CBSE Class 9 – Page 19

NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.1

February 14, 2023

NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.1.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 6
Chapter Name	Coordinate Geometry
Exercise	Ex 6.1
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.1

Question 1.
How will you describe the position of a table lamp on your study table to another person?
Solution:
Let us consider the lamp as a point A and table as a plane.
NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.1 img 1
Take any two perpendicular edges of the table, say OX and OY. Measure the distance of the lamp A from the larger edge OX, let it be 25 cm. Again, measure the distance of the lamp A from the shorter edge OY, let it be 15 cm. Therefore, the position of the lamp A to the edges OX and OY is (15, 25).

Question 2.
(Street Plan) A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1 cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines.
There are many cross-streets in your model. A particular cross-street is made by two street one running in the North-South direction and another in the East-West direction. Each cross-street is referred to in the following manner : If the 2 nd street running in the North-South direction and 5th in the East-West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find
(i) how many cross-streets can be referred to as (4, 3).
(ii) how many cross-streets can be referred to as (3, 4).
Solution:
(i) The street plan is shown by the following figure.

(i) There is only one cros.s-street which can be referred as (4, 3).
(ii) There is only one cross-street which can be referred as (3, 4).
We hope the NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

February 13, 2023

NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 15
Chapter Name	Probability
Exercise	Ex 15.1
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Since, batswoman plays 30 balls, therefore total number of trials is n(S) = 30.
Let E be the event of hitting the boundary.
∴ n(E) = 6
The number of balls not hitting the target
n(E’) = 30-6=24
The probability that she does not hit a boundary = \(\frac { n(E’) }{ n(S) }\) = \(\frac { 24 }{ n(30) }\) = \(\frac { 4 }{ 5 }\)

Question 2.
1500 families with 2 children were selected randomly, and the following data were recorded
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 1
Compute the probability of a family, chosen at random, having
(i) 2 girls (ii) 1 girl (iii) no girl
Also, check whether the sum of these probabilities is 1.
Solution:

Question 3.
In a particular section of class IX, 40 students were asked about the month of their birth and the following graph was prepared for the data so obtained.
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 3
Find the probability that a student of the class was born in August.
Solution:
Total number of students in class IX, n(S) = 40
Number of students bom in the month of August, n(E) = 6
Probability, that the students of the class was born in August = \(\frac { n(E) }{ n(S) }\) = \(\frac { 6 }{ 40 }\) = \(\frac { 3 }{ 20 }\)

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes.
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 4
If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Solution:
In tossing of three coins, getting two heads comes out 72 times,
i.e., n(E) = 72
The total number of tossed three coins n(S) = 200
∴ Probability of 2 heads coming up = \(\frac { n(E) }{ n(S) }\) = \(\frac { 72 }{ 200 }\) = \(\frac { 9 }{ 25 }\)

Question 5.
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below.
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 5
Suppose a family is chosen. Find the probability that the family chosen is
(i) earning ₹ 10000-13000 per month and owning exactly 2 vehicles.
(ii) earning ₹16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not own any vehicle.
(iv) earning ₹13000-16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
Total number of families selected by the organisation, n(S) = 2400
(i) The number of families earning ₹ 10000-13000 per month and owing exactly 2 vehicles, n(E₁) = 29

(ii) The number of families earning ₹ 16000 or more per month and owing exactly 1 vehicle, n(E₂) = 579

(iii) The number of families earning less than ₹ 7000 per month and does not own any vehicle, n(E₃) = 10
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 8
(iv) The number of families earning ₹ 13000-16000 per month and owing more than 2 vehicles, n(E₄) = 25

(v) The number of families owing not more than 1 vehicle,
n(E₅) = (10 + 1 + 2 + 1) + (160 + 305 + 535 + 469 + 579)
=14 + 2048 = 2062

Question 6.
A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows
0-20, 20 – 30, …, 60 – 70, 70 – 100. Then she formed the following table
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 11
(i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Solution:
(i) Total number of students in a class. n(S) = 90
The number of students less than 20% lies in the interval 0-20,
i.e., n(E) = 7
∴ The probability, that a student obtained less than 20% in the Mathematics test = \(\frac { n(E) }{ n(S) }\) = \(\frac { 7 }{ 90 }\)
(ii) The number of students obtained marks 60 or above lies in the marks interval 60-70 and 70-above
i.e., n(F) = 15+ 8 = 23
∴ The probability that a student obtained marks 60 or above = \(\frac { n(E) }{ n(S) }\) = \(\frac { 23 }{ 90 }\)

Question 7.
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 12
Find the probability that a student chosen at random
(i) likes statistics,
(ii) does not like it.
Solution:
Total number of students, n(S) = 200
(i) The number of students who like Statistics, n(E) = 135

(ii) The number of students who does not like Statistics, n(F) = 65
∴ The probability, that the student does not like Statistics

Question 8.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 15
What is the empirical probability that an engineer lives
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within \(\frac { 1 }{ 2 }\) km from her place of work?
Solution:
Total number of engineers lives, n(S) = 40
(i) The number of engineers whose residence is less than 7 km from their place, n(E) = 9
∴ The probability, that an engineer lives less than 7 km from their place of work

(ii) The number of engineers whose residence is more than or equal to 7 km from their place of work, n(F) = 40 – 9 = 31
∴The probability, that an engineer lives more than or equal to 7 km from their place of work = \(\frac { n(F) }{ n(S) }\) = \(\frac { 31 }{ 40 }\)
(iii) The number of engineers whose residence within \(\frac { 1 }{ 2 }\) km from their place of work, i.e., n(G) = 0
∴ The probability, that an engineer lives within \(\frac { 1 }{ 2 }\) km from their place
= \(\frac { n(G) }{ n(S) }\) = \(\frac { 0 }{ 40 }\) = 0

Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler?
Solution:
After observing in front of the school gate in time interval 6:30 to 7:30 am respective frequencies of different types of vehicles are .
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 17
∴ Total number of vehicle, n(S) = 550 + 250 + 80 = 880
Number of two-wheelers, n(E) = 550
∴ Probability of observing two-wheelers = \(\frac { n(E) }{ n(S) }\) = \(\frac { 550 }{ 880 }\) = \(\frac { 5 }{ 8 }\)

Question 10.
Activity : Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digit is divisible by 3.
Solution:
Suppose, there are 40 students in a class.
∴ The probability of selecting any of the student = \(\frac { 40 }{ 40 }\) = 1
A three digit number start from 100 to 999
Total number of three digit numbers = 999 – 99 = 900
∴ Multiple of 3 in three digit numbers = {102,105 ….., 999}
∴ Number of multiples of 3 in three digit number = \(\frac { 900 }{ 3 }\) = 300
i.e., n(E) = 300
∴ The probability that the number written by her/him,is divisible by 3
= \(\frac { n(E) }{ n(S) }\) = \(\frac { 300 }{ 900 }\) = \(\frac { 1 }{ 3 }\)

Question 11.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg)
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags,chosen at random contains more than 5 kg of flour.
Solution:
The total number of wheat flour bags; n(S) = 11
Bags, which contains more than 5 kg of flour, (E)
= {5,05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07}
∴ n(E) = 7
∴ Required probability =\(\frac { n(E) }{ n(S) }\) = \(\frac { 7 }{ 11 }\)

Question 12.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 18
You were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.
Solution:
Now, we prepare a frequency distribution table
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 19
The total number of days for data, to prepare sulphur dioxide, n(S) = 30
The frequency of the sulphur dioxide in the interval 0.12-0.16, n(E) = 2

Question 13.
The blood groups of 30 students of class VIII are recorded as follows
A, B, 0, 0, AB, 0, A, 0, B, A, 0, B, A, 0, 0, A, AB, 0, A, A, 0, 0, AB, B, A, B, 0
You were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Solution:
NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 img 21
The total number of students in class VIII, n(S) = 30
The number of students who have blood group AB, n(E) = 3
∴ The probability that a student has a blood group AB =\(\frac { n(E) }{ n(S) }\) = \(\frac {3 }{ 30 }\) = \(\frac { 1 }{ 10 }\)

We hope the NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

February 13, 2023

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 13
Chapter Name	Surface Areas and Volumes
Exercise	Ex 13.1
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m² costs ₹20.
Solution:
We have a plastic box of
l = length = 15 m
b = width = 1.25 m
h = depth = 65 cm
= \(\frac { 65 }{ 100 }\) m= 0.65 m (∵ 1 m = 100cm)
Surface area of the box = 2 (lb + bh + hl)
= 2(1.5 x 1.25 + 1.25 x 0.65 + 0.65 x 1.5)
= 2(1.875 + 0.8125 + 0.975) = 2 (3.6625)
= 7.325 m²
(i) Area of the sheet required for making the box
= 7.325 – l x b (∵ BOX is opened at the top)
= 7.325 – 1.5 x 1.25
= 7.325 – 1.875 = 5.45 m²
(ii) A sheet measuring 1 m² costs = ₹20
∴ Sheet measuring 5.45 m²costs = ₹20 x 5.45 = ₹109

Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m, respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹17.50 per m².
Solution:
We have a room of l = 5m
b = 4m
h = 3m
Required area for white washing
= Area of the four walls + Area of ceiling
= 2(l+ b) x h+ (l x b)
= 2(5+4) x 3 +(5 x 4)
= 2x 9x 3 + 20
= 54+20
= 74 m2
White washing 1 m² costs = ₹7.50
White washing 74 m² costs = ₹7.50x 74 = ₹555

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹10 per m² is ₹15000, find the height of the hall.
[Hint Area of the four walls = Lateral surface area]
Solution:
Let the rectangular hall of length = l, breadth = b, height = h
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 img 1
Hence, the height of the hall is 6 m.

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container.
Solution:
Given, dimensions of a brick
l = 22.5 cm, b = 10 cm
and b = 7.5cm
Total surface area of bricks = 2 ( l x b + b x h + h x l)
= 2(22.5 x 10 + 10 x 75 + 75 x 225)
= 2(225 + 75 + 168.75)
= 2 x 468.75 cm²
= 9375 cm²

Number of bricks that painted out of this container
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 img 3

Question 5.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Solution:
We have l₁ for cubical box = 10 cm
For cuboidal box l= 12.5 cm
b = 10 cm
h = 8 cm
(i) Lateral surface area of cubical box = 4l² = 4(10)²
= 4 x 100
= 400 cm²
Lateral surface area of cuboidal box = 4 (l + b) x h
= 2 (125 + 10) x 8
= 2 (225) x 8
= 45 x 8 = 360 cm²
(∵ Lateral surface area of cuboidal box) > (Lateral surface area of cuboidal box) (∵ 400 >360)
∴ Required area = (400 – 360) cm2 = 40 cm²
(ii) Total surface area of cubical box = 6l2 = 6(10)2 = 6x 100= 600 cm²
Total surface area of cuboidal box = 2 ( l x b + b x h + h x l)
= 2(125 x 10 + 10 x 8 + 8 x 125)
= 2(125 + 80+ 100)
= 2 x 305
= 610 cm²
∴ (Area of cuboidal box) > (Area of cubical box) (∵ 610 > 600)
Required area = (610 – 600)cm² = 10 cm²

Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Solution:
Dimension for herbarium are
l = 30 cm, b = 25 cm and h = 25 cm
Area of the glass = 2 (l x b + b x h + h x l)
= 2 ( 30 x 25 + 25 x 25 + 25 x 30)
= 2(750 + 625 + 750) = 2 (2125) = 4250cm²
∴ Length of the tape = 4 (l + b + h) = 4(30 + 25 + 25)
[∵ Herbarium is a shape of cuboid length = 4 (1+ b + h)] = 4×80= 320cm

Question 7.
Shanti Sweets Stalll was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
Dimension for bigger box, l = 25 cm, b = 20 cm and b = 5 cm
Total surface area of the bigger size box
=2 ( l x b + b x h + h x l)
= 2(25 x 20 + 20 x 5 + 5 x 25)
= 2(500+ 100+ 125)
= 2(725)= 1450 cm²
Dimension for smaller box, l = 15 cm b = 12 cm and h = 5 cm
Total surface area of the smaller size box = 2(15 x 12 + 12 x 5 + 5 x 15)
= 2 (180 + 60 + 75)= 2 (315)= 630 cm²
Area for all the overlaps = 5% x 2080 cm² = \(\frac { 5 }{ 100 }\) – x 2080 cm² = 104 cm²
Total surface area of both boxes and area of overlaps = (2080 + 104) cm² = 2184 cm²
Total surface area for 250 boxes = 2184 x 250 cm²
Cost of the cardboard for 1000 cm2 = ₹4
Costs of the cardboard for 1 cm = ₹ \(\frac { 4 }{ 1000 }\)
Cost of the cardboard for 2184 x 250 cm²= ₹ \(\frac { 4x 2184 x 250 }{ 1000 }\) = ₹ 2184

Question 8.
Parveen wanted to make a temporary shelter, for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m?
Solution:
Dimension for shetter, l = 4 m, b = 3 m and h = 25 cm
Required area of tarpaulin to make the shelter
= (Area of 4 sides + Area of the top) of the car
= 2(l + b) x h+ ( l x b)
= 2(4+ 3) x 25 + (4 x 3)
= (2 x 7 x 25) + 12 = 35 + 12
= 47 m²

We hope the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.1

February 13, 2023

NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.1.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 11
Chapter Name	Circles
Exercise	Ex 11.1
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.1

Question 1.
Fill in the blanks.
(i) The centre of a circle lies in ___ of the circle. (exterior/interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____ of the circle, (exterior/interior)
(iii) The longest chord of a circle is a ____ of the circle.
(iv) An arc is a ____ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and ____ of the circle.
(vi) A circle divides the plane, on which it lies, in ____ parts.
Solution:
(i) The centre of a circle lies in interior of the circle.
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semi-circle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

Question 2.
Write True or False. Give reason for your answers.
(i) Line segment joining the centre to any point on the circle is a , radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.
Solution:
(i) True. Because all points are equidistant from the centre to the circle.
(ii) False. Because circle has infinitely may equal chords can be drawn.
(iii) False. Because all three arcs are equal, so their is no difference between the major and minor arcs.
(iv) True. By the definition of diameter, that diameter is twice the radius.
(v) False. Because the sector is the region between two radii and an arc.
(vi) True. Because circle is a part of the plane figure.

We hope the NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.1, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.1, drop a comment below and we will get back to you at the earliest.

RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1

February 1, 2023

RD Sharma Class 8 Solutions Chapter 22 Mensuration III (Surface Area and Volume of a Right Circular Cylinder) Ex 22.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1

Other Exercises

Question 1.
Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm.
Solution:
Diameter of the base of cylinder = 7 cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\) cm
Height (h) = 60m
∴ carved surface area = 2πh
= 2 x \(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 2 }\) x 60cm² = 1320cm²
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 1

Question 2.
The curved surface area of a cylindrical road is 132 cm2. Find its length if the radius is 0.35 cm.
Solution:
Curved surface area =132 cm²
Radius (r) = 0.35 cm
Let h be the length of the rod Then 2πrh = 132
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 3

Question 3.
The area of the base of a right circular cylinder is 616 cm2 and its height is 2.5 cm. Find the curved surface area of the cylinder.
Solution:
Let r be the radius of the base of the cylinder, then
Area of the base = πr²
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 4

Question 4.
The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area.
Solution:
Height of the cylinder (h) = 15 cm
Circumference of the base = 88 cm
Let r be the radius of the base, the circumference = 2πr
∴ 2πr = 88 cm …(i)
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 5

Question 5.
A rectangular strip 25 cm x 7 cm is rotated about the longer side. Find the total surface area of the solid thus generated.
Solution:
Dimensions of rectangular strip = 25 cm x 7 cm
By rotating the strip along longer side, a solid is formed whose radius = 7 cm
and height = 25 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 6

Question 6.
A rectangular sheet of paper 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the the total surface area of the cylinder thus generated.
Solution:
By rolling along length wire, we get a cylinder whose circumference of its base = 20 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 8

Question 7.
The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas.
Solution:
Ratio in radii of two cylinders = 2:3
and ratio in their heights = 5:3
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 9

Question 8.
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1: 2. Prove that its height and radius are equal.
Solution:
Let r be the radius and h be the height of a right circular cylinder, then Curved surface area = 2πrh
and total surface area = 2πrh x 2πr² = 2πr (h + r)
But their ratio is 1 : 2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 10
Hence their radius and height are equal.

Question 9.
The curved surface area of a cylinder is 1320 cm² and its base has diameter 21 cm. Find the height of the cylinder.
Solution:
Curved surface area of a cylinder = 1320 cm²
Diameter of its base (d) = 21 cm 21
Radius (r) = \(\frac { 21 }{ 2 }\) cm
Let h be the height of the cylinder
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 11

Question 10.
The height of a right circular, cylinder is 10.5 m. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base.
Solution:
Height of cylinder = 10.5 m
Let r be the radius and h be the height of a right circular cylinder, then Area of its two circular faces = 2π²
and area of curved surface = 2πrh
Now, according to the condition:
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 12

Question 11.
Find the cost of plastering the inner surface of a well at Rs 9.50 per m2, if it is 21 m deep and diameter of its top is 6 m.
Solution:
Diameter of the top of a cylindrical well = 6m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 13
∴ Radius (r) = \(\frac { 6 }{ 2 }\) = 3 m
and depth (h) = 21 m
∴ Curved surface area = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 3 x 21 m² = 396 m²
Rate of plastering = Rs 9.50 per m²
∴ Total cost of plastering = Rs 9.50 x 396 = Rs 3,762

Question 12.
A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of the tin-plating it on the inside at the rate of 50 paise per hundred square centimetre.
Solution:
Diameter of the vessel = 20 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 14

Question 13.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curved surface at Rs 4 per square metre.
Solution:
Diameter of the well = 3.5 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 16

Question 14.
The diameter of a roller is 84 cm and its length is T20 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground ?
Solution:
Diameter of the roller = 84 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 17

Question 15.
Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m, what will be the cost of cleaning them at the rate of Rs 2.50 per square metre ?
Solution:
Diameter of each pillar = 0.50 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 18

Question 16.
The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of the base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.
Solution:
Total surface area of a hollow cylinder opened from both sides = 4620 cm²
Area of base ring = 115.5 cm²
Height of cylinder (h) = 7 cm
Let R be the outer radius and r be the inner radius
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 19

Question 17.
The sum of the radius of the base and height of a solid cylinder is 37 m, if the total surface area of the solid cylinder is 1628 m², find the circumference of its base.
Solution:
Let r be the radius and h be the height of the solid cylinder, then r + h = 37 m …(i)
Total surface area = 1628
⇒ 2πr (r + h) = 1628
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 21

Question 18.
Find the ratio between the total surface area of a cylinder to is curved surface area,given that its height and radius are 7.5 cm and 3.5 cm.
Solution:
Radius (r) of cylinder = 3.5 cm
and height (h) = 7.5 cm
∴ Curved surface area = 2πrh
and total surface area = 2πr (h + r)
∴ Ratio = 2πr (h + r)- 2πrh = (h + r): h = 7.5 + 3.5 : 7.5
⇒ 11 : 7.5
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 22

Question 19.
A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs 3.50 per 1000 cm².
Solution:
Radius of the vessel (r) = 70 cm
and height (h) = 1.4 m = 140 cm
∴ Area of inner and outer curved surfaces and bases = 2 x 2πrh + 2πr²
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 23

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RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2

February 1, 2023

RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2

Other Exercises

Question 1.
The marks obtained by 40 students of class VIII in an examination are given below :
16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13,
21, 13, 15, 19, 24, 16, 3, 23, 5, 12, 18, 8, 12, 6,
8, 16, 5, 3, 5, 0, 7, 9, 12, 20, 10, 2, 23.
Divide the data into five groups, namely 0-5,5-10,10-15,15-20 and 20-25 and prepare a grouped frequency table.
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 1

Question 2.
The marks scored by 20 students in a test are given below :
54, 42, 68, 56, 62, 71, 78, 51, 72, 53, 44, 58, 47, 64, 41, 57, 89, 53, 84, 57.
Complete the following frequency table :
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 2
What is the class interval in which the greatest frequency occurs ?
Solution:

The class in which the greatest frequency is 50-60

Question 3.
The following is the distribution of weights (in kg) of 52 persons :
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 4

(i) What is the lower limit of class 50-60 ?
(ii) Find the class marks of the classes 40-50, 50-60.
(iii) What is the class size ?
Solution:
(i) Lower limit of class 50-60 = 50
(ii) Class marks of 40-50 = \(\frac { 40+50 }{ 2 }\) = \(\frac { 90 }{ 2 }\)
= 45 and of 50-60 = \(\frac { 50+60 }{ 2 }\) = \(\frac { 110 }{ 2 }\) =55
(iii) Class size is 10

Question 4.
Construct a frequency table for the following weights (in gm) of 35 mangoes using the equal class intervals, one of them is 40-45 (45 not included):
30,40,45,32,43,50,55,62,70,70,61,62, 53,52, 50,42,35,37,53,55,65,70, 73, 74,45, 46, 58, 59, 60, 62, 74, 34, 35, 70 ,68.
(i) What is the class mark of the class interval 40-45 ?
(ii) What is the range of the above weights ?
(iii) How many classes are there ?
Solution:
Smallest observation = 30
Greatest observation = 74
Range = 74 – 30 = 44
Now forming the distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 6
(i) Class mark of 40-45
= \(\frac { 40+45 }{ 2 }\) = \(\frac { 85 }{ 2 }\) = 42.5
(ii) Range = 74 – 30 = 44
(iii) Number of classes are 9

Question 5.
Construct a frequency table with class-intervals 0-5 (5 not included) of the following marks obtained by a group of 30 students in an examination :
0, 5, 7,10, 12,15, 20, 22, 25, 27, 8, 11, 17,3, 6, 9,17,19, 21, 29, 31,35,37,40,42, 45, 49, 4, 50, 16.
Solution:
Frequency distribution table.
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 7

Question 6.
The marks scored by 40 students of class VIII in mathematics are given below:
81,55, 68, 79,85,43,29,68,54,73,47, 35, 72,64,95,44,50, 77,64,35,79, 52, 45,54,70,83, 62′, 64,72,92,84,76,63, 43, 54, 38, 73, 68, 52, 54.
Prepare a frequency distribution with class size of 10 marks.
Solution:
Largest marks = 95
Lowest marks = 29
Range = 95 – 29 = 66
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 8

Question 7.
The heights (in cm) of 30 students of class VIII are given below :
155.158.154.158.160.148.149.150.153, 159,161,148,157,153,157,162,159,151, 154,156,152,156,160,152,147,155,163,155,157,153.
Prepare a frequency distribution table with 160-164 as one of the class intervals.
Solution:
Largest height =163
Lowest height =147
Range = 163- 147 = 16
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 9

Question 8.
The monthly wages of 30 workers in a factory are given below :
830,835,890,810,835,836,869,845,898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 806, 840, 890.
Represent the data in the form of a frequency distribution with class size 10.
Solution:
Highest wage = 898
Lowest wage = 804
Range = 898 – 804 = 94
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 10

Question 9.
Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) at 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included):
220, 268, 258, 242, 210, 268, 272,242, 311, 290, 300, 320,319,304,302,318,306,292, 254, 278, 210,240, 280,316,306, 215, 256, 236.
Solution:
Highest wages = 320
Lowest wages = 210
Range = 320-210= 110
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 11

Question 10.
The daily minimum temperatures in degrees Celsius recorded in a certain Arctic region are as follows :
-12.5, -10.8, -18.6, -8.4, -10.8, -4.2, -4.8, -6.7, -13.2, -11.8, -2.3,1.2, 2.6, 0, -2.4, 0, 3.2, 2.7,3.4,0, -2.4, -2.4, 0,3.2, 2.7,3.4, 0,2.4, -5.8, -8.9, -14.6, -12.3, -11.5, -7.8, – 2.9
Represent them as frequency distribution table taking -19.9 to -15 as the first class interval.
Solution:
Lowest temperature = -19.9
Highest temperature = 3.4
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 12

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RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1

February 1, 2023

RD Sharma Class 8 Solutions Chapter 23 Data Handling I (Classification and Tabulation of Data) Ex 23.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1

Other Exercises

Question 1.
Define the following terms :
(i) Observations
(ii) Raw data
(iii) Frequency of an observation
(iv) Frequency distribution
(v) Discrete frequency distribution
(vi) Grouped frequency distribution
(vii) Class-interval
(viii) Class-size
(ix) Class limits
(x) True class limits
Solution:
(i) Observations : Each entry in the given data is called an observation. ‘
(ii) Raw data: A collection of observations by an observer, is called raw data.
(iii) Frequency of an observation : The number of times an observation occurs in the given data is called its frequency.
(iv) Frequency distribution : The presentations of given data in order of magnitude ascending or descending, is called the frequency distribution.
(v) Discrete frequency distribution: When the given data is represented by tally marks after arranging it in an order. This kind of distribution is called discrete frequency distribution.
(vi) Grouped frequency distribution: If the number of data is large, and the difference between the greatest and the smallest observation is large, then we represent then in groups or classes. Such representation of data is called grouped frequency distribution.
(vii) Class intervals: The difference between the upper limit and lower limit of a class is called class interval.
(viii) Class-size : Class intervals are also called the class size. Each size of the same intervals
(ix) Class limits : Every class has two limits : upper limit and lower limit.
(x) True class limits or Exclusive limits : When the upper limit of one is the lower limit of the next interval then these are call true class limits.

Question 2.
The final marks in mathematics of 30 students are as follows :
53, 61, 48, 60, 78, 68, 55, 100, 67, 90, 75, 88,77,37,84,58,60,48,62,56,44,58,52,64, 98, 59, 70, 39, 50, 60
(i) Arrange these marks in the ascending order, 30 to 39 one group, 40 to 49 second group etc.
Now answer the following:
(ii) What is the highest score ?
(iii) What is the lowest score ?
(iv) What is the range ?
(v) If 40 is the pass mark how many have failed ?
(vi) How many have scored 75 or more ?
(vii) Which observations between 50 and 60 have not actually appeared ?
(viii) How many have scored less than 50 ?
Solution:
(i) Arranging the given data in ascending order.
30 to 39 : 37,39
40 to 49 : 44, 48, 48
50 to 59 : 50, 52, 53, 55, 56, 58, 58, 59
60 to 69 : 60, 60, 60, 61, 62, 64, 67, 68
70 to 79 : 70, 75, 77, 78
80 to 89 : 84, 88
90 to 99 : 90, 98
100 to 109 : 100
(ii) Highest score is 100
(iii) Lowest score is 37
(iv) Range is 100 – 37 = 63
(v) If 40 is pass marks then number of failed candidates will be = 2
(vi) Number of students who scored 75 or more = 8
(vii) Between 50 and 60, the observations 51, 54, 57 do not appear.
(viii) Number of students who scored less than 50 = 5

Question 3.
The weights of new born babies (in kg) in a hospital on a particular day are as follows:
2.3, 2.2, 2.1, 2.7, 2.6,3.0, 2.5, 2.9, 2-8,3.1, 2.5, 2.8, 2.7, 2.9, 2.4.
(i) Rearrange the weights in descending order.
(ii) Determine the highest weight:
(iii) Determine the lowest weight.
(iv) Determine the range.
(v) How many babies were born on that day?
(vi) How many babies weigh below 2.5 kg ?
(vii) How many babies weigh more than 2.8 kg ?
(viii) How many babies weigh 2.8 kg ?
Solution:
(i) Arranging the given weights in descending order.
3.1, 3.0, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.5, 2.4, 2.3, 2.2, 2.1
(ii) Highest weight = 3.1 kg
(iii) Lowest weight = 2.1 kg
(iv) Range : 3.1 – 2.1 = 1 kg.
(v) Number of babies born on that day = 15
(vi) Number of babies having weight below 2.5 kg = 4
(vii) Number of babies having weight more than 2.8 kg = 4
(viii) Number of babies weigh 2.8 kg = 2

Question 4.
Following data gives the number of children in 41 families:
1,2,6,5,1,5,1,3,2,6,2,3,4,2,0,0,4,4, 3, 2, 2, 0, 0,1, 2, 2, 4,3, 2,1, 0, 5,1, 2,4, 3, 4, 1, 6, 2, 2.
Represent it in the form of a frequency distribution.
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 1

Question 5.
Prepare a frequency table of the following scores obtained by 50 students in a test:
42, 51, 21, 42, 37. 37, 42, 49, 38, 52, 7, 33, 17, 44, 39, 7, 14, 27, 39, 42, 42, 62, 37, 39, 67, 51, 53, 53, 59, 41, 29, 38, 27, 31, 54,19, 53, 51, 22, 61, 42, 39, 59, 47, 33, 34, 16, 37, 57, 43
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 2

Question 6.
A die was thrown 25 times and following scores were obtained:
1,5,2,4,3,6,1,4,2,5,1,6,2,6,3,5,4,1, 3, 2, 3, 6,1, 5, 2
Prepare a frequency table of the scores.
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 3

Question 7.
In a study of number of accidents per day, the observations for 30 days were obtained as follows:
6,3,5,6,4,3, 2,5,4,2,4,2,1,2, 2,0,5,4,6,1,6,0, 5, 3, 6,1, 5, 5, 2, 6
Prepare a frequency distribution table.
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 4

Question 8.
Prepare a frequency table of the following ages (in years) of 30 students of class VIII in your school:
13,14,13,12,14,13,14,15,13,14,13,14, 16,12,14,13,14,15,16,13,14,13,12,17,13, 12,13,13,13,14
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 5

Question 9.
Following figures relate to the weekly wages (in Rs) of 15 workers in a factory :
300,250,200,250,200,150,350,200,250, 200,150, 300,150, 200, 250 Prepare a frequency table.
(i) What is the range in wages (in Rs) ?
(ii) How many workers are getting Rs 350 ?
(iii) How many workers are getting the minimum wages ?
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 6
(i) Range = 350- 150 = 200
(ii) Number of workers getting Rs 350 = 1
(iii) Number of workers getting minimum wages = 3

Question 10.
Construct a frequency distribution table for the following marks obtained by 25 students in a history test in class Vin of a school:
9,17,12, 20,9,18, 25,17,19,9,12,9,12, 18, 17,19, 20, 25, 9,12,17,19, 19, 20, 9
(i) What is the range of marks ?
(ii) What is the highest mark ?
(iii) Which mark is occurring more frequently ?
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 7
(i) Range = 25 – 9 = 16
(ii) Highest marks = 25
(iii) Marks occurring more frequently = 9

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RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1

February 1, 2023

RD Sharma Class 8 Solutions Chapter 24 Data Handling II (Graphical Representation of Data as Histograms) Ex 24.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1

Question 1.
Given below is the frequency distribution of the heights of 50 students of a class :
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 1
Draw a histogram representing the above data.
Solution:
We represent class intervals along x-axis and frequency along y-axis. Taking suitable intervals along x-axis and y-axis we construct the rectangles as shown in the figure. This is the required histogram.
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 2

Question 2.
Draw a histogram of the following data :
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 3
Solution:
We represent class-intervals along x-axis and frequency along y-axis. Taking suitable intervals along x-axis andy-axis, we construct rectangles as shown in the figure. This is the required histogram.

Question 3.
Number of workshops organized by a school in different areas during the last five years is as follows :
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 5
Draw a histogram representing the above data.
Solution:
We represent years along x-axis and number of workshops along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 6

Question 4.
In a hypothetical sample of 20 people the amounts of money with them were found to be as follows :
114, 108,100, 98, 101,109,117,119, 126, 131, 136, 143, 156, 169, 182, 195, 207, 219, 235, 118.
Draw the histogram of the frequency distribution (taking one of the class intervals as 50-100).
Solution:
Highest sample = 235
Lowest sample = 98
Range = 235-98 = 137
Now frequency distribution table will be as under:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 8
We represent class intervals along x-axis and frequency along j’-axis. Taking suitable intervals, we construct a rectangles as shown in the figure. This is the required histogram.

Question 5.
Construct a histogram for the following data:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 9
Solution:
We represent monthly school fee (in Rs) along x-axis and number of schools along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.

Question 6.
Draw a histogram for the daily earnings of 30 drug stores in the following table :
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 11
Solution:
We represent daily earnings (in Rs) along x-axis and number of stores along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.

Question 7.
Draw a histogram to represent the following data:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 13

Solution:
We represent monthly salary (in Rs) along x-axis and number of teachers along y-axis. Taking suitable intervals we construct rectangles as shown in the figure. This is the required histogram

Question 8.
The following histogram shows the number of literate females in the age group of 10 to 40 years in a town :
(i) Write the age group in which the number of literate female is highest.
(ii) What is the class width ?
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 16
(iii) What is the lowest frequency ?
(iv) What are the class marks of the classes ?
(v) In which age group literate females are least ?
Solution:
(i) The age group in which the number of literate females is 15-20.
(ii) The class width is 5.
(iii) Lowest frequency is 320.
(iv) The class marks of the classes are
\(\frac { 10+15 }{ 2 }\) = \(\frac { 25 }{ 2 }\) =12.5, similarly other class marks will be 17.5,22.5,27.5,32.5,37.5
(v) The least literate females is in the class 10-15

Question 9.
The following histogram shows the monthly wages (in Rs) of workers in a factory:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 17
(i) In which wage-group largest number of workers are being kept ? What is their number ?
(ii) What wages are the least number of workers getting ? What is the number of such workers ?
(iii) What is the total number of workers ?
(iv) What is the factory size ?
Solution:
(i) The largest number of workers are in wage group 950-1000 and is 8.
(ii) The least number of workers are in the wage group 900-950 and is 2.
(iii) Total number of workers is 40 (3 + 7 + 5 + 4 + 2 + 8 + 6 + 5)
(iv) The factory size is 50.

Question 10.
Below is the histogram depicting marks obtained by 43 students of a class :
(i) Write the number of students getting highest marks.
(ii) What is the class size ?
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 18
Solution:
(i) The number of students getting highest marks is 3.
(ii) The class size is 10.

Question 11.
The following histogram shows the frequency distribution of the ages of 22 teachers in a school:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 19
(i) What is the number of eldest and youngest teachers in the school ?
(ii) Which age group teachers are more in the school and which least ?
(iii) What is the size of the classes ?
(iv) What are the class marks of the classes?
Solution:
(i) The number of eldest teacher is 1 and the number of youngest teacher is 2.
(ii) The teachers in age group 35-40 is most.
(iii) Size of classes is 5.
(iv) Class marks of class 20-25 is \(\frac { 20+25 }{ 2 }\)= \(\frac { 45 }{ 2 }\) = 22.5
and similarly others will be 27.5, 32.5, 37.5, 42.5, 47.5, 52.5.

Question 12.
The weekly wages of 30 workers in a factory are given:
830,835,890,810,835,836,869,845,898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Mark a frequency table with intervals as 800-810,810-820 and so on, using tally marks.
Also, draw a histogram and answer the following questions:
(i) Which group has the maximum number of workers ?
(ii) How many workers earn Rs 850 and more ?
(iii) How many workers earn less than Rs 850?
Solution:
The frequency table will be as given below:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 20
We represent wages (in Rs) along x-axis and number of workers along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.
(i) Maximum workers are in the wage group 830-840.
(ii) Number of workers getting Rs 850 and more are 1 + 3 + 1 + 1 + 4 = 10.
(iii) Number of workers getting less than Rs 850 are 3 + 2 + 1 + 9 + 5 = 20
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 21

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RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2

February 1, 2023

RD Sharma Class 8 Solutions Chapter 25 Data Handling III (Pictorial Representation of Data as Pie Charts or Circle Graphs) Ex 25.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2

Other Exercises

Question 1.
The pie-chart given in figure represents the expenditure on different items in constructing a flat in Delhi. If the expenditure incurred on cement is Rs 1,12,500, find the following:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 1
(i) Total cost of the flat,
(ii) Expenditure incurred on labour.
Solution:
Expenditure on cement = Rs 1,12,500
and its central angle = 75°

Question 2.
The pie-chart given in the figure shows the annual agricultural production of an Indian state. If the total production of all the commodities is 81000 tonnes, find the production (in tonnes) of:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 4
(i) Wheat
(ii) Sugar
(iii) Rice
(iv) Maize
(v) Gram
Solution:
Total production = 81000 tonnes

Question 3.
The following pie-chart shows the number of students admitted in different faculties of a college. If 1000 students are admitted in Science answer the following:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 6

(i) What is the total number of students ?
(ii) What is the ratio of students in science and arts ?
Solution:
Students admitted in science = 1000
Central angle = 100°
(i) Total number of students
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 7
∴ Ratio in science and arts = 1000 : 1200 = 5:6

Question 4.
In the figure, the pie-chart shows the marks obtained by a student in an examination. If the student secures 440 marks in all, calculate his marks in each of the given subjects.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 8
Solution:
Total marks secured by a student = 440

Question 5.
In the figure, the pie-charts shows the marks obtained by a student in various subjects. If the student scored 135 marks in mathematics, find the total marks in all the subjects. Also, find his score in individual subjects.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 10
Solution:
Marks obtained in mathematics =135
Central angle = 90°

Question 6.
The following pie-chart shows the monthly expenditure of Shikha on various items. If she spends Rs 16,000 per month, answer the following questions:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 12
(i) How much does she spend on rent ?
(ii) How much does she spend on education ?
(iii) What is the ratio of expenses on food and rent ?
Solution:
Total expenditure per month = Rs 16,000

Question 7.
The pie-chart (as shown in the figure) represents the amount spent on different sports by a sports club in a year. If the total money spent by the club on sports is Rs 1,08,000, find the amount spent on each sport.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 15
Solution:
total amount spent on sports = Rs 1,08,000

Hope given RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data MCQS

January 31, 2023

RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data MCQS

Other Exercises

Question 1.
Which one of the following is not the graphical representation of statistical data:
(a) Bar graph
(b) Histogram
(c) Frequency polygon
(d) Cumulative frequency distribution
Solution:
Cumulative frequency distribution is not a graphical method.

Question 2.
In a frequency distribution, ogives are graphical representation of
(a) Frequency
(b) Relative frequency
(c) Cumulative
(d) Raw data
Solution:
Ogive is the graphical representation of cumulative frequency.

Question 3.
A frequency polygon is constructed by plotting frequency of the class interval and the
(a) upper limit of the class
(b) lower limit of the class
(c) mid value of the class
(d) any values of the class
Solution:
A frequency polygon is constructed by plotting frequency of the class interval and mid value of the class.

Question 4.
In a histogram the area of each rectangle is proportional to
(a) the class mark of the corresponding class interval
(b) the class size of the corresponding class interval
(c) frequency of the corresponding class interval
(d) cumulative frequency of the corresponding class interval
Solution:
In histogram, the area of each rectangle is proportional to frequency of the corresponding class intervals.

Question 5.
In the ‘less than’ type of ogive the cumulative frequency is plotted against
(a) the lower limit of the concerned class interval
(b) the upper limit of the concerned class interval
(c) the mid-value of the concerned class interval
(d) any value of the concerned class interval
Solution:
In the ‘less than’ type of ogive cumulative frequency is plotted against the upper limit of the concerned class interval,

Question 6.
In a histogram the class intervals or the groups are taken along
(a) Y-axis
(b) X-axis
(c) both of X-axis and Y-axis
(d) in between X and Y axis
Solution:
In histogram, the class intervals or the groups are taken along X-axisA

Question 7.
A histogram is a pictorial representation of the grouped data in which class intervals and frequency are respectively taken along
(a) vertical axis and horizontal axis
(b) vertical axis only
(c) horizontal axis only
(d) horizontal axis and vertical axis
Solution:
A histogram is a pictorial representation of the grouped data in which class interval and frequency are respectively taken along horizontal axis and vertical axis.

Question 8.
In a histogram, each class rectangle is constructed with base as
(a) frequency
(b) class-intervals
(c) range
(d) size of the class
Solution:
In a histogram each class rectangle is constructed with base as class intervals.

Hope given RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3

January 31, 2023

RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3

Other Exercises

Question 1.
Construct a histogram for the following data:
RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3 1.1
Solution:

Histogram is given below:

Question 2.
The distribution of heights (in cm) of 96 children is given below. Construct a histogram and a frequency polygon on the same axes.
RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3 2.1
Solution:

Histogram and frequency polygon is given here.
Draw midpoints Of each rectangle at the top and join them to get a frequency polygon.

Question 3.
The time taken, in seconds, to solve a problem by each of 25 pupils is as follows:
16, 20, 26, 27, 28, 30, 33, 37, 38, 40, 42, 43, 46, 46, 46, 48, 49, 50, 53, 58, 59, 60, 64, 52, 20
(a) Construct a frequency distribution for these data, using a class interval of 10 seconds.
(b) Draw a histogram to represent the frequency distribution.
Solution:
Lowest observation = 16,
Highest observation = 64
RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3 3.1
Below is given the histogram

Question 4.
Draw, in the same diagram, a histogram and a frequency polygon to represent the following data which shows the monthly cost of living index of a city in a period of 2 years:
RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3 4.1
Solution:

Question 5.
The following is the distribution of total household expenditure (in Rs.) of manual worker in a city:
RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3 5.1
Draw a histogram and a frequency polygon representing the above data.
Solution:

Here is given histogram and frequency polygon.

Question 6.
The following table gives the distribution of IQ’s (intelligence quotients) of 60 pupils of class V in a school:
RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3 6.1
Draw a frequency polygon for the above data.
Solution:

Now plot the (mid point, frequency) on the graph and join them to get a frequency polygon.

Question 7.
Draw a histogram for the daily earnings of 30 drug stores in the following table:
RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3 7.1
Solution:

Hisotgram is given below:

Question 8.
The monthly profits (in t) of 100 shops are distributed as follows:
RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3 8.1
Draw a histogram for the data and show the frequency polygon for it.
Solution:

Histogram and frequency polygon is given below:

Hope given RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Category: CBSE Class 9

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RD Sharma Class 8 Solutions Chapter 23 Data Handling I (Classification and Tabulation of Data) Ex 23.1

RD Sharma Class 8 Solutions Chapter 24 Data Handling II (Graphical Representation of Data as Histograms) Ex 24.1

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RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data MCQS

RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.3