29
CBSE Sample Paper Mathematics Standard Class X (Term I)
CBSE
QUESTION BANK
Case Study Based Questions
Real Numbers
(iv) 7 ´ 11 ´ 13 ´ 15 + 15 is a
(a) Prime number
(b) Composite number
(c) Neither prime nor composite
(d) None of the above
1. To enhance the reading skills of grade X
students, the school nominates you and two
of your friends to set up a class library.
There are two sections- Section A and
Section B of grade X. There are 32 students in
section A and 36 students in section B.
(v) If p and q are positive integers such that
p = ab 2 and q = a2b, where a and b are
prime numbers, then the LCM ( p , q) is
(a) ab
(c) a 3 b 2
(b) a 2 b 2
(d) a 3 b 3
Sol.
(i) What is the minimum number of books
you will acquire for the class library, so
that they can be distributed equally
among students of Section A or Section B?
(a) 144
(c) 288
(b) 128
(d) 272
(ii) If the product of two positive integers is
equal to the product of their HCF and
LCM is true, then the HCF (32, 36) is
(b) 4
(c) 6
(d) 8
(iii) 36 can be expressed as a product of its
primes as
(a) 22 ´ 3 2
(b) 21 ´ 3 3
(c) 23 ´ 31
(d) 20 ´ 3 0
(iv) (b) 7 ´ 11 ´ 13 ´ 15 + 15 = 15 ´ (7 ´ 11 ´ 13 + 1)
Since, the number is divisible by a number other than
itself and 1.
Hence, it is a composite number.
(v) (b) Given, p = ab 2 and q = a2 b
LCM ( p, q ) = Product of the greatest power of each
prime factor involved in the numbers, with highest
power = a2 ´ b 2
CBSE QUESTION BANK
(a) 2
(i) (c) Given, number of students in Section A = 32
Number of students in Section B = 36
The minimum number of books acquire for the class
library = LCM of (32, 36)
= 2 ´2 ´2 ´2 ´2 ´ 3 ´ 3
= 2 5 ´ 32
= 32 ´ 9 = 288
(ii) (b) Given, product of the two numbers
= LCM ´ HCF
\
32 ´ 36 = LCM (32, 36) ´ HCF (32, 36)
Þ
32 ´ 36 = 288 ´ HCF (32, 36)
32 ´ 36
=4
Þ HCF (32, 36) =
288
(iii) (a) The prime factors of 36 are
36 = 2 ´ 2 ´ 3 ´ 3 = 2 2 ´ 32
30
CBSE Sample Paper Mathematics Standard Class X (Term I)
2. A seminar is being conducted by an
Educational Organisation, where the
participants will be educators of different
subjects. The number of participants in
Hindi, English and Mathematics are 60, 84
and 108 respectively.
(iii) (a) LCM of (60, 84, 108) = Product of the greatest
power of each prime factor involved in the numbers
with highest power
= 2 2 ´ 33 ´ 5 ´ 7 = 4 ´ 27 ´ 35 = 3780
(iv) (d) Now, HCF (60, 84, 108) ´ LCM (60, 84, 108)
= 12 ´ 3780 = 45360
(v) (d) Number 108 can be expressed as a product of its
prime as 2 2 ´ 33 .
3. A
Mathematics Exhibition is being
conducted in your School and one of your
friends is making a model of a factor tree. He
has some difficulty and ask for your help in
completing a quiz for the audience. Observe
the following factor tree and answer the
following :
(i) In each room the same number of
participants are to be seated and all of
them being in the same subject, hence
maximum number participants that can
accommodated in each room are
(a) 14
(b) 12
(c) 16
x
5
y
(d) 18
(ii) What is the minimum number of rooms
required during the event?
(a) 11
(b) 31
(c) 41
(b) 3680
(d) 21
(a) 15005
(c) 56920
(c) 4780 (d) 4680
(a) 23
(b) 35360
(d) 45360
(a) 2 ´ 3
2
(c) 22 ´ 3 2
3
(b) 2 ´ 3
(b) 22
(c) 11
(d) 19
(iii) What will be the value of z ?
(a) 22
(i) (b) Given, number of students in each subject are
Hindi = 60, English = 84 and Mathematics = 108.
The prime factors of each subject students are
The maximum number of participants that can
accommodated in each room
= HCF (60, 84, 108)
= Product of the smallest power of each common
prime factor involved in the numbers
= 2 2 ´ 3 = 12
(ii) (d) The minimum number of rooms required during
the event is
Total number of participants
=
Maximum participants in one room
252
=
= 21rooms.
12
(d) 19
(v) The prime factorisation of 13915 is
(a) 5 ´ 113 ´ 13 2
60 = 2 ´ 2 ´ 3 ´ 5 = 2 2 ´ 3 ´ 5
108 = 2 ´ 2 ´ 3 ´ 3 ´ 3 = 2 2 ´ 33
(c) 17
(a) Composite number
(b) Prime number
(c) Neither prime nor composite
(d) Even number
(d) 22 ´ 3 3
84 = 2 ´ 2 ´ 3 ´ 7 = 2 2 ´ 3 ´ 7
(b) 23
(iv) According to Fundamental Theorem of
Arithmetic 13915 is a
3
Sol.
CBSE QUESTION BANK
(b) 13915
(d) 17429
(ii) What will be the value of y ?
(v) 108 can be expressed as a product of its
primes as
3
z
(i) What will be the value of x ?
(iv) The product of HCF and LCM of 60, 84
and 108 is
(a) 55360
(c) 45500
253
11
(iii) The LCM of 60, 84 and 108 is
(a) 3780
2783
2
(c) 5 ´ 11 ´ 23
(b) 5 ´ 113 ´ 23 2
(d) 5 ´ 112 ´ 13 2
Sol.
(i) (b) x = 5 ´ 2783 = 13915
(ii) (c) We have, 2783 = y ´ 253
2783
y=
= 11
Þ
253
(iii) (b) We have, 253 = 11 ´ z
253
z=
= 23
Þ
11
(iv) (a) Here, 13915 = 5 ´ 2783
Since, 13915 has factor other than 1 and the number
itself. It is a composite number.
(v) (c) 13915 = 5 ´ 11 ´ 11 ´ 23 = 5 ´ 112 ´ 23
31
CBSE Sample Paper Mathematics Standard Class X (Term I)
Sol.
Polynomials
1. The below picture are few natural examples
of parabolic shape which is represented by a
quadratic polynomial. A parabolic arch is an
arch in the shape of a parabola.
(i) (c) In the standard form of quadratic polynomial
ax2 + bx + c; ‘a’ is a non-zero real number, and b
and c are any real number.
(ii) (d) In a quadratic polynomial, if roots are equal, then
discriminant, D = 0.
1
(iii) (b) Given, a and are the zeroes of quadratic
a
polynomial 2 x2 - x + 8k.
Now, product of zeroes,
1
Constant term
a´ =
a Coefficient of x2
8k
2 1
Þk= =
1=
Þ
2
8 4
2
(iv) (c) Given equation is x + 1 = 0.
x2 = - 1
y = x2 +1
In structures, their curve represents an
efficient method of load, and so can be
found in bridges and in architecture in a
variety of forms.
(i) In the standard form of quadratic
polynomial, ax 2 + bx + c , a, b and c are
(a) All are real numbers
(b) All are rational numbers.
(c) a is a non-zero real number, b and c are
any real numbers.
(d) All are integers.
(ii) If the roots of the quadratic polynomial
are equal, where the discriminant
D = b 2 - 4 ac, then
(a) D > 0
(c) D ³ 0
(b) D < 0
(d) D = 0
1
are the zeroes of the quadratic
a
polynomial 2 x 2 - x + 8 k, then k is
(iii) If a and
(a) 4
(b)
1
4
(c)
-1
4
(0, 1)
Hence, it is neither touches nor intersects X-axis.
(v) (c) Given, sum of roots = - p
1
and product of roots = p
\ Required quadratic polynomial
= k [ x2 - (Sum of roots) x + Product of roots]
é
1ö
æ 1 öù
æ
= k ê x2 - (- p)x + ç - ÷ ú = k ç x2 + px - ÷
pø
è p øû
è
ë
2. An asana is a body posture, originally and
still a general term for a sitting meditation
pose, and later extended in hatha yoga and
modern yoga as exercise, to any type of pose
or position, adding reclining, standing,
inverted, twisting, and balancing poses.
In the figure, one can observe that poses
can be related to representation of quadratic
polynomial.
(d) 2
(iv) The graph of x 2 + 1 = 0
(v) If the sum of the roots is -p and product
1
of the roots is - , then the quadratic
p
polynomial is
æ
ö
x
(a) k çç - px 2 + + 1÷÷
p
è
ø
æ 2
1ö
(c) k çç x + px - ÷÷
pø
è
æ
ö
x
(b) k çç px 2 - - 1÷÷
p
è
ø
æ 2
1ö
(d) k çç x - px + ÷÷
pø
è
CHAKRASANA
TRIKONASANA
CBSE QUESTION BANK
(a) Intersects X-axis at two distinct points.
(b) Touches X-axis at a point.
(c) Neither touches nor intersects X-axis.
(d) Either touches or intersects X-axis.
32
CBSE Sample Paper Mathematics Standard Class X (Term I)
(v) (b) Let p( x) = 4 3 x2 + 5 x - 2 3
(i) The shape of the poses shown is
= 4 3 x2 + (8 - 3)x - 2 3
[by splitting middle term]
(a) Spiral
(b) Ellipse
(c) Linear
(d) Parabola
= 4 3 x2 + 8 x - 3 x - 2 3
(ii) The
graph
of
parabola
downwards, if ………… .
(a) a ³ 0
(c) a < 0
= 4 x( 3 x + 2 ) -
opens
= (4 x -
(4 x -
\
(iii) In the graph, how many zeroes are there
for the polynomial?
Þ
Þ
(b) 1
(c) 2
4x -
3 )( 3 x + 2 ) = 0
3 = 0 and 3 x + 2 = 0
3
2
and x = x=
4
3
3. Basketball and soccer are played with a
4
1
(a) 0
3 )( 3 x + 2 )
For finding the zeroes, put p( x) = 0
(b) a = 0
(d) a > 0
–2
3( 3 x + 2 )
(d) 3
(iv) The two zeroes in the below shown graph
are
spherical ball. Even though an athlete
dribbles the ball in both sports, a basketball
player uses his hands and a soccer player
uses his feet. Usually, soccer is played
outdoors on a large field and basketball is
played indoor on a court made out of wood.
The projectile (path traced) of soccer ball
and basketball are in the form of parabola
representing quadratic polynomial.
Y
X¢
–2 –1
1 2 3 4
X
Y¢
(a) 2, 4
(c) - 8, 4
(b) - 2, 4
(d) 2, - 8
(v) The zeroes of the quadratic polynomial
4 3 x 2 + 5 x - 2 3 are
2
3
,
3 4
2
3
(b) ,
3 4
2
3
(c)
,4
3
2
3
(d) ,4
3
CBSE QUESTION BANK
(a)
Sol.
(i) The shape of the path traced shown is
(i) (d) The shape of given poses are parabolic.
(ii) (c) The graph of parabola opens downwards, if a < 0.
(iii) (c) Number of zeroes is equal to number of times
intersects parabola on the X-axis.
\ Number of zeroes = 2.
(iv) (b) The curve intersect X-axis at points x = - 2
and x = 4.
Hence, two zeroes in the given graph are - 2 and 4.
(a) Spiral
(b) Ellipse
(c) Linear
(d) Parabola
(ii) The
graph
of
parabola
downwards, if ………… .
(a) a = 0
(c) a > 0
(b) a < 0
(d) a ³ 0
opens
33
CBSE Sample Paper Mathematics Standard Class X (Term I)
(iii) Observe the following graph and answer.
Pair of Linear Equations in
Two Variables
1. A test consists of ‘True’ or ‘False’ questions.
6
One mark is awarded for every correct
1
answer while mark is deducted for every
4
wrong answer. A student knew answers to
some of the questions. Rest of the questions
he attempted by guessing. He answered 120
questions and got 90 marks.
2
–4
–3
–2
–1
1
2
3
4
–2
–6
Type of
question
In the above graph, how many zeroes are
there for the polynomial?
(a) 0
(c) 2
(b) 1
(d) 3
(iv) The three zeroes in the above shown
graph are
(a) 2 , 3 , - 1
(c) - 3 , - 1, 2
(b) - 2, 3, 1
(d) -2, - 3 , - 1
(v) What will be the expression of the
polynomial?
True/False
Marks given for Marks deducted
correct answer for wrong answer
1
0.25
(i) If answer to all questions he attempted by
guessing were wrong, then how many
questions did he answer correctly?
(ii) How many questions did he guess?
(iii) If answer to all questions he attempted by
guessing were wrong and answered 80
correctly, then how many marks he got?
(c) x3 + 2x2 + 5 x - 6
(iv) If answer to all questions he attempted by
guessing were wrong, then how many
questions answered correctly to score 95
marks?
(d) x3 + 2x2 + 5 x + 6
Sol. Let the number of questions whose answer is known
(a) x3 + 2x2 - 5 x - 6
(b) x3 + 2x2 - 5 x + 6
Sol.
= x3 - (Sum of zeroes) x2
+ (Sum of product of two zeroes at a time) x
- (Product of three zeroes)
= x3 - (-2 ) x2 + (-5)x - (6)
= x3 + 2 x2 - 5 x - 6
(i)
(ii)
(iii)
(iv)
CBSE QUESTION BANK
(i) (d) The shape of the path traced shown in the given
figure is the form of parabola.
(ii) (b) The graph of parabola opens downwards, if a < 0.
(iii) (d) In the given graph, we see that curve intersect the
X-axis at three points. Hence, number of zeroes in the
given polynomial are 3.
(iv) (c) The given curve intersect the X-axis at points
x = - 3, -1and 2.
Hence, three zeroes in the given graph are -3, - 1, 2.
(v) (a) Since, given polynomial has three zeroes.
So, it will be a cubic polynomial.
Now, sum of zeroes = - 3 - 1 + 2 = - 2
Sum of product of two zeroes at a time
= - 3 ´ (-1) + (-1) ´ 2 + 2 ´ (-3)
= 3-2 - 6= - 5
and product of all zeroes = - 3 ´ - 1 ´ 2
=6
\ Required cubic polynomial
to the student be x and questions attempted by
guessing be y.
Then,
… (i)
x + y = 120
1
and
… (ii)
x - y = 90 Þ 4 x - y = 360
4
On adding Eqs. (i) and (ii), we get
480
= 96
5 x = 480 Þ x =
5
Put x = 96 in Eq. (i), we get
96 + y = 120 Þ y = 120 - 96 = 24
He answered 96 questions correctly.
He guesses only 24 questions.
In out of 120 questions attempted 80 answered are
correct and 40 guessing answered are wrong.
1
Then, he got the marks = 80 - of 40
4
1
= 80 - ´ 40 = 80 - 10 = 70
4
According to the given condition,
1
1
x - of (120 - x) = 95 Þ x - ´ (120 - x) = 95
4
4
Þ
4 x - 120 + x = 380
Þ
5 x = 500
Þ
x = 100
Hence, he answered correctly 100 questions to score
95 marks.
34
CBSE Sample Paper Mathematics Standard Class X (Term I)
2. Amit is planning to buy a house and the
layout is given below. The design and the
measurement has been made such that
areas of two bedrooms and kitchen together
is 95 sq m.
5m
x
2
y
Bedroom 1
Bath
room
Kitchen
2m
Living room
5m
Bedroom 2
15 m
But, it is also given, the cost of laying tiles in
kitchen at the rate of `50 per m 2 .
\Total cost of laying tiles in the kitchen = 35 ´ 50
= ` 1750
3. It is common that Governments revise travel
fares from time to time based on various
factors such as inflation ( a general increase
in prices and fall in the purchasing value of
money) on different types of vehicles like
auto, Rickshaws, taxis, Radio cab etc. The
auto charges in a city comprise of a fixed
charge together with the charge for the
distance covered. Study the following
situations
Based on the above information, answer the
following questions:
(i) Form the pair of linear equations in two
variables from this situation.
(ii) Find the length of the outer boundary of
the layout.
(iii) Find the area of each bedroom and
kitchen in the layout.
Name of
the city
Distance travelled
(km)
Amount
paid (in `)
(iv) Find the area of living room in the layout.
City A
10
75
15
110
8
91
14
145
(v) Find the cost of laying tiles in kitchen at
the rate of ` 50 per sq m.
City B
CBSE QUESTION BANK
Sol.
(i) From the given figure we see that area of two
bedrooms = 2(5 x) = 10x m 2
Area of kitchen = 5 ´ y = 5 y m2
According to the question,
Area of the two bedrooms and area of kitchen
= 95
\
10 x + 5 y = 95
Þ
2 x + y = 19 [divide both sides by 5] … (i)
Also, length of the home = 15 cm
… (ii)
\
x + 2 + y = 15 Þ x + y = 13
Hence, pair of linear equations is
2 x + y = 19 and x + y = 13
(ii) The length of the outer boundary of the layout
= 2(l + b ) = 2(15 + 12 )
= 2(27 ) = 54 m
(iii) On solving Eqs. (i) and (ii), we get
x = 6 and y = 7
\Area of each bedroom = 5 ´ x = 5 ´ 6 = 30 m2
and area of kitchen = 5 ´ y = 5 ´ 7 = 35 m 2
(iv) Area of living room = 15 ´ (5 + 2 ) - Area of bedroom 2
= 15 ´ 7 - 5 ´ 6
= 105 - 30 = 75 m 2
(v) Since, area of kitchen = 5 ´ y
= 5 ´ 7 = 35 m2
Situation 1 In city A, for a journey of 10 km,
the charge paid is ` 75 and for a journey of 15
km, the charge paid is ` 110.
Situation 2 In a city B, for a journey of 8 km,
the charge paid is ` 91 and for a journey of 14
km, the charge paid is ` 145.
Refer situation 1
(i) If the fixed charges of auto rickshaw be
`x and the running charges be ` y km/h,
the pair of linear equations representing
the situation is
(a) x + 10y = 110, x + 15 y = 75
(b) x + 10y = 75 , x + 15 y = 110
(c) 10x + y = 110, 15 x + y = 75
(d) 10x + y = 75, 15 x + y = 110
(ii) A person travels a distance of 50 km.
The amount he has to pay is
(a) ` 155
(b) ` 255
(c) ` 355
(d) ` 455
35
CBSE Sample Paper Mathematics Standard Class X (Term I)
Refer situation 2
(iii) What will a person have to pay for
travelling a distance of 30km ?
(a) ` 185
(c) ` 275
(b) ` 289
(d) ` 305
(iv) The graph of lines representing the
conditions are: (situation 2)
Y
25
(20, 25)
20
15
(a)
10
(30, 5)
5 (0, 5)
X¢
0
–5
5 10 15 20 25 30 35
X
–10
Y¢
Y
25
20
15
(b)
10 (0, 10)
5
X¢
(20, 10)
(12.5, 0)
0
–5
5 10 15 20 25 30 35
(5, –10)
–10
Y¢
X
1. Vijay is trying to find the average height of a
tower near his house. He is using the
properties of similar triangles.The height of
Vijay’s house, if 20 m when Vijay’s house
casts a shadow 10m long on the ground.
20
15
(11, 10) (19, 9)
10
5
X¢
- 5 y = - 35 Þ y = 7
\
x + 10 ´ 7 = 75
Þ
x = 75 - 70 = 5
To travel a distance of 50 km, a person has to pay
amount = x + 50 y = 5 + 50 ´ 7
= 5 + 350 = ` 355
(iii) (b) As per the situation 2, the pair of linear is
… (i)
x + 8 y = 91
and
… (ii)
x + 14 y = 145
On solving Eqs. (i) and (ii), we get
x + 8 y = 91
x + 14 y = 145
- - 6 y = - 54
Þ
y=9
Put y = 9 in Eq. (i), we get
x + 8 ´ 9 = 91
Þ
x = 91 - 72 = 19
To travel a distance of 30 km, a person has to pay
amount = x + 30 ´ y
= 19 + 30 ´ 9 = 19 + 270 = ` 289
(iv) (c) In situation 2, the intersection point of two lines is
(19, 9), which is shown in figure (c).
Similar Triangles
(25, –10)
Y
25
(c)
(ii) (c) On solving the above equations, we get
x + 10 y = 75
x + 15 y = 110
- -
(5, 10)
0
–5
5 10 15 20 25 30 35
X
At the same time, the tower casts a shadow
50 m long on the ground and the house of
Ajay casts 20 m shadow on the ground.
Y¢
Y
25
15
(d)
10
(0, 10)
(15, 15)
(35, 10)
Vijay’s
house
5
X¢
0
–5
5 10 15 20 25 30 35
(15, 5)
X
Y¢
Sol.
(i) (b) As per the situation 1, the pair of linear equation
representing the situation is
x + 10 y = 75
and
x + 15 y = 110
Tower
Ajay’s
house
(i) The height of the tower is
(a) 20 m
(c) 100 m
(b) 50 m
(d) 200 m
(ii) What will be the length of the shadow of
the tower when Vijay’s house casts a
shadow of 12 m?
(a) 75 m
(c) 45 m
(b) 50 m
(d) 60 m
CBSE QUESTION BANK
20
36
CBSE Sample Paper Mathematics Standard Class X (Term I)
(iii) What is the height of Ajay’s house?
(a) 30 m
(c) 50 m
2. Rohan wants to measure the distance of a
(b) 40 m
(d) 20 m
pond during the visit to his native. He marks
points A and B on the opposite edges of a
pond as shown in the figure below. To find
the distance between the points, he makes a
right-angled triangle using rope connecting
Bwith another point C are a distance of 12 m,
connecting C to point D at a distance of 40 m
from point C and the connecting D to the
point A which at distance of 30 m from D
such the Ð ADC = 90°.
(iv) When the tower casts a shadow of 40 m,
same time what will be the length of the
shadow of Ajay’s house?
(a) 16 m
(c) 20 m
(b) 32 m
(d) 8 m
(v) When the tower casts a shadow of 40 m,
same time what will be the length of the
shadow of Vijay’s house?
(a) 15 m
(c) 16 m
(b) 32 m
(d) 8 m
12 m
B
A
C
Sol.
30 m
(i) (c) Let CD = h m be the height of the tower. Let
BE = 20 m be the height of Vijay’s house and GF be
the height of Ajay’s house.
40 m
D
D
E
G
A 10 m B
50 m
CBSE QUESTION BANK
(iii)
(iv)
(v)
(a) Similarity of triangles
(b) Thales Theorem
(c) Pythagoras Theorem
(d) Area of similar triangles
h
20 m
(ii)
(i) Which property of geometry will be used
to find the distance AC ?
C
F 20 m H
50 m
DACD ~ ABE
AC CD
\
=
AB EB
h
50
=
Þ h = 100 m
Þ
10 20
(d) Given AB = 12 m, let AC = h
In similar DABE and DACD,
AB BE 12
20
=
Þ
=
AC CD
h 100
12 ´ 100
h=
Þ
= 12 ´ 5 = 60 m
20
(b) Let height of Ajay’s house be GF = h1
Since, DHFG ~ DHCD
HF FG
=
\
HC CD
h
20
= 1
Þ
50 100
20 ´ 100
h1 =
= 40 m
Þ
50
(a) Given, HC = 40 cm
Let length of the shadow of Ajay’s hour be HF = l m
Since, DHFG ~ DHCD
HF FG
\
=
HC CD
40 ´ 40
l
40
= 16 m
=
Þl=
Þ
100
40 100
(d) Given, AC = 40 cm
Let length of the shadow of Vijay’s house be AB = l m
Since,
DABE ~ ACD
AB EB
\
=
AC CD
20 ´ 40
l
20
=
Þ h=
= 8m
Þ
100
40 100
(ii) What is the distance AC ?
(a) 50 m
(c) 100 m
(b) 12 m
(d) 70 m
(iii) Which is the following does not form a
Pythagoras triplet?
(a) (7,24,25)
(c) (5,12,13)
(b) (15,8,17)
(d) (21,20,28)
(iv) Find the length AB?
(a) 12 m
(c) 50 m
(b) 38 m
(d) 100 m
(v) Find the length of the rope used.
(a) 120 m
(c) 82 m
(b) 70 m
(d) 22 m
Sol.
(i) (c) To find the distance AC in the given figure, we
use Pythagoras theorem.
(ii) (a) In right DADC, use Pythagoras theorem,
AC = ( AD)2 + (CD)2 = (30)2 + (40)2
= 900 + 1600 = 2500 = 50 m
(iii) (d) (a) Now, 242 + 7 2 = 576 + 49 = 625 = (25)2 ,
which forms a Pythagoras triplet
(b) (15)2 + (8)2 = 225 + 64 = 289 = (17 )2 ,
which forms a Pythagoras triplet
(c) (12 )2 + (5)2 = 144 + 25 = 169 = (13)2 ,
which form a Pythagoras triplet.
(d) (20)2 + (21)2 = 400 + 441 = 881¹ (28)2 ,
which does not form a Pythagoras triplet.
(iv) (b) Since, AC = 50 m
\
AB = AC - BC = 50 - 12 = 38 m
(v) (c) The length of the rope used = BC + CD + DA
= 12 + 40 + 30 = 82 m
37
CBSE Sample Paper Mathematics Standard Class X (Term I)
(v) The length of AB in the given figure
3. A scale drawing of an object is the same
shape at the object but a different size. The
scale of a drawing is a comparison of the
length used on a drawing to the length it
represents. The scale is written as a ratio.
The ratio of two corresponding sides in
similar figures is called the scale factor.
Scale factor
Length in range
=
Corresponding length in object
If one shape can become another using
revising, then the shapes are similar. Hence,
two shapes are similar when one can
become the other after a resize, flip, slide or
turn. In the photograph below showing
the side view of a train engine. Scale factor is
1 : 200.
This means that a length of 1 cm on the
photograph above corresponds to a length
of 200 cm or 2 m, of the actual engine. The
scale can also be written as the ratio of two
lengths.
(i) If the length of the model is 11cm, then
the overall length of the engine in the
photograph above, including the
couplings(mechanism used to connect) is
(a) 22 cm
(c) 220 m
(b) 220 cm
(d) 22 m
(ii) What will affect the similarity of any two
polygons?
(a) They are flipped horizontally
(b) They are dilated by a scale factor
(c) They are translated down
(d) They are not the mirror image of one
another.
(a) 0.7 m
(c) 0.07 cm
(b) 0.7 cm
(d) 0.07 m
(iv) If two similar triangles have a scale factor
5 : 3 which statement regarding the two
triangles is true?
(a) The ratio of their perimeters is 15 : 1
(b) Their altitudes have a ratio 25 :15
(c) Their medians have a ratio 10 : 4
(d) Their angle bisectors have a ratio 11 : 5
x cm
B
C
3 cm
4 cm
D
E
6 cm
(a) 8 cm
(c) 4 cm
(b) 6 cm
(d) 10 cm
Sol. Given, scale factors = 1 : 200
It means that length of 1 cm on the photograph
above corresponds to a length of 200 cm (or 2 m) of
the actual engine.
(i) (d) Since, length of the model is 11 cm.
Therefore, the overall length of the engine
= 11 ´ 200
= 2200 cm
= 22 m
(ii) (d) The similarity of any two polygons will affect that
they are not the mirror image of one another.
(iii) (a) The actual width of the door = 0.35 ´ 200 cm
= 70 cm
= 07
. m
(iv) (b) If two similar triangles have a scale factor 5 : 3,
then their altitudes have a ratio 25 : 15.
(v) (c) In the given BC || DE.
\
Þ
Þ
Þ
Þ
Þ
DABC ~ ADE,
AB BC
=
AD DE
x
3
=
x+ 4 6
x
1
=
x+ 4 2
2x = x + 4
x = 4 cm
Coordinate Geometry
1. In order to conduct Sports Day activities in
your School, lines have been drawn with
chalk powder at a distance of 1 m each, in a
rectangular shaped ground ABCD, 100
flowerpots have been placed at a distance of
1 m from each other along AD, as shown in
given figure below. Niharika runs 1/4 th the
distance AD on the 2nd line and posts a
1
green flag. Preet runs th distance AD on
5
the eighth line and posts a red flag.
CBSE QUESTION BANK
(iii) What is the actual width of the door, if the
width of the door in photograph is 0.35
cm?
A
38
CBSE Sample Paper Mathematics Standard Class X (Term I)
D
C
(iii) (c) Distance between these flags = GR
= (8 - 2 )2 + (25 - 20)2
= (6)2 + (5)2
= 36 + 25 = 61 m
(iv) (a) The point at which Rashmi should post her blue
flag is the mid-point of the line joining these points.
Let this point be A( x, y).
2+ 8
25 + 20
, y=
x=
2
2
10
45
x=
= 5, y =
= 22.5
2
2
Hence, A( x, y) = (5, 22.5)
(v) (a) Let the point at which Joy post his flag be B( x, y).
G
R
1:3
2
1
G(2, 25)
B
A
1
2
3
4
5
6
7
8
9
10
(i) Find the position of green flag
(a) (2, 25)
(c) (25, 2)
(b) (2, 0.25)
(d) (0, - 25)
B
R(8, 20)
1´ 8 + 3 ´2
1 ´ 20 + 3 ´ 25
, y=
1+ 3
1+ 3
14
95
x=
= 3.5, y =
= 23.75 » 24
4
4
Hence, B( x, y) = (3.5, 24)
Then, x =
(ii) Find the position of red flag
(a) (8, 0)
(c) (8, 20)
(b) (20, 8)
(d) (8, 0.2)
(iii) What is the distance between both the
flags?
(a) 41 m
(c) 61 m
(b) 11 m
(d) 51 m
(iv) If Rashmi has to post a blue flag exactly
halfway between the line segment joining
the two flags, where should she post her
flag?
(a) (5, 22.5)
(c) (2, 8.5)
(b) (10, 22)
(d) (2.5, 20)
(v) If Joy has to post a flag at one-fourth
distance from green flag ,in the line
segment joining the green and red flags,
then where should he post his flag?
(a) (3.5, 24)
(c) (2.25, 8.5)
(b) (0.5, 12.5)
(d) (25, 20)
CBSE QUESTION BANK
Sol.
(i) (a) It can be observed that Niharika posted
1
the green flag at th distance of AD
4
1
i.e. ´ 100 = 25 m from the starting point
4
of 2nd line.
Therefore, the coordinates of this point G is
(2, 25).
1
(ii) (c) Preet posted red flag at th distance of AD,
5
1
i.e. ´ 100 = 20 m from the starting point of 8th line.
5
Therefore the coordinates of this point R is
(8, 20).
Areas Related to Circles
1. Pookalam is the flower bed or flower
pattern designed during Onam in Kerala.
It is similar as Rangoli in North India
and Kolam in Tamil Nadu. During the
festival of Onam , your school is planning to
conduct a Pookalam competition. Your
friend who is a partner in competition,
suggests two designs given below. Observe
these carefully.
A
B
A
B
D
C
C
Design I This design is made with a circle of
radius 32 cm leaving equilateral triangle
ABC in the middle as shown in the given
figure.
Design II This Pookalam is made with
9 circular design each of radius 7 cm.
Refer Design I
(i) The side of equilateral triangle is
(a) 12 3 cm
(c) 48 cm
(b) 32 3 cm
(d) 64 cm
(ii) The altitude of the equilateral triangle is
(a) 8 cm
(c) 48 cm
(b) 12 cm
(d) 52 cm
39
CBSE Sample Paper Mathematics Standard Class X (Term I)
Refer Design II
(iii) The area of square is
(a) 1264 cm 2
(c) 1830 cm 2
(b) 1764 cm 2
(d) 1944 cm 2
(iv) Area of each circular design is
(a) 124 cm 2
(c) 144 cm 2
(b) 132 cm 2
(d) 154 cm 2
(v) Area of the remaining portion of the
square ABCD is
(a) 378 cm 2
(c) 340 cm 2
(b) 260 cm 2
(d) 278 cm 2
Sol.
Directions (i-ii)
Given, radius of the circle = 32 cm
Let the side of the equilateral DABC be a cm.
Let h be the height of the triangle.
Now, area of one circle = pr 2 = p (7 )2
22
=
´ (7 )2 = 154 cm 2
7
\Area of nine circles = 9 ´ 154 = 1386 cm 2
Area of square ABCD = (Side)2 = (42 )2 = 1764 cm2
Hence, area of the remaining portion of the handkerchief
= Area of square - Area of nine circles
= 1764 - 1386 = 378 cm 2
(iii) (b) 1764 cm 2
(iv) (d) 154 cm 2
(v) (a) 378 cm 2
2. A brooch is a small piece of jewellery which
has a pin at the back so it can be fastened on a
dress, blouse or coat. Designs of some
brooch are shown below. Observe them
carefully.
A
a
a
O
h
A
B
a/2
D
a/2
C
We know that in an equilateral triangle, centroid and
circumcentre coincide.
2
\
AO = h cm
3
[Q centroid divides the median in the ratio 2 : 1]
which is equal to the radius of circle.
2
…(i)
\
h = 32 Þ h = 48 cm
3
Now, we draw a perpendicular from vertex A to
side BC which bisects BC at D.
In right angled DADB,
AB2 = BD2 + AD2
[by Pythagoras theorem]
2
Þ
Þ
Þ
a2
3a2
æaö
a2 = ç ÷ + h2 Þ h2 = a2 =
4
4
è2 ø
2
a
3
[from Eq. (i)]
(48)2 =
4
a2 = 3072
a = 3072
= 32 3 cm
(i) (b) 32 3 cm
(ii) (c) 48 cm
[taking positive square root]
Directions (iii-v)
Given, radius of each circle, r = 7 cm
[Q diameter = 2 ´ radius]
\Diameter of circle, d = 14 cm
In the given figure, horizontal three circles touch each other.
\Length of a side of square = 3 ´ Diameter of one circle
= 3 ´ 14 = 42 cm
Design A Brooch A is made with silver wire
in the form of a circle with diameter 28 mm.
The wire used for making 4 diameters which
divide the circle into 8 equal parts.
Design B Brooch b is made two colours_
Gold and silver. Outer part is made with
Gold. The circumference of silver part is
44 mm and the gold part is 3mm wide
everywhere.
Refer to Design A
(i) The total length of silver wire required is
(a) 180 mm
(c) 250 mm
(b) 200 mm
(d) 280 mm
(ii) The area of each sector of the brooch is
(a) 44 mm 2
(c) 77 mm 2
(b) 52 mm 2
(d) 68 mm 2
Refer to Design B
(iii) The circumference of outer part (golden) is
(a) 48.49 mm
(c) 72.50 mm
(b) 82.2 mm
(d) 62.86 mm
CBSE QUESTION BANK
Þ
B
40
CBSE Sample Paper Mathematics Standard Class X (Term I)
(iv) The difference of areas of golden and
silver parts is
(a) 18p mm 2
(c) 51p mm 2
(b) 44p mm 2
(d) 64p mm 2
(v) A boy is playing with brooch B. He makes
revolution with it along its edge. How
many complete revolutions must it take
to cover 80 p mm ?
(a) 2
(c) 4
Probability
1. On a weekend Rani was playing cards with
her family. The deck has 52 cards.If her
brother drew one card.
(b) 3
(d) 5
Sol.
(i) (b) Given, diameter of circle, d = 28 mm
[Qd = 2 r]
\Circumference of circle = pd
22
=
´ 28 = 88 mm
7
Now, length of 4 diameters = 4 ´ 28 = 112 mm
Total length of the silver wire = pd + 4 d
= 88 + 112 = 200 mm
(ii) (c) Here, we see that total circle is divided into
8 parts.
1
\ Area of each sector = ´ Area of circle
8
1
= ´ pr 2
8
1 22
= ´
´ 14 ´ 14
8 7
= 77 mm 2
(i) Find the probability of getting a king of
red colour.
(a)
(b)
1
13
(c)
1
52
(d)
1
4
(ii) Find the probability of getting a face card.
(a)
1
26
(b)
1
13
(c)
2
13
(d)
3
13
(iii) Find the probability of getting a jack of
hearts.
(a)
1
26
(b)
1
52
(c)
3
52
(d)
3
26
(iv) Find the probability of getting a red face
card.
(a)
Directions (iii-iv)
1
26
3
13
(b)
1
13
(c)
1
52
(d)
1
4
(v) Find the probability of getting a spade.
(a)
r
3m
CBSE QUESTION BANK
R
(b)
1
13
(c)
1
52
(d)
1
4
Sol. Total number of cards in one deck of cards is 52.
Silver
m
1
26
Gold
We have, circumference of silver part = 44 mm
\
2 pr = 44
44
r=
= 7 mm
Þ
22
2´
7
\
R = r + 3= 7 + 3
= 10 mm
(iii) (d) Circumference of golden part = 2pR
22
=2 ´
´ 10
7
= 62.86 mm
(iv) (c) Difference of areas = pR 2 - pr 2 = p(R 2 - r 2 )
= (102 - 7 2 ) p = 51p mm 2
(v) (c) Required number of revolutions
Distance covered
=
Circumference
80p
80p
=
=
=4
2 pR 2 p ´ 10
\ Total number of outcomes = 52
(i) (a) Let E1 = Event of getting a king of red colour
\ Number of outcomes favourable to E1 = 2
[Q there are four kings in a deck of playing
cards out of which two are
red and two are black]
Hence, probability of getting a king of red colour,
2
1
=
P(E1 ) =
52 26
(ii) (d) Let E2 = Event of getting a face card
\ Number of outcomes favourable to E2 = 12
[Q in a deck of cards, there are
12 face cards, namely 4 kings,
4 jacks, 4 queens]
Hence, probability of getting a face card,
12
3
=
P(E2 ) =
52 13
(iii) (b) Let E3 = Event of getting a jack of heart
\ Number of outcomes favourable to E3 = 1
[Q there are four jack cards in a deck,
namely 1 of heart, 1 of club,
1 of spade and 1 of diamond]
Hence, probability of getting a jack of heart,
41
CBSE Sample Paper Mathematics Standard Class X (Term I)
1
52
(iv) (*) Let E4 = Event of getting a red face card.
\Number of outcomes favourable to E4 = 6
[Q in a deck of cards, there are 12 face cards
out of which 6 are red cards]
Hence, probability of getting a red face card,
6
3
=
P(E4 ) =
52 26
(v) (d) Let E5 = Event of getting a spade
\ Number of outcomes favourable to E5 = 13
[Q in a deck of cards, there are 13 spades,
13 clubs, 13 hearts and 13 diamonds]
Hence, probability of getting a spade,
13 1
=
P(E5 ) =
52 4
the top face of the dice is less than or
equal to 12 ?
P(E3 ) =
2. Rahul and Ravi planned to play Business
(board game) in which they were supposed
to use two dice.
(a) 1
(b)
5
36
(c)
1
18
(d) 0
(iv) Rahul got next chance. What is the
probability that he got the sum of the two
numbers appearing on the top face of the
dice is equal to 7 ?
(a)
5
9
(b)
5
36
(c)
1
6
(d) 0
(v) Now it was Ravi’s turn. He rolled the
dice. What is the probability that he got
the sum of the two numbers appearing on
the top face of the dice is greater than 8 ?
(a) 1
(b)
5
36
(c)
1
18
(d)
5
18
Sol. Total number of out comes = 6 ´ 6 = 36
(i) (b) Let E1 = Event of getting sum 8
\Number of favourable outcomes to E1 = 5
i.e. (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
5
P(E1 ) =
\
36
(ii) (d) Let E2 = Event of getting sum 13.
Since, we can’t get sum more than 12.
\
P(E2 ) =
36
=0
0
(iii) (a) Let E3 = Event of getting sum less than or equal
to 12.
\Number of favourable outcomes = 36
(i) Ravi got first chance to roll the dice. What
is the probability that he got the sum of
the two numbers appearing on the top
face of the dice is 8?
(a)
1
26
(b)
5
36
(c)
1
18
(d) 0
(ii) Rahul got next chance. What is the
probability that he got the sum of the two
numbers appearing on the top face of the
dice is 13?
1
(c)
18
(d) 0
(iii) Now it was Ravi’s turn. He rolled the
dice. What is the probability that he got
the sum of the two numbers appearing on
\ P(E3 ) =
36
=1
36
(iv) (c) Let E4 = Event of getting sum 7.
\Number of favourable outcomes to E2 = 6
i.e. (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) (6, 1)
6
1
\
=
P(E4 ) =
36 6
(v) (d) Let E5 = Event of getting sum greater than 8 i.e.
getting sum equal to 8, 9, 10, 11, 12
\Favourable outcomes = 10 i.e. (3, 6), (4, 5), (5, 4),
(6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6).
10
5
=
P(E5 ) =
\
36 18
CBSE QUESTION BANK
(a) 1
5
(b)
36
As sum of all the out comes is less than or equal to 12.