Class 9 Maths RD Sharma Solutions

RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.2

Other Exercises

• RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1
• RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.2
• RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data MCQS

Question 1.
Define cumulative frequency distribution.
Solution:
Cumulative frequency : In a discrete frequency distribution, the cumulative frequency of a particular value of the variable is the total of all the frequencies of the values of the variable which are less than or equal to the particular value.

Question 2.
Explain the difference between a frequency distribution and a cumulative frequency distribution.
Solution:
The number of times an observation occurs in the given data, is called frequency of the
observation while a cumulative frequency of a particular value of the variable is the total of all the frequencies of the values of the variable which are less than or equal to the particular value.

Question 3.
The marks scored by 55 students in a test are given below:

Prepare a cumulative frequency table.
Solution:
Cumulative frequency table is given below:

Question 4.
Following are the ages of 360 patients getting medical treatment in a hospital on a day:

Construct a cumulative frequency distribution.
Solution:
Cumulative frequency distribution table (less than) is given below:

Question 5.
The water bills (in rupees) of 32 houses in a certain street for the period 1.1.98 to 31.3.98 are given below:
56, 43, 32, 38, 56, 24, 68, 85, 52, 47, 35, 58, 63, 74, 27, 84, 69, 35, 44, 75, 55, 30, 54, 65, 45, 67, 95, 72, 43, 65, 35, 59.
Tabulate the data and present the data as a cumulative frequency table using 70-79 as one of the class intervals.
Solution:
Highest bill = 95 Lowest bill = 24
Range = 95 – 24 = 71
Cumulative frequency table is given below:

Question 6.
The number of books in different shelves of a library are as follows:
30, 32, 28, 24, 20, 25, 38, 37, 40, 45, 16, 20, 19, 24, 27, 30, 32, 34, 35, 42, 27, 28, 19, 34,
38, 39, 42, 29, 24, 27, 22, 29, 31, 19, 27, 25, 28, 23, 24, 32, 34, 18, 27, 25, 37, 31, 24, 23,
43, 32, 28, 31, 24, 23, 26, 36, 32, 29, 28, 21.
Prepare a cumulative frequency distribution table using 45-49 as the last class interval.
Solution:
Greatest number = 43
Lowest number =16
Range = 43 – 16 = 27
Cumulative frequency table is given below:

Question 7.
Given below are the cumulative frequencies showing the weights of 685 students of a school. Prepare a frequency distribution table.

Solution:
The frequency table of the given cumulative frequency is given below:

Question 8.
The following cumulative frequency distribution table shows the daily electricity consumption (in kW) of 40 factories in an industrial state:

(i) Represent this as a frequency distribution table.
(ii) Prepare a cumulative frequency table.
Solution:
(i) The frequency distribution table is given below:

(ii) Now cumulative frequency table (more then)

Question 9.
Given below is a cumulative frequency distribution table showing the ages of people living in a locality.

Prepare a frequency distribution table.
Solution:
Frequency distribution table is given below:

Hope given RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.2 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 22 Tabular Representation of Statistical Data Ex 22.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1

Other Exercises

• RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1
• RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.2
• RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data MCQS

Question 1.
What do you understand by the word ‘statistics’ in (i) singular form, (ii) plural forms.
Solution:
The word ‘statistics’ is used in both its singular as well as its plural senses.
In singular sense, statistics piay be defined as the science of collection, presentation, analysis and interpretation of numerical data.
In plural sense, statistics means numerical facts or observations collected with definite purpose.
For example, income and expenditure of persons in a particular locality, number of males and females in a particular town are statistics.

Question 2.
Describe some fundamental characteristics of statistics.
Solution:
Statistics in plural sense have the following characteristics:
(i) A single observation does not form statistics. Statistics are a sum total of observations.
(ii) Statistics are expressed quantitatively and not qualitatively.
(iii) Statistics are collected with a definite purpose.
(iv) Statistics in an experiment are comparable and can be classified into various groups.

Question 3.
What are (i) primary data? (ii) secondary data? Which of the two – the primary or the secondary data – is more reliable and why?
Solution:
(i) Primary data : When an investigator collects the data himself with a definite plan Or design in his mind, it is called primary data.
(ii) Secondary data : Data which are not originally collected rather obtained from the published or unpublished, sources are called secondary data.
Primary data are reliable and relevent because they are original in character and are collected by some individuals or by some institutions or by research bodies.

Question 4.
Why do we group data?
Solution:
When the number of observations is large, then arranging data in ascending or descending order is tedius job and it does not tell us much except minimum or maximum(s) of data. So, to make it easily understandable and clear, we tabulate data in the form of a table.

Question 5.
Explain the meaning of the following terms:
(i) variate
(ii) class intervals
(iii) class size
(iv) class mark
(v) frequency
(vi) class limits
(vii) true class limits.
Solution:
(i) Variate : The observations of a data are called variate.
(ii) Class intervals : When the presentations of data in classes, or groups, then groups are called classes or class intervals.
(iii) Class size : The difference between upper limit and lower limit is called class size.
(iv) Class mark : The mean of lower limit and upper limit is called class mark or mid value. Therefore class mark
= \(\frac { lower limit + upper limit }{ 2 }\)
(v) Frequency : The number of times an observation occurs in the given data, is called frequency of that observation.
(vi) Class limits : Every class has two limits : lower limit and upper limit.
(vii) True class limits : Whenever inclusive method is used, it is necessary tp make an adjustment to determine the correct class intervals, and to have continuity. If a-b is a class in inclusive method, then in order to change it into exclusive method, it becomes
a – \(\frac { h }{ 2 }\) – b + \(\frac { h }{ 2 }\) where h = \(\frac { 1 }{ 2 }\) [lower limit of a class – upper limit of previous class]

Question 6.
The ages of ten students of a group are given below. The ages have been recorded in years and months.
8-6, 9-0, 8-4, 9-3, 7-8, 8-11, 8-7, 9-2, 7¬10, 8-8
(i) What is the lowest age?
(ii) What is the highest age?
(iii) Determine the range?
Solution:
From the given data
(i) Lowest age is 7 years 8 months
(ii) The highest age is 9 years, 3 months
(iii) Range = Highest term – Lowest term
= 9 years 3 months – 7 years 8 months
= 1 years 7 months

Question 7.
The monthly pocket money of six friends is given below:
₹45, ₹30, ₹40, ₹50, ₹25, ₹45
(i) What is the highest pocket money?
(ii) What is the lowest pocket money?
(iii) What is the range?
(iv) Arrange the amounts of pocket money in ascending order.
Solution:
From the given data
(i) Highest pocket money = ₹50
(ii) Lowest pocket money = ₹25
(iii) Range = Highest term – Lowest term = ₹50-₹25 = ₹25
(iv) In ascending order : ₹25, ₹30, ₹40, ₹45, ₹45, ₹50

Question 8.
Write the class-size in each of the following:
(i) 0-4,5-9,10-14
(ii) 10-19, 20-29, 30-39
(iii) 100-120, 120-140, 160-180
(iv) 0-0.25, 0.25-0.50, 0.50-0.75
(v) 5-5,01, 5.01-5.02, 5.02-5.03
Solution:
(i) In 0-4, 5-9, 10-14
0-4 means from 0 to 4, similarly 5-9 means 5 to 9 and 10-14 means 10 to 14
class-size is 5
(ii) In 10-19, 20-29, 30-39
Here 10-19, means 10 to 19, 20-29 means 20 to 29 and 30-39 means 30 to 39
Class-size = 10
(iii) 100-120, 120-140, 160-180
Here 100-120, 120-140, 160-180
Then 120 – 100 = 20, 140 – 120 = 20, 180 – 160 = 20 is class-size = 20
(iv) 0-0.25, 0.25-0.50, 0.50-0.75
Here 0.25 – 0 = 0.25, 0.50 – 0.25 = 0.25 and 0.75 – 0.50 = 0.25
∴ Class-size = 0.25
(v) 5-5.01, 5.01-5.02, 5.02-5.03
Here 5.01 – 5 = 0.01, 5.02 – 5.01 = 0.01 and 5.03 – 5.02 = 0.01
∴ Class-size = 0.01

Question 9.
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 in ascending order:
37, 39, 44, 48, 48, 50, 52, 53, 55, 56, 58, 58, 59, 60, 60, 60, 61, 62, 64, 67, 68, 70, 75, 77, 78, 84, 88, 90, 98, 100

(ii) Highest score =100
(iii) Lowest score = 37
(iv) Range =  Highest score – Lowest score = 100 – 37 = 63
(v) If 40 is marks, then the number of students who failed = 2
(vi) No. of students who scored 75 or more = 8
(vii) Between 50 and 60, the scores which are missing 51, 54, 57
(viii) Number of students who scored less then 50 = 2 + 3 = 5

Question 10.
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 bom 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) Weights in descending order are
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
(iii) Lowest weight = 2.1
(iv) Range = Highest weight – Lowest wight = 3.1 – 2.1 = 1.0
(v) No. of babies who bom = 15
(vi) No. of babies whose weights are below 2.5 kg = 4
(vii) No. of babies whose weight are more than 2.8 kg = 4.
(viii) No. of babies whose weight is 2.8 kg = 2

Question 11.
The number of runs scored by a cricket player in 25 innings are as follows:
26, 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47.
(i) Rearrange these runs in ascending order.
(ii) Determine the player, is highest score.
(ii) How many times did the player not score a run?
(iv) How many centuries did he score?
(v) How many times did he score more than 50 runs?
Solution:
(i) Arranging in ascending order
0, 0, 0, 15, 26, 34, 35, 39, 42, 42, 47, 48, 48, 53, 54, 64, 64, 67, 71, 82, 84, 94, 105, 124, 139
(ii) Highest score is 139
(iii) 3 times when his score is 0
(iv) No. of times, he made century = 3
(v) No. of times his score more then 50 runs = 12

Question 12.
The class size of a distribution is 25 and the first class-interval is 200-224. There are seven class-intervals.
(i) Write the class-intervals.
(ii) Write the class-marks of each interval
Solution:
Class size = 25
First class is 200-224
∴ Number of class = 7
∴ Class interval will be
(i) 200-224, 225-249, 250-274, 275-299, 300-324, 325-349, 350-374

Question 13.
Write the class size and class limits in each of the following:
(i) 104, 114, 124, 134, 144, 154 and 164
(ii) 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97 and 102
(iii) 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5
Solution:
(i) 104, 114, 124, 134, 144, 154 and 164
Here class size = 114 – 104 = 10
Here in first class, 104 – \(\frac { 10 }{ 2 }\)
= 104 – 5 = 99 and 104 + 5 = 109
∴ Class will be 99-109
In this way other classes will be 109-119, 119-129, 129-139, 139-149, 149-159, 159¬169
(ii) In 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97, 102
Here class size = 52 – 47 = 5
∴ In first class, 47 – \(\frac { 5 }{ 2 }\) and 47 + \(\frac { 5 }{ 2 }\)
= 44.5-49.5
In this way other classes will be = 49.5-54.5, 54.5-59.5, 59.5-64.5, 64.5-69.5,69.5-74.5, 74.5-79.5, 79.5-84.5, 84.5-89.5, 89.5-94.5,94.5-99.5, 99.5-104.5 070 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5
Here class interval (size) = 17.5 – 12.5 = 5
In the class = 12.5 – \(\frac { 5 }{ 2 }\) , 12.5 + \(\frac { 5 }{ 2 }\)
⇒ 12.5 – 2.5, 12.5 + 2.5
⇒ 10, 15
∴ First class will be 10-15
In this way other classes 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50

Question 14.
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 is given below:

Question 15.
The marks scored by 40 students of class IX 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:
Maximum marks = 95,
minimum marks = 29,
range = 95 – 29 = 66
Frequency distribution is given below

Question 16.
The heights (in cm) of 30 students of class IX 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:
Maximum height = 163 cm,
minimum height = 147 cm,
range 163 – 147 = 16
Frequency distribution of the given data is given below

Question 17.
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 = ₹890,
lower wage = ₹804,
range = 890 – 804 = 86
Frequency distribution table is given below

Question 18.
The daily maximum temperatures (in degree Celsius) recorded in a certain city during the month of November are as follows:
25.8, 24.5, 25.6, 20.7, 21.8, 20.5, 20.6, 20.9, 22.3, 22.7, 23.1, 22.8, 22.9, 21.7, 21.3, 20.5, 20.9, 23.1, 22.4, 21.5, 22.7, 22.8, 22.0, 23.9, 24.7, 22.8, 23.8, 24.6, 23.9, 21.1
Represent them as a frequency distribution table with class size 1°C.
Solution:
Maximum temperature = 25.8
Minimum temperature = 20.5
Range = 25.8 – 20.5 = 5.3
Frequency distribution table is given below:

Question 19.
Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 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:
Maximum wages = ₹320
Minimum wages = ₹210
Range = 320 – 210 = 110
Required frequency table is given below:

Question 20.
The blood groups of 30 students of class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
Represent this data in the form of a frequency distribution table. Find out which is the most common and which is the rarest blood group among these students. (NCERT)
Solution:
Frequency distribution table is given below:

Question 21.
Three coins were tossed 30 times. Each time the number of heads occuring was noted down as follow:
0 1 2 2 1 2 3 1 3 0
1 3 1 1 2 2 0 1 2 1
3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above. (NCERT)
Solution:
Frequency distribution table is given below:

Question 22.
Thirty children were asked about the number of hours they watched T.V. programmes in the previous week. The results were found as follows:
1 6 2 35 12 5 8 4 8
10 3 4 12 2 8 15 1 17 6
3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
(ii) How many children watched television for 15 or more hours a week? (NCERT)
Solution:
Highest number of hours = 17
Lowest number of hours = 1
(i) The frequency table is given below:

(ii) No. of children watching T.V. for 15 or more hours a week = 2

Question 23.
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:
Maximum temperature = -18.6
Minimum temperature = 3.4
Range = 3.4 – (-18.6) = 3.4 + 18.6 = 22
The required frequency distribution table is given below:

Hope given RD Sharma Class 9 Solutions Chapter 22 Tabular Representation of Statistical Data Ex 22.1 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 21 Surface Areas and Volume of a Sphere MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1
• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2
• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS
• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS

Mark the correct alternative in each of the following:
Question 1.
In a sphere, the number of faces is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
Number of faces of a sphere is 1 (a)

Question 2.
The total surface area of a hemisphere of radius r is
(a) πr²
(b) 2πr²
(c) 3πr²
(d) 4πr²
Solution:
Total surface area of a hemisphere is 37πr² (c)

Question 3.
The ratio of the total surface area of a sphere and a hemisphere of same radius is
(a) 2 : 1
(b) 3 : 2
(c) 4 : 1
(d) 4 : 3
Solution:
Total surface area of a sphere = 4πr²
and total surface area of a hemisphere = 3m²
∴ Ratio 4πr²: 3πr²
= 4 : 3 (d)

Question 4.
A sphere and a cube are of the same height. The ratio of their volumes is
(a) 3 :4
(b) 21 : 11
(c) 4 : 3
(d) 11 : 21
Solution:
Let r be the height of a sphere and cube

Question 5.
The largest sphere is cut off from a cube of side 6 cm. The volume of the sphere will be
(a) 27π cm³
(b) 36π cm³
(c) 108π cm³
(d) 12π cm³
Solution:
Side of cube = 6 cm
∴ Diameter of sphere cut off from it = 6 cm

Question 6.
A cylmderical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius. The number of balls will be
(a) 4
(b) 3
(c) 6
(d) 8
Solution:
Let r be the radius of a cylindrical rod = r
Then its height (h) = 8r
Volume = πr²h = πr² x 8r = 8πr³
Radius of spherical ball = r

Question 7.
If the ratio of volumes of two spheres is 1 : 8, then the ratio of their surface areas is
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16
Solution:
Let r₁ and r₂ be the radius of two spheres

Question 8.
If the surface area of a sphere is 144π m² then its volume (in. m3) is
(a) 288π
(b) 316π
(c) 300π
(d) 188π
Solution:
Surface area of a sphere = 144π m²
Let r be the radius, then
4πr² = 144π

Question 9.
If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball (in sq. cm) is
(a) 100π
(b) 75π
(c) 60π
(d) 50π
Solution:
Radius of a sphere (r) = 10 cm

Question 10.
If a sphere is inscribed in a cube, then the ratio of the volume of the sphere to the volume of the cube is
(a) π : 2
(b) π : 3
(c) π : 4
(d) π : 6
Solution:
Let side of a cube = a
Then volume of cube = a³
The diameter of inscribed sphere = a

Question 11.
If a solid sphere of radius r is melted and cast into the shape of a solid cone of height r, then the radius of the base of the cone is
(a) 2r
(b) 3r
(c) r
(d) 4r
Solution:
Radius of a sphere = r

Question 12.
A sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder. If the radius of the sphere is r, then the volume of the cylinder is

Solution:
Radius of sphere = r

Question 13.
The ratio between the volume of a sphere and volume of a circumscribing right circular cylinder is
(a) 2 : 1
(b) 1 : 1
(c) 2 : 3
(d) 1 : 2
Solution:
Let r be the radius of the sphere, then 4
Volume = \(\frac { 4 }{ 3 }\)πr³
Diameter of circumscribed cylinder = 2r
∴ Radius = r
and height (h) = 2r

Question 14.
A cone and a hemisphere have equal bases and equal volumes the ratio of their heights is
(a) 1 : 2
(b) 2 : 1
(c) 4 : 1
(d) \(\sqrt { 2 } \) : 1
Solution:
Let radius of hemisphere and a cone be r

Question 15.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is
(a) 1 : 2 : 3
(b) 2 : 1 : 3
(c) 2 : 3 : 1
(d) 3 : 2 : 1
Solution:
∵ Bases of a cone, hemisphere and a cylinder are same
Let radius of each = r
and height of each = r

Hope given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere 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 21 Surface Areas and Volume of a Sphere VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1
• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2
• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS
• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS

Question 1.
Find the surface area of a sphere of radius 14 cm.
Solution:
Radius of a sphere (r) = 14 cm
∴ Surface area = 4πr² = 4 x \(\frac { 22 }{ 7 }\) x 14 x 14 cm²
= 2464 cm³

Question 2.
Find the total surface afea of a hemisphere of radius 10 cm.
Solution:
Radius of hemisphere (r) = 10 cm
∴ Total surface area = 3πr²

Question 3.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
Surface area of a sphere = 154 cm²

Question 4.
The hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7 m. Find the area available to the motor cyclist for riding.
Solution:
Diameter of hollow sphere = 7 m

Question 5.
Find the volume of a sphere whose surface area is 154 cm².
Solution:
Surface area of a sphere = 154 cm²

Question 6.
How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?
Solution:
Edge of a solid cube = 44 cm
∴ Volume = a² = (44)² cm²
= 44 × 44 × 44 cm³
Diameter of a spherical bullet = 4 cm

Question 7.
If a sphere of radius 2r has the same volume as that of a cone with circular base of radius r, then find the height of the cone.
Solution:
Radius of a sphere (R) = 2r

Question 8.
If a hollow sphere of intefnal and external diameters 4 cm and 8 cm respectively melted into a cone of base diameter 8 cm, then find the height of the cone.
Solution:
Internal diameter of a hollow sphere = 4cm
∴ Internal radius = \(\frac { 4 }{ 2 }\) = 2 cm
Similarly the outer radius (R) = \(\frac { 8 }{ 2 }\) = 4 cm
∴ Volume of melted used in hollow sphere

Question 9.
The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height of the cone.
Solution:
Radius of a sphere (r) = 5 cm
∴ Surface area = 4πr²
= 4π x 5 x 5 = 100π cm²
Radius of cone (r₁) = 4 cm

Question 10.
If a sphere is inscribed in a cube, find the ratio of the volume of cube to the volume of the sphere.
Solution:
Let edge of a cube = a
Then its volume = a³
∵ A sphere is inscribed in the cube
∴ Diameter of sphere = a

Hope given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 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 21 Surface Areas and Volume of a Sphere Ex 21.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2

Other Exercises

• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1
• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2
• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS
• RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS

Question 1.
Find the volume of a sphere whose radius is
(i) 2 cm
(ii) 3.5 cm
(iii) 10.5 cm
Solution:
(i) Radius of sphere (r) = 2 cm

Question 2.
Find the volume of a sphere whose diameter is,
(i) 14 cm
(ii) 3.5 dm
(iii) 2.1 m
Solution:
(i) Diameter of a sphere = 14 cm

Question 3.
A hemspherical tank has inner radius of 2.8 m. Find its capacity in litres.
Solution:
Radius of hemispherical tank (r) = 2.8 m

Question 4.
A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.
Solution:
Thickness of steel = 0.25 cm = \(\frac { 1 }{ 4 }\)cm
Inside radius of the hemispherical bowl (r) = 5 cm
∴ Outer radius (R) = 5 + 0.25 = 5.25 cm
∴ Volume of the steel used = \(\frac { 1 }{ 4 }\)π(R³ – r³)

Question 5.
How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?
Solution:
Edge of cube (r) = 22 cm
∴ Volume = a³ = (22)³ cm³
= 22 x 22 x 22 = 10648 cm³
Diameter of a bullet = 2 cm

Question 6.
A shopkeeper has one laddoo of radius 5 cm. With the same material how many laddoos of radius 2.5 cm can be made?
Solution:
Radius of bigger laddoo (R) = 5 cm

Question 7.
A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. It the diameters of two balls be \(\frac { 3 }{ 2 }\) cm and 2 cm, find the diameter of the third ball.
Solution:
Diameter of a spherical ball of lead = 3 cm

Question 8.
A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises \(\frac { 5 }{ 3 }\) cm. Find the radius of the cylinder.
Solution:
Radius of sphere (r1) = 5 cm

Level of water rises in the cylinder after immersing the sphere in it
∴ Height of water level = \(\frac { 5 }{ 3 }\) cm
Let r be radius of the cylinder, then Volume of water = Volume of the sphere

Question 9.
If the radius of a sphere is doubled, what is the ratio of the volumes of the first sphere to that of the second sphere?
Solution:
Let r2 be the radius of the given sphere
then volume = \(\frac { 4 }{ 3 }\) πr³
By doubling the radius the radius of the new sphere = 2r

Question 10.
A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.
Solution:
Radius of hemispherical bowl (r) = 3.5 cm

Question 11.
A cylinder whose height is two thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.
Solution:
Radius of a sphere (r) = 4 cm

Question 12.
A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.
Solution:
Radius of hemispherical bowl (r) = 6 cm

Question 13.
The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.
Solution:
Diameter of a copper sphere = 18 cm

Question 14.
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.
Solution:
Diameter of a sphere = 6 cm

Question 15.
The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height 2 \(\frac { 2 }{ 3 }\) cm. Find the diameter of the cylinder.
Solution:
Internal radius of the hollow spherical shell (r) = 3 cm
and external radius (R) = 5 cm

Question 16.
A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Solution:
Radius of hemisphere (r) = 7 cm

Question 17.
A hollow sphere of internal and external radius 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.
Solution:
Internal radius of a hollow sphere (r) = 2 cm
and external radius (R) = 4 cm
∴ Volume of the metal used

Question 18.
A metallic sphere of radius 10.5 cm is melted and thus recast into small cones each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.
Solution:
Radius of a metallic sphere (R) = 10.5 cm

Question 19.
A cone and a hemisphere have equal bases and equal volumes. Find the ratio Of their heights.
Solution:
Let r be the radius and h be the height of the cone, hemisphere

Question 20.
The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.
Solution:
By carving a largest sphere out of the cube, the diameter of the sphere = 10.5

Question 21.
A cube, of side 4 cm, contains a sphere touching its sides. Find the volume of the gap in between.
Solution:
Side of cube = 4 cm
∴ Volume = (side)³ = 4x4x4 = 64 cm³
Diameter of the largest sphere touching its sides = 4 cm

Question 22.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank. (NCERT)
Solution:
Thickness of hemispherical tank = 1 cm
Inner radius (r) = 1 m = 100 cm

Question 23.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule? (NCERT)
Solution:
Diameter of a medicine spherical capsule = 3.5 mm

Question 24.
The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? (NCERT)
Solution:

Question 25.
A cone and a hemisphere have equal bases and equal volumes. Find the ratio in their heights.
Solution:
Let r be the radius of cone and hemisphere and let h be the height of the cone then

Question 26.
A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?
Solution:
Radius of cylinderical tub (r) = 16 cm
Height of water in it (h) = 30 cm

Question 27.
A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use π = 22/7).
Solution:
Radius of cylinder (r) = 12 cm
Depth of water in it (h) = 20 cm
By dropping a ball, the water level rose by 6.75 cm

Question 28.
A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres?
Solution:
Radius of cylinderical jar (r) = 6 cm
Level of oil in it (h) = 2 cm

Question 29.
A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm eacfy are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?
Solution:
Diameter of measuring jar = 10 cm

Now after swing the ball in the water of jar Let volume of water raised, by h cm

Question 30.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2:3.
Solution:
∵ Bases and heights of a cones hemisphere and a cylinder are equal
Let r be the radius and h be their heights

Question 31.
A cylinderical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?
Solution:
Radius of the cylinderical tub (r) = 12 cm
Depth of water in it (h) = 20 cm
By dropping a spherical ball in it, the water raised by 6.75 cm

Question 32.
A sphere, a cylinder and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.
Solution:
Diameter of a sphere, cylinder and a cone are equal
Let each as diameter = 2r
Then radius of each = r
Height of cylinder = diameter = 2r
and height of cone = 2r
Now volume of sphere = \(\frac { 4 }{ 3 }\)πr³
Volume of cylinder = πr²h

Hope given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.2 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.