours Maximum Marks: 80
General Instructions:
1. This Question Paper has 5 Sections A, B, C, D, and E.
2. Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
3. Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
4. Section C has 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
5. Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
6. Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the
values of 1, 1 and 2 marks each respectively.
7. All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and
2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks
questions of Section E.
8. Draw neat figures wherever required. Take π =22/7 wherever required if not stated.
SECTION A
SECTION A CONSIST OF 20 QUESTIONS OF 1 MARKS EACH
S.NO. MARKS
1. The smallest number by which √27 should be multiplied so as to get a rational
number is
(a) √27 (b) 3√3 (c) √3 (d) 3
1
2. The ratio of LCM and HCF of the least composite number and least prime
number is
(a) 1: 2(𝑏) 2 ∶ 1 (𝑐) 1 ∶ 1 (𝑑) 1 ∶ 3
1
3. If P and Q are zeroes of 3𝑋2
+2X-9 then the value of P-Q is
(a) −3 (𝑏) −
2
3
(𝑐)
4√7
3
(𝑑)None of these
1
4. The value of K for which the system of equations 2X + 3 Y =5 and 4X+KY =10
has infinite number of solution is
(a) 1 (𝑏) 3 (𝑐) 6 (𝑑) 0
1
5. If one root of the equation 𝑎𝑥2
+ b x+ c =0 is three times the other, then 𝑏2
∶ 𝑎𝑐
(a) 3 : 1 (b) 3 : 16 (c) 16 :3 (d) 16 : 1
1
6. The ratio in which x- axis divides the line segment joining (3 ,6 ) and (12 , -3 ) is
(a) 2 : 1 (b) 1 : 2 (c) 3 :2 (d) 1 : 3
1
7. In ∆ ABC , D and E are the points on sides AB and AC
respectively such that
DE ∥ 𝐵𝐶 And
𝐴𝐷
𝐷𝐵
=
3
1
if EA= 4.2 c m then AC is equal to
(a) 3.3 𝑐 𝑚 (𝑏)12.6 𝑐 𝑚 (𝑐)5.6 𝑐 𝑚 (𝑑) 4.2 𝑐 𝑚
1
8. Which of the following is true for the following two triangles? 1

48
(a) ∆ 𝐴𝐵𝐶 ~ ∆ 𝐷𝐸𝐹 (𝑏) ∆ 𝐴𝐵𝐶~ 𝐷𝐹𝐸 (𝑐) ∆ 𝐵𝐴𝐶~∆ 𝐷𝐸𝐹 (𝑑)∆𝐴𝐵𝐶 𝑖𝑠 𝑛𝑜𝑡 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜∆ 𝐷𝐸
9. 𝐼𝑛 𝑡ℎ𝑖𝑠 𝑓𝑖𝑔𝑢𝑟𝑒 𝑡𝑤𝑜 𝑙𝑖𝑛𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝐴𝐶 𝑎𝑛𝑑 𝐵𝐷 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟 𝑎𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡
𝑃 such that
𝑃𝐴 = 6 𝑐 𝑚, 𝑃𝐵 = 3𝑐 𝑚, 𝑃𝐶 = 2.5 𝑐 𝑚 𝑃𝐷 = 5𝑐𝑚 , ∠𝐴𝑃𝐵 = 50° 𝑎𝑛𝑑
∠𝐶𝐷𝑃 = 30° 𝑡ℎ𝑒𝑛 ∠ 𝑃𝐵𝐴 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜
(a)500 (b)300 (c) 600 (d)1000
1
10. If 7𝑐𝑜𝑠30°
+ 5 𝑡𝑎𝑛 𝑡𝑎𝑛300
+ 6 𝑐𝑜𝑡 𝑐𝑜𝑡 𝑖𝑠60°
is
(a)
43
2√3
(b)
41√3
2
(c)
47
2√3
(d)
49√3
2
1
11. If a cos𝜃 + 𝑏𝑠𝑖𝑛 𝜃 =12 and a sin𝜃 − 𝑏𝑐𝑜𝑠𝜃=5 then 𝑎2
+ 𝑏2
is equal to
(a) 13 (𝑏)12 (𝑐)14 4 (𝑑)169
1
12. A bag contains 24 balls of which x are red and 2x are white and 3x are blue, a
ball is selected at random the probability that it is white ball is
(a)
1
3
(b)
5
6
(c)
3
7
(d)
7
24
1
13.
If sin𝜃 =
𝑎2−𝑏2
𝑎2+𝑏2
then the value of tan𝜃 is
(a)
𝑎2+𝑏2
2𝑎𝑏
(𝑏)
𝑎2−𝑏2
2𝑎𝑏
(𝑐)
2𝑎𝑏
𝑎2−𝑏2
(𝑑)
2𝑎𝑏
𝑎2+𝑏2
1
14. The length of minute hand of a clock is 14 c m find the area swept by the minute
hand in 5 minutes.
(a)
154
3
𝑐𝑚2
(b)
160
3
𝑐𝑚2
(c)
154
6
𝑐𝑚2
(d)
181
6
𝑐𝑚2
1
15. Mode for the following distribution is 22 if x< 𝑦 < 10 𝑡ℎ𝑒𝑛 the value of y is
C I 0 -10 10 -20 20-30 30-40 40-50 Total
Frequency 5 8 10 x y 30
(a) 2 (b) 5 (c) 3 (d) 4
1
16. If for a data mean : median = 9 : 8 then median : mode =
(a) 8 : 9 (b) 4 :3 (c) 7 :6 (d) 5 : 4
1
17. Find the area of quadrant of a circle whose circumference is 22 cm.
(a)
22
8
𝑐𝑚2(𝑏)
77
8
𝑐𝑚2(𝑐)
77
22
𝑐𝑚2(𝑑)
8
77
𝑐𝑚2
1
18. If the perimeter of the bases of two right circular cone are in the ratio 3:4 and
their volumes are in the ratio 9:32 then the ratio of their heights is-
(a) 1:3 (b) 2:1 (c) 1:2 (d) 1:3
1
Direction for questions 19 & 20: In question numbers 19 and 20, a
statement of Assertion (A) is followed by a statement of Reason (R). Choose
the correct option.

49
19. Assertion: If HCF (90,144)= 18 then LCM(90,144) = 720
Reason: HCF (a,b) x LCM (a,b)= axb
(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
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
1
20. Assertion: if the coordinate of the midpoints of sides AB and AC of triangle ABC
are D(3,5) and E(-3,-3) respectively then BC= 20 units
Reason:
𝑇ℎ𝑒 𝑙𝑖𝑛𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑗𝑜𝑖𝑛𝑖𝑛𝑔 𝑡ℎ𝑒 𝑚𝑖𝑑 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑡𝑤𝑜 𝑠𝑖𝑑𝑒𝑠 𝑜𝑓 𝑎 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑠𝑖𝑑𝑒
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct
1
Section B
Section B consists of 5 questions of 2 marks each.
21. Solve for X and Y: 8X+5Y=9 3X +2Y=4 2
22. Diagonal AC and BD of trapezium ABCD with AB∥ 𝐷𝐶 intersect each other at the
point O. using a similarity criterion for two triangles.
Show that
𝐴𝑂
𝐶𝑂
=
𝐵𝑂
𝐷𝑂
2
23. Prove that the tangents drawn at the ends of a diameter of a circle are parallel. 2
24. If cot𝜃 =
7
8
then evaluate
(1+𝑠𝑖𝑛𝜃) (1−𝑠𝑖𝑛𝜃)
(1+𝑐𝑜𝑠𝜃)(1−𝑐𝑜𝑠𝜃)
OR
Evaluate
5𝑐𝑜𝑠2
600
+ 4𝑠𝑒𝑐2
300
− 𝑡𝑎𝑛2
450
𝑠𝑖𝑛2300 + 𝑐𝑜𝑠2300
2
25. In a circle of radius 21 c m. an arc subtends an angle of
600
at the centre. find
(a) Length of an arc
(b)Area of sector
OR
Find the area of shaded region (in this figure ) if radii of
two concentric circles with centre o. are 7 c m and 14
cm respectively and ∠𝐴𝑂𝐶 = 400
2
Section C
Section C consists of 6 questions of 3 marks each.
26. Prove that7√2 is an irrational number. 3
27. Find the zeroes of the polynomial 𝑋2
+
1
6
X -2 and verify the relationship between
the zeroes and the coefficient.
3
28. If the equation (1+𝑚2
) 𝑥2
+2mcx + 𝑐2
- 𝑎2
= 0 has equal roots. Show that 𝑐2
=
𝑎2
(1 + 𝑚2
)
OR
Find two consecutive positive integers. Sum of whose squares is 365.
3

50
29. Two tangents TP and TQ are drawn to a circle with centre o from external point
T. prove that ∠𝑃𝑇𝑄 = 2∠𝑂𝑃𝑄
OR
In this figure XY and 𝑋1
𝑌1
are two parallel tangents to a circle with centre o.
and another tangent AB with point of contact C intersect XY at A and 𝑋1
𝑌1
at B
prove that ∠𝐴𝑂𝐵 = 900
3
30.
Prove that : √
1+𝑠𝑖𝑛𝐴
1−𝑠𝑖𝑛𝐴
= 𝑠𝑒𝑐𝐴 + 𝑡𝑎𝑛𝐴
3
31. A die is thrown twice. What is the probability that
(a ) 5 will come up at least once .
(b ) 5 will not come up either time .
3
Section D
Section D consists of 4 questions of 5 marks each.
32. State and prove that the Basic proportional theorem.
Using this theorem ( in this figure ) If DE∥ OQ and DF ∥ OR then show that 𝐸𝐹 ∥ 𝑄𝑅
5
33. The median of the following data is 525. find the value of X and Y. if total frequency
is 100.
C I Frequency
0 - 100 2
100 - 200 5
200 - 300 X
300 - 400 12
400 - 500 17
500 - 600 20
600 - 700 Y
700 - 800 9
800 - 900 7
900 - 1000 4
5

51
34. RS 9000 were divided equally among a certain number of persons. Had there been
20 more persons each would have got RS 160 less. Find the original number of
persons.
OR
Two water taps together can fill the tank in 9
3
8
hrs. The tap of larger diameter takes
10 hrs less the smaller one to fill the tank separately. Find the time in which each
tap can separately fill the tank.
5
35. A tent is in the shape of cylinder surmounted by a conical top. If the height and
diameter of the cylindrical part are 2.1 m and 4 m respectively and slant height of
top is 2.8 m. find the area of canvas used for making the tent .Also find the cost of
the canvas of tent at the rate of Rs 500 per 𝑚2
.
OR
A Gulabjamun contains sugar syrup upto about 30% of its volume. Find
approximately how much syrup would be found in 45 gulabjamuns, each shaped
like a cylinder with two hemi spherical ends with length 5cm and diameter 2.8 cm.
5
Section E
Case study based questions are compulsory.
36 As observed from the top of a 60 M high light house from the sea level the angle of
depression of two ships are 280
and 45°
one ship exactly behind the other on the
same side of the light house . (use tan 28°
= 0.5317)
(i) Find the distance between the two ships.
(ii) Find the distance between the foot of the light house and first ship from the
light house .
(iii)Find the distance from the top of light house (A) to the first ship(D) .
OR
Find the value of cosec B
1
1
2
37. A triangular plot is marked with parallel and vertical lines. A
pair of such lines are taken as x- axis and y- axis respectively
as shown in graph P ,Q ,R are mid points of AB, BC and CA
respectively where three pillars for makings were erected in
the ground and are joined with ropes.
(i) Find coordinates of mid point of rope PQ .
(ii) Find coordinates of R.
(iii) Find the ratio in which BC is divided by y-axis.
OR
Find the ratio in which y-axis divides AC .
1
1
2

52
38. Rishi wants to buy a car and plans to take loan from a bank to
buy the car. He pays his total loan of Rs 1180000 by paying
every month starting with the first instalment of RS 10000. If he
increases the instalment by RS 1000 every month answer the
following.
(i) Find the amount paid by Rishi in 30𝑡ℎ
instalment.
(ii) Find the amount paid by Rishi in 30 instalment.
(iii)If the loan is to be repaid in 40 instalments, thenfind the amount paid in the
last instalment .
OR
Find the ratio of the first instalment to the last instalment.
1
1
2

53
MARKING SCHEME Sample Paper 1
Q.No SECTION – A Marks
1. (c ) √3 1
2. (b) 2 : 1 1
3. (c )
4√7
3
1
4. (c) k =6 1
5. (c)
16
3
1
6. (a) 2:1 1
7. ( c) AC =5.6 1
8. (d) ΔABC is not similar to ΔDEF 1
9. (d) 100֯ 1
10. (a)
43
√3
2
1
11. (d) 169 1
12. (a)
1
3
1
13. (b)
𝑎2 −𝑏2
2𝑎𝑏
1
14. (a)
154
3
𝑐𝑚2 1
15. (b) y = 5 1
16. (b)4:3 1
17. (b)
77
8
1
18. (c) 1:2 1
19. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct
1
20 (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the
correct explanation of Assertion (A).
1
21. 8x +5y = 9
3x + 2y = 4
X = -2
And y =
7
5
1
1
22. In ΔAOB and ΔDOC
∠1 = ∠2(alternate interior)
∠3 = ∠4 ( alternate interior)
Δ AOB ~ Δ COD (by aa)
𝐴𝑂
𝐶𝑂
=
𝑂𝐵
𝑂𝐷
1
1
23. ∠1 + ∠2 = 180֯ so l ∥ 𝑚 2
24. Sin 𝜃=
8
√113
, Cos 𝜃=
7
√113
(1 +
8
√113
) (1 -
8
√113
)
(1 +
7
√113
) (1 -
7
√113
)
49
113
64
113
=
49
64
OR
1
1
1
1

54
5
12
2
+ 4
22
√3
− 12
12
2
+
√3
2
2
5
4
+
16
3
−1
1
4
+
3
4
=
67
12
1+1
25. (i)Length =
𝜋𝑟𝜃
180
= 22 cm
(ii) Area of sector=
𝜃
360
𝜋𝑟2
11 x 21= 231 𝑐𝑚2
OR
Area of shaded region =
𝜃
360
𝜋(𝑅2
− 𝑟2
)
308
3
𝑐𝑚2
1
1
1
1
26. Assume, √2is a rational number, it can be written as p/q, in which p and q are
co-prime integers and q≠0,
i.e. √2= p/q. where, p and q are coprime numbers, and q≠0.
On squaring both sides of the above equation;
2= (p/q)2
2 = p2/q2
2q2 = p2 ...(i)
p2 is a multiple of 2 so p is a multiple of 2 ...(ii)
Since, p is a multiple of two.
So p = 2m
p² = 4m² …(iii)
Using equation(i) into the equation (iii), we get;
2q² = 4m²
So q² = 2m²
q2 is a multiple of 2 so q is a multiple of 2 ...(iv)
Equation (ii) and (iv), implies that p and q have a common factor 2.
It contradicts the fact that they are co-primes which lead from our wrong
assumption that 2is a rational number.
Multiplication of rational and irrational number is always irrational number so
7 √2 is an irrational number.
1
1
1
27. 1
6
{ 6𝑥2
+ 𝑥 − 12} = 0
(3x -4) (2x+3)
X =
4
3
or−
3
2
verify
𝛼 + 𝛽=
4
3
−
3
2
= −
1
6
∝ 𝛽 = -2
1
1
1
28. 𝑏2
- 4ac = 0
2𝑚𝑐2
– 4 (1 + 𝑚)2
(𝑐2
- 𝑎2
)= 0
𝑐2
= 𝑎2
( 1 + 𝑚2
)
OR
𝑥2
+ (𝑥 + 1)2
= 365
𝑥2
+ 𝑥 − 182 = 0
1
1
1
1

55
(x+14) (x-13) = 0
So x = 13
1 positive integer = 13
2 positive integer = 14
1
1
29. Consider the problem
Let us join point O to C
In ΔOPA and ΔOCA
OP=OC (Radii of the same circle)
AP=AC (Tangent from point A)
AO=AO (Common side)
ΔOPA≅ΔOCA (SSS congruence criterion)
Therefore, P↔C,A↔A,O↔O
∠POA=∠COA.........(1)
Similarly,
∠QOB≅∠OCB
∠QOB=∠COB.........(2)
Since, POQ is a diameter of the circle, it is a straight line.
Therefore, ∠POA+∠COA+∠COB+∠QOB=180°
So, from equation (1) and equation (2)
2∠COA+2∠COB=180°∠COA+∠COB=90°∠AOB=90°
OR
We know that length of tangents drawn from an external point to a circle are
equal
∴ TP=TQ−−−(1)
4∴ ∠TQP=∠TPQ (angles of equal sides are equal)−−−(2)
Now, PT is tangent and OP is radius.
∴ OP⊥TP (Tangent at any point of circle is perpendicular to the radius through
point of cant act)
∴ ∠OPT=90°
or, ∠OPQ+∠TPQ=90°
or, ∠TPQ=90°−∠OPQ−−−(3)
In △PTQ
∠TPQ+∠PQT+∠QTP=180° (∴ Sum of angles triangle is 180°)
or, 90°−∠OPQ+∠TPQ+∠QTP=180°
or, 2(90°−∠OPQ)+∠QTP=180° [from (2) and (3)]
or, 180°−2∠OPQ+∠PTQ=180°
∴ 2∠OPQ=∠PTQ−−−− proved
1
1
1
1
1
1
30.
LHS √
𝟏+𝒔𝒊𝒏𝑨
𝟏−𝒔𝒊𝒏𝑨
√
𝟏 + 𝒔𝒊𝒏𝑨( 𝟏 + 𝒔𝒊𝒏𝑨)
𝟏 − 𝒔𝒊𝒏𝑨( 𝟏 + 𝒔𝒊𝒏𝑨)
= √
(𝟏 + 𝒔𝒊𝒏𝑨)𝟐
𝟏 − 𝒔𝒊𝒏𝟐𝑨
1

56
=√
(𝟏+𝒔𝒊𝒏𝑨)𝟐
𝒄𝒐𝒔𝟐 𝑨
=
𝟏+𝒔𝒊𝒏𝑨
𝒄𝒐𝒔𝑨
=
𝒔𝒊𝒏𝑨
𝒄𝒐𝒔𝑨
++ +
𝟏
𝒄𝒐𝒔𝑨
= tan A + sec A = RHS
1
1
31. Throwing a die twice and throwing two dice simultaneously are treated as the
same experiment.
Sample space ={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
i) P(5 will not come up either time)= Number of times 5 does not show
divided by total number of outcomes
P(5 will not come up either time)=
25
36
ii) P(5 will come up at least once)= Number of times 5 shows up at least once
divided by total number of outcomes
P(5 will come up at least once)=
11
36
1
1
1
32. Computation of Median
Class interval Frequency
(f)
Cumulative frequency (cf)
0-100 2 2
100-200 5 7
200-300 x 7+x
300-400 12 19+x
400-500 17 36+x
500-600 20 56+x
600-700 y 56+x+y
700-800 9 65+x+y
800-900 7 72+x + y
900-1000 4 76+x + y
Total = 100
We have,
N=∑fi=100
⇒76+x+y=100⇒x+y=24
It is given that the median is 525. Clearly, it lies in the class 500−600
∴l=500, h=100,f=20,F=36+x and N=100
Now,Median= l +
𝑛
2
−𝑐𝑓
𝑓
×h
⇒525=500+
100
2
−36−𝑥
20
×100
⇒525−500=(14−x)×5
⇒25=70−5x⇒5x=45⇒x=9
Putting x=9 in x + y=24, we get y=15.
Hence, x=9 and y=15.
1
1
1
1
1
33. Given, to prove, figure
Correct proof of bpt theorem
In ΔPOQ, we have
1
2

57
DE∥ OQ(Given)
By Basic proportionality Theorem, we have
𝑬 𝑸
𝑷𝑬
=
𝑭𝑹
𝑷𝑭
(i)
Similarly,
In ΔPOR, we have
DF∥ OR(Given)
∴
𝑫𝑶
𝑷𝑫
=
𝑭𝑹
𝑷𝑭
(ii)
Now, from (i) and (ii), we have
⇒
𝑫𝑶
𝑷𝑫
=
𝑬𝑸
𝑷𝑬
𝒔𝒐EF∥QR
[Applying the converse of Basic proportionality Theorem in ΔPQR]
2
34. Let there be n persons and each get p rupees
Hence, p=
9000
𝑛
9000
𝑛
-
9000
𝑛+20
= 160
𝑛2
+20n−1125=0
𝑛2
+45n−25n−1125=0
(n+45)(n−25)=0
n=25,−45
Thus, number of persons are 25
OR
It is given that the tank is filled in 875 hours that is, the taps fill 758 part of the
tank in 1 hour. Then,
1
𝑥
+
1
𝑥+10
=
8
75
4𝑥2
−115x+375=0
(4x−15)(x−25)=0
4x−15=0
x=415
Or,
x−25=0
x=25
When x=415, then, x−10=415−10
=415−40
=−425
This cannot be possible because time can never be negative.
When x=25, then,
x−10=25−10
x=25
Therefore, the tap of smaller diameter can separately fill the tank in 25 hours.
1
2
1
1
2
1
1
1
35. CSA of tent = CSA of cylinder+ CSA of cone
= 2𝜋𝑟ℎ + 𝜋rl
=𝜋𝑟(2ℎ + 𝑙)
=
22
7
× 2(4.2 + 2.8)
= 44𝑚2
= cost of canvas= 44× 500 = 𝑅𝑆22000
OR
Diameter of cylinder = Diameter of hemisphere =2.8 cm
Radius = 1.4 cm, height of cylinder= 2.2 cm
Volume of gulabjamun = volume of cylinder +2(𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒)
1
2
1
1
1
2

58
Volume of 1 gulabjamun = 𝜋𝑟2
h+
4
3
𝜋𝑟3
=
22
7
× 1.4 × 1.4 (2.2 +
4
3
× 1.4)
= 25.06𝑐𝑚3
Volume of 45 gulabjamun = 25.06× 45 = 1127.25𝑐𝑚3
30 percentage volume of sugar syrup = 1127.25×
30
100
=338𝑐𝑚3
1
1
36. 1.) 52.85M 2. ) 60 M 3. )60√2 M
OR
Cosec B = 0
1+1
2
37 1) (1 , 4 )
2) Coordinate of point R are ( 4 , - 3 )
3 ) Ratio 1: 5
OR
Ratio 1 : 5
1
1
2
38 1) 39000
2) 735000
3 )49000
OR
10 : 49
1
1
2

59
KENDRIYA VIDYALAYA SANGHTHAN, JAIPUR REGION
Class - X Session 2022-23
Subject - Mathematics STANDARD (041)
SAMPLE PAPER 2
Time Allowed: 3 Hours Maximum Marks: 80
General Instructions: As per the previous question paper
Section A
Section A consists of 20 questions of 1 mark each.
S No. Marks
1 If one zero of the polynomial f(x) = (k2 + 4)x2 + 13x +4k is
reciprocal of the other, then k
(a)2 (b)-2 (c)1 (d)-1
1
2 If the LCM of two prime number p and q (p>q) is 221 then the value
of 3p – q is
(a) 4 (b) 28 (c) 38 (d) 48
1
3 If 217x + 131y = 913 and 131x + 217y = 827, then x + y is equal to
(a)5 (b)6
(c)7 (d)8
1
4 If 2 is a root of the equation x2 + bx + 12 = 0 and the equation x2 +
bx + q = 0 has equal roots, then q is
(a)8 (b)-8 (c)16 (d) -16
1
5 If ABC and DEF are similar triangles such that A = 470 and E =
830, then C is
(a) 500(b)600 (c) 700 (d) 800
1
6 In adjoining figure, the value of x for
which DE ll BC is
(a)4 (b)1
(c)3 (d)2
1
7 A vertical stick 20 m long casts a shadow 10 m long on the ground.
At the same, a tower casts a shadow 50 m long on the ground. The
height of the tower is
(a)100 m (b)120 m
(c) 25 m (d) 200 m
1
8 If the centroid of the triangle formed by the points (3, -5), (-7, 4),
(10,-k) is at the point (k, -1), then k is
(a)3 (b)1
(c)2 (d) 4
1
9 If x = 2sin2 and y = 2cos2 +1, then x + y is equal to
(a)3 (b)2
(c)1 (d)4
1
10 If 𝑘 + 1 = sec2
𝜃(1 + sin 𝜃)(1 − sin 𝜃), 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑘 𝑖𝑠 1

60
(a)1 (b)2 (c) 0 (d)-1
11 If  is an acute angle such that tan2 =
8
7
, then the value of
(1+𝑠𝑖𝑛𝜃)(1−𝑠𝑖𝑛𝜃)
(1+𝑐𝑜𝑠𝜃)(1−𝑐𝑜𝑠𝜃)
is
(a)
7
8
(b)
8
7
(c)
7
4
(d)
64
49
1
12 In adjoining figure, APB is a tangent to a
circle with centre O at point P. if QPB =
500, then the measure of POQ is
(a)1000 (b) 1400
(c) 1200 (d)1500
1
13 The minute hand of a clock is 12 cm long. The area swept by minute
hand in 35 minutes is
(a) 156 cm2 (b)264 cm2
(c)164 cm2 (d) 120 cm2
1
14 If an arc of a circle of radius 14 cm subtends an angle of 45o at the
centre of the circle, then its is
(a) a minor arc of length of 5.5 cm (b)a major arc of length 77 cm
(c) a major arc of length 38.5 cm (d) a minor arc of length 11cm
1
15 The maximum volume of a cone that can be carved out of a solid
hemisphere of radius r is
(a)3𝜋r2 (b)
𝜋𝑟3
3
(c)3𝜋𝑟3 (d)
𝜋𝑟2
3
1
16 Two dice are thrown together. The probability of getting the
difference of numbers on their upper faces equal to 2, is
(a)
5
9
(b)
4
9
(c)
1
3
(d)
2
9
1
17 If the sum of 15 observations of a data is (434 + x) and the mean of
the observation is x, then x is equal to
(a)25 (b)27
(c)31 (d) 33
1
18 If the difference of mode and median of a data is 24, then the
difference of median and mean is
(a)12 (b)24
(c)8 (d) 36
1
Direction for question 19 & 20: In question numbers 19 and 20, a
statement of Assertion (A) is followed by a statement of reason(R).
Choose the correct option.
19 Assertion: If HCF(a, b) = 4 and ab = 96 x 404, then LCM(a, b) = 9696
Reason: LCM of two number a and b = HCF(a, b) x ab
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is
not the correct explanation of Assertion (A).
1

61
20 Assertion: The perimeter of ∆OAB where O is origin, A(3,0), B(0, 4)
is 7 units.
Reason: Perimeter of a triangle is the sum of all three sides of the
triangle.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not
the correct explanation of Assertion (A).
1
Section B
Section B consists of 5 questions of 2 marks each.
21 For what value of k, will the following system of equation has no
solution
(3k + 1)x + 3y – 2 = 0 (k2 + 1)x + (k - 2)y – 5 = 0
2
22 In the adjoining figure,
𝑄𝑅
𝑄𝑆
=
𝑄𝑇
𝑃𝑅
and PQR =
PRQ. Prove that
∆𝑃𝑄𝑆 ~ ∆𝑇𝑄𝑅.
2
23 Two tangents PA and PB are drawn to a circle with centre O from an
external point P. Prove that APB = 2 OAB
2
24 The perimeter of a sector of a circle of radius 5.8 m is 27.2 m. Find
the area of sector.
OR
Area of a sector of central angle 200o of a circle is 770 cm2. Find the
length of the corresponding arc of this sector.
2
25 Given that cos(A + B) = cosAcosB – sinAsinB
Find the value of cos1050
OR
If sin𝜃 + sin2𝜃 = 1, then prove that cos2𝜃 + cos4𝜃= 1
2
Section C
Section C consists of 6 questions of 3 marks each.
26 Prove that √7 is an irrational number. 3

62
27 Find the value of k such that the polynomial x2 – (k + 6)x + 2(2k +
1) has sum of its zeroes equal to half of their product.
3
28 The area of a rectangle gets reduced by 9 sq units if its length is
reduced by 5 units and breadth is increased by 3 units. If we
increase the length by 3 units and breadth by 2 units the area
increased by 67 units. Find the dimensions of the rectangle?
OR
The sum of numerator and denominator of a fraction is 4 more than
twice the numerator. If the numerator and denominator are
increased by 3 they are in the ratio 2:3. Find the fraction.
3
29 Prove that opposite sides of a quadrilateral circumscribing a circle
subtend supplementary angles at the centre of the circle.
OR
A circle is touching the side BC of ∆ABC at P and touching AB and AC
produced at Q and R respectively. Prove that
AQ =
1
2
(Perimeter of ∆ABC)
3
30 If tan𝜃 + sin𝜃 = m and tan𝜃 – sin𝜃 = n show that m2 – n2 = 4√𝑚𝑛 3
31 A card is drawn at random from a well shuffled deck of playing
cards. Find the probability the card drawn is
(a) A card of spade and an ace
(b)Either a king or a queen
(c) A red king
3
Section D
Section D consists of 4 questions of 5 marks each.
32 A fast train takes 3 hours less than a slow train for a journey of 600
km. If the speed of the slow train is 10 km/hr less than that of the
fast train, find the speeds of the two train.
OR
A two digit number is such that the product of its digit is 14. If 45 is
added to the number, the digits interchange their places. Find the
number.
5
33 Sides AB and AC and median AD of a triangle ABC are respectively
proportional to sides PQ and PR and median PM of another triangle
PQR. Show that Δ ABC ~ Δ PQR.
5
34 The interior of a building is in the form of a right circular cylinder of
diameter 4.2 m and height 4 m, surmounted by a cone. the vertical
height of the cone is 2.4 m. find the outer surface area and the
volume of the building.
OR
A solid is in the form of a cylinder with hemispherical ends, the total
height of the solid is 19 cm and diameter of the cylinder is 7 cm. Find
the volume and total surface area of solid.
5
35 If the mean of the following frequency distribution is 188. Find the
missing frequency f1 and f2
Classes 0-80 80-
160
160-240 240-
320
320-
400
Total
frequency 20 25 f1 f2 10 100
5

Here are the answers to the questions on polynomials from Chapter 2:1. (b) -3, -22. (a) 6 3. (a) x2 - 6x - 4 = 04. (a) m, m + 35. x2 - 5x + 66. The zeroes of the polynomial f(x) = x2 - 5x + 6 are:(a) 3, 2 (b) 2, 3 (c) -3, -2 (d) -2, -36. (c) -3, -2

Here are the answers to the questions on polynomials from Chapter 2:1. (b) -3, -22. (a) 6 3. (a) x2 - 6x - 4 = 04. (a) m, m + 35. x2 - 5x + 66. The zeroes of the polynomial f(x) = x2 - 5x + 6 are:(a) 3, 2 (b) 2, 3 (c) -3, -2 (d) -2, -36. (c) -3, -2

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Similar to Here are the answers to the questions on polynomials from Chapter 2:1. (b) -3, -22. (a) 6 3. (a) x2 - 6x - 4 = 04. (a) m, m + 35. x2 - 5x + 66. The zeroes of the polynomial f(x) = x2 - 5x + 6 are:(a) 3, 2 (b) 2, 3 (c) -3, -2 (d) -2, -36. (c) -3, -2 (20)

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Here are the answers to the questions on polynomials from Chapter 2:1. (b) -3, -22. (a) 6 3. (a) x2 - 6x - 4 = 04. (a) m, m + 35. x2 - 5x + 66. The zeroes of the polynomial f(x) = x2 - 5x + 6 are:(a) 3, 2 (b) 2, 3 (c) -3, -2 (d) -2, -36. (c) -3, -2