Maths RS Aggarwal Solutions Class 9

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H

Question 1.
Solution:
(i) First eight natural numbers are 1, 2, 3, 4, 5, 6, 7, 8
∴ Mean = \(\frac { 1+2+3+4+5+6+7+8 }{ 8 } \) = \(\frac { 36 }{ 8 } \) = \(\frac { 9 }{ 2 } \) = 4.5
(ii) First ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
∴ Mean = \(\frac { 1+3+5+7+9+11+13+15+17+179 }{ 10 } \) = \(\frac { 100 }{ 10 } \) = 10
(iii) First five prime numbers are 2, 3, 5, 7, 11.
∴ Mean = \(\frac { 2+3+5+7+11 }{ 5 } \) = \(\frac { 28 }{ 5 } \) = 5.6
(iv) First six even numbers are 2, 4, 6, 8, 10, 12
∴ Mean = \(\frac { 2+4+6+8+10+12 }{ 10 } \) = \(\frac { 42 }{ 6 } \) = 7
(v) First seven multiples of 5 are 5, 10, 15, 20, 25, 30, 35
∴ Mean = \(\frac { 5+10+15+20+25+30+35 }{ 7 } \) = \(\frac { 140 }{ 7 } \) = 20
(vi) All the factors of 20 are, 1, 2, 4, 5, 10, 20
∴ Mean = \(\frac { 1+2+4+5+10+20 }{ 6 } \) = \(\frac { 42 }{ 6 } \) = 7

Question 2.
Solution:
No. of families (n) = 10
Sum of children (∑x1) = 2 + 4 + 3 + 4 + 2 + 0 + 3 + 5 + 1 + 6 = 30
∴ Mean \(\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 30 }{ 10 } =3\)

Question 3.
Solution:
Here number of days (n) = 7
Number of books (∑x1) = 105 + 216 + 322 + 167 + 273 + 405 + 346 = 1834
∴ Mean \(\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 1864 }{ 7 } =262 books\)

Question 4.
Solution:
Number of days (n) = 6
Sum of temperature (∑x) = 35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9 = 179.4°F
∴ Mean temperature = \(\frac { \sum { x } }{ n } =\frac { 179.4 }{ 6 } \)=29.9°F

Question 5.
Solution:
Number of students (n) = 12
Sum of marks (∑x) = 64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1 = 474
= \(\frac { \sum { x1 } }{ n } =\frac { 474 }{ 12 } \) = 39.5

Question 6.
Solution:
Here, n = 6
and arithmetic mean =13
Total sum = 13 x 6 = 78.
But sum of 7 + 9+ 11 + 13 + 21=61
Value of x = 78 – 61 = 17
Hence x = 17 Ans.

Question 7.
Solution:
Let x₁, x₂, x₃, … x₂₄ be the 24 numbers
\(\frac { x1+x2+x3+…..+x24 }{ 24 } =35\)
=> x₁ + x₂ + x₃ +….+ x₂₄ = 35 x 24

Question 8.
Solution:
Let x₁ + x₂ + x₃……..x₂₀ be the 20 numbers

Question 9.
Solution:
Let x₁, x₂, x₃ … x₁₅ be the numbers
Mean = \(\frac { x1+x2+x3+…..+x15 }{ 15 } = 27\)

Question 10.
Solution:
Let x₁, x₂, x₃ … x₁₂ be the numbers
Mean = \(\frac { x1+x2+x3+…..+x12 }{ 12 }\) = 40
=> x1+x2+x3+….+x12 = 40 X 12 =480
Now,new numbers are \(\frac { x1 }{ 8 } ,\frac { x2 }{ 8 } ,\frac { x3 }{ 8 } ,..\frac { x12 }{ 8 } \)
Mean = \(\frac { \frac { x1 }{ 8 } +\frac { x2 }{ 8 } +\frac { x3 }{ 8 } +…..+\frac { x12 }{ 8 } }{ 12 } \)
= \(\frac { 1 }{ 8 } \frac { \left( x1+x2+x3+….x12 \right) }{ 12 } \)
= \(\frac { 480 }{ 8X12 } \) = 5
Mean of new numbers =5

Question 11.
Solution:
Let x₁, x₂, x₃,…..x₂₀ are the numbers
Mean = \(\frac { x1+x2+x3+…x20 }{ 20 }\) = 18
=> x₁ + x₂ + x₃ +….+ x₂₀ = 18 X 20 =360

Question 12.
Solution:
Mean weight of 6 boys = 48 kg
Their total weight = 48 x 6 = 288 kg
Weights of 5 boys among them, are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg
Sum of weight of 5 boys = (51 + 45 + 49 + 46 + 44) kg = 235 kg
Weight of 6th boy = 288 kg – 235 kg = 53 kg Ans.

Question 13.
Solution:
Mean of 50 students = 39
Total score = 39 x 50 = 1950
Now correct sum of scores = 1950 – Wrong item + Correct item= 1950 – 23 + 43
= 1950 + 20 = 1970
Correct mean = \(\frac { 1970 }{ 50 } \) = 39.4 Ans.

Question 14.
Solution:
Mean of 100 items = 64
The sum of 100 items = 64 x 100 = 6400
New sum of 100 items = 6400 + 36 + 90 – 26 – 9 = 6526 – 35 = 6491,
Correct mean = \(\frac { 6491 }{ 100 }\) =64.91 = 64.91 Ans.

Question 15.
Solution:
Mean of 6 numbers = 23
Sum of 6 numbers = 23 x 6 = 138
Excluding one number, the mean of remaining 5 numbers = 20
Total of 5 numbers = 20 x 5 = 100
Excluded number = 138 – 100 = 38 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D 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.

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H

Question 1.
Solution:
It is clear that the given frequency distribution is in exclusive form, we represent the daily wages (in rupees) along x-axis and no. of workers along y-axis. Then we construct rectangles with class intervals as bases and corresponding frequencies as heights as shown given below:

Question 2.
Solution:
We shall take daily earnings along x-axis and number of stores along y-axis. Then we construct rectangles with the given class intervals as bases and corresponding frequency as height as shown in the graph.

Question 3.
Solution:
We shall take heights along x-axis and number of students along y-axis. Then we shall complete the rectangles with the given class intervals as bases and frequency as height and complete the histogram as shown below :

Question 4.
Solution:
We shall take class intervals along x-axis and frequency along y-axis Then we shall complete the rectangles with given class interval as bases and frequency as heights and complete the histogram as shown below.

Question 5.
Solution:
The given frequency distribution is in inclusive form first we convert it into exclusive form at given below.


Now, we shall take class intervals along x-axis and frequency along y-axis and draw rectangles with class intervals as bases and frequency as heights and complete the histogram as shown.

Question 6.
Solution:

Question 7.
Solution:
In the given distribution class intervals are different size. So, we shall calculate the adjusted frequency for each class. Here minimum class size is 4. We know that adjusted
frequency of the class is \(\frac { max\quad class\quad size }{ class\quad size\quad of\quad this\quad class } \)  x its frequency

Question 8.
Solution:
We shall take two imagined classes one 0-10 at the beginning with zero frequency a id other 70-80 at the end with zero frequency.


Now, plot the points” (5, 0), (15, 2), (25, 5), (35, 12), (45, 19), (55, 9), (65, 4), (75, 0) on the graph and join them its order to get a polygon as shown below.

Question 9.
Solution:
We take Age (in years) along x-axis and number of patients along y-axis. First we complete the histogram and them join the midpoints of their tops in order

We also take two imagined classes. One 0-10 at the beginning and other 70-80 at the end and also join their midpoints to complete the polygon as shown.

Question 10.
Solution:
We take class intervals along x-axis and frequency along y-axis.
First we complete the histogram and then join the midpoints of the tops of adjacent rectangles in order. We shall take two imagined classes 15-20 at the beginning and 50-55 at the end.
We also join their midpoints to complete the polygon as shown

Question 11.
Solution:
We represent class interval on x-axis and frequency on y-axis. First we construct the histogram and then join the midpoints of the tops of each rectangle by line segments. We shall take two imagined class i.e. 560-600 at the beginning and 840-900 at the end. Join their midpoints also to complete

Question 12.
Solution:
We will take the imagined class 11-0 at the beginning and 61-70 at the end. Each with frequency zero. Thus, we have,


Now we will mark the points A (-5.5, 0), B(5.5, 8), C(15.5, 3) D(25.5, 6), E(35.5, 12), F(45.5, 2), G(55.5, 7) and H(65.5, 0) on the graph and join them in order to get the required frequency polygon.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H

Question 1.
Solution:
We shall take the game along x-axis and number of students along y-axis.

Question 2.
Solution:
We shall take the time on x-axis and temperature (in °C) on y-axis.

Question 3.
Solution:
We shall take name of vehicle on x-axis and velocity (in km/hr) on y-axis

Question 4.
Solution:
We shall take sports on x-axis and number of students on y-axis.

Question 5.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 6.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 7.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 8.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 9.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 10.
Solution:
We shall take years on x-axis and number of students on y-axis.

Question 11.
Solution:
We shall take week on x-axis and Rate per 10g (in Rs.) on y-axis.

Question 12.
Solution:
We shall take mode of transport on x-axis and number of students on y-axis.

Question 13.
Solution:
We see from the graph that
(i) It shows the marks obtained by a student in various subjects.
(ii) The student is very well in mathematics.
(iii) The student is very’ poor, in Hindi.
(iv) Average marks
= \(\frac { 60+35+75+50+60 }{ 5 } \) (Here x = 5)
= \(\frac { 280 }{ 5 } \)
= 56 marks

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B 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.

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H

Question 1.
Solution:
Statistics is a science which deals with the collection, presentation, analysis and interpretations of numerical data.

Question 2.
Solution:
(i) Numerical facts alone constitute data
(ii) Qualitative characteristics like intelligence, poverty etc. which cannot be measured, numerically, don’t form data.
(iii) Data are an aggregate of facts. A single observation does not form data.
(iv) Data collected for a definite purpose may not be suited for another purpose.
(v) Data is different experiments are comparable.

Question 3.
Solution:
(i) Primary data : The data collected by the investigator himself with a definite plan in mind are called primary data.
(ii) Secondary data : The data collected by some one other than the investigator are called secondary data. The primary data is more reliable and relevant.

Question 4.
Solution:
(i) Variate : Any character which is capable of taking several different values is called a variate or variable.
(ii) Class interval : Each group into which the raw data is condensed, is called a class interval
(iii) Class size : The difference between the true upper limit and the true lower limit of a class is called class size.
(iv) Class Mark : \(\frac { upper\quad limit+lower\quad limit }{ 2 } \) is called a class mark
(v) Class limits : Each class is bounded by two figures which are called class limits which are lower class limit and upper class limit.
(vi) True class limits : In exclusive form, the upper and lower limits of a class are respectively are the true upper limit and true lower limit but in inclusive form, the true lower limit of a class is obtained by subtracting O.S from lower limit of the class and for true limit, adding 0.5 to the upper limit.
(vii) Frequency of a class : The number of times an observation occurs in a class is called its frequency.
(viii) Cumulative frequency of a class : The cumulative frequency corresponding to a class is the sum of all frequencies upto and including that class.

Question 5.
Solution:
The given data can be represent in form of frequency table as given below:

Question 6.
Solution:
The frequency distribution table of the given data is given below :

Question 7.
Solution:
The frequency distribution table of the

Question 8.
Solution:
The frequency table is given below :

Question 9.
Solution:
The frequency table of given data is given below :

Question 10.
Solution:
The frequency distribution table of the given data in given below :

Question 11.
Solution:
The frequency table of the given data:

Question 12.
Solution:
The cumulative frequency of the given table is given below:

Question 13.
Solution:
The given table can be represented in a group frequency table in given below :

Question 14.
Solution:
Frequency table of the given cumulative frequency is given below :

Question 15.
Solution:
A frequency table of the given cumulative frequency table is given below :

 

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A 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.

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13A
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13B
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13C
• RS Aggarwal Solutions Class 9 Chapter 13 Volume and Surface Area Ex 13D

Question 1.
Solution:
(i) Radius of sphere = 3.5cm
(a) Volume = \(\frac { 4 }{ 3 } \) πr³

Question 2.
Solution:
Let r be the radius of the sphere and volume = 38808 cm³
∴\(\frac { 4 }{ 3 } \) πr³ = 38803
=> \(\frac { 4 }{ 3 } \) x \(\frac { 22 }{ 7 } \) r³ = 38803

Question 3.
Solution:
Let r be the radius of the sphere
∴ Volume = \(\frac { 4 }{ 3 } \) πr³

Question 4.
Solution:
Surface area of a sphere = 394.24 m²
Let r be the radius, then 4πr² = 394.24

Question 5.
Solution:
Surface area of sphere = 576π cm²
Let r be the radius, then 4r² = 576π

Question 6.
Solution:
Outer diameter of shell = 12cm,
Outer radius (R) = \(\frac { 12 }{ 2 } \) = 6cm
and inner diameter = 8cm

Question 7.
Solution:
Length of cuboid of (l) = 12cm
Breadth (b) = 11cm
and height (h) = 9cm

Question 8.
Solution:
Radius of sphere (r) = 8cm
Volume = \(\frac { 4 }{ 3 } \)πr³

Question 9.
Solution:
Radius of solid sphere (R) = 3cm.
Volume = \(\frac { 4 }{ 3 } \)π(R)³ = \(\frac { 4 }{ 3 } \)π(3)³ cm³

Question 10.
Solution:
Radius of metallic sphere (R) = 10.5cm

Question 11.
Solution:
Diameter of a cylinder = 8cm
Radius (r) = \(\frac { 8 }{ 2 } \) = 4cm

Question 12.
Solution:
Diameter of sphere = 6cm
Radius (R) = \(\frac { 6 }{ 2 } \) = 3cm

Question 13.
Solution:
Diameter of sphere = 18cm
Radius (R) = \(\frac { 18 }{ 2 } \) = 9cm.

Question 14.
Solution:
Diameter of the sphere = 15.6 cm
Radius (R) = \(\frac { 15.6 }{ 2 } \) = 7.8 cm

Question 15.
Solution:
Diameter of the canonball = 28cm
Radius (R) = \(\frac { 28 }{ 2 } \) = 14 cm

Question 16.
Solution:
Given,
Radius of spherical big ball (R) = 3cm

Question 17.
Solution:
Ratio in the radii of two spheres = 1:2
Let radius of smaller sphere = r then,
radius of bigger sphere = 2r

Question 18.
Solution:
Let r1 and r2 be the radii of two spheres

Question 19.
Solution:
Radius of the cylindrical tub = 12cm.
First level of water = 20cm
Raised water level = 6.75cm.

Question 20.
Solution:
Radius of the ball (r) = 9cm.
Volume of ball = \(\frac { 4 }{ 3 } \)πr³

Question 21.
Solution:
Given,
Radius of hemisphere of lead (r) = 9cm.

Question 22.
Solution:
Given,
Radius of hemispherical bowl (r) = 9cm

Question 23.
Solution:
External radius of spherical shell (R) = 9cm

Question 24.
Solution:
Inner radius (r) = 4 cm
Thickness of steel used = 0.5

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D 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.