NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 14 Statistics, are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 14 Statistics.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1

Question 1.
Give five examples of data that you collect from your day-to-day life.
Solution:
Five examples of data that we can gather from our day-to-day life are
(i) number of students in our class.
(ii) number of fans in our class.
(iii) electricity bills of our house for last two years.
(iv) election results obtained from television or news paper.
(v) literacy rate figures obtained from educational survey. Of course, remember that there can be many more different answers.

Question 2.
Classify the data in Q.1 above as primary or secondary data.
Solution:
We know that, when the information was collected by the investigator herself or himself with a definite objective in her or his mind, the data obtained is called primary data.
∴ In the given data (in Q.1) examples (i), (ii) and (iii) are called primary data and when the information was gathered from a source which already had the information stored, the data obtained is called secondary data.
∴ In the given data (in Q.1) examples (iv) and (v) are called secondary data.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The blood groups of 30 students of class VIII are recorded as follows
A, B, 0, 0, AB, 0, A, 0, B, A, 0, B, A, 0, 0,
A, AB, 0, A, A, 0, 0, AB, B, A, 0, B, A, B, 0
Represent this data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students?
Solution:
The number of students who have a certain type of blood group is called the frequency of those blood groups. To make data more easily undrestandable, we write it in a table, as given below

From table, we observe that the higher frequency blood group i.e., most common blood group is O and the lowest frequency blood group i.e., rarest blood group is AB.

Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows

Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Solution:
To present such a large amount of data, so that a reader can make sense of it easily, we condense it into groups like 0-5, 5-10,…. 30-35 (since, our data is from 5 to 32). These grouping are called ‘classes’ or ‘class-intervals’ and their size is called the class size or class width which is 5 in this case. In each of these classes, the least number is called the iower class limit and the greatest number is called the upper class limit, e.g., in 0-5, 0 is the ‘lower class limit’ and 5 is the‘upper class limit’.
Now, using tally marks, the data (given) can be condensed in tabular form as follows

Presenting data in this form simplifies and condenses data and enables us to observe certain important feature at a glance. This is called a grouped frequency distribution table. We observe that the classes in the table above are non-overlapping.

Question 3.
The relative humidity (in %) of a certain city for a month of 30 days was as follows

(i) Construct a grouped frequency distribution table with classes 84-86, 86-88 etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution:
(i) We condense the given data into groups like 84 – 86, 86 – 88, 98-100. (since, our data is from 86.5 to 99.2) the class width in this case is 2. Now, the given data can be condensed in tabular form as follows

(ii) From table, we observe that the data appears to be taken in the rainy season as the relative humidity is high.
(iii) We know that, Range = Upper limit of data – Lower limit of data
∴ Range = 99.2 – 84.9 =14.3

Question 4.
The heights of 50 students, measured to the nearest centimeters have been found to be as follows

(i) Represent the data given above by a grouped frequency distribution table, taking class intervals as 160-165, 165-170 etc.
(ii) What can you conclude about their heights form the table?
Solution:
(i) We condense the given data into groups like 150-155, 155-160, …,170-175. (since, our data is from 150 to 172) The class width in this case is 5.
Now, the given data can be condensed in tabular form as follows

(ii) From the table, our conclusion is that more than 50% of students (i.e.,12 + 9+ 14 = 35) are shorter than 165 cm.

Question 5.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00-0.04, 0.04-0.08 and so on.
(ii) For how many day’s was the the concentration of sulphur dioxide more than 0.11 parts per million (ppm)?
Solution:
(i) We condense the given data into groups like 0.00-0.04, 0. 04-0.08,…, 0.20-0.24. (since, our data is from 0.01 to 0.22). The class width in this case is 0.04.
Now, the given data can be condensed in tabular form as follows

(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 2 + 4 + 2 = 8 days (by table).

Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows

Prepare a frequency distribution table for the data given above.
Solution:
Firstly, we write the data in a table

In above table, we observe that the repeatition of ‘0’ in given data is 6 times, 1 as to 10 times, 2 as 9 times and 3 as 5 times. Also, the above table is called an ungrouped frequency distribution table or simply a frequency distribution table.

Question 7.
The value of π upto 50 decimal places is given below 
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
Solution:
Firstly, we write the data i.e., digits from 0 to 9 after the decimal point in a table below

(i) From the table, we observe that the digit’s after the decimal points i.e., 0,1, 2, 3, 4, 5, 6, 7, 8, 9 repeated 2, 5, 5, 8, 4, 5, 4, 4, 5, 8 times, respectively.
(ii) From the table, we observe that the digits after the decimal point 3 and 9 are most frequently occurring i.e„ 8 times. The digit ‘0’ is the least occurring i.e., only 2 times.

Question 8.
Thirty children were asked about the number of hours they watched TV programmes in the previous week.
The results were found as follows

(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?
Solution:
(i) We condense the given data into groups like 0-5, 5-10 15-20.
(since, our data is from 1 to 17). The class width in this case is 5.
Now, the given data can be condensed in tabular form as follow

(ii) From table, we observe that the number of children is 2, who watched television for 15 or more hours a week.

Question 9.
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2-2.5.
Solution:
We condense the given data into groups like 2.0-2.5, 2.5-3.0 ,4.5- 5.0. (since, our data is from 2.2 to 4.6). The class width in this case is 5.
Now, the given data can be condensed in tabular form as follows

The above table is called a grouped frequency distribution table.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15-44 (in years) worldwide, found the following figures (in %)

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution:
(i) We draw the bar graph of this data in the following steps. Note that, the unit in the second column is percentage.

1. We represent the causes (variable) on the horizontal axis choosing any scale, since width of the bar is not important but for clarity, we take equal widths for all bars and maintain equal gaps in between. Let on cause be represented by one unit.
2. We represent the female fatality rate (value) on the vertical axis. Here, we can choose the scale as 1 unit = 4%.
3. To represent our first cause i.e, reproductive health conditions, we draw & rectangle bar with width 1 unit and height 31.8 units.
4. Similarly, other heads are represented leaving a gap of 1 unit in between two consecutive bars.

Now the graph is drawn in figure.

(ii) From graph, we observe that ‘reproductive health conditions’ is the major cause of women’s ill health and death world wide because it has maximum percentage among the causes i.e.,31.8%.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss, what conclusions can be arrived at from the graph.
Solution:
(i) We draw the bar graph of this data, note that the unit in the second column is number of girls per thousand boys.

We represent the section on the horizontal axis choosing any scale, since width of the bar is not important but for clarity, we take equd widths for all bars and maintain equal gaps in between. Let on section be represented by one unit.
We represent the number of girls per thousand boys on the vertical axis.
Now, the graph is drawn in figure.
We represent the number of girls per thousand boys on the vertical axis. Here, we can choose the scale as 1 unit = 100.
Now, the graph is drawn in figure

(ii) From graph, we observe that scheduled tribe (ST) number of girls is major section in different sections of Indian society, because it has maximum number of girls per thousand boys i.e., 970.

Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution:
We draw the bar graph of this data, note that the unit in the second column is seats won by political party.
(i) We represent the political party on the horizontal axis choosing any scale, since width of the bar is not important but for clarity, we take equal widths for all bars and maintain equal gaps in between. Let on political party be represented by one unit.
(ii) We represent the seats won on the vertical axis. Here, we can choose the scale as 1 unit = 10
Now, the graph is drawn in figure

Party ‘A’ Won the maximum number of seats. i.e, 75.

Question 4.
The length of 40 leaves of a plant measured correct to one millimetre and the obtained data is represented in the following table

(i) Draw a histogram to represent the given data.
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves 153 mm long and Why?
Solution:
(i) We know that, the areas of the rectangles are proportional to the frequencies in a histogram. Now, we get the following modified table by given data

Now, we draw the histogram for given data

(ii) Frequency polygon.
(iii) No, because the maximum number of leaves have their lengths lying in
the interval 145-153.

Question 5.
The following table gives the lifetimes of 400 neon lamps

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a lifetime of more 700 h?
Solution:
(i) Here, we will make a modified table by given data.

Now, we draw the histogram for above table

(ii) 184 lamps have a life time of more than 700 h i.e.,74 + 62 + 48 = 184.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution:
Here, we make modified tables by given data.

Now, required frequency polygon curves are

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below

Represent the data of both the teams on the same graph by frequency polygons.
Solution:
First make the class intervals continuous.

Now, draw a frequency polygon curve

Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows

Draw a histogram to represent the data above.
Solution:
We know that, the areas of the rectangles are proportional to the frequencies in a histogram. Here, the widths of the rectangles are varying. So, we need to make certain modifications in the lengths of the rectangles, so that the areas are again proportional to the frequencies.

1. Select a class interval with the minimum class size. The minimum class size is 1.
2. The lengths of the rectangles are then modified to be proportionate to the class size

Now, we get the following table

So, the correct histogram with varying width is given below.

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Solution:
(i) We know that, the areas of the rectangles are proportional to the frequencies in a histogram. Here, the widths of the rectangles are varying. So, we need to make certain modifications in the lengths of the rectangles. So that the areas are again proportional to the frequencies.

1. Select a class interval with the minimum class size. The minimum class size is 2.
2. The lengths of the rectangles are then modified to be proportionate to the class size 2.

Since we have calculated these lengths for interval of 2 letters in each case, we may call these lengths as ‘proportion of surnames per 2 mark ‘ interval’.
So, the correct histogram with varying width is given below.
Here, we make a modified table by given data with minimum class size 2.


(ii) The class interval in which the maximum number of surnames lie is 6-8.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following number of goals were scored by a team in a series of 10 matches
2, 3, 4, 5, 0, 1, 3, 3, 4, 3.
Find the mean, median and mode of these scores.
Solution:

(iii) Mode Arranging the given data in ascending order, we have 0, 1,2, 3, 3, 3, 3, 4, 4, 5.
Here, 3 occurs most frequently (4 times)
∴ Mode = 3

Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Solution:
(i) Mean

(ii) Median Arranging the given data in descending order, we have 98, 96, 62, 60, 54, 52, 52, 52, 48, 46, 42, 41,40, 40, 39
Number of observations (n) = 15 which is odd.

(iii) Mode Arranging the data in descending order, we have 98, 96, 62, 60, 54, 52, 52, 52, 48, 46, 42, 41,40, 40, 39.
Here, 52 occurs most frequently (3 times).
∴ Mode = 52

Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
Number of observations (n) = 10which is even.


According to question, median = 63
∴ x + 1 = 63 ⇒ x = 63 – 1 = 62
Hence, the value of x is 62.

Question 4.
Find the mode of 14, 25,14, 28,18,17,18,14, 23, 22,14 and 18.
Solution:
The given data is,
14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
Arranging the data in ascending order, we have
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Here, 14 occurs most frequently (4 times).
∴ Mode = 14

Question 5.
Find the mean salary of 60 workers of a factory from the following table

Solution:


Hence, the mean salary is ₹ 5083.33.

Question 6.
Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Solution:
(i) Mean marks in a test in mathematics,
(ii) Average beauty

We hope the NCERT Solutions for Class 9 Maths Chapter 14 Statistics, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 14 Statistics, drop a comment below and we will get back to you at the earliest.

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

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

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

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

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

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹10 per m² is ₹15000, find the height of the hall.
[Hint Area of the four walls = Lateral surface area]
Solution:
Let the rectangular hall of length = l, breadth = b, height = h

Hence, the height of the hall is 6 m.

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

Number of bricks that painted out of this container

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

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

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

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

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Question 1.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.
Solution:
We have, height = 14 cm
Curved surface area Of a right circular cylinder = 88 cm²

r = 1 cm
Diameter = 2 x Radius = 2 x 1 = 2 cm

Question 2.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?
Solution:
Let r be the radius and h be the height of the cylinder.
Given, r = \(\frac { 140 }{ 2 }\) = 70 cm = 0.70 m
and h = 1 m
Metal sheet required to make a closed cylindrical tank
= Total surface area
= 2πr (h + r)
= 2 x \(\frac { 22 }{ 7 }\) x 0.7(1 + 0.70)
= 2 x 22 x 0.1 x 170 = 7.48m²
Hence, the sheet required to make a closed cylindrical tank = 7.48m²

Question 3.
A metal pipe is 77 cm long. The inner ft diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its
(i) inner curved surface area.
(ii) outer curved surface area.
(iii) total surface area.

Solution:
We have, h = 77cm
Outer diameter (d₁) = 4.4 cm
and inner diameter (d₂) = 4 cm
Outer radius (r₁) = 2.2 cm
Inner radius (r₂) = 2cm
(i) Inner curved surface area = 2πr²h = 2x \(\frac { 22 }{ 7 }\) x 2 x 77 = 88 x 11 = 968 cm²
(ii) Outer curved surface area = 2πr₁h
= 2 x \(\frac { 22 }{ 7 }\) x 2.2 x 77
= 44 x 2.2 x 11= 1064.8cm²
(iii) Total surface area = Inner curved surface area + Outer curved surface area + Areas of two bases
= 968 + 1064.8 + 2\(\frac { 22 }{ 7 }\) (r₁² – r²)
= 968 + 1064.8 + 2 x \(\frac { 22 }{ 7 }\) [(2.2) – r²]
= [2032.8 + 2 x \(\frac { 22 }{ 7 }\) (4.84 – 4)]
= 2032.8 +\(\frac { 44 }{ 7 }\) x 0.84 = 2032.8+ 44x 0.12
= 2032.8 + 5.28 cm² = 2038.08 cm²

Question 4.
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².
Solution:
We have, diameter of a roller = 84 cm
r = radius of a roller = 42 cm
h = 120 cm
To cover 1 revolution = Curved surface area of roller
= 2πrh
= 2 x \(\frac { 22 }{ 7 }\) x 42 x 120
= 44 x 720 cm²
= 31680 cm²

Area of the playground = Takes 500 complete revolutions = 500 x 3.168 m²
= 1584 m²

Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m².
Solution:
Given, Diameter = 50 cm

Curved surface area of the pillar = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 0.25 x 35
= 2 x 22 x 0.25 x 0.5 = 55 m²
Cost of painting per m2 = ₹ 12.50
Cost of painting 5.5 m2 = ₹ 12.50×5.5 = ₹ 68.75

Question 6.
Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.
Solution:
We have, curved surface area of a right circular cylinder = 4.4m²
∴ 2πrh = 4.4

Hence, the height of the right circular cylinder is 1 m

Question 7.
he inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area.
(ii) the cost of plastering this curved surface at the rate of ₹40 per m².
Solution:
We have, inner diameter = 3.5 m
∴ inner radius = \(\frac { 3.5 }{ 2 }\) m
and h= 10m
(i) Inner curved surface area = 2πrh = 2x \(\frac { 22 }{ 7 }\) x \(\frac { 3.5 }{ 2 }\) x10 = 22 x 5 = 110m²
(ii) Cost of plastering perm2 = ₹40
Cost of plastering 110 m2 = ₹40x 110= ₹4400

Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
We have, h = 28 m
Diameter = 5 cm

Total radiating surface in the system = Curved surface area of the cylindrical pipe
= 2πrh =2 x \(\frac { 22 }{ 7 }\) x 0.025 x 28 = 4.4 m²

Question 9.
Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if \(\frac { 1 }{ 12 }\) of the steel actually used was wasted in making the tank?
Solution:
(i) We have, diameter = 4.2 m

Since, \(\frac { 1 }{ 2 }\) of the actual steel used was wasted, therefore the area of the steel which was actually used for making the tank

Question 10.
In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Solution:
Given r = \(\frac { 20 }{ 2 }\) cm = 10cm
h = 30 cm
Since, a margin of 2.5 cm is used for folding it over the top and bottom so the total height of frame,
h1 = 30 + 25 + 25
h1 = 35 cm
∴ Cloth required for covering the lampshade = Its curved surface area

Question 11.
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Solution:
Cardboard required by each competitor
= Base area + Curved surface area of one penholder
= πr2 + 2πrh [Given, h = 10.5 cm, r = 13cm]
= \(\frac { 22 }{ 7 }\) x (3)2 + 2 x \(\frac { 22 }{ 7 }\) x 3 x 105
= (28.28 + 198) cm²
= 22828 cm²
For 35 competitors cardboard required = 35 x 22628 = 7920 cm²
Hence, 7920 cm² of cardboard was required to be bought for the competition.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
We have, diameter = 10.5 cm
Radius (r) = \(\frac { 10.5 }{ 2 }\) = 5.25 cm
and slant height l= 10 cm
Curved surface area = πrl= \(\frac { 22 }{ 7 }\) x 5.25 x 10 = 165 cm²

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
We have, slant height l = 21 m
and diameter = 24 cm

Question 3.
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone.
Solution:
We have, slant height, l= 14cm
Curved surface area of a cone = 308 cm²

Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹70.
Solution:
We have, h = 10 m

Hence, the slant height of the canvas tent is 26 m.
(ii) Canvas required to make the tent = Curved surface area of tent
= πrl
= π x 24 x 26 = 624π m²
∵ Cost of 1 m2 canvas = ₹ 70
Cost of 624n m2 canvas = ₹ 70 x 624π
= ₹ 70x 624 x \(\frac { 22 }{ 7 }\)
= ₹ 10 x 624 x 22
= ₹ 137280
Hence, the cost of the canvas is ₹ 137280.

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14)
Solution:
Let r, h and l be the radius, height and slant height of the tent, respectively.

The extra material required for stitching margins and cutting = 20 cm = Q² m Hence, the total length of tarpaulin required = 62.8 + Q² = 63 m

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m².
Solution:
We have, slant height, l = 25m
and diameter = 14m
∴ Radius, r= 7m

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:

∵ The sheet required to make 1 cap = 550 cm²
∴ The sheet required to make 10 caps = 550 x 10= 5500 cm²

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take \( \sqrt{104} \) = 102)
Solution:

Curved surface area of a cone = πrl = 3.14 x 0.2 x 1.02 = 0.64056 m²
Cost of painting per m2 = ₹12
Cost of painting 0.64056 m2 = ₹12 x 0.64056= ₹ 7.68672
Cost of painting for 1 cone = ₹ 7.68672
Cost of painting 50 cones = ₹ 7.68672x 50= ₹ 384.336= ₹ 384.34

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Question 1.
Find the surface area of a sphere of radius
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution:
(i) We have, r = 105 cm

Question 2.
Find the surface area of a sphere of diameter
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
Solution:

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
We have, r = 10 cm
Total surface area of a hemisphere = 3πr²
= 3 x 3.14 x (10)²
= 9.42 x 100
= 942 cm²

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Let initial radius, r₁ = 7 cm
After increases, r₂ = 14 cm
Surface area for initial balloon = 4πr₁² = 4 x \(\frac { 22 }{ 7 }\) x 7 x 7 = 88 x 7
A₁ = 616 cm²
Surface area for increasing balloon = 4πr₂² = 4x \(\frac { 22 }{ 7 }\) x 14 x 14 = 88 x 28
A₂ = 2464 cm²
∴ Required ratio = A₁ : A₂ = 616 : 2464 = 1 : 4

Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².
Solution:
We have, inner diameter = 10.5 cm
Inner radius = \(\frac { 10.5 }{ 2 }\) cm = 5.25 cm
Curved surface area of hemispherical bowl of inner side = 2πr²

Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
Surface area of a sphere = 154 cm²


Hence, the radius of the sphere is 3.5 cm.

Question 7.
The diameter of the Moon is approximately one-fourth of the diameter of the Earth. Find the ratio of their surface areas.
Solution:
Let diameter of the Earth = d₁

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
Outer radius of the bowl = (Inner radius + Thickness)
= ( 5 + 0.25) cm = 5.25 cm

Question 9.
A right circular cylinder just encloses a sphere of radius r (see figure). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
Solution:

The radius of the sphere = r
Radius of the cylinder = Radius of the sphere = r
Height of the cylinder = Diameter = 2r
(i) Surface area of the sphere A₁ = 4πr²
(ii) Curved surface area of the cylinder = 2πrh
A₂ = 2π x r x 2r
A₂ = 4πr²
(iii) Required ratio = A₁ :A₂ = 4πr² : 4πr² = 1 : 1

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A matchbox measures 4 cm x 2.5 cm x 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Solution:
Volume of a match box = 4 cm x 2.5 cm x 1.5 cm= 15 cm³
Volume of a packet = 12 x 15 cm³ = 180 cm³

Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? ( 1 m³ = 1000L)
Solution:
Volume of a cuboidal water tank = 6 m x 5 m x 4.5 m
= 30 x 4.5 m³ = 135 m³
= 135 x 1000L = 135000L (∵ 1 m³ = 1000L)

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Solution:
Volume of a cuboidal vessel = hold liquid
∴ l x b x h = 380 m3
⇒ 10 x 8 x h = 380
⇒ h = \(\frac { 380 }{ 80 }\)
⇒ h= 4.75m
Hence, the cuboidal vessel must be made 4.75 m high.

Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m3.
Solution:
Volume of a cuboidal pit = l x bx h= (8 x 6 x 3)m³ =144 m³
∵ Cost of digging per m³ = ₹ 30
∴ Cost of digging 144m³ = ₹ 30x 144 = ₹ 4320

Question 5.
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are 2.5 m and 10 m, respectively.
Solution:
Given, l = 2.5 m and h = 10m
Let breadth of tank be b.

Hence, breadth of the cuboidal tank is 2 m.

Question 6.
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m x 15 m x 6 m. For how many days will the water of this tank last?
Solution:

Question 7.
A go down measures 40 m x 25 m x 10 m. Find the maximum number of wooden crates each measuring 15 m x 125 m x 0.5 m that can be stored in the go down.
Solution:
Dimension for godown, l = 40m b = 25 m and c = 10 m
Volume of the godown = l x b x h= 40 m x 25 m x 10 m
Dimension for wooden gate l = 1.5 m, b = 1.25 m, b = 1.25 m, h = 0.5 m
Volume of wooden gates = l x b x h = 1.5 m x 1.25 m x 0.5m

Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution:
Volume of a solid cube = 12 cm x 12 cm x 12 cm = 1728 cm³
The solid cube is cut into eight cubes of equal volume.

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Solution:
Given, l = 2 km= 2 x 1000 m = 2000m
b = 40 m
and h = 3 m
Since, the water flows at the rate of 2 km h^-1,
i.e., the water from 2 km of river flows into the sea in one hour.
The volume of water flowing into the sea in one hour = Volume of the cuboid
= l x b x h
= (2000 x 40 x 3) m³
= 240000 m³
∴ The volume of water flowing into the sea in one minute
= \(\frac { 240000 }{ 60 }\) m³
= 4000 m³

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

Question 1.
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000cm3 =1 L.)
Solution:
We have, circumference of the base = 132 cm
∴ 2πr = 132

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.
Solution:
Given, inner diameter = 24 cm
Inner radius (r₁) = 12cm
Outer diameter = 28 cm
Outer radius (r₂) = 14cm
h = 35cm

Question 3.
A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm.
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Solution:
(i) We have, h = 15cm
l = 5cm, b = 4cm, h = 15 cm
Volume of cuboidical body =l x b x h= 5 x 4 x 15 = 300cm3 …(i)
(ii) We have, Diameter = 7cm
Radius, r = \(\frac { 7 }{ 2 }\) cm
Height, h = 10cm

∴ From Eqs. (i) and (ii), we see that volume of a plastic cylinder has greater capacity and its capacity is 385 – 300 = 85 cm3 is more than the tin can.

Question 4.
If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find
(i) radius of its base,
(ii) its volume. (Use π = 3.14)
Solution:
We have, lateral surface of a cylinder = 94.2 cm² and h = 5 cm
∴ 2πrh = 94.2
⇒ 2 x 3.14 x r x 5 = 94.2
⇒ r = \(\frac { 94.2 }{ 31.4 }\)
⇒ r = 3 cm
(i) Hence, radius of base, r = 3cm
(ii) Volume of a cylinder = πr2h= 3.14(3)²x 5
= 3.14 x 9 x 5
= 141.3 cm³

Question 5.
It costs ₹2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹20 per m², find
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.
Solution:
We have, cost to paint the inner curved surface = ₹2200
Cost to paint per m² = ₹20

(i) Inner curved surface area of the vessel = 110m²
(ii) Hence, radius of the base is 1.75 m.
(iii) Capacity of the vessel = Volume of the vessel

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Solution:
Capacity of a closed cylindrical vessel = 15.4 L
= 15.4 x 1000cm3 (∵ 1L = 1000 cm²)
∴ πr²h = 15400 cm³
πr² x 100 = 15400
r² = 154 x \(\frac { 7 }{ 22 }\) (∵ h= 1m= 100cm)
⇒ r² = 7 x 7
⇒ r = 7 cm
Total surface area of closed cylindrical vessel = 2πr(r + h)

Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of. the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution:

Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
We have, diameter = 7cm
Radius, r = \(\frac { 7 }{ 22 }\) cm
h= 4 cm
Capacity of the cylindrical bowl = Volume of the cylinder = πr²h

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7

Question 1.
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
Solution:
(i) We have, r = 6 cm and h = 7cm

Question 2.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
Solution:
(i) We have, r = 7 cm and l = 25 cm

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π = 3.14)
Solution:

Question 4.
If the volume of a right circular cone of height 9 cm is 48 cm3, find the diameter of its base.
Solution:
We have, volume of a right circular cone = 48 π cm3

Question 5.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
We have, diameter = 3.5 m

Question 6.
The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
Solution:
We have, d = 28cm
⇒ r = 14 cm
∵ Volume of a right circular cone = 9856 cm³

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
On revolving the right ∆ ABC about the side AB, we get a cone as shown in the adjoining figure.

∴ Volume of the solid so obtained = \(\frac { 1 }{ 3 }\) πr²h
= \(\frac { 1 }{ 3 }\) x π x 5 x 5 x 4 = 100π cm³
Hence, the volume of the solid so obtained is 100π cm³.

Question 8.
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
On revolving the right ∆ ABC about the side BC( = 5 cm),
we get a cone as shown in the adjoining figure.

In above Question 7, the volume V₂ obtained by revolving the ∆ ABC about the side 12 cm i.e., V₂ = 100π cm³.

Hence, the required ratio =5:12

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Solution:

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 cm
Solution:
(i) We have, r = 7 cm

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Solution:
(i) We have, d = 28 cm
⇒ r = 14 cm

Question 3.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?
Solution:
We have, d = 4.2 cm ⇒ r = 2.1 cm

The density of the metal per cm³ = 8.9 g
The density of the metal 38.808 cm³ = 38.808 x 8.9 g = 345.39 g
Hence, the mass of the ball = 345.39 g

Question 4.
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?
Solution:
Let diameter of the Earth is d₃.
We have, diameter of Moon (d₁) = \(\frac { 1 }{ 4 }\) diameter of the Earth

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:
Diameter = 10.5 cm
⇒ Radius, r = 5.25 cm

Question 6.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Solution:
Inner radius (r₁) = 1 m -100 cm
Outer radius (r₂) = (100 + 1)cm ( ∵ Inner radius + Thickness)
= 101 cm

Question 7.
Find the volume of a sphere whose surface area is 154 cm².
Solution:
We have, surface area of a sphere = 154 cm²

Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹498.96. If the cost of white washing is ₹2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Solution:
Cost of white washed = ₹ 498.96
Cost of white washing per square metre = ₹ 2.00

Question 9.
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the
(i) radius r’ of the new sphere,
(ii) ratio of S and S’.
Solution:

Question 10.
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?
Solution:
We have, diameter = 3.5 mm

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows : Height = 110cm, Depth = 25cm, Breadth = 85cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing-is 20 paise per cm² and the rate of pointing is 10 paise per cm², find the total expenses required for palishing and painting the surface of the bookshelf.

Solution:
Here, l = 85 cm b = 25 cm and h = 110 cm
Area of the bookshelf of outer surface = 2 lb + 2bh + hl
= [2(85 x 25)+ 2(110 x 25)+ 85×110] cm2 = (4250 + 5500 + 9350) cm² = 19100 cm²
Cost of polishing of the outer surface of bookshelf
= 19100 x \(\frac { 20 }{ 100 }\) = ₹ 3820
Thickness of the plank = 5 cm
Internal height of bookshelf = (110 – 2 x 5) = 100 cm
Internal depth of bookshelf = (25 – 5) = 20 cm
Internal breadth of bookshelf = 85 – 2 x 5 = 75 cm
Hence, area of the internal surface of bookshelf
= 2(75 x 20)+ 2(100x 20)+ 75×100
= 3000 + 4000 + 7500 = 14500 cm2
So, cost of painting of internal surface of bookshelf
= 14500 x \(\frac { 10 }{ 100 }\) = ₹1450
Hence, total costing of polishing and painting = 3820 + 1450= ₹ 5270

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are-used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Solution:
It is obvious, we have to subtract the cost of the sphere that is resting on the supports while calculating the cost of silver paint.
Surface area to be silver paint

Question 3.
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Solution:
Let d be the diameter of the sphere

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

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Question 1.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.
Solution:
We have, height = 14 cm
Curved surface area Of a right circular cylinder = 88 cm²

r = 1 cm
Diameter = 2 x Radius = 2 x 1 = 2 cm

Question 2.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?
Solution:
Let r be the radius and h be the height of the cylinder.
Given, r = \(\frac { 140 }{ 2 }\) = 70 cm = 0.70 m
and h = 1 m
Metal sheet required to make a closed cylindrical tank
= Total surface area
= 2πr (h + r)
= 2 x \(\frac { 22 }{ 7 }\) x 0.7(1 + 0.70)
= 2 x 22 x 0.1 x 170 = 7.48m²
Hence, the sheet required to make a closed cylindrical tank = 7.48m²

Question 3.
A metal pipe is 77 cm long. The inner ft diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its
(i) inner curved surface area.
(ii) outer curved surface area.
(iii) total surface area.

Solution:
We have, h = 77cm
Outer diameter (d₁) = 4.4 cm
and inner diameter (d₂) = 4 cm
Outer radius (r₁) = 2.2 cm
Inner radius (r₂) = 2cm
(i) Inner curved surface area = 2πr²h = 2x \(\frac { 22 }{ 7 }\) x 2 x 77 = 88 x 11 = 968 cm²
(ii) Outer curved surface area = 2πr₁h
= 2 x \(\frac { 22 }{ 7 }\) x 2.2 x 77
= 44 x 2.2 x 11= 1064.8cm²
(iii) Total surface area = Inner curved surface area + Outer curved surface area + Areas of two bases
= 968 + 1064.8 + 2\(\frac { 22 }{ 7 }\) (r₁² – r²)
= 968 + 1064.8 + 2 x \(\frac { 22 }{ 7 }\) [(2.2) – r²]
= [2032.8 + 2 x \(\frac { 22 }{ 7 }\) (4.84 – 4)]
= 2032.8 +\(\frac { 44 }{ 7 }\) x 0.84 = 2032.8+ 44x 0.12
= 2032.8 + 5.28 cm² = 2038.08 cm²

Question 4.
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².
Solution:
We have, diameter of a roller = 84 cm
r = radius of a roller = 42 cm
h = 120 cm
To cover 1 revolution = Curved surface area of roller
= 2πrh
= 2 x \(\frac { 22 }{ 7 }\) x 42 x 120
= 44 x 720 cm²
= 31680 cm²

Area of the playground = Takes 500 complete revolutions = 500 x 3.168 m²
= 1584 m²

Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m².
Solution:
Given, Diameter = 50 cm

Curved surface area of the pillar = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 0.25 x 35
= 2 x 22 x 0.25 x 0.5 = 55 m²
Cost of painting per m2 = ₹ 12.50
Cost of painting 5.5 m2 = ₹ 12.50×5.5 = ₹ 68.75

Question 6.
Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.
Solution:
We have, curved surface area of a right circular cylinder = 4.4m²
∴ 2πrh = 4.4

Hence, the height of the right circular cylinder is 1 m

Question 7.
he inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area.
(ii) the cost of plastering this curved surface at the rate of ₹40 per m².
Solution:
We have, inner diameter = 3.5 m
∴ inner radius = \(\frac { 3.5 }{ 2 }\) m
and h= 10m
(i) Inner curved surface area = 2πrh = 2x \(\frac { 22 }{ 7 }\) x \(\frac { 3.5 }{ 2 }\) x10 = 22 x 5 = 110m²
(ii) Cost of plastering perm2 = ₹40
Cost of plastering 110 m2 = ₹40x 110= ₹4400

Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
We have, h = 28 m
Diameter = 5 cm

Total radiating surface in the system = Curved surface area of the cylindrical pipe
= 2πrh =2 x \(\frac { 22 }{ 7 }\) x 0.025 x 28 = 4.4 m²

Question 9.
Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if \(\frac { 1 }{ 12 }\) of the steel actually used was wasted in making the tank?
Solution:
(i) We have, diameter = 4.2 m

Since, \(\frac { 1 }{ 2 }\) of the actual steel used was wasted, therefore the area of the steel which was actually used for making the tank

Question 10.
In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Solution:
Given r = \(\frac { 20 }{ 2 }\) cm = 10cm
h = 30 cm
Since, a margin of 2.5 cm is used for folding it over the top and bottom so the total height of frame,
h1 = 30 + 25 + 25
h1 = 35 cm
∴ Cloth required for covering the lampshade = Its curved surface area

Question 11.
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Solution:
Cardboard required by each competitor
= Base area + Curved surface area of one penholder
= πr2 + 2πrh [Given, h = 10.5 cm, r = 13cm]
= \(\frac { 22 }{ 7 }\) x (3)2 + 2 x \(\frac { 22 }{ 7 }\) x 3 x 105
= (28.28 + 198) cm²
= 22828 cm²
For 35 competitors cardboard required = 35 x 22628 = 7920 cm²
Hence, 7920 cm² of cardboard was required to be bought for the competition.

We hope the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6, are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

Question 1.
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000cm3 =1 L.)
Solution:
We have, circumference of the base = 132 cm
∴ 2πr = 132

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.
Solution:
Given, inner diameter = 24 cm
Inner radius (r₁) = 12cm
Outer diameter = 28 cm
Outer radius (r₂) = 14cm
h = 35cm

Question 3.
A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm.
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Solution:
(i) We have, h = 15cm
l = 5cm, b = 4cm, h = 15 cm
Volume of cuboidical body =l x b x h= 5 x 4 x 15 = 300cm3 …(i)
(ii) We have, Diameter = 7cm
Radius, r = \(\frac { 7 }{ 2 }\) cm
Height, h = 10cm

∴ From Eqs. (i) and (ii), we see that volume of a plastic cylinder has greater capacity and its capacity is 385 – 300 = 85 cm3 is more than the tin can.

Question 4.
If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find
(i) radius of its base,
(ii) its volume. (Use π = 3.14)
Solution:
We have, lateral surface of a cylinder = 94.2 cm² and h = 5 cm
∴ 2πrh = 94.2
⇒ 2 x 3.14 x r x 5 = 94.2
⇒ r = \(\frac { 94.2 }{ 31.4 }\)
⇒ r = 3 cm
(i) Hence, radius of base, r = 3cm
(ii) Volume of a cylinder = πr2h= 3.14(3)²x 5
= 3.14 x 9 x 5
= 141.3 cm³

Question 5.
It costs ₹2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹20 per m², find
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.
Solution:
We have, cost to paint the inner curved surface = ₹2200
Cost to paint per m² = ₹20

(i) Inner curved surface area of the vessel = 110m²
(ii) Hence, radius of the base is 1.75 m.
(iii) Capacity of the vessel = Volume of the vessel

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Solution:
Capacity of a closed cylindrical vessel = 15.4 L
= 15.4 x 1000cm3 (∵ 1L = 1000 cm²)
∴ πr²h = 15400 cm³
πr² x 100 = 15400
r² = 154 x \(\frac { 7 }{ 22 }\) (∵ h= 1m= 100cm)
⇒ r² = 7 x 7
⇒ r = 7 cm
Total surface area of closed cylindrical vessel = 2πr(r + h)

Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of. the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution:

Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
We have, diameter = 7cm
Radius, r = \(\frac { 7 }{ 22 }\) cm
h= 4 cm
Capacity of the cylindrical bowl = Volume of the cylinder = πr²h

We hope the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3, are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
We have, diameter = 10.5 cm
Radius (r) = \(\frac { 10.5 }{ 2 }\) = 5.25 cm
and slant height l= 10 cm
Curved surface area = πrl= \(\frac { 22 }{ 7 }\) x 5.25 x 10 = 165 cm²

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
We have, slant height l = 21 m
and diameter = 24 cm

Question 3.
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone.
Solution:
We have, slant height, l= 14cm
Curved surface area of a cone = 308 cm²

Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹70.
Solution:
We have, h = 10 m

Hence, the slant height of the canvas tent is 26 m.

(ii) Canvas required to make the tent = Curved surface area of tent
= πrl
= π x 24 x 26 = 624π m²
∵ Cost of 1 m2 canvas = ₹ 70
Cost of 624n m2 canvas = ₹ 70 x 624π
= ₹ 70x 624 x \(\frac { 22 }{ 7 }\)
= ₹ 10 x 624 x 22
= ₹ 137280
Hence, the cost of the canvas is ₹ 137280.

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14)
Solution:
Let r, h and l be the radius, height and slant height of the tent, respectively.

The extra material required for stitching margins and cutting = 20 cm = Q² m Hence, the total length of tarpaulin required = 62.8 + Q² = 63 m

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m².
Solution:
We have, slant height, l = 25m
and diameter = 14m
∴ Radius, r= 7m

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:

∵ The sheet required to make 1 cap = 550 cm²
∴ The sheet required to make 10 caps = 550 x 10= 5500 cm²

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take \( \sqrt{104} \) = 102)
Solution:

Curved surface area of a cone = πrl = 3.14 x 0.2 x 1.02 = 0.64056 m²
Cost of painting per m2 = ₹12
Cost of painting 0.64056 m2 = ₹12 x 0.64056= ₹ 7.68672
Cost of painting for 1 cone = ₹ 7.68672
Cost of painting 50 cones = ₹ 7.68672x 50= ₹ 384.336= ₹ 384.34

We hope the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3, drop a comment below and we will get back to you at the earliest.