## Online Education for RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles MCQS

These Solutions are part of Online Education RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is
(a) 15 cm
(b) 16 cm
(c) 17 cm
(d) 34 cm
Solution:
Length of chord AB of circle = 16 cm
Distance from the centre OL = 15 cm

Let OA be the radius, then in right ∆OAL,
OA2 = OL2 + AL2
16
= (15)2 + $$\frac { 16 }{ 2 }$$ = 152 + 82
= 225 + 64 = 289 = (17)2
∴ OA = 17 cm
Hence radius of the circle = 17 cm (c)

Question 2.
The radius of a circle Js 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is

Solution:
Radius of the cirlce (r) = 6 cm
Perpendicular distance from centre = ?
Length of chord = 8 cm

Let AB be chord, OL is the distance
In right ∆OAL
OA2 = AL2 + OL2

Question 3.
If O is the centre of a circle of radius r and AB is a chord of the circle at a distance $$\frac { r }{ 2 }$$ from O, then ∠BAO =
(a) 60°
(b) 45°
(c) 30°
(d) 15°
Solution:
r is the radius of the circle with centre O
AB is the chord, at a distance of $$\frac { r }{ 2 }$$ from the centre

Question 4.
ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB=
(a) 70°
(b) 100°
(c) 125°
(d) 150°
Solution:
ABCD is a cyclic quadrilateral ∠DCA = 80° and ∠ADB = 30°

∵∠ADB = ∠ACB (Angles in the same segment)
∴ ∠ACB = 30°
∴ ∠BCD = 80° + 30° = 110°
∵ ABCD is a cyclic quadrilateral
⇒ ∠BAD = 180°- 110° = 70°
or ∠DAB = 70° (a)

Question 5.
A chord of length 14 cm is at a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is
(a) 12 cm
(b) 14 cm
(c) 16 cm
(d) 18 cm
Solution:
In a circle AB chord = 14 cm
and distance from centre OL = 6 cm
Let r be the radius of the circle, then OA2 = AL2 + OL2
⇒ r2 = (7)2 + (6)2 = 49 + 36 = 85

In the same circle length of another chord CD = ?
Distance from centre = 2 cm
∴ r2 = OM2 + MD2
⇒ 85 = (2)2 + DM2
⇒ 85 = 4 + DM2
⇒ DM2 = 85-4 = 81 = (9)2
∴ DM = 9
∴ CD = 2 x DM = 2 x 9 = 18 cm
∴Length of another chord = 18 cm (d)

Question 6.
One chord of a circle is known to be 10 cm. The radius of this circle must be
(a) 5 cm
(b) greater than 5 cm
(c) greater than or equal to 5 cm
(d) less than 5 cm
Solution:
Length of chord of a circle = 10 cm
Length of radius of the circle greater than half of the chord
More than $$\frac { 10 }{ 2 }$$ = 5 cm (b)

Question 7.
ABC is a triangle with B as right angle, AC = 5 cm and AB = 4 cm. A circle is drawn with O as centre and OC as radius. The length of the chord of this circle passing through C and B is
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
Solution:
In right ∆ABC, ∠B = 90°
AC = 5 cm, AB = 4 cm

∴ BC2 = AC-AB2
= 52 – 42 = 25 – 16
= 9 = (3)2
∴ BC = 3 cm
∴ Length of chord BC = 3 cm (a)

Question 8.
If AB, BC and CD are equal chords of a circle with O as centre, and AD diameter then ∠AOB =
(a) 60°
(b) 90°
(c) 120°
(d) none of these
Solution:
In a circle chords AB = BC = CD

O is the centre of the circle
∴ ∠AOB = cannot be found (d)

Question 9.
Let C be the mid-point of an arc AB of a circle such that m $$\breve { AB }$$ = 183°. If the region bounded by the arc ACB and line segment AB is denoted by S, then the centre O of the circle lies
(a) in the interior of S
(b) in the exterior of S
(c) on the segment AB
(d) on AB and bisects AB
Solution:
$$\breve { AB }$$ = 183°
∴ AB is the diameter of the circle with centre O and C is the mid point of arc AB

Line segment AB = S
∴ Centre will lie on AB (c)

Question 10.
In a circle, the major arc is 3 ti.nes the minor arc. The corresponding central angles and the degree measures of two arcs are
(a) 90° and 270°
(b) 90° and 90°
(d) 60° and 210°
Solution:
In a circle, major arc is 3 times the minor arc i.e. arc ACB = 3 arc ADB
∴ Reflex ∠AOB = 3∠AOB
But angle at O = 360°

and let ∠AOB = x
x + 3x – 360°
⇒ 4x = 360°
⇒ x = $$\frac { { 62 }^{ \circ } }{ 2 }$$  = 90°
∴ 3x = 90° x 3 = 270°
Here angles are 270° and 90° (c)

Question 11.
If A and B are two points on a circle such that m($$\breve { AB }$$) = 260°. A possible value for the angle subtended by arc BA at a point on the circle is
(a) 100°
(b) 75°
(c) 50°
(d) 25°
Solution:
A and B are two points on the circle such that reflex ∠AOB = 260°

∴ ∠AOB = 360° – 260° = 100°
C is a point on the circle
∴ By joining AC and BC,
∠ACB = $$\frac { 1 }{ 2 }$$∠AOB = $$\frac { 1 }{ 2 }$$ x 100° = 50° (c)

Question 12.
An equilateral triangle ABC is inscribed in a circle with centre O. The measures of ∠BOC is
(a) 30°
(b) 60°
(c) 90°
(d) 120°
Solution:
∆ABC is an equilateral triangle inscribed in a circle with centre O

∴ Measure of ∠BOC = 2∠BAC
= 2 x 60° = 120° (d)

Question 13.
If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a
(a) rhombus
(b) rectangle
(c) parallelogram
(d) square
Solution:
Two diameter of a circle AB and CD intersect each other at right angles

∵ The diagonals are equal and bisect each other at right angles
∴ ACBD is a square (d)

Question 14.
In ABC is an arc of a circle and ∠ABC = 135°, then the ratio of arc $$\breve { AB }$$ to the circumference is
(a) 1 : 4
(b) 3 : 4
(c) 3 : 8
(d) 1 : 2
Solution:
Arc ABC of a circle and ∠ABC = 135°
Join OA and OC

∴ Angle subtended by arc ABC at the centre = 2 x ∠ABC = 2 x 135° = 270°
Angle at the centre of the circle = 360°
∴ Ratio with circumference = 270° : 360° = 3:4 (b)

Question 15.
The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is
(a) 60°
(b) 75°
(c) 120°
(d) 150°
Solution:
The chord of a circle = radius of the circle In the figure OA = OB = AB

∴ ∠AOB = 60°
(Each angle of an equilateral = 60°) (a)

Question 16.
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
(a) 41°
(b) 23°
(c) 67°
(d) 18°
Solution:
PQRS is a cyclic quadrilateral with centre O and ∠QPR = 67°
∠SPR = 72°

∴ ∠QPS = 67° + 72° = 139°
∵ ∠QPS + ∠QRS = 180° (Sum of opposite angles of a cyclic quad.)
⇒ 139° + ∠QRS = 180°
⇒ ∠QRS = 180° – 139° = 41° (a)

Question 17.
If A, B, C are three points on a circle with centre O such that ∠AOB = 90° and ∠BOC = 120°, then ∠ABC =
(a) 60°
(b) 75°
(c) 90°
(d) 135°
Solution:
A, B and C are three points on a circle with centre O
∠AOB = 90° and ∠BOC = 120°

∴ ∠AOC = 360° – (120° + 90°)
= 360° -210°= 150°
But ∠AOC is at the centre made by arc AC and ∠ABC at the remaining part of the circle
∴ ∠ABC = $$\frac { 1 }{ 2 }$$ ∠AOC
= $$\frac { 1 }{ 2 }$$ x 150° = 75° (b)

Question 18.
The greatest chord of a circle is called its
(b) secant
(c) diameter
(d) none of these
Solution:
The greatest chord of a circle is called its diameter. (c)

Question 19.
Angle formed in minor segment of a circle is
(a) acute
(b) obtuse
(c) right angle
(d) none of these
Solution:
The angle formed in minor segment of a circle is obtuse angle. (b)

Question 20.
Number of circles that can be drawn through three non-collinear points is
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
The number of circles that can pass through three non-collinear points is only one. (a)

Question 21.
In the figure, if chords AB and CD of the circle intersect each other at right angles, then x + y =
(a) 45°
(b) 60°
(c) 75°
(d) 90°

Solution:
In the circle, AB and CD are two chords which intersect each other at P at right angle i.e. ∠CPB = 90°

∠CAB and ∠CDB are in the same segment
∴ ∠CDB = ∠CAB = x
Now in ∆PDB,
Ext. ∠CPB = ∠D + ∠DBP
⇒ 90° = x + y (∵ CD ⊥ AB)
Hence x + y = 90° (d)

Question 22.
In the figure, if ∠ABC = 45°, then ∠AOC=
(a) 45°
(b) 60°
(c) 75°
(d) 90°

Solution:
∵ arc AC subtends
∠AOC at the centre of the circle and ∠ABC
at the remaining part of the circle

∴ ∠AOC = 2∠ABC
= 2 x 45° = 90°
Hence ∠AOC = 90° (d)

Question 23.
In the figure, chords AD and BC intersect each other at right angles at a point P. If ∠D AB = 35°, then ∠ADC =
(a) 35°
(b) 45°
(c) 55°
(d) 65°

Solution:
Two chords AD and BC intersect each other at right angles at P, ∠DAB = 35°
AB and CD are joined
In ∆ABP,
Ext. ∠APC = ∠B + ∠A

⇒ 90° = ∠B + 35°
∠B = 90° – 35° = 55°
∵ ∠ABC and ∠ADC are in the same segment
∴ ∠ADC = ∠ABC = 55° (c)

Question 24.
In the figure, O is the centre of the circle and ∠BDC = 42°. The measure of ∠ACB is
(a) 42°
(b) 48°
(c) 58°
(d) 52°

Solution:
In the figure, O is the centre of the circle
∠BDC = 42°
∠ABC = 90° (Angle in a semicircle)
and ∠BAC and ∠BDC are in the same segment of the circle.
∴ ∠BAC = ∠BDC = 42°
Now in ∆ABC,
∠A + ∠ABC + ∠ACB = 180° (Sum of angles of a triangle)

⇒ 42° + 90° + ∠ACB = 180°
⇒ 132° + ∠ACB – 180°
⇒ ∠ACB = 180° – 132° = 48° (b)

Question 25.
In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is

Solution:
AB and CD are two diameters of a circle with centre O

Question 26.
Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circles is

Solution:
Two equal circles pass through the centre of the other and intersect each other at A and B
Let r be the radius of each circle

Question 27.
If AB is a chord of a circle, P and Q are the two points on the circle different from A and B,then
(a) ∠APB = ∠AQB
(b) ∠APB + ∠AQB = 180° or ∠APB = ∠AQB
(c) ∠APB + ∠AQB = 90°
(d) ∠APB + ∠AQB = 180°
Solution:
AB is chord of a circle,
P and Q are two points other than from points A and B

∵ ∠APB and ∠AQB are in the same segment of the circle
∴ ∠APB = ∠AQB (a)

Question 28.
AB and CD are two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm. The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is

Solution:
AB and CD are two parallel chords of a circle with centre O
Let r be the radius of the circle AB = 6 cm, CD = 12 cm
and distance between them = 3 cm
Join OC and OA, LM = 3 cm

Let OM = x, then OL = x + 3

Question 29.
In a circle of radius 17 cm, two parallel chords are drawn on opposite side of a diameter. This distance between the chords is 23 cm. If the length of one chord is 16 cm then the length of the other is
(a) 34 cm
(b) 15 cm
(c) 23 cm
(d) 30 cm
Solution:
Radius of a circle = 17 cm
The distance between two parallel chords = 23 cm
AB || CD and LM = 23 cm
Join OA and OC,
∴ OA = OC = 17 cm
Let OL = x, then OM = (23 – x) cm
AB = 16 cm
Now in right ∆OAL,
OA2 = OL22 + AL2
⇒ (17)2 = x2 + AL2
⇒ 289 = x2 + AL2

Question 30.
In the figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
(a) 130°
(b) 115°
(c) 65°
(d) 165°

Solution:
O is the centre of the circle and ∠AOC = 130°
Reflex ∠AOC = 360° – 130° = 230°

Now arc ADB subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle
∴ ∠ABC = $$\frac { 1 }{ 2 }$$reflex ∠AOC
= $$\frac { 1 }{ 2 }$$ x 230°= 115° (b)

Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles MCQS are helpful to complete your math homework.

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## 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.

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## 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:

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.

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

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:

Solution:

Question 5.
The following is the distribution of total household expenditure (in Rs.) of manual worker in a city:

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:

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:

Solution:

Hisotgram is given below:

Question 8.
The monthly profits (in t) of 100 shops are distributed as follows:

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.

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## RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.2

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 23.2

Other Exercises

Question 1.
Explain the reading and interpretation of bar graphs.
Solution:
Reading and interpretation of the bar graph:
The first step in reading a bar graph is to know what it represents or what is the information given by it. For this, we read the captions which are generally written just below the horizontal line (x-axis) and adjacent to vertical line (y-axis). After knowing that what does a bar graph represent we read the scale so that we can know the precise values in the given data.
After reading a bar graph one must be able to draw certain conclusions from it. Drawing some conclusions from a given bar graph means interpretation of the bar graph.

Question 2.
(i) What information is given by the bar graph?
(ii) In which year the export is minimum?
(iii) In which year the import is maximum?
(iv) In which year the difference of the values of export and import is maximum?

Solution:
After reading the bar graph. We find:
(i) Export and import in 100 crores of rupees year wise.
(ii) In the year 1982-83, export is minimum i.e. 100 crores.
(iii) In the year 1986-87, import is maximum i.e. 22 hundred crores.
(iv) In the year 1986-87, the difference of the values of import and export is maximum i.e. 22 – 12 = 10 x 100 crore.

Question 3.
The following bar graph show’s the results of an annual examination in a secondary school.
Read the bar graph and choose the correct alternative in each of the following:
(i) The pair of classes in which the results of boys and girls are inversely proportional are:
(a) VI, VIII (b) VI, IX (c) VIII, IX (d) VIII, X
(ii) The class having the lowest failure rate of girls is
(a) VII (b) X (c) IX (d) VIII
(iii) The class having the lowest pass rate of students is
(a) VI (b) VII (c) VIII (d) IX

Solution:
After reading the given bar graph, we find that
(i) The pair of classes in which the results of boys and girls are inversely proportional is VI and IX as in VI, 80% boys and 70% girls and in IX, 70% boys and 80% girls are the results.
(ii) Lowest failure of girls of students is VII as pass % is 100%.
(iii) Class having the lowest pass rate is VII boys as it is 50% only.

Question 4.
The following data gives the number (in thousands) of applicants registered with an Employment Exchange during, 1995-2000:

Construct a bar graph to represent the above data.

Solution:

Bar graph of the above is given below:

Question 5.
The production of saleable steel in some of the steel plants of our country during 1999 is given below:

Construct a bar graph to represent the above data on a graph paper by using the scale 1 big divisions = 20 thousands tonnes.
Solution:

Bar graph of the above data is given below:

Question 6.
The following table gives the route length (in thousand kilometres) of the Indian Railways in some of the years:

Represent the above data with the help of a bar graph.
Solution:

Question 7.
The following data gives the amount of loans (in crores of rupees) disbursed by a bank during some years:

(i) Represent the above data with the help of a bar graph.
(ii) With the help of the bar graph, indicate the year in which amount of loan is not increased over that of the preceding year.
Solution:

Question 8.
The following table shows the interest paid by a company (in lakhs):

Draw the bar graph to represent the above information.
Solution:

Question 9.
The following data shows the average age of men in various countries in a certain year:

Represent the above information by a bar graph.
Solution:

Question 10.
The following data gives the production of foodgrains (in thousand tonnes) for some years:

Represent the above data with the help of a bar graph.
Solution:

Question 11.
The following data gives the amount of manure (in thousand tonnes) manufactured by a company during some years:

(i) Represent the above data with the help of a bar graph.
(ii) Indicate with the help of the bar graph the year in which the amount of manufactured by the company was maximum.
(in) Choose the correct alternative:
The consecutive years during which there was maximum decrease in manure production are:
(a) 1994 and 1995
(b) 1992 and 1993
(c) 1996 and 1997
(d) 1995 and 1996
Solution:

(ii) The maximum amount of manure manufactured in the year 1994, i.e. 45 thousand tonnes.
(iii) There was maximum decrease in manure production in 1996 and 1997.

Question 12.
The following data gives the demand estimates of the Government of India, Department of Electronics for the personnel in the Computer sector during the Eighth Plan period (1990-95):

Represent the data with the help of a bar graph. Indicate with the help of the bar graph the course where estimated requirement is least.
Solution:

The course where requirement is least is DCE.

Question 13.
The income and expenditure for 5 years of a family is given in the following data:

Represent the above data by a bar graph.
Solution:

Question 14.
The investment (in ten crores of rupees) of Life Insurance Corporation of India in different sectors are given below:

Represent the above data with the help of a bar graph.
Solution:

Question 15.
The following data gives the value (in crores of rupees) of the Indian export of cotton textiles for different years:

Represent the above data with the help of a bar graph. Indicate with the help of a bar graph the year in which the rate of increase in exports is maximum over the preceding year.
Solution:

Question 16.
The following table gives the quantity of good (in crore tonnes)

Represent this information with the help of a bar graph.
Explain through the bar graph if the quantity of goods carried by the Indian Railways in 1965-66 is more than double the quantity of goods carried in the year 1950-51.
Solution:

Question 17.
The production of oil (in lakh tonnes) in some of the refineries in India during 1982 was given below:

Construct a bar graph to represent the above data so that the bars are drawn horizontally.
Solution:

Question 18.
The expenditure (in 10 crores of rupees) on health by the Government of India during the various five year plans is shown below:

Construct a bar graph to represent the above data.
Solution:

Hope given RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.2 are helpful to complete your math homework.

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## RD Sharma Class 9 Solutions Chapter 23 Graphical Representation of Statistical Data Ex 23.1

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 23.1

Other Exercises

Question 1.
The following table shows the daily production of T.V. sets in an industry for 7 days of a week:

Represent the above information by a pictograph.
Solution:

Question 2.
The following table shows the number of Maruti cars sold by five dealers in a particular month:

Represent the above information by a pictograph.
Solution:

Question 3.
The population of Delhi State in different census years is as given below:

Represent the above information with the help of a bar graph.
Solution:

Question 4.
Read the bar graph shown in Figure and answer the following questions:

(i) What is the information given by the bar graph?
(ii) How many tickets of Assam State Lottery were sold by the agent?
(iii) Of which state, were the maximum number of tickets sold?
(iv) State whether true or false.
The maximum number of tickets sold is three times the minimum number of tickets sold.
(v) Of which state were the minimum number of tickets sold?
Solution:
(i) Sale of lottery tickets if different states.
(ii) 40 tickets of Assam state lottery were sold.
(iii) Maximum number of lottery tickets were sold of Haryana state.
(iv) Maximum tickets were 100 and minimum = 20 which are not three time. False.
(v) Minimum Lottery tickets of 10 Rajasthan will sold.

Question 5.
Study the bar graph representing the number of persons in various age groups in a town shown in figure. Observe the bar graph and answer the following questions:
(i) What is the percentage of the youngest age-group persons over those in the oldest age group?
(ii) What is the total population of the town?
(iii) What is the number of persons in the age-group 60-65?
(iv) How many persons are more in the age-group 10-15 than in the age group 30-35?
(v) What is the age-group of exactly 1200 persons living in the town?
(vi) What is the total number of persons living in the town in the age-group 50-55?
(vii) What is the total number of persons living in the town in the age-groups 10-15 and 60-65?
(viii) Whether the population in general increases, decreases or remains constant with the increase in the age-group.

Solution:
(i) Youngest age group = 300 and oldest age group = 1400

(ii) Total population = 1400 + 1200 + 1100 + 1000 + 900 + 800 + 300 – 6700
(iii) In age group of 60-65, number of persons = 800
(iv) Difference of person in the age group 10-15 and of 30-35 = 1400 – 1100 = 300
(v) 1200 persons are in the age group 20-25
(vi) In age group 50-55 number of persons are 900
(vii) Total number of person of the age group 10-15 and
60-65 = 1400 + 800 = 2200
(viii)Population is decreasing with the age-group.

Question 6.
Read the bar graph shown in figure and answer the following questions:
(i) What is the information given by the bar graph?
(ii) What was the number of commercial banks in 1977?
(iii) What is the ratio of the number of commercial banks in 1969 to that in 1980?
(iv) State whether true or false:
The number of commercial banks in 1983 is less than double the number of commercial banks in 1969.

Solution:
(i) The information is number of commercial banks in India during same years.
(ii) In 1977, commercial banks were 130.
(iii) Ratio of banks in 1969 to 1980 = 90 : 150 = 3:5
(iv) Number of banks in 1983 = 230 and number of banks in 1969 = 90
∴ Number of banks in 1983 is less than double the number of banks in 1969 – False.

Question 7.
Given below is the bar graph indicating the marks obtained out of 50 in mathematics paper by 100 students. Read the bar graph and answer the following questions:

(i) It is decided to distribute work books on mathematics to the students obtaining less than 20 marks, giving one workbook to each of such students. If a work book costs T5, what sum is required to buy the work books?
(ii) Every student belonging to the highest mark group is entitled to get a prize of ? 10. How much amount of money is required for distributing the prize money?
(iii) Every student belonging to the lowest mark-group has to solve 5 problems per day. How many problems, in all, will be solved by the students of this group per day?
(iv) State whether true or false. .
(a) 17% students have obtained marks ranging from 40 to 49.
(b) 59 students have obtained marks ranging from 10 to 29.
(v) What is the number of students getting less than 20 marks?
(vi) What is the number of students getting more than 29 marks?
(vii) What is the number of students getting marks between 9 and 40?
(viii)What is the number of students belonging to the highest mark group?
(ix) What is the number of students obtaining more than 19 marks?
Solution:
(i) Students getting less than 20 marks = 27 + 12 = 39
Cost of 1 work book = ₹5
∴ Total cost = 39 x 5 = ₹195
(ii) No. of students getting highest mark group = 17
∴ Amount paid at the rate of Rs. 10 to each = 10 x 17 = Rs. 170
(iii) Lowest mark group = 27
∴ Total number of problems given = 27 x 5 = 135
(iv) (a) True (b) False
(v) No. of students getting less then 20 marks = 27 + 12 = 39
(vi) Students getting more than 29 marks = 24 + 17 = 41
(vii) Students getting marks between 9 and 40 = 12 + 20 + 24 = 56
(viii) Students to the highest mark group = 17
(ix) Students setting more than 19 marks = 20 + 24 + 17 = 61

Question 8.
(i) What is the information given by the bar graph?
(ii) State each of the following whether true or false.
(a) The number of government companies in 1957 is that of 1982 is 1 : 9.
(b) The number of government companies have decreased over the year 1957 to 1983.

Solution:
(i) Number of government companies in India during some years.
(ii) (a) False (b) False
(a) 50 : 400 = 1:8 and (b) companies are increasing not decreasing.

Question 9.

(i) What information is given by the bar graph?
(ii) Which state is the largest producer of rice?
(iii) Which state is the largest producer of wheat?
(iv) Which state has total production of rice and wheat as its maximum?
(v) Which state has the total production of wheat and rice minimum?
Solution:
(i) Production of rice and wheat in different states of India.
(ii) West Bengal is the largest state of production of rice.
(iii) Largest producer of wheat is U.P.
(iv) Maximum production of wheat and rice is U.P.
(v) Minimum production of wheat and rice is Maharashtra.

Question 10.
The following bar graph represents the heights (in cm) of 50 students of Class XI of a particular school. Study the graph and answer the following questions:

(i) What percentage of the total number of students have their heights more than 149 cm?
(ii) How many students in the class are in the range of maximum height of the class?
(iii) The school wants to provide a particular type of tonic to each student below the height of 150 cm to improve his height. If the cost of the tonic for each student comes out to be Rs. 55, how much amount of money is required?
(iv) How many students are in the range of shortest height of the class?
(v) State whether true or false:
(a) There are 9 students in the class whose heights are in the range of 155-159 cm.
(b) Maximum height (in cm) of a student in the class is 17.
(c) There are 29 students in the class whose heights are in the range of 145-154 cm.
(d) Minimum height (in cm) of a student is the class is in the range of 140-144 cms.
(e) The number of students in the class having their heights less than 150 cm is 12.
(f) There are 14 students each of whom has height more than 154 cm.
Solution:
(i) No. of students having more than height of 149 cm = 17 + 9 + 5 = 31
Percentage = $$\frac { 31 }{ 50 }$$ x 100 = 62%
(ii) No. of students in the range of maximum height = 5
(iii) No. of students below the height of 150 cm = 7 + 12 = 19
Cost of tonic for each student = ₹55
∴ Total cost = ₹55 x 19 = ₹1045
(iv) No. of students in the shortest height = 7
(v) (a) True
(b) False
(c) True
(d) True
(e) False
(f) True

Question 11.
(i) What information is given by the bar graph?
(ii) What was the production of cement in the year 1980-81?
(iii) What is the minimum and maximum productions of cement and corresponding years?

Solution:
(i) Industrial production of cement in different years in India.
(ii) In the year 1980-81, the production was 186 lakh tonnes.
(iii) Minimum production was 30 lakh tonnes in 1950-51 and maximum production was 232 lakh tonnes in 1982-83.

Question 12.
The bar graph shown in the figure represents the circulation of newspapers in 10 languages. Study the bar graph and answer the following questions:

(i) What is the total number of newspapers published in Hindi, English, Urdu, Punjabi and Bengali?
(ii) What percent is the number of news papers published in Hindi of the total number of newspapers?
(iii) Find the excess of the number of newspapers published in English over those published in Urdu,
(iv) Name two pairs of languages which publish the same number of newspapers.
(v) State the languages in which the smallest number of newspapers are published.
(vi) State the language in which the largest number of newspapers are published.
(vii) State the language in whih the number of newspapers published is between 2500 and 3500.
(viii)State whether true or false:
(a) The number of newspapers published in Malayalam and Marathi together is- less than those published in English.
(b) The number of newspapers published in Telugu is more than those published in Tamil.
Solution:
(i) Total number of newspapers in Hindi, English, Urdu, Punjabi, and Bengali
= 3700 + 3400 + 700 + 200 + 1100 = 9100

(iii) Difference in English and Urdu = 3400 – 700 = 2700
(iv) Malayalam and Marathi, Bengali and Gujrati
(v) Punjabi (200)
(vi) Hindi (3700)
(vii) English (3400)
(viii)(a) True (b) False

Question 13.
Read the bar graph given in figure and answer the following questions:
(i) What information is given by the bar graph?
(ii) What was the crop-production if rice in 1970-71?
(iii) What is the difference between the maximum and minimum production of rice?

Solution:
(i) Production of rice in India in different years.
(ii) Crop production in 1970-71 was 42.5 lakh tonnes.
(iii) Differences in maximum and minimum production of rice = 55 – 22 lakh tonnes = 33 lakh tonnes

Question 14.
Read the bar graph given in figure and answer the following questions:

(i) What information does it give?
(ii) in which part the expenditure on education is maximum in 1980?
(iii) In which part the expenditure has gone up from 1980 to 1990?
(iv) In which part the gap between 1980 and 1990 is maximum?
Solution:
(i) Public expenditure on education by vertices sub-continents in 1980 and 1990.
(ii) Maximum expenditure on education in 1980 was in Africa.
(iii) The expenditure from 1980 to 1990 was gone up in East Africa.
(iv) Maximum gap between 1980 and 1990 was in Africa.

Question 15.
Read the bar graph given in the figure and answer the following questions:

(i) What information is given by the bar graph?
(ii) In whicli years the areas under the sugarcane crop were the maximum and the minimum?
(iii) State whether true or false:
The area under the sugarcane crop in the year 1982-83 is three times that of the year 1950-51.
Solution:
(i) Areas under sugarcane crop during different years in India.
(ii) Maximum sugarcane crop was in 1982-83 and minimum in 1950-51.
(iii) False.

Question 16.
Read the bar graph given in the figure and answer the following questions:

(i) What information is given by the bar graph?
(ii) What was the expenditure on health and family planning in the year 1982-83?
(iii) In which year is the increase in expenditure maximum over the expenditure in previous year? What is the maximum increase?
Solution:
(i) Expenditure on health and family planning sixth five year plan in India.
(ii) Expenditure during 1982-83 was 700 crore of rupees.
(iii) Maximum increase in 1984-85 was them in previous year and difference 10.2 – 8 = 220 crores.

Question 17.
Read the bar graph given in the figure and answer the following questions:
(i) What is the information given by the bar graph?
(ii) What is the number of families having 6 members?
(iii) How many members per family are there in the maximum number of families? Also tell the number of such families.
(iv) What are the number of members per family for which the number of families are equal? Also, tell the number of such families?

Solution:
(i) Number of families with different number of members in a locality.
(ii) Number of families having 6 members are 50.
(iii) Maximum number of families are of 3 members family and there are 120.
(iv) Number of families having 9 members and having 10 members are equal and each is 5.

Question 18.
Read the bar graph given in the figure and answer the following questions:

(i) What information is given by the bar graph?
(ii) Which Doordarshan centre covers maximum area? Also tell the covered area.
(iii) What is the difference between the areas covered by the centres at Delhi and Bombay?
(iv) Which Doordarshan centres are in U.P. State? What are the areas covered by them?
Solution:
(i) Coverage of some Doordarshan centres of India.
(ii) Doordarshan kender Kolkata covers maximum area which is 36000 sq. km.
(iii) Difference of areas covered by Delhi and Bombay centres = 32000 – 18000 = 14000 sq. km.
(iv) In U.R Doordarshan centres are Kanpur (32000 sq. km) and Lucknow (25000 sq. km).

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## RD Sharma Class 9 Solutions Chapter 22 Tabular 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 22 Tabular Representation of Statistical Data MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
Tally marks are used to find
(a) class intervals
(b) range
(c) frequency
(d) upper limits
Solution:
Frequency (c)

Question 2.
The difference between the highest and lowest values of the observations is called
(a) frequency
(b) mean
(c) range
(d) class intervals
Solution:
range (c)

Question 3.
The difference between the upper limit and the lower class limits is called
(a) mid-points
(b) class size
(c) frequency
(d) mean
Solution:
class size (b)

Question 4.
In the class intervals 10-20, 20-30, 20 is taken in
(a) the interval 10-20
(b) the interval 20-30
(c) both interval 10-20, 20-30
(d) none of the intervals
Solution:
the interval 20-30 (b)

Question 5.
In a frequency distribution, the mid value of a class is 15 and the class intervals is 4. The lower limit of the class is
(a) 10
(b) 12
(c) 13
(d) 14
Solution:
Mid value = 15 and class interval is 4
∴ Lower limit = 15 – $$\frac { 4 }{ 5 }$$ = 15 – 2 = 13 (c)

Question 6.
The mid-value of a class interval is 42. If the class size is 10, then the upper and lower limits of the class are
(a) 47 and 37
(b) 37 and 47
(c) 37.5 and 47.5
(d) 47.5 and 37.5
Solution:
Mid value = 42 and class size =10
∴ Upper limit = 42 + $$\frac { 10 }{ 2 }$$ = 42 + 5 = 47
and lower limit = 42 – $$\frac { 10 }{ 2 }$$ = 42 – 5 = 37
Upper class limit and lower class limits are 47, 37 (a)

Question 7.
The number of times a particular item occurs in a given data is called its
(a) variation
(b) frequency
(c) cumulative frequency
(d) class-size
Solution:
frequency (b)

Question 8.
The width of each of nine classes in a frequency distribution is 2.5 and the lower class boundary of the lowest class 10.6. Then the upper class boundary of the highest class is
(a) 35.6
(b) 33.1
(c) 30.6
(d) 28.1
Solution:
Width of each class = 2.5
No. of classes = 9
Lower class boundary of the lowest class = 10.6
∴ Upper class limit of highest class = 10.6 + 9 x 2.5
= 10.6 + 22.5 = 33.1 (b)

Question 9.
The following marks were obtained by the students in a test:
81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62
The range of the marks is
(a) 9
(b) 17
(c) 27
(d) 33
Solution:
Marks are 81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62
Here highest marks = 95
and lowest marks = 62
∴ Range = highest marks – lowest marks = 95 – 62 = 33 (d)

Question 10.
Tallys are usually marked in a bunch of
(a) 3
(b) 4
(c) 5
(d) 6
Solution:
4 i.e. IIII (b)

Question 11.
Let l be the lower class limit of a class- interval in a frequency distribution and m be the mid-point of the class. Then, the upper class limit of the class is

Solution:
l is the lower class limit of a class interval
m is the mid point of the class
m = $$\frac { u+l }{ 2 }$$
⇒ 2m = u+ l ⇒ u = 2m – l
Then, upper class limit = 2 m – l (c)

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## 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

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.

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## 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

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.
(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.

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## 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

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:

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
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

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.

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## 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

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.
Solution:

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.