Category: Class 9

MP Board Class 9th Maths Solutions Chapter 14 Statistics Ex 14.2

Question 1.
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. Which is the most common, and which is the rarest, blood group among these students?
Solution:
Frequency distribution table:

Common blood group – O
Rarest blood group – 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:
Frequency distribution table:

We can observe that residence of most of the engineers are at a distance of 5 – 20 km from their place of work.

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

1. Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
2. Which month or season do you think this data is about?
3. What is the range of this data?

Solution:
1. Frequency distribution table.

2. Rainy season as the relative humidity is high.
3. 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:

1. Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 -165,165 -170, etc.
2. What can you conclude about their heights from the table?

Solution:
1. Frequency distribution table

2. We observe that most of the students have their heights between 160 – 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:

1. Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
2. For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

Solution:
1.

2. The concentration of sulphur dioxide was more than 0.11 ppm for 8 days.

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:

Question 7.
The value of n upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510

1. Make a frequency distribution of the digits from 0 to 9 after the decimal point.
2. What are the most and the least frequently occurring digits?

Solution:
1.

2. The most frequently occurring digits are 3 and 9. The least occurring is 0.

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:

1. Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 – 10.
2. How many children watched television for 15 or more hours a week?

Solution:
1.

2. 2 children.

Question 9.
A company manufactures car batteries of a particular type.ffhe 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:

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 14 Statistics Ex 14.1

1. Give five examples of data that you can collect from your day – to – day life.
2. Classify the data in –

Question 1.
Above as primary or secondary data
Solution:
Do it yourself.

Presentation of Data:
The data collected in original form is called raw data. These data are arranged in ascending or descending order. This arrangement is called array. The data are divided or grouped into classes in condensed form. This is known as presentation of data.

Range of Data:
The difference between highest and lowest value of data is called the range of data.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 15 Probability Ex 15.1

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Total number of trials = 30
Number of balls in which boundary hit = 6
Number of balls in which boundary not hit = 30 – 6 = 24
Let E denotes the event that she did not hit a boundary.
P(E) = \(\frac{24}{30}\) =0.8.

Question 2.
1500 families with 2 children were selected randomly, and the following data were recorded.

Compute the probability of a family, chosen at random, having

1. 2 girls
2. 1 girl
3. No girl

Also check whether the sum of these probabilities is 1.
Solution:
Total number of families = 1500
Let E , E2 and E} denote the event that the families having 2 girls, 1 girl and no girl.
P(E₁) = \(\frac{475}{1500}\) = 0.317
P(E₂) = \(\frac{814}{1500}\) = 0.543
P(E₃) = \(\frac{211}{1500}\) = 0.140
P(E₁) + P(E₂) + P(E₃) = 0.317 + 0.543 + 0.140 = 1.

Question 3.
In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained:

Find the probability that a student of the class was born in August.
Solution:
Total number of students = 40
Number of students that were bom in the month of August = 6 Let E denotes the event that a student of the class was bom in August
P(E) = \(\frac{6}{40}\) = 0.15

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are tossed simultaneously tossed again, compute the probability of 2 heads coming up.
Solution:
Total number of trials = 200
Frequency of 2 heads coming up = 72
Let E denotes the event that 2 heads coming up
P(E) = \(\frac{72}{200}\) = 0.36

Question 5.
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

Suppose a family is chosen. Find the probability that the family chosen is

1. earning ₹ 10000 -13000 per month and owning exactly 2 vehicles.
2. earning ₹ 16000 or more per month and owning exactly 1 vehicle.
3. earning less than ₹ 7000 per month and does not own any vehicle.
4. earning ₹ 13000 – 16000 per month and owning more then 2 vehicles.
5. owning not more than 1 vehicle.

Solution:
Total number of families = 2400

1. P (familiy chosen is earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles)
= \(\frac{29}{2400}\)

2. P (that the family chosen is earning ₹ 16000 or more and owning exactly 1 vehicle)
= \(\frac{579}{2400}\)

3. P (that the family chosen is earning less than ₹ 7000 per month and does not own any vehicle)
= \(\frac{10}{2400}\) = \(\frac{1}{240}\)

4. P (that the family chosen is earning ₹ 13000 – 16000 per month and owning more than 2 vehicles)
\(\frac{25}{2400}\) = \(\frac{1}{96}\)

5. Number of families owning not more than 1 vehicle
= 10 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579
= 2062
P (that the family chosen is owning not more than 1 vehicle)
\(\frac{2062}{2400}\) = \(\frac{1031}{1200}\)

Question 6.
Refer the table below:

1. Find the probability that a student obtained less then 20% in the mathematics test.
2. Find the probability that a student obtained marks 60 or above.

Solution:
Total number of students = 90

1. Number of students getting less that 20% marks = 7
Let E denotes the event th.-t a student obtained less than 20% in the mathematics test
P(E) = \(\frac{7}{90}\) = 0.078

2. Number of students getting marks 60 or above = 15 + 8 = 23
Let E denote the event that a student obtained marks 60 or above.
P(E) = \(\frac{23}{90}\) = 0.255

Question 7.
To know the opinion of the student about the subject statistics, a survey of200 students was conducted. The data is recorded in the following table.

Find the probability that a student chosen at random

1. likes statistics
2. does not like it.

Total number of students on which survey was conducted = 200
1. Let E₁ denotes the event that the student chosen likes statistics.
P(E₁) = \(\frac{135}{200}\) = 0.675

2. Let E₂ denotes the event that the student chosen dislikes statistics.
P(E₂) = \(\frac{65}{200}\) = 0.325

Question 8.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows:

What is the empirical probability that an engineer lives:

1. less than 7 km from her place of work?
2. more than or equal to 7 km from her place of work?
3. within 1/2 km from her place of work?

Solution:
1. Number of engineers residing less than 7 km from their place of work = 9
Total number of engineers = 40
Let E₁ denotes the event that an engineer lives less than 7 km from her place of work.
P(E₁) = \(\frac{9}{40}\)

2. Number of engineers residing more than or equal to 7 km from their place of work = 40 – 9 = 31
Let E₂ denotes the event that an engineer lives more than or equal to 7 km from their place of work
P(E₂) = \(\frac{31}{40}\)

3. Number of engineers residing within \(\frac{1}{2}\) km = 0
Let E₃ denotes the event that an engineer lives within \(\frac{1}{2}\) km from their place of work.
P(E₃) = \(\frac{0}{40}\)

Question 9.
Activity: Note the frequency of two-wheelers, three wheelers and four wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Solution:
Do if yourself.

Question 10.
Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Solution:
Do it with the help of your class mates.

Question 11.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg);
4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Total number of bags = 11
Number of bags containing more than 5 kg of flour = 7
Let E denotes the event that any of these bags chosen at random
contains more then 5 kg of flour.
p(E) = \(\frac{7}{11}\)

Question 12.
You were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphar dioxide in the interval 0.12 -0.16 on any of these days.

Solution:
Frequency distribution table

Total number of days = 30
Let£ denotes the event of the concentration of sulphar dioxide in the interval 0.12 – 0.16 on any of these days.
p(E) = \(\frac{2}{30}\) = \(\frac{1}{15}\)

Question 13.
You were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AS.
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.
Solution:
Frequency distribution table
Number of students having blood group AB = 3
Total number of students = 30
Let E denotes the event that a student of this class selected at random has blood group AB.
p(E) = \(\frac{3}{10}\) = \(\frac{1}{10}\)

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.8

Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
Find the volume of a sphere whose radius is

(i) 7 cm
(ii) 0.63 m

Solution:
(i) Here, radius (r) = 7 cm
∴ Volume of the sphere

Thus, the required volume is 1.05 m³ (approx.)

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm
(ii) 0.21 m

Solution:
(i) Diameter of the ball = 28 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:
d = 4.2 cm
⇒ r = 2.1 cm
Density (D) = 8.9 gm/cm³
Volume of metallic ball = \(\frac{4}{3}\) x \(\frac{22}{7}\) x 2.1 x 2.1 x 2.1 = 38.808 cm³
Mass = D x V
= 8.9 x 38.808
= 345.3912 gm.

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 volume of the moon?
Solution:
Let d and d be the diameter of moon and earth respectively.

Question 5.
How many liters of milk can hemispherical bowl of diameter 10 J cm hold?
Solution:
d = 10.5 cm
r = 5.25 cm
Volume of the hemisphere = \(\frac{2}{3}\) x \(\frac{22}{7}\) x 5.25 x 5.25 x 0.25
= 303.18 cm³
= 0.303 l

Question 6.
A hemispherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Solution:
t = 1 cm
r₁ = 1 m = 100 cm
r₂ = r₁ + t = 100 + 1 = 101 cm
Outer volume of tank (V₂) = \(\frac{2}{3}\) πr₂²
Inner volume of tank (V₁) = \(\frac{2}{3}\) πr₁³
Volume of iron = Outer volume – inner volume

= 0.06348 ml

Question 7.
Find the volume of a sphere whose surface area is 154 cm2.
Solution:
Surface area = 154 cm²
Surface area of sphere = 4πr²

Question 8.
A dome of a building is in the from 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 meter, find the

(i) inside surface area of the dome.
(ii) volume of the air inside the dome.

Solution:
(i) Cost = ₹ 498.96
Rate = ₹ 2 per m²
Inside surface area = \(\frac{498.96}{2}\) = 249.48 m²
Inside curved surface area = 2πr²

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:
Let V and be the volume of old and new sphere respectively
(i) Volume of new sphere = 27 x volume of old sphere
V₁ = 27 x V

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:
d = 3.5 mm
r = 1.75 mm
Volume of medicine needed to fill the capsule
= \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\) x \(\frac{22}{7}\) x 1.75 x 1.75 x 1.75
= 22.458 mm³
= 22.46 mm³ (Approx.)

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7

Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
Find the volume of the right circular cone with

1. radius 6 cm, height 7 cm
2. radius 3.5 cm, height 12 cm.

Solution:
1. Here, radius of the cone r = 6 cm
height (h) = 7 cm
Volume = \(\frac{1}{3}\) x πr²h
= \(\frac{1}{3}\) x \(\frac{22}{7}\) x 6 x 6 x 7 cm³
= 22 x 2 x 6 cm³
= 264 cm³

2. Here, radius of the cone (r) ;
= 3.5 cm = \(\frac{35}{10}\) cm
Height (h) = 12 m
Volume of the cone

Question 2.
Find the capacity in litres of a conical vessel with

1. radius 7 cm, slant height 25 cm
2. height 12 cm, slant height 13 cm

Solution:
1. Here, r = 7 and l = 25 cm

Thus, the required capacity of the conical vessel is 1.232 l.

2. Here, height (h) – 12 cm and l = 13 cm

Thus, the required capacity of the conical vessel is \(\frac{11}{35}\) l.

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use 71 = 3.14)
Solution:
Here, height of the cone (h) = 15 cm
Volume of the cone (v) = 1570 cm³
Let the radius of the base be ‘r’ cm.

Question 4.
If the volume of a right circular cone of height 9 cm is 48 JI cm3, find the diameter of its base.
Solution:
Volume of cone = \(\frac{1}{3}\) πr²h
\(\frac{1}{3}\) x πr² x 9 = 48π
r² = \(\frac{48π}{9π}\) x 3
r² = 16
r = 4 cm
Diameter = 2 x 4 = 8 cm

Question 5.
A conical pit of top diameter 3,5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
d = 3.5 m
r = 1.75 m
h = 12m
Volume of the pit = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 1. 75 x 1.75 x 12
= 38.5 m³ = 38.5 kl (1 kl= 1 m³)

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base 28 cm, find.

(i) height of the cone.
(ii) slant height of the cone.
(iii) curved surface area of the cone.

Solution:
V = 9856 cm²
d =28 cm
r = 14 cm

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the sides 12 cm. Find the volume of the solid 30 obtained.
Solution:
h = 12cm
r = 5cm

Volume of solid, V₁ = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) x π x 5 x 5 x 12
= 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:
h = 5 cm
r = 12 cm
Volume of solid, V₂ = \(\frac{1}{3}\) x π x 12 x 12 x 5
= 240π

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 by covered canvas to protect it from rain. Find the area of the canvas required.
Solution:
d = 10.5 m ⇒ r = 5.25 m
h = 3m
Volume of heap of wheat = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 5.25 x 5.25 x 3
= 86.625 m³
l² = 3² + (5.25)²
l = \(\sqrt{9+27.56}\) = \(\sqrt{36.56}\) = 6.04 cm
Area of canvas required = CSA of cone –
= \(\frac{22}{7}\) x 5.25 x 6.04 = 99.66 m²

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6

Assume = \(\frac{22}{7}\) unless stated otherwise.

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? (1000 cm³ = 1l)
Solution:
Circumference of base, C =132 cm ,
h = 25 cm
C = 2πr
132 = 2 x \(\frac{22}{7}\) x r
r = \(\frac{132×7}{2×22}\) = 21 cm
Volume of cylinder =πr²h
= \(\frac{22}{7}\) x 21 x 21 x 25 = 34650 cm³
= \(\frac{34650}{1000}\)

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:

Inner diameter
d₁ = 24 cm ⇒ r₁ = 12 cm
Outer diameter, d₂ = 28 cm ⇒ r₂ = 14 cm
Volume of wood in the piple = πr₂²h – πr₁²h
πh(r₂² – r₁²) = \(\frac{22}{7}\) x 35(14² – 12²)
= \(\frac{22}{7}\) x 35 x 52 = 5720 cm³
Mass of pipe = 0.6 x 5720 = 3432 gm
= 3.432 kg.

Question 3.
A soft drink is available in two packs –

1. a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
2. a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Solution:
1. Volume of cuboidal can = (5 x 4 x 15) = 300 cm³

2. Volume of cylindrical can = πr²h = \(\frac{22}{7}\) x 3.5 x 3.5 x 10
= 385 cm²
Capacity of cylindrical can is more than the cuboidal can by 85 cm2.

Question 4.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find

1. radius of its base
2. its volume. (Use π = 3.14)

Solution:
1. CSA of cylinder = 94.2 cm²
2πrh = 94.2 cm²
2 x 3.14 x 5 = 94.2 cm²
r = \(\frac{94.2}{2×3.14×5}\) = 3 cm
Volume of the cylinder = πr²h
= 3.14 x 3 x 3 x 5
= 3.14 x 45 = 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

1. inner curved surface area of the vessel
2. radius of the base
3. Capacity of the Vessel.

1. Inner curved surface area =

2. Let r be the radius of the base ICSA = 2πrh
110 = 2 x \(\frac{22}{7}\) x r x 10 r
= 1.75 m

3. Capacity of vessel = πr²h = \(\frac{22}{7}\) x 1.75 x 1.75 x 10
= 96.25 m³.

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:
h = 1 m
V = 15.4l
= \(\frac{15.4}{1000}\)m³ = 0.0154 m³
Volume of the vessel = πr²h
0. 0154 = \(\frac{22}{7}\) x r² x 1
∴ r² = \(\frac{0.0154×7}{22}\)
⇒ r² = 0.0049
∴ r = 0.07 m
Area of metal sheet required = 2nr (h + r)
= 2x \(\frac{22}{7}\) x 0.07(1 + 0.07)
= 0.4708 m².

Question 7.
A lead pencil consists of a cylinder ofvyood 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:
Volume of the pencil = πr²h
= \(\frac{22}{7}\) x 3.5 x 3.5 x 140 = 5390 mm³
Volume of graphite = πr²h
= \(\frac{22}{7}\) x 0.5 x 0.5 x 140 = 110 mm³
Volume of wood = Volume of pencil – Volume of graphite
= 5390 – 110
= 5280 mm³
=5.28 cm³.

Question 8.
A patient in a hospital is given soup daily in a cylindrical biftvl 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:
d = 7cm
r = 3.5 cm
h = 4cm
Volume of the bowl = πr²h
= \(\frac{22}{7}\) x 3.5 x 3.5 x 4 = 154 cm³
Volume of soup needed for 250 patients = 154 x 250
= 38500 cm³
= 38.5l.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Assume π = \(\frac{22}{7}\), unless stated otherwise.

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:
Measures of matchbox (cuboid) is 4 cm x 2.5 cm x 1.5 cm
l = 4 cm,
b = 2.5 cm and
h = 1.5 cm
∴ Volume of matchbox = (l x b) x h
= [4 cm x 2.5 cm] x 1.5 cm³
= 4 x \(\frac{25}{10}\) x \(\frac{25}{10}\) cm³ = 15 cm³
Volume of 12 boxes =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 = 1000 l)
Solution:
Here, Length (l) = 6 m
Breadth (b) = 5 m
Depth (h) = 4.5 m
Capacity = l x b x h = 6 x 5 x 4.5 m³
= 6 x 5 x \(\frac{45}{10}\) m³ = 3 x 45
= 135 m³
∴ 1 m³ can hold 1000 l.
∴ 135 m³ can hold (135 x 1000 l = 135000 l) of water.
∴ The required amount of water in the tank = 135000 l.

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

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 cuboidal pit = 8 x 6 x 3 = 144 m³
Cost of digging = 144 x 30 = ₹ 4320.

Question 5.
The capacity of a cuboidal tank is 50000 liters of water. Find the breadth of the tank, it its length and depth are respectively 2.5 m and 10 m.
Solution:
Capacity of cuboidal tank = \(\frac{50000}{1000}\) m³ = 50 m³
Volume of cuboidal tank = l x b x h
50 = 2.5 x 6 x 10
50 = 25b
Breadth of tank, b = \(\frac{50}{25}\) = 2 m.

Question 6.
A village, having a population of4000, requires 150 liters of water per head per day. It has a tank measures 20 m x 15 m x 6m. For how many days will the water of this tank last?
Solution:
Volume of water required for the village per day = \(\frac{4000×150}{1000}\)
Volume of tank = 20 x 15 x 6 = 1800 m³
No. of days = \(\frac{1800}{600}\) = 3.

Question 7.
A godown measures 60 m x 25 m x 10 m. Find the maxium number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown.
Solution:
Volume of godown = (60 x 25 x 10) m³ = 15000 m³
Volume of one wooden crate = (1.5 x 1.25 x 0.5) m³ = 0.9375 m³
No. of wooden crates which can be stored in the godown

= 16000.

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 cubk? Also, find the ratio between their surface area.
Solution:
Volume of cube of side 12 cm = (12 x 12 x 12) cm³
Let V be volume of new cube.
Volume of cube of side 12 cm = 8 x volume of new cube
12 x 12 x 12 = 8 x V
V = \(\frac{12x12x12}{8}\) = \(\frac{12x12x12}{2x2x2}\) = 216 cm³
Let a be the side of new cube
a³ = 216 = 6³
a = 6 cm
TSA of cube of side 12 = 6 x (12)²
TSA of new cube = 6 x (6)²

Question 9.
Anver 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:
b = 40 m,
h = 3 m
V = 2 km/hr
= \(\frac{2×1000}{60}\) = \(\frac{100}{3}\) m/min
Volume of water coming out of the river per min = b x h x V
= \(\frac{100}{3}\) x 3 x 40 = 4000 m³.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
Find the surface area of a sphere of radius:

(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm

Solution:
(i) Here r = 10.5 cm
∴ Surface area of the sphere = 4πr²

Question 2.
Find the surface area of a sphere of diameter:

(i) 14 cm
(ii) 21 cm
(Hi) 3.5 m

Solution:
(i) Here, Diameter = 14 cm

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
Here, radius (r) = 10 cm

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:
Initial radius of the balloon, r₁ = 7 cm
Final radius of the balloon, r₂ = 14 cm
Initial surface area of the balloon = S₁
Final surfacg area of the balloon = S₂

S₁ : S₂ = 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:
d = 10.5 cm
r =5.25 cm
Rate of tinplating = \(\frac{16}{100}\) = ₹ 0.16 per cm²
Inner curved surface area of bow l = 2 πr²
ICSA = 2 x \(\frac{22}{7}\) x 5.25 x 5.25 = 173.25 cm²
Cost of tin-plating = Rate x ICSA
= 0.16 x 173.25 = ₹ 27.72

Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
Surface area of sphere = 4 πr2
154 = 4 x \(\frac{22}{7}\) x r²
\(\frac{154×7}{4×22}\) = r²
⇒ r² = 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:
d₁ = \(\frac{1}{4}\)d₂

Let .S₁, and .S₂, be the surface area of moon and earth respectively.
S₁ = 4 πr₁²
S₂ = 4 πr₂²

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:
r₁ = 5 cm
t (thickness) = 0.25 cm
r₂ = r₁ + t
= 5 + 0.25 = 5.25 cm
OCSA of bowl = 2x \(\frac{22}{7}\) x 5.25 x 5.25
= \(\frac{44}{7}\) x 5.25 x 5.25 = 173.25 cm²

Question 9.
A right circular cylinder just encloses a sphere of radius r (see Fig.). Find

1. surface area of the sphere.
2. curved surface area of the cylinder.
3. ratio of the areas obtained in (i) and (ii).

Solution:
As the sphere just encloses a right circular cylinder, the height of the cylinder is equal to the diameter of the sphere.
∴ Height of cylinder h = 2r
where r is the radius of sphere

1. Surface area of sphere S₁ = 4πr²

2. CSA of a cylinder,= 2πrh
= 2πr x 2r = 4πr²

3.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Assume π = \(\frac{22}{7}\), unless stated otherwise.

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:
Here, diameter of the base = 10.5 cm
Radius (r) = \(\frac{10.5}{2}\)
Slant height (l) = 10 cm
Curved surface area of the cone = πrl
= \(\frac{22}{7}\) x \(\frac{10.5}{2}\) x 10 cm²
= \(\frac{22}{7}\) x \(\frac{10.5}{20}\) x 10 cm²
= 11 x 15 x 1 cm² = 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:
Here, diameter = 24 m
Radius (r) = \(\frac{24}{2}\) m = 12m
Slant height (l) = 21 m
Total surface area = πr (r + l)

Question 3.
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find:

1. radius of the base and
2. total surface area of the cone.

Solution:
Here, curved surface area = 308 cm²
Slant height (l) = 14 cm

1. Let the radius of the base be ‘r’ cm.
πrl = 308
\(\frac{22}{7}\) x r x 14 = 308
r = \(\frac{308×7}{22×14}\) = 7 cm
Thus, the required radius of the cone is 7 cm.

2. base area = πr² = \(\frac{22}{7}\) x 7² cm.
and curved surface area = 308 cm² given
∴ Total surface area = [Curved surface area] + [Base area]
= 308 cm² + 154 cm² = 462 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:
h = 10 m
r =24m

Cost of canvas = CSA x Rate of canvas
= 1961.14 x 70
= ₹ 137279.8
= ₹ 137280.

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? 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:

h = 8m
r = 6m
Width of tarpaulin = 3 m
Extra length required = 20 cm = 0.2 m

CSA of conical tent = πrl
= 3.14 x 6 x 10 = 188.4 m²
CSA of conical tent = Area of tarpaulin
188.4 = l x 3 (Here l = length of tarpaulin)
l = \(\frac{188.4}{3}\)
Total length of tarpaulin required = 62.8 + 0.2 = 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:
l = 25m
d = 14m
r = 7m
Rate of white washing = ₹ 210 per 100 m²
= \(\frac{210}{100}\) = ₹ 2.10 per m²
CSA of conical tomb = πrl
= \(\frac{22}{7}\) x 7 x 25 = 550 m²
Cost of white washing = Rate x CSA
= 2.10 x 550 (Rate per m² = \(\frac{220}{100}\))
= ₹ 1155

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:
r = 7 cm
h = 24 cm

Area of sheet required = CSA of 10 such 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 cost of painting all these cones? (Use π = 3.14 and take \(\sqrt{1.04}\) =1.02).
Solution:
d = 40 cm
r = 20 cm = 0.20 m
h = l m
Rate = ₹ 12 per m²

= 1.02
CSA of cone = 3.14 x 0.2 x 1.02 = 0.64m²
Total area of 50 hollow cones= 50 x 0.64 = 32 m²
Cost of painting = 32 x 12 = ₹ 384
(Rate of painting = ₹ 12/m²)

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2

Assume π = \(\frac{22}{7}\) unless stated otherwise.

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:
h = 14 cm
CSA of cylinder = 88 cm²
CSA of cylinder = 2 πrh
88 = 2x \(\frac{22}{7}\) xr x 14
\(\frac{88}{4}\) = 2r
r = \(\frac{2}{2}\) = 1
Diameter = 2 r
= 1 x 2 = 2 cm.

Question 2.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140cm from a metal sheet. How many square metres of the sheet are required for the same?
h = 1 m
d = 140 cm
r = 7o cm = 0.7 m
Area of metal required = TSA of cylinderical tank
= 2 πr (r + h)
= 2 x \(\frac{22}{7}\) x 0.7 (0.7 + 1)
= 4.4 x 1.7 = 7.4m²

Question 3.
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm. the outer diameter being 4.4 cm. (see Fig.). Find its

(i) inner curved surface area.
(ii) outer curved surface area.
(iii) total surface area.

(iii) TSA = ICSA + OCSA + 2 x area of ring
= 968 + 1064.8 + 2 x 2.64
= 2032.8 + 5.28
= 20.38.08 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:
d = 84cm
∴ r = 42 cm
h = 120 cm
No. of revolution = 500
CSA of the roller = 2πrh
= 2 x \(\frac{22}{7}\) x 42 x 120
= 44 x 720
= 31680 cm²
Area of the playground = 31680 x 500
= 15840000 cm²
Area in m² \(\frac{ 15840000}{100 x 100}\)
= 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.S0 per m².
Solution:
d = 50cm
r = 25 cm²
h = 3.5 m = 350 cm
CSA of pillar = 2πrh
= 2x \(\frac{22}{7}\) x 25 x 350
= 44 x 1250
= 55000 cm²
CSA in m² = \(\frac{55000}{100×100}\)
Cost of painting the cylindrical pillar = ₹ 12.50 x 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.
Solution:
CSA = 4.4m².
r = 0.7 m
CSA of cylinder =2πrh
4.4 = 2x \(\frac{22}{7}\) x 0.7 x h
4.4 = 44 x 0.1 x h
4.4 = 4.4 x h
h = \(\frac{44}{4.4}\) = 1m

Question 7.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find

1. its inner curved surface area
2. the cost of plastering this curved surface at the rate of ₹ 40 per m².

Solution:
d = 3.5m
r = 1.75 m
h = 10 m

1. ICSA of the well = 2πrh
= 2 x \(\frac{22}{7}\) x 1.75 x 10
= 110 m²

2. Cost of plastering = ₹ 40 x 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:
h = 28m
d = 5cm
r = 2.5cm = 0.025m
Total radiating surface = CSA of pipe = 2πrh
= 2 x \(\frac{22}{7}\) x 0.025 x 28
= 4.4 m².

9. Find

1. 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.

2. How much steel was actually used, if \(\frac{1}{2}\) of the steel actually used was wasted in making the tank.

Solution:
1. d = 4.2 m
r = 2.1m
h = 4.5 m
CSA of cylinderical tank = 2πrh
= 2x \(\frac{22}{7}\) x 2.1 x 4.5
= 59.4 m².
TSA = 2πrr(r + h)
= 2 x \(\frac{22}{7}\) x 2.1 (2.1 + 4.5)
= 2 x \(\frac{22}{7}\) x 2.1 x 6.6
= 44 x 1.98
= 87.12 m²

2. Let A be the area of sheet actually used
TSA = A – \(\frac{1}{12}\)

A = 95.04m².

Question 10.
In Fig. 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. Amargin 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:

d = 20 cm
r = 10 cm
h = 30 + 2.5 + 2.5 cm = 35 cm
Area of cloth required = CSA of cyliner of height 35 cm
= 2πrh
= 2 x \(\frac{22}{7}\) x 10 x 35
= 44 x 50 = 2200 cm².

Question 11.
The students of a Vidyalaya were asked to participate in a competi-tion 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:
r = 3 cm
h = 10.5 cm
Cardboard required for one penholder = CSA of penholder + Area of base
= 2πrh + πr²

MP Board Class 9th Maths Solutions