RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2

RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2

Other Exercises

Question 1.
In which of the following tables x and y vary inversely :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 1
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 2
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 3
Solution:
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 4
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 5
We see that it in 15 x 4 and 3 x 25 are not equal to 36 others are 72
In it x and y do not vary.

Question 2.
It x and y vary inversely, fill in the following blanks :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 6
Solution:
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 7
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 8
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 9

Question 3.
Which of the following quantities vary inversely as each other ?
(i) The number of x men hired to construct a wall and the time y taken to finish the job.
(ii) The length x of a journey by bus and price y of the ticket.
(iii) Journey (x km) undertaken by a car and the petrol (y litres) consumed by it.
Solution:
(i) Here x and’y var inversely
More men less time and more time less men.
(ii) More journey more price, less journey less price
x and y do not vary inversely.
(iii) More journey more petrol, less journey, less petrol
x and y do not vary inversely.
In (i) x and y, vary inversely.

Question 4.
It is known that for a given mass of gas, the volume v varies inversely as the pressure p. Fill in the missing entries in the following table :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 10
Solution:
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 11
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 12
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 13

Question 5.
If 36 men can do a piece of work in 25 days, in how many days will 15 men do it ?
Solution:
Here less men, more days.
Let in x days, 15 men can finish the work
Therefore.
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 14

Question 6.
A work force of 50 men with a contractor can finish a piece of work in 5 months. In how many months the same work can be completed by 125 men.
Solution:
Let in x months, the work will be completed by 125 men
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 15

Question 7.
A work-force of 420 men with contractor can finish a certain piece of work in 9 months. How many extra men must he employ to complete the job in 7 months?
Solution:
Let total x men can finish the work in 7 months.
Therefore,
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 16
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 17
Total men = 540
Number of men already employed = 420
Extra men required = 540 – 420 = 120

Question 8.
1200 men can finish a stock of food in 35 days. How many more men should join them so that the same stock may last for 25 days ?
Solution:
Let x men can finish the stock, then
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 18
Total men required = 1680
Already men working = 1200
More men required = 1680 – 1200 = 480

Question 9.
In a hostel of 50 girls, there are food provisions for 40 days. If 30 more girls join the hostel. How long will these provisions last ?
Solution:
Number of girls in the beginning = 50
More girls joined = 30
Total number of girls = 50 + 30 = 80
Let the provisions last for x days.
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 19
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 20

Question 10.
A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance ?
Solution:
Let x km/hr be the speed. Then
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 21
Speed required = 60 km/hr.
Already speed = 48 km/hr
Speed to be increase = 60 – 48 = 12 km/hr

Question 11.
1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 more days. How many soldiers left the fort ?
Solution:
Period = 28 days
After 4 day, the remaining period = 28 – 4 = 24 days
In the beginning number of soldiers in the fort = 1200
Period for which the food lasted = 32 days
Let for x soldier, the food was sufficient, then
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 22
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 23

Question 12.
Three spraying machines working together can finish painting a house in 60 minutes. How long will it take 5 machines of the same capacity to do the same job ?
Solution:
Let in x minutes, 5 machines can do the work
Now
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 24

Question 13.
A group of 3 friends staying together, consume 54 kg of wheat every month. Some more friends join this group and they find that the same amount of wheat lasts for 18 days. How new many numbers are there in this group now ?
Solution:
Let x members can finish the wheat in 18 day.
Therefore :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 25
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 26
5 member can consume the wheat
Number of members already = 3
5 – 3 = 2 more member joined them.

Question 14.
55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days ?
Solution:
Let number of cows required = x
Therefore :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 27

Question 15.
18 men can reap a field in 35 days. For reaping the same field in 15 days, how many men are required ?
Solution:
Let x men are required,
Therefore,
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 28

Question 16.
A person has money to buy 25 cycles worth Rs. 500 each. How many cycles he will be able to buy if each cycle is costing Rs. 125 more ?
Solution:
Price of one cycle = Rs. 500
Number of cycle purchased = 25
New price of the cycle = Rs. 500 + Rs. 125 = Rs. 625
Let number of cycle will be purchase = x
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 29

Question 17.
Raghu has enough money to buy 75 machines worth Rs. 200 each. How many machines can he buy if he gets a discount of Rs. 50 on each machine ?
Solution:
Price of each machine = Rs. 200
Price after given discount of Rs. 50 = Rs. 200 – 50 = Rs. 150
Let machine can be purchase = x
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 30
Number of machines can be purchased = 100

Question 18.
If x and y vary inversely as each other and
(i) x = 3 when y = 8, find y when x = 4
(ii) x = 5 when y = 15, find x when y = 12
(iii) x = 30, find y when constant of variation = 900.
(iv) y = 35, find x when constant of variation = 7.
Solution:
x and y vary inversely
x x y is constant of variation
(i) x = 3, y = 8
Constant = xy = 3 x 8 = 24
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 31
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.2 32

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RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1

RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1

Other Exercises

Question 1.
Explain the concept of direct variation.
Solution:
If two quantifies a and b vary with each other in such a way that the ratio \(\frac { a }{ b }\) remains constant and is positive, then we say that a and b vary directly with each other or a and b are in direct variation.

Question 2.
Which of the following quantities vary directly with each other ?
(i) Number of articles (x) and their price (y).
(ii) Weight of articles (x) and their cost (y).
(iii) Distance x and time y, speed remaining the same.
(iv) Wages (y) and number of hours (x) of work.
(v) Speed (x) and time (y) (distance covered remaining the same).
(vi) Area of a land (x) and its cost (y).
Solution:
(i) It is direct variation because more articles more price and less articles, less price.
(ii) It is direct variation because, more weight more price, less weight, less price.
(iii) It is not direct variation. The distance and time vqry indirectly or inversely.
(iv) It is direct variation as more hours, more wages, less hours, less wages.
(v) It is not direct variation, as more speed, less time, less speed, more time.
(vi) It is direct variation, as more area more cost, less area, less cost.
Hence (i), (ii), (iv) and (vi) are in direct variation.

Question 3.
In which of the following tables x and y vary directly ?
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 1
Solution:
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 2
All are different.
It is not in direct variation.
Hence (i) and (ii) are in direct variation.

Question 4.
Fill in the blanks in each of the following so as to make the statement true :
(i) Two quantities are said to vary ……….. with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.
(ii) x and y are said to vary directly with each other if for some positive number k = k.
(iii) If u = 3v, then u and v vary ……….. with each other.
Solution:
(i) Two quantities are said to vary directly with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.
(ii) x and y are said to vary directly with each other if for some positive number k, \(\frac { x }{ y }\) = k.
(iii) If u = 3v, then u and v vary directly with each other.

Question 5.
Complete the following tables given that x varies directly as y.
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 3
Solution:
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 4
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 5
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 6
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 7
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 8

Question 6.
Find the constant of variation from the table given below :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 9
Solution:
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 10

Set up a table and solve the following problems. Use unitary method to verify the answer.
Question 7.
Rohit bought 12 registers for Rs. 156, find the cost of 7 such registers.
Solution:
Price of 12 registers = Rs. 156
Let cost of 7 registers = Rs. x. Therefore
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 11

Question 8.
Anupama takes 125 minutes in walking a distance of 100 metre. What distance would she cover in 315 minutes.
Solution:
For walking 100 m, time is taken = 125 minutes
Let in 315 minutes, distance covered = m
Therefore,
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 12
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 13

Question 9.
If the cost of 93 m of a certain kind of plastic sheet is Rs. 1395, then what would it cost to buy 105 m of such plastic sheet.
Solution:
Cost of 93 m of plastic sheet = Rs. 1395
Let cost of 105 m of such sheet = Rs. x
Therefore,
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 14

Question 10.
Suneeta types 1080 words in one hour. What is GWAM (gross words a minute rate) ?
Solution:
1080 words were typed in = 1 hour = 60 minutes
Let x words will be typed in 1 minute
Therefore,
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 15

Question 11.
A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 12 minutes.
Solution:
Speed of car = 50 km/hr = 50 km in 60 minutes
Let it travel x km in 12 minutes. Therefore
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 16

Question 12.
68 boxes of a certain commodity require a shelf length of 13.6 m. How many boxes of the same commodity would occupy a shelf of 20.4 m ?
Solution:
For 68 boxes of certain commodity is required a shelf length of 13.6 m
Let x boxes are require for 20.4 m shelf Then
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 17

Question 13.
In a library 136 copies of a certain book require a shelf length of 3.4 metre. How many copies of the same book would occupy a shelf-length of 5.1 metres ?
Solution:
For 136 copies of books require a shelf of length = 3.4 m
For 5.1 m shelf, let books be required = x Therefore :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 18
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 19

Question 14.
The second class railway fare for 240 km of journey is Rs. 15.00. What would be the fare for a journey of 139.2 km ?
Solution:
Fare of second class for 240 km = Rs. 15.00
Let fare for 139.2 km journey = Rs. x
Therefore :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 20

Question 15.
If the thickness of a pile of 12 cardboards is 35 mm, find the thickness of a pile of 294 cardboards.
Solution:
Thickness of a pile of 12 cardboards = 35 mm.
Let the thickness of a pile of 294 cardboards = x mm
Therefore :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 21

Question 16.
The cost of 97 metre of cloth is Rs. 242.50. What length of this can be purchased for Rs. 302.50 ?
Solution:
Cost of 97 m of cloth = Rs. 242.50
Let x m can be purchase for Rs. 302.50
Therefore :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 22

Question 17.
men can dig 6\(\frac { 3 }{ 4 }\) metre long trench in one day. How many men should be employed for digging 27 metre long trench of the same type in one day ?
Solution:
11 men can dig a trench = 6\(\frac { 3 }{ 4 }\) m long
Let x men will dig a trench 27 m long.
Therefore,
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 23

Question 18.
A worker is paid Rs. 210 for 6 days work. If his total income of the month is Rs. 875, for how many days did he work ?
Solution:
Payment for 6 day’s work = Rs. 210
Let payment for x day’s work = Rs. 875
Therefore :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 24

Question 19.
A worker is paid Rs. 200 for 8 days work. If he works for 20 days, how much will he get ?
Solution:
Labour for 8 days work = Rs. 200
Let x be the labour for 20 days work, then
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 25

Question 20.
The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm ?
Solution:
150 gm of weight produces an extension = 2.9 cm
Let x gm of weight will produce an extension of 17.4 cm
Therefore :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 26
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 27

Question 21.
The amount of extension in an elastic spring varies directly with the weight hung on it. If a weight of 250 gm produces an extension of 3.5 cm, find the extension produced by the weight of 700 gm.
Solution:
A weight of 250 gm produces an extension of 3.5 cm.
Let a weight of 700 gm will produce an extension of x cm. Therefore :
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 28

Question 22.
In 10 days, the earth picks up 2.6 x 108 pounds of dust from the atmosphere. How much dust will it pick up in 45 days.
Solution:
In 10 days dust is picked up = 2.6 x 108 pounds
Let x pounds of dust is picked up in = 45 days
Therefore,
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 29

Question 23.
In 15 days, the earth picks up 1.2 x 108 kg of dust from the atmosphere. In how many days it will pick up 4.8 x 10s kg of dust ?
Solution:
Dust of 1.2 x 108 kg is picked up in = 15 days
Let the dust of 4.8 x 108 will be picked up in x days
Therefore,
RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 30

Hope given RD Sharma Class 8 Solutions Chapter 10 Direct and Inverse variations Ex 10.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4

RD Sharma Class 8 Solutions Chapter 21 Mensuration II (Volumes and Surface Areas of a Cubiod and a Cube) Ex 21.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4

Other Exercises

Question 1.
Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.
Solution:
Length of room (l) = 12m
Breadth (b) = 9 m
Height (h) = 8 m
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 1
Longest rod to be kept in the room
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 2

Question 2.
If V is the volume of the cuboid of dimensions a, b, c and S its the surface area then prove that
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 3
Solution:
∵ a, b, c are the dimensions of a cuboid
∴ Volume (V) = abc
Surface area (S) = 2(ab + bc + ca)
Now
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 4

Question 3.
The areas of three adjacent faces of a cuboid are .v, y and z. If the volume is V1 prove that V2 = xyz.
Solution:
Let length of cuboid = l
Breadth = b
and height = h
Volume = Ibh
∴ x = lb,y = bh and z = hl
Now x.y.z = lb.bh.hl
= l2 b2 h2 = (Ibh)2 = V2
∴ V2 = xyz Hence proved

Question 4.
A rectangular water reservoir contains 105 m3 of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m.
Solution:
Volume of the water in reservoir = 105 m2
Length (l)= 12 m
and breadth (b) = 3.5 m
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 5

Question 5.
Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and moulded into a new cube D. Find the edge of the bigger cube D.
Solution:
Edge of cube A = 18 cm
∴ Volume = a2 = (18)3 cm3 = 5832 cm3
Edge of cube B = 24 cm
∴ Volume = (24)3 = 13824 cm3
Edge of cube C = 30 cm
∴Volume = (30)3 = 27000 cm3
Volume of A, B, C cubes
= 5832+ 138-24+ 27000 = 46656 cm3
Volume of cube D = 46656 cm3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 6

Question 6.
The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu.dm. Find its dimensions.
Solution:
Volume of room = 512 cu.dm
Let height of the room (h) = x
Then breadth (b) = 2x
and length (l) = 2x x 2 = 4x.
∴ Volume = l x b x h = 4x x 2x x x = 8×3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 7

Question 7.
A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs 5 per metre sheet, sheet being 2 m wide.
Solution:
Length of iron tank (l) = 12 m
Breadth (b) = 9 m
Depth (h) = 4 m
∴ Surface area of the tank = 2(l x b + b x h + h x l)
= 2(12 x 9 + 9 x 4 + 4 x 12) m2
= 2(108 + 36 + 48) = 2 x 192 m2
= 384 m2
Width of sheet used = 2 m
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 8

Question 8.
A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12 m x 8 m x 6 m, find the cost of iron sheet at Rs 17.50 per metre.
Solution:
Dimensions of the open iron tank = 12mx 8m.x 6m
∴ Surface area (without top)
= 2(1 x b) x h + lb
= 2(12 + 8) x 6+12 x 8m2
= 2 x 20 x 6 + 96 = 240 + 96 m2 = 336 m2
Width of sheet used = 4 m
∴ Length of sheet = \(\frac { Area }{ b }\) = \(\frac { 336 }{ 4 }\) m = 84 m b 4
Rate of sheet = Rs 17.50 per m.
∴ Total cost = Rs 17.50 x 84 = Rs 1470

Question 9.
Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Let edge of each equal cubes = x
Then, surface area of one cube = 6x2
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 9
and surface area of three cubes = 3 x 6x2 = 18x2
By placing the cubes in a row,
The length of newly formed cuboid (l) = 3x
Breadth (b) = x
and height (h) = x
∴ Surface area of the cuboid so formed
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 10

Question 10.
The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs 3.50 per square metre.
Solution:
Dimensions of a room = 12.5 m x 9 m x 7 m
∴ Total surface area of the walls = 2(1 + b) x h = 2(12.5 + 9) x 7 m2
= 2 x 21.5 x 7 = 301 0 m2
Area of 2 doors = 2 x (2.5 x 1.2) m2 = 2 x 3.00 = 6 m2
Area of 4 windows = 4 x (1.5 x 1) m2
4 x 1.5 = 6 m2
∴ Remaining area of the walls = 301 -(6 + 6) m2
= 301 – 12 = 289 m2
∴ Rate of painting the walls = Rs 3.50 per m2
∴ Total cost = Rs 3.50 x 289 = Rs 1011.50

Question 11.
A field is 150 m long and 100 m wide. A plot (outside the field) 50 m long and 30 m wide is dug to a depth of 8 m and the earth taken out from the plot is spread evenly in the field. By how much the level of field is raised ?
Solution:
Length of the plot (l) = 50 m
Width (b) = 30 m
and depth (h) = 8 m
∴ Volume of the earth dug out = l x b x h = 50 x 30 x 8 = 12000 m3
Length of the field = 150 m
and breadth = 100 m
∴ Height of the earth spread out on the field
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 11

Question 12.
Two cubes, each of volume 512 cm3 are joined end to end, find the surface area of the resulting cuboid.
Solution:
Volume of each cube = 512 cm3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 12
Now by joining the two equal cubes of side 8 cm, the length of so formed cuboid (l)
= 2 x 8 = 16 cm
Breadth (b) = 8 cm
and height (h) = 8 cm
∴ Surface area = 2( l x b + b x h + h x l)
= 2(16 X 8 + 8 X 8 + 8X16) cm2
= 2(128 + 64 + 128) cm2
= 2 x 320 = 640 cm2
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 13

Question 13.
Three cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.
Solution:
Edge of first cube = 3 cm
∴ Volume = a3 = (3)3 27 cm3
Edge of second cube = 4 cm
∴Volume = a3 = (4)3 = 64 cm3
Edge of third cube = 5 cm
∴ Volume = a3 = (5)3 = 125 cm3
Volume of three cubes together = 27 + 64+ 125 = 216 cm3
∴ Volume of the new cube = 216 cm3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 14

Question 14.
The cost of preparing the’walls of a room 12 m long at the rate of Rs 1.35 per square metre is Rs 340.20 and the cost of matting the floor at 85 paise per square metre is Rs 91.80. Find the height of the room.
Solution:
Length of the room (l) = 12 m
Rate of matting the floor = 85 paise per m2
Total cost of matting = Rs 91.80
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 15

Question 15.
The length of a hall is 18 m and width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the wall.
Solution:
Length of hall (l) = 18 m
and breadth (b) = 12 m
∴ Area of floor = l x b = 18 x 12 = 216 m2
and area of roof = 216 m2
Total area of floor and roof
= (216 + 216) m2 = 432 m2
∴ Area of four walls = 432 m2
But area of 4 walls = 2(l + b) x h
∴ 2h (l + b) = 432
⇒ 2h (18 + 12) = 432
⇒ 2h x 30 = 432 432
⇒ h = \(\frac { 432 }{ 60 }\) = 7.2m
∴ Height of the wall = 7.2 m

Question 16.
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.
Solution:
Edge of metal bigger cube = 12 cm
∴ Volume = (12)3 = 1728 cm3
∴ Sum of volumes of 3 smaller cubes = 1728 cm3
Edge of first smaller cube = 6 cm
∴ Volume = (6)3 = 216 cm3
Edge of second smaller cube = 8 cm
∴ Volume = (8)3 = 512 cm3
Sum of volumes of two smaller cubes = 216+ 512 = 728 cm3
∴ Volume of third smaller cube = 1728-728 cm3 = 1000 cm3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 16

Question 17.
The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall if each person requires 150 m3 of air ?
Solution:
Length of cinema hall (l) = 100 m
Breadth (b) = 50 m
and height (h) = 18 m
∴ Volume of air of the hall = l x b x h
= 100 x 50 x 18 m3
= 90000 m3
Each person requires air = 150 m3
∴ Number of persons = \(\frac { 90000 }{ 150 }\)= 600

Question 18.
The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm x 3 cm x 0.75 cm can be put in this box ?
Solution:
Outer dimensions of a closed wooden box = 48 cm x 36 cm x 30 cm
Thickness of wood = 1.5 cm.
∴ Inner length (l) = 48 – 2 x 1.5 cm = 48 – 3 = 45 cm
Breadth(b) = 36-2 x 1.5 = 36-3 = 33 cm
Height (h) = 30 – 2 x 1.5 = 30 – 3 = 27 cm
∴ Volume of inner box = l x b x h = 45 x 33 x 27 cm3 = 40095 cm3
Volume of one brick of size 6 cm x 3 cm x 0.75 cm
= 6 x 3 x 0.75 = 6 x 3 x \(\frac { 3 }{ 4 }\) cm3 = \(\frac { 27 }{ 2 }\) cm3
∴ Number of bricks = \(\frac { 40095 x 2 }{ 27 }\)
= 1485 x 2 = 2970 bricks

Question 19.
The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m2 is Rs 1,248. Find the dimensions of the box.
Solution:
Ratio in the dimensions of a box =2:3:4
Difference in total cost = Rs 1,248
Difference in rates = Rs 9.50 – Rs 8 = Rs 1.50
Let length (l) = 2x
Then breadth (b) = 3x
and height (h) = 4x
∴ Surface area = 2 (l x b + b x h + h x l)
= 2(2x 3x  + 3x x 4x + 4x x 2x)
= 2(6x2 + 12x2 + 8 x2) = 2 x 26x2 = 52x2
First rate of paper = Rs 9.50 per m2
and second rate = 8.00 per m2
∴ First cost = Rs 52x2 x 9.50
and second cost = Rs 52x2 x 8
∴ 52x2 x 9.50 – 52x2 x 8= 1248
⇒ 52x2 (9.50 – 8) = 1248
⇒ 52x2(1.50) = 1248
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 17

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RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3

RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3

Other Exercises

Question 1.
Find the surface area of a cuboid whose :
(i) length = 10 cm, breadth = 12 cm and height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2 m, breadth = 4 m and height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.
Solution:
(i) Length of cuboid (l) = 10 cm
Breadth (b) = 12 cm
Height (h) = 14 cm
∴ Surface area = 2(1 × b + b × h + h × l)
= 2(10 x 12 + 12 x 14 + 14 x 10) cm2
= 2(120+ 168 + 140) cm2
= 2 x 428 = 856 cm2
(ii) Length of cuboid (l) = 6 dm
Breadth (b) = 8 dm
Height (h) = 10 dm
∴ Surface area = 2 ( l × b + b x h + h× l)
= 2(6 x 8 + 8 x 10 + 10 x 6) dm2
= 2(48 + 80 + 60) dm2 = 2 x 188 = 376 dm2
(iii) Length of cuboid (l) = 2 m
Breadth (b) = 4 m
Height (h) = 5 m
∴ Surface area = 2(l × b + b × h + h × l)
= 2(2 x 4 + 4 x 5 + 5 x 2) m2
= 2(8 + 20 + 10) m2 = 76 m2
(iv) Length of cuboid (l) = 3.2 m = 32 dm
Breadth (b) = 30 dm
Height (h) = 250 cm = 25 dm
∴ Surface area = 2(1 x b + b x h + h x l)
= 2(32 x 30 + 30 x 25 + 25 x 32) dm2
= 2(960 + 750 + 800) dm2
= 2 x 2510 = 5020 dm2

Question 2.
Find the surface area of a cube whose edge is
(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1m
Solution:
(i) Edge of the cube (a) = 1.2 m
∴ Surface area = 6a2= 6 x (1,2)2 m2
= 6 x 1.44 = 8.64 m2
(ii) Edge of cube (a) = 27 cm
∴ Surface area = 6a2 = 6 x (27)2 m2
= 6 x 729 = 4374 m2
(iii) Edge of cube (a) = 3 cm
Surface area = 6a2 = 6 x (3)2 m2
= 6×9 cm2 = 54 cm2
(iv) Edge of cube (a) = 6 m
∴ Surface area = 6a2 = 6 x (6)2 m2
= 6 x 6 x 6 = 216 m2
(v) Edge of the cube (a) = 2.1 m
∴ Surface area = 6a2 = 6 x (2.1)2 m2
= 6 x 4.41 = 26.46 m2

Question 3.
A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area.
Solution:
Length of cuboid box (l) = 5 cm
Breadth (b) = 5 cm
and height (h) = 4 cm
∴ Surface area = 2 (l x b + b x h + h x l)
= 2 (5 x 5 + 5 x 4 + 4 x 5) cm2
= 2 (25 + 20 + 20)
= 2 x 65 cm2
= 130 cm2

Question 4.
Find the surface area of a cube whose volume is :
(i) 343 m3
(ii) 216 dm3.
Solution:
(i) Volume of a cube = 343 m3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3 1

Question 5.
Find the volume of a cube whose surface area is
(i) 96 cm2
(ii) 150 m2.
Solution:
(i) Surface area of a cube = 96 cm2
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3 2

Question 6.
The dimensions of a cuboid are in the ratio 5:3:1 and its total surface area is 414 m2. Find the dimensions.
Solution:
Ratio in .dimensions = 5 : 3 : 1
Let length (l) = 5x
breadth (b) = 3x
and height (h) = x
∴ Surface area = 2(1 x b + b x h + h x l)
= 2(5x x 3x + 3x x x + x x 5x)
= 2(15×2 + 3×2 + 5×2) = 2 x 23×2 = 46×2
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3 3

Question 7.
Find the area of the cardboard required to make a closed box of length 25 cm, 0.5 m and height 15 cm.
Solution:
Length of cardboard (l) = 25 cm
Breadth (b) = 0.5 m = 50 cm
Height (h)= 15 cm.
∴ Surface area of cardboard = 2 (l x b + b x h + h x l)
= 2(25 x 50 + 50 x 15 + 15 x 25) cm2
= 2(1250+ 750+ 375) cm2
= 2(2375)
= 4750 cm2

Question 8.
Find the surface area of a wooden box whose shape is of a cube and if the edge of the box is 12 cm.
Solution:
Edge of cubic wooden box = 12 cm
∴ Surface area = 6a2 = 6(12)2 cm2
= 6 x 144 = 864 cm2

Question 9.
The dimensions of an oil tin are 26 cm x 26 cm x 45 cm. Find the area of the tin sheet required for making 20 such tins. If 1 square metre of the tin sheet costs Rs 10, find the cost of the tin sheet used for these 20 tins.
Solution:
Length of tin (l) = 26 cm = 0.26 m
Breadth (b) = 26 cm = 0.26 m
Height (h) = 45 cm = 0.45 m
∴ Surface area = 2(l x b + b x h +h xl)
= 2(0.26 x 0.26 + 0.26 x 0.45 + 0.45 x 0.26) m2
= 2(0.0676 + 0.117 + 0.117) m2
= 2(0.3016) = 0.6032 m2
Sheet required for such 20 tins
= 0.6032 x 20= 12.064 m2
Cost of 1 m2 tin sheet = 10 m
∴ Total cost = Rs 12.064 x 10 = Rs 120.64
and area of sheet = 12.064 m2 = 120640 cm2

Question 10.
A classroom is 11 m long, 8 m wide and 5 m high. Find the sum of the areas of its floor and the four walls (including doors, windows etc.)
Solution:
Length of room (l) = 11 m
Width (b) = 8 m
and height (h) = 5 m
Area of floor = l x b = 11 x8 = 88m2
Area of four walls = 2 (l + b) x h
= 2(11 + 8) x 5 m2 = 2 x 19×5 = 190 m2
∴ Total area = 88 m2 + 190 m2 = 278 m2

Question 11.
A swimming pool is 20 m long, 15 m wide and 3 m deep. Find the cost of repairing the floor and wall at the rate of Rs 25 per square metre.
Solution:
Length of pool (l) = 20 m
Breadth (b) = 15 m
and Depth (h) = 3 m.
Area of floor = l x b = 20 x 15 = 300 m2
and area of its walls = 2(l + b) x h
= 2(20 + 15) x 3 = 2 x 35 x 3 m2 = 210 m2
∴ Total area = 300 + 210 = 510 m2
Rate of repairing it = Rs 25 per sq. metre
∴ Total cost = Rs 25 x 510 = Rs 12750

Question 12.
The perimeter of a floor of a room is 30 m and its height is 3 m. Find the area of four walls of the room.
Solution:
Perimeter of floor = 30 m
i.e. 2(1 + b) = 30 m
Height = 3 m
∴ Area of four walls = Perimeter x height = 30 x 3 = 90 m2

Question 13.
Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume.
Solution:
Let length of the room = l
and breadth = b
and height = h
Volume = l x b x h
Area of floor = l x b = lb.
Area of two adjacent walls = hl x bh.
∴ Product of areas of floor and two adjacent walls of the room = lb (hi x bh)
= l2b2h2 = (l.b.h)2 = (Volume)2
Hence proved

Question 14.
The walls and ceiling of a room are to be plastered. The length, breadth and height of the room are 4.5, 3m and 350 cm, respectively. Find the cost of plastering at the rate of Rs 8 per square metre.
Solution:
Length of room (l) = 4.5 m
Width (b) = 3 m
and height (h) = 350 cm = 3.5 m
∴ Area of walls = 2(l + b) x h
= 2(4.5 + 3) x 3.5 m2 = 2 x 7.5 x 3.5 m2 = 52.5 m2
Area of ceiling = l x b = 4.5 x 3 = 13.5 m2
∴ Total area = 52.5 + 13.5 m2 = 66 m2
Rate of plastering = Rs 8 per sq. m
∴ Total cost = Rs 8 x 66 = Rs 528

Question 15.
A cuboid has total surface area of 50 m2 and lateral surface area its 30 m2. Find the area of its base.
Solution:
Total surface area of cuboid = 50 m2
Lateral surface area = 30 m2
∴ Area of floor and ceiling = 50 – 30 = 20 m2
But area of floor = area of ceiling
∴ Area of base (floor) = \(\frac { 20 }{ 2 }\) = 10 m2

Question 16.
A classroom is 7 m long, 6 m broad and 3.5 m high. Doors and windows occupy an area of 17 m2. What is the cost of white-washing the walls at the rate of Rs 1.50 per m2.
Solution:
Length of room (l) = 7 m
Breadth (b) = 6 m
and height (h) = 3.5 m
∴ Area of four walls = 2(1 + b) x h
= 2(7 + 6) x 3.5 m2 = 2 x 13 x 3.5 = 91 m2
Area of doors and windows = 17 m2
∴ Remaining area of walls = 91 – 17 = 74 m2
Rate of whitewashing = Rs 1.50 per m2
∴ Total cost = 74 x Rs 1.50 = Rs 111

Question 17.
The central hall of a school is 80 m long and 8 m high. It has 10 doors each of size 3 m x 1.5 m and 10 windows each of size 1.5 m x l m. If the cost of the white-washing the walls of the hall at the rate of Rs 1.20 per m2 is Rs 2385.60, find the breadth of the hall.
Solution:
Length of hall (l) = 80 m
Height (h) = 8 m
Size of each door = 3 m x 1.5 m
∴ Area of 10 doors = 3 x 1,5 x 10 m2
= 45 m2
A size of each windows = 1.5 m x 1 m
∴ Area of 10 windows = 1.5 m x 1 x 10= 15 m2
Total cost of whitewashing the walls = Rs 2385.60
Rate of whitewashing = Rs 1.20 per m2
∴ Area of walls which are whitewashed
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3 4

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RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2

RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2

Other Exercises

Question 1.
Find the volume in cubic metres (cu.m) of each of the cuboids whose dimensions are :
(i) length = 12 cm, breadth = 10 m, height = 4.5 m
(ii) length = 4 m, breadth = 2.5 m, height = 50 cm
(iii) length = 10 m, breadth = 25 dm, height = 25 cm.
Solution:
(i) Length of cuboid (l) = 12 m
Breadth (b) = 10m
and height (h) = 4.5 m
∴Volume = l x b x h = 12 x 10 x 4.5 m3
= 540 m3
(ii) Length of cuboid (l) = 4 m
Breadth (b) = 2.5m
Height (h) = 50 cm = 0.5 m
∴ Volume = l x b x h = 4 x 2.5 x 0.5 = 5 m3
(iii) Length of cuboid (l) = 10 m
Breadth (b) = 25 dm = 2.5 m
Height (h) = 25 cm 0.25 m
∴ Volume = l x b x h = 10 x 2.5 x 0.25 m3 = 6.25 m3

Question 2.
Find the volume in cubic decimetre of each of the cubes whose side is
(i) 1.5 m
(it) 75 cm
(iii) 2 dm 5 cm
Solution:
(i) Side of cube (a) = 1.5 m
∴ Volume = a3 = (1.5)3 m3
= 1.5 x 1.5 x 1.5 m3 = 3.375 m3
= 3.375 x 1000 = 3375 dm3
(ii) Side of cube (a) = 75 cm = 7.5 dm
∴ Volume = a3 = (7.5)3 dm3
= 421.875 dm3
(iii) Side of cube (a) = 2 dm 5 cm = 2.5 dm
∴ Volume = (a)3 = (2.5)3 dm3
= 15.625 dm3

Question 3.
How much clay is dug out in digging a well measuring 3m by 2m by 5m?
Solution:
Length of well (l) = 3m
breadth (b) = 2 m
and height (depth) (h) = 5 m
Volume of earth dug out = l x b x h = 3 x 2 x 5 = 30m3

Question 4.
What will be the height of a cuboid of volume 168 m3, if the area of its base is 28 m2 ?
Solution:
Volume of a cuboid = 168 m3
Area of its base l.e., l x b = 28 m3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 1

Question 5.
A tank is 8 m long, 6 m broad and 2 m high. How much water can it contain ?
Solution:
Length of tank (l) = 8 m
Breadth (b) = 6 m
Height (h) = 2 m
∴ Volume of water in the tank = l x b x h = 8 x 6 x 2 = 96 m3
= 96 x 1000 = 96000litres (∵1m3 = 1000litre)

Question 6.
The capacity of a certain cuboidal tank is 50000 litres of water. Find the breadth of the tank if its height and length are 10 m and 2.5 m respectively.
Solution:
Capacity of water in the tank = 50000 litres
∴ Volume of water = 50000 x \(\frac { 1 }{ 1000 }\) = 50 m3 (1000 l = 1 m3)
Height of tank (h)= 10 m
and length (l) = 2.5 m
Volume 50
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 2

Question 7.
A rectangular diesel tanker is 2 m long, 2 m wide and 40 cm deep. How many litres of diesel can it hold ?
Solution:
Length of tanker (l) = 2 m
Breadth (b) = 2m
Depth (h) = 40 cm = 0.4 m
∴ Volume = l x bx h = 2 x 2 x 0.4=1.6m3
Quantity of diesel = 1.6 x 1000 litres (1 m3= 1000 l)
= 1600 litres

Question 8.
The length, breadth and height of a room are 5 m, 4.5 m and 3 m, respectively. Find the volume of the air it contains.
Solution:
Length of room (l) = 5 m
Breadth (6) = 4.5 m
and height (h) = 3 m
∴ Volume of air it contains
= l x b x h = 5 x 4.5 x 3 m3
= 67.5 m3

Question 9.
A water tank is 3 m long, 2 m broad and 1 m deep. How many litres of water can it hold ?
Solution:
Length of tank (l) = 3 m
Breadth (b) = 2 m
and depth (h) = 1 m
∴ Volume of tank = l x b x h
= 3 x 2 x 1 = 6 m3
∴ Quantity of water it can contains
= 6 x 1000 litres = 6000 litres (1 m3= 1000 litres)

Question 10.
How many planks each of which is 3 m long, 15 cm broad and 5 cm thick can be prepared from a wooden block 6 m long, 75 cm broad and 45 cm thick ?
Solution:
Length of wooden block (l) = 6 m
Width (b) = 75 cm = 0.75 m
Thickness (h) = 45 cm = 0.45 m
∴ Volume = l x b x h = 6 x 0.75 x 0.45 m3
Length of plank (l) = 3 m
Breadth (b) = 15 cm = 0.15 m
Thickness (h) = 5 cm = 0.05 m
∴ Volume = 3 x 0.15 x 0.05 m3
Number of planks
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 3

Question 11.
How many bricks each of size 25 cm x 10 cm x 8 cm will be required to build a wall 5 m long, 3 m high and 16 cm thick assuming that the volume of sand and cement used in the construction is negligible ?
Solution:
Size of one brick = 25 cm x 10 cm x 8 cm
∴ Volume of one brick = 25 x 10 x 8 cm3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 4
Length of wall (l) = 5 m
Width (b) = 0.16 m
Height (h) = 3 m
∴ Volume of wall = l x b x h
= 5 x 0.16 x 3 m3 = 2.4 m3
∴ Number of bricks required
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 5

Question 12.
A village, having a population of 4000 requires 150 litres water per head per day. It has a tank which is 20 m long, 15 m broad and 6 m high. For how many days the water of this tank will last ?
Solution:
Total population of a village = 4000
Water required for each person for one day = 150 litres
∴ Water required for 4000 persons for one day = 150 x 4000 = 600000 litres
Length of tank (l) = 20 m
Breadth (b) = 15 m
Height (h) = 6 m
∴ Volume of tank = l x b x h = 20 x 15 x 6 m3 = 1800 m3
Capacity of water in the tank = 1800 x 1000 l= 1800000l (1 m3 = 1000 l)
∴ Number of days, the water will last
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 6

Question 13.
A rectangular field is 70 m long and 60 m broad. A well of dimensions 14 m x 8 m x 6 m is dug outside the field and the earth dugout from this well is spread evenly on the field. How much will the earth level rise ?
Solution:
Length of well (l) = 14 m
Breadth (A) = 8m
Depth (A) = 6m
∴ Volume of earth dugout = l x bx h
= 14 x 8 x 6 = 672 m3 Length of field = 70 m
and breadth = 60 m
Let h be the height of earth spread over
Then 70 x 60 x h = 672
⇒ h = \(\frac { 672 }{ 70×60 }\) = 0.16m
∴ Height of earth = 0.16 m = 16 cm

Question 14.
A swimming pool is 250 m long and 130 m wide. 3250 cubic metres of water is pumped into it. Find the rise in the level of water.
Solution:
Volume of water = 3250 m3
Length of pool (l) = 250 m
Breadth (b)= 130 m
∴ Height of water level
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 7

Question 15.
A beam 5 m long and 40 cm wide contains 0.6 cubic metres of wood. How thick is the beam?
Solution:
Volume of wood of the beam = 0.6 m3 = 600000
Length of beam (l) = 5 m = 500 cm
Breadth (b) = 40 cm
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 8

Question 16.
The rainfall on a certain day was 6 cm. How many litres of water fell on 3 hectares of field on that day ?
Solution:
Area of the field = 3 hectares
= 3 x 10000 square metres
= 30000 square metres
Height of rainfall = 6 cm = m3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 9

Question 17.
An 8 m long cuboidal beam of wood when sliced produces four thousand 1 cm cubes and there is no wastage of wood in this process. If one edge of the beam is 0.5 m, find the third edge.
Solution:
Length of cuboidal beam (l) = 8 m = 800 cm
Number of cubical sliced = 4000
Edge of each cube = 1 cm
Volume of beam = 4000 (1)3 cm3 = 4000 cm3
One edge of the beam = 0.5 m = 50 cm.
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 10

Question 18.
The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm. It is melted and recast into cubes, each of the side 45 cm. How many cubes are formed ?
Solution:
Dimensions of metal block = 2.25 m x 1.5 m x 27 cm
∴ Volume = 2.25 x 1.5 x 0.27 m3
= 225 x 150 x 27 cm3 = 911250 cm3
Side of each cube (a) = 45 cm
∴ Volume of one cube = a3 = (45)3 cm3 = 91125 cm3
∴ Number of cubes = \(\frac { 911250 }{ 91125 }\) = 10

Question 19.
A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm. Find the weight of this piece if 1 cm3 of iron weighs 8 gm.
Solution:
Dimensions of a piece of rectangular iron = 6m x 6cm x 2cm
∴ Volume = 600 x 6 x 2 cm3 = 7200 cm3
Weight of 1 cm3 = 8 gm
∴ Total weight of the piece = 7200 x 8 gm
= 57600 gm = \(\frac { 57600 }{ 1000 }\) kg = 57.6 kg

Question 20.
Fill in the blanks in each of the following so as to make the statement true :
(i) 1 m3 = ……… cm3
(ii) 1 litre = …….. cubic decimetre
(iii) 1 kl = …… m3
(iv) The volume of a cube of side 8 cm is …….. .
(v) The volume of wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm3. The height of the cuboid is …….. cm
(vi) 1 cu.dm = …….. cu.mm
(vii) 1 cu.km = ……cu.m
(viii) 1 litre =…….. cu.cm
(ix) 1 ml = ……… cu.cm
(x) 1 kl = ……… cu.dm = ……. cu.cm.
Solution:
(i) 1 m3 = 1000000 or 10cm3
(ii) 1 litre = 1 cubic decimetre
(iii) 1 kl = 1 m3
(iv) The volume of a cube of side 8 cm is 512 cm3 (V = a3 = 8 x 8 x 8 = 512 cm3)
(v) The volume of a wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm3. The height of the cuboid is 50 cm
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 11
(vi) 1 cu.dm = 1000000 cu mm = 106 cu.mm
(vii) 1 cu.km = 1000 x 1000 x 1000 cu.m = 109 cu.m
(viii) 1 litre = 1000 cu.cm = 103 cu.cm
(ix) 1 ml = 1 cu.cm
(x) 1 kl = 1000 cu.dm = 100 x 100 x 100 cu.cm = 106 cu.cm

Hope given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1

RD Sharma Class 8 Solutions Chapter 21 Mensuration II (Volumes and Surface Areas of a Cubiod and a Cube) Ex 21.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1

Other Exercises

Question 1.
Find the volume of a cuboid whose
(i) length = 12 cm, breadth = 8 cm, height = 6 cm
(ii) length = 1.2 m, breadth = 30 cm, height = 15 cm
(iii) length = 15 cm, breadth = 2.5 dm, height = 8 cm
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 1
Solution:
In a cuboid,
(i) Length (l) = 12 cm
Breadth (b) = 8 cm
Height (h) = 6 cm
∴ Volume = Ibh = 12 x 8 x 6 cm3 = 576 cm3
(ii) Length (l) = 1.2 m = 120 cm
breadth (6) = 30 cm
Height (h) = 15 cm
∴ Volume = Ibh = 120 x 30 x 15 cm3 = 54000 cm3
(iii) Length (l) = 15 cm
Breadth (b) = 2.5 dm = 25 cm
Height (h) = 8 cm
∴ Volume = Ibh
= 15 x 25 x 8 cm3 = 3000 cm2

Question 2.
Find the volume of the cube whose side is
(i) 4 cm
(ii) 8 cm
(iii) 1.5 dm
(iv) 1.2 m
(v) 25 mm.
Solution:
(i) Side of a cube (a) = 4 cm
∴ Volume = a3 = (4)3 cm3 = 4 x 4 x 4 = 64 cm3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 2
(ii) Side of cube (a) = 8 cm
∴ Volume = a3 = (8)3 4 cm
= 8 x 8 x 8 cm3 = 512 cm3
(iii) Side of cube (a) = 1.5 dm = 15 cm
∴ Volume = a3 = (1.5)3 dm2 = (15)3 cm3
= 15 x 15 x 15 = 3375 cm3
(iv) Side of cube (a) = 1.2 m = 120 cm
∴ Volume = a3 = (120)3 cm3
= 120 x 120 x 120 = 1728000 cm3
(v) Side of cube (a) = 25 mm = 2.5 cm.
∴ Volume = a3 = (2.5)3 cm3
= 2.5 x 2.5 x 2.5 cm3 = 15.625 cm3

Question 3.
Find the height of a cuboid of volume 100 cm3 whose length and breadth are 5 cm and 4 cm respectively.
Solution:
Volume of a cuboid =100 cm3
Length (1) = 5 cm
and breadth (b) = 4 cm
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 3

Question 4.
A cuboidal vessel is 10 cm long and 5 cm wide, how high it must be made to hold 300 cm3 of a liquid ?
Solution:
Volume of the liquid in the vessel = 300 cm3
Length (l)= 10 cm
Breadth (b) = 5 cm
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 4

Question 5.
A milk container is 8 cm long and 50 cm wide. What should be its height so that it can hold 4 litres of milk ?
Solution:
Capacity of milk = 4 litres
∴ Volume of the container = 4 x 1000 cm3 = 4000 cm3
Length (l) = 8 cm
Width (b) = 50 cm
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 5

Question 6.
A cuboidal wooden block contains 36 cm3 wood. If it be 4 cm long and 3 cm wide, find its height.
Solution:
Volume of wooden cuboid block = 36 cm3
Length (l) = 4 cm
Breadth (b) = 3 cm
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 6

Question 7.
What will happen to the volume of a cube, if its edge is (i) halved (ii) trebled ?
Solution:
Let side of original cube = a cm
∴ Volume = a3 cm3
(i) In first case,
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 7
(ii) In second case, when side (edge) is trebled, then side = 3a
∴ Volume = (3a)3 = 27a3
∴ It will be 27 times

Question 8.
What will happen to the volume of a cuboid if its (i) Length is doubled, height is same and breadth is halved ? (ii) Length is doubled, height is doubled and breadth is same ?
Solution:
Let l, b and h be the length, breadth and height of the given cuboid respectively.
∴ Volume = lbh.
(i) Length is doubled = 21
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 8
∴ The volume will be the same.
(ii) Length is doubled = 21
breadth is same = b height is doubled = 2h
∴ Volume = 2l x b x 2h = 4 lbh
∴ Volume will be 4 times

Question 9.
Three cuboids of dimensions 5 cm x 6 cm x 7 cm, 4 cm x 7 cm * 8 cm and 2 cm x 3 cm x 13 cm are melted and a cube is made. Find the side of cube.
Solution:
Dimensions of first cuboid = 5 cm x 6 cm x 7 cm
∴ Volume = 5 x 6 x 7 = 210 cm3
Dimensions of second cuboid = 4 cm x 7 cm x 8 cm
∴ Volume = 4x 7 x 8 = 224 cm3
Dimensions of third cuboid = 2 cm x 3 cm x 13 cm
∴ Volume = 2 x 3 x 13 = 78 cm3
Total volume of three cubes = 210 + 224 + 78 cm3 = 512 cm3
∴ Volume of cube = 512 cm3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 9

Question 10.
Find the weight of solid rectangular iron piece of size 50 cm x 40 cm x 10 cm, if 1 cm3 of iron weighs 8 gm.
Solution:
Dimension of cuboidal iron piece = 50 cm x 40 cm x 10 cm
∴ Volume = 50 x 40 x 10 = 20000 cm3
Weight of 1 cm3 = 8 gm
∴ Total weight of piece = 20000 x 8 gm
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 10

Question 11.
How many wooden cubical blocks of side 25 cm can be cut from a log of wood of size 3 m by 75 cm by 50 cm, assuming that there is no wastage ?
Solution:
Length of log (l) = 3 m = 300 cm.
Breadth (b) = 75 cm
and height (h) = 50 cm
∴ Volume of log = lbh = 300 x 75 x 50 cm3 = 1125000 cm3
Side of cubical block = 25 cm
∴ Volume of one block = a2 = 25 x 25 x 25 cm3 = 15625 cm3
∴ Number of blocks to be cut out
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 11

Question 12.
A cuboidal block of silver is 9 cm long, 4 cm broad and 3.5 cm in height. From it, beads of volume 1.5 cm2 each are to be made. Find the number of beads that can be made from the block ?
Solution:
Length of block (l) = 9 cm
Breadth (b) = 4 cm
and height (h) = 3.5 cm
∴ Volume = l x b x h = 9 x 4 x 3.5 cm3 = 126 cm3
Volume of one bead = 1.5 cm3
∴ Number of beads = \(\frac { 126 }{ 105 }\) = 84

Question 13.
Find the number of cuboidal boxes measuring 2 cm by 3 cm by 10 cm which can be stored in a carton whose dimensions are 40 cm, 36 cm and 24 cm.
Solution:
Length of cuboidal box (l) = 2 cm
breadth (b) = 3 cm
and height (h) = 10 cm
∴ Volume = lx b x h = 2 x 3 x 10 = 60 cm3
Volume of carton = 40 x 36 x 24 cm3
= 34560 cm3
∴ Number of boxes to be height in the carton
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 12

Question 14.
A cuboidal block of solid iron has dimensions 50 cm, 45 cm and 34 cm. How many cuboids of size 5 cm by 3 cm by 2 cm can be obtained from the block ? Assume cutting causes no wastage.
Solution:
Dimensions of block = 50 cm, 45 cm, 34 cm
∴ Volume = 50 x 45 x 34 = 76500 cm3
Size of cuboid = 5 cm x 3 cm x 2 cm
∴ Volume of cuboid =  5 x 3 x 2 = 30 cm3
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 13

Question 15.
A cube A has side thrice as long as that of cube B ? What is the ratio of the volume of cube A to that of cube B ?
Solution:
Let side of cube B = a
Then Volume = a3
and side of cube A = 3a
Volume = (3a)3 = 3a x 3a x 2a = 27a3
∴ Ratio of volume’s A and B = 27a3 : a3
= 27 : 1

Question 16.
An ice-cream brick measures 20 cm by 10 cm by 7 cm. How many such bricks can be stored in a deep fridge whose inner dimensions are 100 cm by 50 cm by 42 cm ?
Solution:
Dimensions of ice cream brick = 20 cm x 10 cm x 7 cm
∴ Volume = 20 x 10 x 7 cm3 = 1400 cm3
Dimensions of inner of fridge = 100 cm x 50 cm x 42 cm = 210000 cm3
∴ Number of bricks to be kept in the fridge
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 14

Question 17.
Suppose that there are two cubes, having edges 2 cm and 4 cm, respectively. Find the volume V1 and V2 of the cubes and compare them.
Solution:
Side of first cube (a) = 2 cm
∴ Volume (V1) = a3 = (2) = 8 cm3
Similarly side of second cube = 4 cm
and volume (V2) = (4)3 = 64 cm3
Now V2 = 64 cm3 = 8 x 8 cm3
= 8 x V1
⇒ V2 = 8V1

Question 18.
A tea-packet measures 10 cm x 6 cm x 4 cm.How many such tea-packets can be placed in a cardboard box of dimensions 50 cm x 30 cm x 0.2 m ?
Solution:
Dimensions of tea-packet = 10cm x 6cm x 4 cm
∴ Volume =10 x 6 x 4 = 240 cm3
Dimensions of box = 50 cm x 30 cm x 0.2 m
= 50 cm x30 cm x20 cm
∴ Volume = 50 x 30 x 20 = 30000 cm3
∴ Number of tea-packets to be kept = \(\frac { 30000 }{ 240 }\)

Question 19.
The weight of a metal block of size 5 cm by 4 cm by 3 cm is 1 kg. Find the weight of a block of the same metal of size 15 cm by 8 cm by 3 cm.
Solution:
Dimensions of a metal block = 5 cm x 4 cm x 3 cm = 5 x 4 x 3 = 60 cm3
Dimensions of a second block = 15 cm x 8 cm x 3 cm = 15 x 8 x 3 = 360 cm3
But weight of first block = 1 kg
∴ Weight of second block
= \(\frac { 1 }{ 16 }\) x 360 = 6 kg

Question 20.
How many soap cakes can be placed in a box of size 56 cm x 0.4 m x 0.25 m, it the size of soap cake is 7 cm x 5 cm x 2.5 cm ?
Solution:
Size of box = 56 cm x 0.4 m x 0.25 m = 56 cm x 40 cm x 25 cm
∴ Volume = 56 x 40 x 25 cm3 = 56000 cm3
Size of a soap cake = 7 cm x 5 cm x 2.5 cm
∴ Volume = 7 x 5 x 2.5 cm3 = 87.5 cm3
∴ Number of cakes to be kept in the box
= \(\frac { 56000 }{ 87.5 }\) = 640

Question 21.
The volume of a cuboid box is 48 cm3. If its height and length are 3 cm and 4 cm respectively, find its breadth.
Solution:
Volume of cuboid box = 48 cm3
Length (l) = 4 cm
Height = (h) = 3 cm
RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 15

Hope given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2

RD Sharma Class 8 Solutions Chapter 22 Mensuration III (Surface Area and Volume of a Right Circular Cylinder) Ex 22.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2

Other Exercises

Question 1.
Find the volume of a cuboid whose
(i) r = 3.5 cm, h = 40 cm
(ii) r = 2.8 m, h = 15 m
Solution:
(i) Radius (r) = 3.5 cm
Height (h) = 40 cm
Volume of cylinder = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 1

Question 2.
Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h) are:
(i) d= 21 cm, h = 10 cm
(ii) d = 7 m, h = 24 m.
Solution:
(i) Diameter (d) = 21 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 2

Question 3.
The area of the base of a right circular cylinder is 616 cm3 and its height is 25 cm. Find the volume of the cylinder.
Solution:
Base area of cylinder = 616 cm2
Height (h) = 25 cm.
∴ Volume = Area of base x height
= 616 x 25 cm3 = 15400 cm3

Question 4.
The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find the volume of the cylinder.
Solution:
Circumference of the base of cylinder = 88 cm
Let r be the radius
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 3

Question 5.
A hollow cylindrical pipe is 21 dm long. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of the copper used in making the pipe.
Solution:
Length (Height) of hollow cylindrical pipe = 21 dm = 210 cm
Inner diameter = 6 cm
Outer diameter = 10 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 4
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 5

Question 6.
Find the (i) curved surface area (ii) total surface area and (iii) volume of a right circular cylinder whose height is 15 cm and the radius of the base is 7 cm
Solution:
Radius of the cylider (r) = 7 cm
and height (h) = 15 cm
(i) Curved surface area = 2πrh
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 6

Question 7.
The diameter of the base of a right circular cylinder is 42 cm and its height is 10 cm. Find the volume of the cylinder.
Solution:
Diameter of the base of cylinder = 42 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 7

Question 8.
Find the volume of a cylinder, the diameter of whose base is 7 cm and height being 60 cm. Also, find the capacity of the cylinder in litres.
Solution:
Diameter of cylinder = 7 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 8

Question 9.
A rectangular strip 25 cm x 7 cm is rotated about the longer side. Find the volume of the solid, thus generated.
Solution:
Size of rectangular strip = 25 cm x 7 cm
By rotating, about the longer side, we find a right circular cylinder,
then the radius of cylinder (r) = 7 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 9

Question 10.
A rectangular sheet of paper 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.
Solution:
Size of sheet of paper = 44 cm x 20 cm
By rolling along length, a cylinder is formed in which circumference of its base = 20 cm
and height (h) = 44 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 10

Question 11.
The volume and the curved surface area of a cylinder are 1650 cm3 and 660 cm2 respectively. Find the radius and height of the cylinder.
Solution:
Volume of cylinder = 1650 cm3
and curved surface area = 660 cm2
Let r be the radius and h be the height of the cylinder, then
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 11
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 12

Question 12.
The radii of two cylinders are in the ratio 2 :3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes.
Solution:
Ratio in radii = 2:3
and in heights = 5:3
Let r1,h1 and r2, h2 are the radii and heights of two cylinders respectively.
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 13

Question 13.
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder, if its total surface area is 616 cm2.
Solution:
Ratio in curved surface are and total surface area =1 : 2
Let r be the radius and h be the height.
Then 2πrh : 2πr (h + r)= 1 : 2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 14
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 15

Question 14.
The curved surface area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the volume of the cylinder.
Solution:
Curved surface area of a cylinder = 1320 cm2
Diameter of base = 21 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 16

Question 15.
The ratio between the radius of the base and the height of a cylinder is 2 : 3. Find the total surface area of the cylinder, if its volume is 1617 cm3.
Solution:
Ratio between radius and height of a cylinder = 2:3
Volume = 1617 cm3
Let r be the radius and A be the height of the cylinder, then
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 17

Question 16.
The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar.
Solution:
Let r be the radius and A be the height of the cylinder, then
2πrh = 264 …(i)
and πr2h= 924 …(ii)
Dividing (ii) by (i),
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 18
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 19

Question 17.
Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of their radii.
Solution:
Volumes of two cylinders are equal.
Let r1, r2 are the radii and h1, h2 are their heights, then
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 20
Now, volume of first cylinder = πr12h1
and Volume of second cylinder = πr22h2
Their volumes are equal
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 21

Question 18.
The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder.
Solution:
Height of cylinder (A) = 10.5 m.
Let r be the radius, then
Sum of areas of two circular faces = 2π2
and curved surface area = 2πrh
According to the condition,
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 22

Question 19.
How many cubic metres of earth must be dug-out to sink a well 21 m deep and 6 m diameter ?
Solution:
Diameter of well = 6 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 23

Question 20.
The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m, find the volume of the timber that can be obtained from the trunk.
Solution:
Circumference of cylindrical trunk = 176 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 24

Question 21.
A well is dug 20 m deep and it has a diameter of 7 m. The earth which is so dug out is spread out on a rectangular plot 22 m long and 14 m broad, what is the height of the platform so formed ?
Solution:
Diameter of the well = 7 m
∴ Radius (r) = \(\frac { 7 }{ 2 }\) m
Depth (h) = 20 m
= Volume of earth dug out = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 25

Question 22.
A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 14 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 26
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 27

Question 23.
A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.
Solution:
Diameter of the base of a cylindrical container = 56 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 28
Dimensions of the rectangular solid = 32 cm x 22 cm x 14 cm
∴ Volume of solid = 32 x 22 x 14 cm3 = 9856 cm3
∴Volume of water rose up = 9856 cm3
Let h be the height of water, then πr2h = 9856
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 29

Question 24.
A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways Le., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.
Solution:
Size of paper = 30 cm and 18 cm.
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 30
(i) By rolling length wise,
The circumference of base = 30 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 31
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 32
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 33

Question 25.
The rain which falls on a roof 18 m long and 16.5 m wide is allowed to be stored in a cylindrical tank 8 m in diameter. If it rains 10 cm on a day, what is the rise of water level in the tank due to it.
Solution:
Length of roof (l) = 18 m
Breadth (b) = 16.5 m
Height of water on the roof = 10 cm
∴ Volume of water collected
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 34

Question 26.
A Piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 5 cm. It is drawnout into a wire of diameter 1 mm. What will be the length of the wire so formed ?
Solution:
Diameter of ductile metal = 1 cm
and length (h) = 5 cm
and radius (r) = \(\frac { 1 }{ 2 }\) cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 35

Question 27.
Find the length of 13.2 kg of copper wire of diameter 4 mm when 1 cubic cm of copper weighs 8.4 gm.
Solution:
Weight of wire = 13.2 kg
Diameter of wire = 4 mm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 36
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 37

Question 28.
2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.
Solution:
Volume of brass = 2.2 cu.dm
= 2.2 x 10 x 10 x 10 = 2200 cm2
Diameter of wire = 0.25 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 38

Question 29.
The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube.
Solution:
Let r and R be the radii of inner and outer surfaces of a cylindrical tube,
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 39
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 40

Question 30.
Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes.
Solution:
Diameter of pipe = 2 cm
∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1 cm = 0.01 m
Length of flow of water in 1 second = 6 m Length of flow in 30 minutes = 6 x 30 x 60 m = 10800 m
∴ Volume of water = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 41

Question 31.
A cylindrical tube, open at both ends, is to made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal tube is 8 mm everywhere. Calculate the volume of the metal.
Solution:
Length of metal tube (h) = 25 cm
Internal diameter = 10.4 cm
∴ Internal radius (r) = \(\frac { 10.4 }{ 2 }\) = 5.2 cm
Thickness of metal = 8 mm = 0.8 cm
∴ Outer radius = 5.2 + 0.8 = 6 cm
Now volume of the metal
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 42

Question 32.
From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour.
Solution:
Inner radius of pipe (r) = 0.75 cm
Rate of water flow = 7 m per second
∴ Length of water flow in 1 hour (h)
= 7 x 3600 m = 25200 m
∴ Volume of water in 1 hour
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 43

Question 33.
A cylindrical water tank of diameter 1. 4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled ?
Solution:
Diameter of cylindrical tank = 1.4 m
∴ Radius (r) = \(\frac { 1.4 }{ 2 }\) = 0.7 m
Height (h) = 2.1m.
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 44

Question 34.
A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways Le. either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinder thus formed.
Solution:
In first case,
By rolling the paper along its length,
the circumference of the base = 30 cm
and height (h) = 18 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 45
Note: See Q.No. 24 this exercise

Question 35.
How many litres of water flow out of a pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec ?
Solution:
Speed of water = 30 cm/sec
∴ Water flow in 1 minute = 30 cm x 60 = 1800 cm
Area of cross-section = 5 cm2
∴ Volume of water = 1800 x 5 = 9000 cm3
Capacity of water in litres = 9000 x l ml
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 46

Question 36.
A solid cylinder has a total surface area of 231 cm2. Its curved surface area is \(\frac { 2 }{ 3 }\) of the total surface area. Find the volume of the cylinder.
Solution:
Total surface area of a solid cylinder = 231 cm2
Curved surface area = \(\frac { 2 }{ 3 }\) of 231 cm2 = 2 x 77 = 154 cm2
and area of two circular faces = 231-154 = 77 cm2
Let r be the radius, then
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 47

Question 37.
Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic metre. Find also the cost of cementing its inner curved surface at Rs 2.50 per square metre.
Solution:
Diameter of well = 3 m
∴ Radius (r) = \(\frac { 3 }{ 2 }\) m
and depth (h) = 280 m
(i) Volume of earth dug out = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 48
Rate of sinking the well = Rs 3.60 per m3
∴ Total cost of sinking = Rs 1980 x 3.60 = Rs 7,128
(ii) Inner curved surface area = 2πrh
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 49
Rate of cementing = Rs 2.50 per m2
∴ Total cost of cementing = Rs 2.50 x 2640
= Rs 6,600

Question 38.
Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.
Solution:
Note : See Q.No. 27 of this exercise

Question 39.
2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.
Solution:
Note: See Q.No. 28 of this exercise.

Question 40.
A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 10 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 50
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 51

Question 41.
A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.
Solution:
Width of roller (h) = 63 cm.
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 52
Outer circumference of roller = 440 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 53
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 54

Question 42.
What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm ?
Solution:
Length of hollow cylinder (h) = 16 cm
External diameter = 20 cm
∴ External radius (R) = \(\frac { 20 }{ 2 }\) = 10 cm
Thickness of iron = 2.5 mm
∴ Internal radius (r) = 10 – 0.25 = 9.75 cm
∴ Volume of iron = πh (R2 – r2)
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 55

Question 43.
In the middle of rectangular field measuring 30 m x 20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.
Solution:
Diameter of well = 7 m
∴ Radius (r) = \(\frac { 7 }{ 2 }\) m
Depth (h) = 10 m.
∴ Volume of earth dug out = πr2h
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 56
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 57

Hope given RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1

RD Sharma Class 8 Solutions Chapter 22 Mensuration III (Surface Area and Volume of a Right Circular Cylinder) Ex 22.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1

Other Exercises

Question 1.
Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm.
Solution:
Diameter of the base of cylinder = 7 cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\) cm
Height (h) = 60m
∴ carved surface area = 2πh
= 2 x \(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 2 }\) x 60cm2 = 1320cm2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 1
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 2

Question 2.
The curved surface area of a cylindrical road is 132 cm2. Find its length if the radius is 0.35 cm.
Solution:
Curved surface area =132 cm2
Radius (r) = 0.35 cm
Let h be the length of the rod Then 2πrh = 132
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 3

Question 3.
The area of the base of a right circular cylinder is 616 cm2 and its height is 2.5 cm. Find the curved surface area of the cylinder.
Solution:
Let r be the radius of the base of the cylinder, then
Area of the base = πr2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 4

Question 4.
The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area.
Solution:
Height of the cylinder (h) = 15 cm
Circumference of the base = 88 cm
Let r be the radius of the base, the circumference = 2πr
∴ 2πr = 88 cm …(i)
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 5

Question 5.
A rectangular strip 25 cm x 7 cm is rotated about the longer side. Find the total surface area of the solid thus generated.
Solution:
Dimensions of rectangular strip = 25 cm x 7 cm
By rotating the strip along longer side, a solid is formed whose radius = 7 cm
and height = 25 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 6
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 7

Question 6.
A rectangular sheet of paper 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the the total surface area of the cylinder thus generated.
Solution:
By rolling along length wire, we get a cylinder whose circumference of its base = 20 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 8

Question 7.
The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas.
Solution:
Ratio in radii of two cylinders = 2:3
and ratio in their heights = 5:3
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 9

Question 8.
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1: 2. Prove that its height and radius are equal.
Solution:
Let r be the radius and h be the height of a right circular cylinder, then Curved surface area = 2πrh
and total surface area = 2πrh x 2πr2 = 2πr (h + r)
But their ratio is 1 : 2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 10
Hence their radius and height are equal.

Question 9.
The curved surface area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the height of the cylinder.
Solution:
Curved surface area of a cylinder = 1320 cm2
Diameter of its base (d) = 21 cm 21
Radius (r) = \(\frac { 21 }{ 2 }\) cm
Let h be the height of the cylinder
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 11

Question 10.
The height of a right circular, cylinder is 10.5 m. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base.
Solution:
Height of cylinder = 10.5 m
Let r be the radius and h be the height of a right circular cylinder, then Area of its two circular faces = 2π2
and area of curved surface = 2πrh
Now, according to the condition:
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 12

Question 11.
Find the cost of plastering the inner surface of a well at Rs 9.50 per m2, if it is 21 m deep and diameter of its top is 6 m.
Solution:
Diameter of the top of a cylindrical well = 6m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 13
∴ Radius (r) = \(\frac { 6 }{ 2 }\) = 3 m
and depth (h) = 21 m
∴ Curved surface area = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 3 x 21 m2 = 396 m2
Rate of plastering = Rs 9.50 per m2
∴ Total cost of plastering = Rs 9.50 x 396 = Rs 3,762

Question 12.
A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of the tin-plating it on the inside at the rate of 50 paise per hundred square centimetre.
Solution:
Diameter of the vessel = 20 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 14
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 15

Question 13.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curved surface at Rs 4 per square metre.
Solution:
Diameter of the well = 3.5 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 16

Question 14.
The diameter of a roller is 84 cm and its length is T20 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground ?
Solution:
Diameter of the roller = 84 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 17

Question 15.
Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m, what will be the cost of cleaning them at the rate of Rs 2.50 per square metre ?
Solution:
Diameter of each pillar = 0.50 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 18

Question 16.
The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of the base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.
Solution:
Total surface area of a hollow cylinder opened from both sides = 4620 cm2
Area of base ring = 115.5 cm2
Height of cylinder (h) = 7 cm
Let R be the outer radius and r be the inner radius
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 19
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 20

Question 17.
The sum of the radius of the base and height of a solid cylinder is 37 m, if the total surface area of the solid cylinder is 1628 m2, find the circumference of its base.
Solution:
Let r be the radius and h be the height of the solid cylinder, then r + h = 37 m …(i)
Total surface area = 1628
⇒ 2πr (r + h) = 1628
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 21

Question 18.
Find the ratio between the total surface area of a cylinder to is curved surface area,given that its height and radius are 7.5 cm and 3.5 cm.
Solution:
Radius (r) of cylinder = 3.5 cm
and height (h) = 7.5 cm
∴ Curved surface area = 2πrh
and total surface area = 2πr (h + r)
∴ Ratio = 2πr (h + r)- 2πrh = (h + r): h = 7.5 + 3.5 : 7.5
⇒ 11 : 7.5
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 22

Question 19.
A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs 3.50 per 1000 cm2.
Solution:
Radius of the vessel (r) = 70 cm
and height (h) = 1.4 m = 140 cm
∴ Area of inner and outer curved surfaces and bases = 2 x 2πrh + 2πr2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 23

 

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RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2

RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2

Other Exercises

Question 1.
The marks obtained by 40 students of class VIII in an examination are given below :
16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13,
21, 13, 15, 19, 24, 16, 3, 23, 5, 12, 18, 8, 12, 6,
8, 16, 5, 3, 5, 0, 7, 9, 12, 20, 10, 2, 23.
Divide the data into five groups, namely 0-5,5-10,10-15,15-20 and 20-25 and prepare a grouped frequency table.
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 1

Question 2.
The marks scored by 20 students in a test are given below :
54, 42, 68, 56, 62, 71, 78, 51, 72, 53, 44, 58, 47, 64, 41, 57, 89, 53, 84, 57.
Complete the following frequency table :
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 2
What is the class interval in which the greatest frequency occurs ?
Solution:
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 3
The class in which the greatest frequency is 50-60

Question 3.
The following is the distribution of weights (in kg) of 52 persons :
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 4
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 5
(i) What is the lower limit of class 50-60 ?
(ii) Find the class marks of the classes 40-50, 50-60.
(iii) What is the class size ?
Solution:
(i) Lower limit of class 50-60 = 50
(ii) Class marks of 40-50 = \(\frac { 40+50 }{ 2 }\) = \(\frac { 90 }{ 2 }\)
= 45 and of 50-60 = \(\frac { 50+60 }{ 2 }\) = \(\frac { 110 }{ 2 }\) =55
(iii) Class size is 10

Question 4.
Construct a frequency table for the following weights (in gm) of 35 mangoes using the equal class intervals, one of them is 40-45 (45 not included):
30,40,45,32,43,50,55,62,70,70,61,62, 53,52, 50,42,35,37,53,55,65,70, 73, 74,45, 46, 58, 59, 60, 62, 74, 34, 35, 70 ,68.
(i) What is the class mark of the class interval 40-45 ?
(ii) What is the range of the above weights ?
(iii) How many classes are there ?
Solution:
Smallest observation = 30
Greatest observation = 74
Range = 74 – 30 = 44
Now forming the distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 6
(i) Class mark of 40-45
= \(\frac { 40+45 }{ 2 }\) = \(\frac { 85 }{ 2 }\) = 42.5
(ii) Range = 74 – 30 = 44
(iii) Number of classes are 9

Question 5.
Construct a frequency table with class-intervals 0-5 (5 not included) of the following marks obtained by a group of 30 students in an examination :
0, 5, 7,10, 12,15, 20, 22, 25, 27, 8, 11, 17,3, 6, 9,17,19, 21, 29, 31,35,37,40,42, 45, 49, 4, 50, 16.
Solution:
Frequency distribution table.
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 7

Question 6.
The marks scored by 40 students of class VIII in mathematics are given below:
81,55, 68, 79,85,43,29,68,54,73,47, 35, 72,64,95,44,50, 77,64,35,79, 52, 45,54,70,83, 62′, 64,72,92,84,76,63, 43, 54, 38, 73, 68, 52, 54.
Prepare a frequency distribution with class size of 10 marks.
Solution:
Largest marks = 95
Lowest marks = 29
Range = 95 – 29 = 66
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 8

Question 7.
The heights (in cm) of 30 students of class VIII are given below :
155.158.154.158.160.148.149.150.153, 159,161,148,157,153,157,162,159,151, 154,156,152,156,160,152,147,155,163,155,157,153.
Prepare a frequency distribution table with 160-164 as one of the class intervals.
Solution:
Largest height =163
Lowest height =147
Range = 163- 147 = 16
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 9

Question 8.
The monthly wages of 30 workers in a factory are given below :
830,835,890,810,835,836,869,845,898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 806, 840, 890.
Represent the data in the form of a frequency distribution with class size 10.
Solution:
Highest wage = 898
Lowest wage = 804
Range = 898 – 804 = 94
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 10

Question 9.
Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) at 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included):
220, 268, 258, 242, 210, 268, 272,242, 311, 290, 300, 320,319,304,302,318,306,292, 254, 278, 210,240, 280,316,306, 215, 256, 236.
Solution:
Highest wages = 320
Lowest wages = 210
Range = 320-210= 110
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 11

Question 10.
The daily minimum temperatures in degrees Celsius recorded in a certain Arctic region are as follows :
-12.5, -10.8, -18.6, -8.4, -10.8, -4.2, -4.8, -6.7, -13.2, -11.8, -2.3,1.2, 2.6, 0, -2.4, 0, 3.2, 2.7,3.4,0, -2.4, -2.4, 0,3.2, 2.7,3.4, 0,2.4, -5.8, -8.9, -14.6, -12.3, -11.5, -7.8, – 2.9
Represent them as frequency distribution table taking -19.9 to -15 as the first class interval.
Solution:
Lowest temperature = -19.9
Highest temperature = 3.4
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.2 12

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RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1

RD Sharma Class 8 Solutions Chapter 23 Data Handling I (Classification and Tabulation of Data) Ex 23.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1

Other Exercises

Question 1.
Define the following terms :
(i) Observations
(ii) Raw data
(iii) Frequency of an observation
(iv) Frequency distribution
(v) Discrete frequency distribution
(vi) Grouped frequency distribution
(vii) Class-interval
(viii) Class-size
(ix) Class limits
(x) True class limits
Solution:
(i) Observations : Each entry in the given data is called an observation. ‘
(ii) Raw data: A collection of observations by an observer, is called raw data.
(iii) Frequency of an observation : The number of times an observation occurs in the given data is called its frequency.
(iv) Frequency distribution : The presentations of given data in order of magnitude ascending or descending, is called the frequency distribution.
(v) Discrete frequency distribution: When the given data is represented by tally marks after arranging it in an order. This kind of distribution is called discrete frequency distribution.
(vi) Grouped frequency distribution: If the number of data is large, and the difference between the greatest and the smallest observation is large, then we represent then in groups or classes. Such representation of data is called grouped frequency distribution.
(vii) Class intervals: The difference between the upper limit and lower limit of a class is called class interval.
(viii) Class-size : Class intervals are also called the class size. Each size of the same intervals
(ix) Class limits : Every class has two limits : upper limit and lower limit.
(x) True class limits or Exclusive limits : When the upper limit of one is the lower limit of the next interval then these are call true class limits.

Question 2.
The final marks in mathematics of 30 students are as follows :
53, 61, 48, 60, 78, 68, 55, 100, 67, 90, 75, 88,77,37,84,58,60,48,62,56,44,58,52,64, 98, 59, 70, 39, 50, 60
(i) Arrange these marks in the ascending order, 30 to 39 one group, 40 to 49 second group etc.
Now answer the following:
(ii) What is the highest score ?
(iii) What is the lowest score ?
(iv) What is the range ?
(v) If 40 is the pass mark how many have failed ?
(vi) How many have scored 75 or more ?
(vii) Which observations between 50 and 60 have not actually appeared ?
(viii) How many have scored less than 50 ?
Solution:
(i) Arranging the given data in ascending order.
30 to 39 : 37,39
40 to 49 : 44, 48, 48
50 to 59 : 50, 52, 53, 55, 56, 58, 58, 59
60 to 69 : 60, 60, 60, 61, 62, 64, 67, 68
70 to 79 : 70, 75, 77, 78
80 to 89 : 84, 88
90 to 99 : 90, 98
100 to 109 : 100
(ii) Highest score is 100
(iii) Lowest score is 37
(iv) Range is 100 – 37 = 63
(v) If 40 is pass marks then number of failed candidates will be = 2
(vi) Number of students who scored 75 or more = 8
(vii) Between 50 and 60, the observations 51, 54, 57 do not appear.
(viii) Number of students who scored less than 50 = 5

Question 3.
The weights of new born babies (in kg) in a hospital on a particular day are as follows:
2.3, 2.2, 2.1, 2.7, 2.6,3.0, 2.5, 2.9, 2-8,3.1, 2.5, 2.8, 2.7, 2.9, 2.4.
(i) Rearrange the weights in descending order.
(ii) Determine the highest weight:
(iii) Determine the lowest weight.
(iv) Determine the range.
(v) How many babies were born on that day?
(vi) How many babies weigh below 2.5 kg ?
(vii) How many babies weigh more than 2.8 kg ?
(viii) How many babies weigh 2.8 kg ?
Solution:
(i) Arranging the given weights in descending order.
3.1, 3.0, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.5, 2.4, 2.3, 2.2, 2.1
(ii) Highest weight = 3.1 kg
(iii) Lowest weight = 2.1 kg
(iv) Range : 3.1 – 2.1 = 1 kg.
(v) Number of babies born on that day = 15
(vi) Number of babies having weight below 2.5 kg = 4
(vii) Number of babies having weight more than 2.8 kg = 4
(viii) Number of babies weigh 2.8 kg = 2

Question 4.
Following data gives the number of children in 41 families:
1,2,6,5,1,5,1,3,2,6,2,3,4,2,0,0,4,4, 3, 2, 2, 0, 0,1, 2, 2, 4,3, 2,1, 0, 5,1, 2,4, 3, 4, 1, 6, 2, 2.
Represent it in the form of a frequency distribution.
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 1

Question 5.
Prepare a frequency table of the following scores obtained by 50 students in a test:
42, 51, 21, 42, 37. 37, 42, 49, 38, 52, 7, 33, 17, 44, 39, 7, 14, 27, 39, 42, 42, 62, 37, 39, 67, 51, 53, 53, 59, 41, 29, 38, 27, 31, 54,19, 53, 51, 22, 61, 42, 39, 59, 47, 33, 34, 16, 37, 57, 43
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 2

Question 6.
A die was thrown 25 times and following scores were obtained:
1,5,2,4,3,6,1,4,2,5,1,6,2,6,3,5,4,1, 3, 2, 3, 6,1, 5, 2
Prepare a frequency table of the scores.
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 3

Question 7.
In a study of number of accidents per day, the observations for 30 days were obtained as follows:
6,3,5,6,4,3, 2,5,4,2,4,2,1,2, 2,0,5,4,6,1,6,0, 5, 3, 6,1, 5, 5, 2, 6
Prepare a frequency distribution table.
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 4

Question 8.
Prepare a frequency table of the following ages (in years) of 30 students of class VIII in your school:
13,14,13,12,14,13,14,15,13,14,13,14, 16,12,14,13,14,15,16,13,14,13,12,17,13, 12,13,13,13,14
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 5

Question 9.
Following figures relate to the weekly wages (in Rs) of 15 workers in a factory :
300,250,200,250,200,150,350,200,250, 200,150, 300,150, 200, 250 Prepare a frequency table.
(i) What is the range in wages (in Rs) ?
(ii) How many workers are getting Rs 350 ?
(iii) How many workers are getting the minimum wages ?
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 6
(i) Range = 350- 150 = 200
(ii) Number of workers getting Rs 350 = 1
(iii) Number of workers getting minimum wages = 3

Question 10.
Construct a frequency distribution table for the following marks obtained by 25 students in a history test in class Vin of a school:
9,17,12, 20,9,18, 25,17,19,9,12,9,12, 18, 17,19, 20, 25, 9,12,17,19, 19, 20, 9
(i) What is the range of marks ?
(ii) What is the highest mark ?
(iii) Which mark is occurring more frequently ?
Solution:
Frequency distribution table
RD Sharma Class 8 Solutions Chapter 23 Data Handling I Ex 23.1 7
(i) Range = 25 – 9 = 16
(ii) Highest marks = 25
(iii) Marks occurring more frequently = 9

 

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RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1

RD Sharma Class 8 Solutions Chapter 24 Data Handling II (Graphical Representation of Data as Histograms) Ex 24.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1

Question 1.
Given below is the frequency distribution of the heights of 50 students of a class :
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 1
Draw a histogram representing the above data.
Solution:
We represent class intervals along x-axis and frequency along y-axis. Taking suitable intervals along x-axis and y-axis we construct the rectangles as shown in the figure. This is the required histogram.
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 2

Question 2.
Draw a histogram of the following data :
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 3
Solution:
We represent class-intervals along x-axis and frequency along y-axis. Taking suitable intervals along x-axis andy-axis, we construct rectangles as shown in the figure. This is the required histogram.
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 4

Question 3.
Number of workshops organized by a school in different areas during the last five years is as follows :
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 5
Draw a histogram representing the above data.
Solution:
We represent years along x-axis and number of workshops along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 6
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 7

Question 4.
In a hypothetical sample of 20 people the amounts of money with them were found to be as follows :
114, 108,100, 98, 101,109,117,119, 126, 131, 136, 143, 156, 169, 182, 195, 207, 219, 235, 118.
Draw the histogram of the frequency distribution (taking one of the class intervals as 50-100).
Solution:
Highest sample = 235
Lowest sample = 98
Range = 235-98 = 137
Now frequency distribution table will be as under:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 8
We represent class intervals along x-axis and frequency along j’-axis. Taking suitable intervals, we construct a rectangles as shown in the figure. This is the required histogram.

Question 5.
Construct a histogram for the following data:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 9
Solution:
We represent monthly school fee (in Rs) along x-axis and number of schools along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 10

Question 6.
Draw a histogram for the daily earnings of 30 drug stores in the following table :
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 11
Solution:
We represent daily earnings (in Rs) along x-axis and number of stores along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 12

Question 7.
Draw a histogram to represent the following data:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 13
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 14
Solution:
We represent monthly salary (in Rs) along x-axis and number of teachers along y-axis. Taking suitable intervals we construct rectangles as shown in the figure. This is the required histogram
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 15

Question 8.
The following histogram shows the number of literate females in the age group of 10 to 40 years in a town :
(i) Write the age group in which the number of literate female is highest.
(ii) What is the class width ?
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 16
(iii) What is the lowest frequency ?
(iv) What are the class marks of the classes ?
(v) In which age group literate females are least ?
Solution:
(i) The age group in which the number of literate females is 15-20.
(ii) The class width is 5.
(iii) Lowest frequency is 320.
(iv) The class marks of the classes are
\(\frac { 10+15 }{ 2 }\) = \(\frac { 25 }{ 2 }\) =12.5, similarly other class marks will be 17.5,22.5,27.5,32.5,37.5
(v) The least literate females is in the class 10-15

Question 9.
The following histogram shows the monthly wages (in Rs) of workers in a factory:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 17
(i) In which wage-group largest number of workers are being kept ? What is their number ?
(ii) What wages are the least number of workers getting ? What is the number of such workers ?
(iii) What is the total number of workers ?
(iv) What is the factory size ?
Solution:
(i) The largest number of workers are in wage group 950-1000 and is 8.
(ii) The least number of workers are in the wage group 900-950 and is 2.
(iii) Total number of workers is 40 (3 + 7 + 5 + 4 + 2 + 8 + 6 + 5)
(iv) The factory size is 50.

Question 10.
Below is the histogram depicting marks obtained by 43 students of a class :
(i) Write the number of students getting highest marks.
(ii) What is the class size ?
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 18
Solution:
(i) The number of students getting highest marks is 3.
(ii) The class size is 10.

Question 11.
The following histogram shows the frequency distribution of the ages of 22 teachers in a school:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 19
(i) What is the number of eldest and youngest teachers in the school ?
(ii) Which age group teachers are more in the school and which least ?
(iii) What is the size of the classes ?
(iv) What are the class marks of the classes?
Solution:
(i) The number of eldest teacher is 1 and the number of youngest teacher is 2.
(ii) The teachers in age group 35-40 is most.
(iii) Size of classes is 5.
(iv) Class marks of class 20-25 is \(\frac { 20+25 }{ 2 }\)= \(\frac { 45 }{ 2 }\) = 22.5
and similarly others will be 27.5, 32.5, 37.5, 42.5, 47.5, 52.5.

Question 12.
The weekly wages of 30 workers in a factory are given:
830,835,890,810,835,836,869,845,898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Mark a frequency table with intervals as 800-810,810-820 and so on, using tally marks.
Also, draw a histogram and answer the following questions:
(i) Which group has the maximum number of workers ?
(ii) How many workers earn Rs 850 and more ?
(iii) How many workers earn less than Rs 850?
Solution:
The frequency table will be as given below:
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 20
We represent wages (in Rs) along x-axis and number of workers along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.
(i) Maximum workers are in the wage group 830-840.
(ii) Number of workers getting Rs 850 and more are 1 + 3 + 1 + 1 + 4 = 10.
(iii) Number of workers getting less than Rs 850 are 3 + 2 + 1 + 9 + 5 = 20
RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 21

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