CBSE Class 8

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

• 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 10.2

Question 1.
In which of the following tables x and y vary inversely :



Solution:


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 :

Solution:

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 :

Solution:

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.

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

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,


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

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.

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

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

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

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 :


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 :

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,

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

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

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

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

• 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 10.2

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 ?

Solution:

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.

Solution:

Question 6.
Find the constant of variation from the table given below :

Solution:

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

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,

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,

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,

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

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

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 :

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 :

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 :

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 :

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,

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 :

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

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 :

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 :

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,

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,

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 (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

• RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1
• RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2
• RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3
• RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4

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

Longest rod to be kept in the room

Question 2.
If V is the volume of the cuboid of dimensions a, b, c and S its the surface area then prove that

Solution:
∵ a, b, c are the dimensions of a cuboid
∴ Volume (V) = abc
Surface area (S) = 2(ab + bc + ca)
Now

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

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

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

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

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

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

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

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

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

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

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

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

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

Hope given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 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 21.3

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

• RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1
• RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2
• RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3
• RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4

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

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

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

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

Hope given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3 are helpful to complete your math homework.

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