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.
(i) Length of cuboid = 12 cm
Breadth of cuboid = 8 cm
Height of cuboid = 6 cm
Hence,
Volume of cuboid = length × breadth × height = 12 × 8 ×6 = 576 cm3
(ii) Length = 1.2 m = 120 cm
Breadth = 30 cm
Height = 15 cm
Hence,
Volume of cuboid = 120 × 30 × 15 = 54000 cm3
(iii) Length = 15 cm
Breadth = 2.5 dm = 25 cm
Height = 8 cm
Hence,
Volume of cuboid = 15 × 25 × 8 = 3000 cm3
Find the volume of a cube whose side is
(i) 4 cm
(ii) 8 cm
(iii) 1.5 dm
(iv) 1.2 m
(v) 25 mm
(i) Given,
side of cube = 4 cm
volume of cube = (side)3 = 43 = 64 cm3
(ii) side of cube = 8 cm
volume of cube =(side)3 = 83 = 512 cm3
(iii) side of cube = 1.5 dm
volume of cube = (side)3 = 1.53 = 3.375 dm3 = 3375 cm3
(iv) side of cube = 1.2 m
volume of cube = 1.23 = 1.728 m3
(v) side of cube = 25 mm
volume of cube = 253 = 15625 mm3= 15.625 cm3
Find the height of a cuboid of volume 100 cm3, whose length and breadth are 5 cm and 4 cm respectively.
Given,
Volume of cuboid = 100 cm3
Length = 5 cm
Breadth = 4 cm
Let height of cuboid = h cm
So,
=
=
A cuboidal vessel is 10 cm long and 5 cm wide. How high it must be made to hold 300 cm3 of a liquid?
Given,
Length of cuboidal vessel = 10 cm
Width = 5 cm
Volume of liquid in it = 300 cm3
Let height of vessel = h cm
So,
=
=
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?
Given,
Length of milk container = 8 cm
Width = 50 cm
Volume to hold = 4 litre = 4000 cm3
Let height of container = h cm
So,
=
=
A cuboidal wooden block contains 36 cm3 wood. If it be 4 cm long and 3 cm wide, find its height.
Given,
Volume of wood in cuboidal block = 36 cm3
Length of block = 4 cm
Breadth of block = 3 cm
Let height of block = h cm
So,
=
=
What will happen to the volume of a cube, if its edge is
(i) halved (ii) trebled?
Given,
Let edge of cube = a
So volume of cube = a3
Case (i)
Edge become =
Volume become = =
Case (ii)
Edge becomes = 3a
Volume become =
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?
(i) Let ,
Length of cuboid = l
Breadth = b
Height = h
Volume of cuboid = lbh
Now,
Case(i)
Length become = 2l
Height = h
Breadth =
Volume of cuboid =
(ii)
Case(ii)
Length become = 2l
Breadth = b
Height = 2h
Volume of cuboid =
Three cuboids of dimemsions 5 cm × 6cm × 7cm, 4cm × 7cm × 8cm and 2 cm×3 cm× 13 cm are melted and a cube is made. Find the side of cube.
Volume of First cuboids = 5 × 6 × 7 = 210 vm3
Volume of second cuboids = 4 × 7 × 8 = 224 cm3
Volume of third cuboids = 2 × 3 × 13 = 78 cm3
Volume of cube = 210 + 224 + 78 = 512
Let side of cube = a
⇒ a3 = 512
a = 8 cm
Find the weight of solid rectangular iron piece of size 50 cm×40cm × 10 cm, if 1 cm3 of iron weights 8 gm.
Given,
Dimension of rectangular iron piece = 50cm×40cm×10cm
Volume of solid rectangular =
Weight of 1 cm3 iron = 8 gm
∴ weight of 20000 cm3 iron =
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?
Given,
Dimensions of log of wood = 3m × 75 cm ×50cm
Side of cubical block = 25 cm
Hence,
No. of cubical block that can be made from wooden log =
=
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 cm3 each are to be made. Find the number of beads that can be made from the block.
Given,
Dimensions of cuboidal block of silver = 9cm×4cm×3.5cm
Volume of beads made = 1.5 cm3
So,
Number of beads can be made from cuboidal block =
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.
Given,
Dimensions of cuboidal boxes = 2cm×3 cm×10 cm
Dimesions of carton = 40cm×36cm×24cm
So,
Number of boxes can be stored in carton =
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 this block? Assume cutting causes no wastage.
Given,
Dimensions of cuboidal block of iron = 50cm×45cm×34cm
Size of small cuboids cutting from it = 5cm×3cm×2cm
So,
Number of small cuboids can be cut =
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?
Given,
Let side of cube B = X cm
Then, side of cube A = 3X cm
So,
=
An ice-cream brick measures 20 cm by 10 cm by 7 cm. How many such bricks can be stored in deep fridge whose inner dimensions are 100 cm by 50 cm by 42 cm?
Given,
Dimensions of ice cream brick = 20 cm× 10cm× 7cm
Dimensions of fridge = 100 cm×50cm×42 cm
So,
Number of bricks can be put in fridge =
Suppose that there are two cubes, having edges 2 cm and 4 cm, respectively. Find the volumes V1 and V2 of the cubes and compare them.
Given,
Edge of one cube = 2 cm
Edge of second cube = 4 cm
Hence ,
=
=
=
A tea-packet measures 10 cm×6 cm× 4 cm. How many such tea-packets can be placed in a cardboard box of dimensions 50 cm× 30cm× 0.2 m?
Given,
Dimensions of tea packet = 10 cm×6 cm×4cm
Dimension of cardboard box = 50cm×30cm×0.2 m
So,
Number of tea packets can be put in cardboard box =
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.
Given,
Dimensions of metal block = 5cm×4cm×3cm
Weight of block = 1 kg
Volume of box = 5×4×3 = 60 cm3
Dimension of new block = 15cm×8cm×3cm
Volume of new box = 15×8×3 = 360 cm3
We know that,
=
=
How many soap cakes can be placed in a box of size 56 cm×0.4 m× 0.25 m, if the size of a soap cake is 7 cm× 5cm× 2.5 cm?
Given,
Dimensions of box = 56cm×0.4m×0.25m
Dimensions of soap cake = 7cm×5cm×2.5cm
So,
Number of soap cakes can be placed in box =
The volume of acuboidal box is 48 cm3. If its height and kength are 3 cm and 4 cm respectively, find its breadth.
Given,
Volume of cuboidal box = 48 cm3
Height of box = 3 cm
Length of box = 4 cm
Let height of box = h cm
So,
=
=
Find the volume in cubic metres (cu.m) of each of the cuboids whose dimensions are :
(i) length = 12 m, breadth = 10 m, height = 4.5 m
(ii) length = 14 m, breadth = 2.5m, height = 50 cm
(iii) length = 10m, breadth = 25 dm, height = 25 cm.
(i) Given,
Length of cuboid = 12 m
Breadth of cuboid = 10m
Height of cuboid = 4.5 m
So,
Volume of cuboid =
(ii) Given,
length of cuboid = 14 m
breadth of cuboid = 2.5 m
height of cuboid = 50 cm = .50m
so.
Volume of cuboid =
(iii) Given,
length of cuboid = 10m
breadth of cuboid = 25 dm = 2.5 m
height of cuboid = 25 cm = .25 m
so,
volume of cuboid =
Find the volume in cubic decimetre of each of the cubes whose side is
(i) 1.5 m
(ii) 2 dm 5 cm
(i) Given,
Side of cube = 1.5m = 15 dm
Volume of cube = 153 = 3375 dm3
(ii) Given,
side of cube = 2dm 5 cm = 2.5 dm
volume of cube = 2.53 = 15.625 dm3
How much clay is dug out in digging a well measuring 3 m by 2 m by 5 m?
Given
Dimensions of well = 3m×2m×5m
So,
Volume of clay dug out from it =
What will be the height of a cuboid of volume 168 m3, if the area of its base is 28 m2?
Given,
Volume of cuboid = 168 m3
Area of base =
Let height of cuboid = h m
So,
=
=
=
So height of cuboid = 6 m
A tank is 8 m long, 6 m broad and 2 m high. How much water can it contain?
Given,
Dimensions of a tank = 8m×6m×2m
So,
Capacity of tank = volume of tank = = 96000 litre
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.
Given,
Capacity of cuboidal tank = 50000 litre = 50 m3
Height of tank = 10 m
Length of tank = 2.5 m
Let breadth of tank = b m
So,
=
= 2 m
Breadth of tank = 2 m
A rectangular diesel tanker is 2 m long, 2 m wide and 40 cm deep. How many litres of diesel can it hold?
Given,
L = 2m
B = 2m
H = 40cm
Dimensions of rectangular diesel tank = 2m × 2m × 40 cm
So,
Amount of diesel it can hold = volume of tank = = 1600 litre
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.
Given,
L = 5m
B = 4.5m
H = 3m
Dimensions of a room = 5m × 4.5m × 3m
So,
Volume of air it contains = l × b × h = 5 × 4.5 × 3 = 67.5m3
A water tank is 3 m long, 2 m broad and 1 m deep. How many litres of water can it hold?
Given,
L = 2m
B = 2m
H = 40cm
Dimensions of water tank = 3m × 2m ×1m
So,
Capacity of water it can hold =
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?
Given,
Dimensions of one plank = 3m × 15cm × 5cm = 300cm ×15cm×5cm
Dimensions of wooden block = 6m × 75cm ×45 cm = 600 cm×75 cm×45cm
So,
No. of planks can be made = = = 90 planks
How many bricks each of size 25 cm× 10 cm× 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?
Given,
Size of one brick = 25cm × 10cm × 8cm
Dimensions of wall = 5m × 3m × 16cm = 500 cm ×300 cm ×16cm
So,
Number of bricks needed to make the wall =
A village, having a population of 4000, required 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 will the water of this tank last?
Given,
Dimensions of water tank = 20m × 15m ×6m
Population of village = 4000
Water require per head per day = 150 litre
Total requirement of water per day = 150 ×4000 = 600000 litre
Volume of water tank =
So,
=
A rectangular field is 70 m long and 60 m broad. A well of dimensions 14 m × 8 m × 6 m is dug outside the field and the earth dug-out from this well is spread evenly on the field. How much will the earth level rise?
Given,
Dimension of rectangular field = 70m ×60m
Dimension of well = 14m × 8m ×6m
Amount of earth dug out from well =
So,
Rise in earth level of rectangular field =
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.
Given,
Dimensions of swimming pool = 250 m ×130m
Volume of water pumped in it = 3250 m3
So,
Rise in water level in pool =
A beam 5 m long and 40 cm wide contains 0.6 cubic metre of wood. How thick is the beam?
Given,
Length of beam = 5 m
Width of beam = 40 cm = 0.4 m
Volume of wood in beam = 0.6 m3
Let thickness of beam = h m
So,
=
=
The rainfall on a certain day was 6 cm. How many litres of water fell on 3 hectares of field on that day?
Given,
Area of field = 3 hectare = 3×10000 m2 = 30000 m2
Depth of water on the field = 6 cm =
∴ volume of water = area of field × depth of water
= 30000 × 0.06 = 1800 m3
=
∴
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.
Given,
Length of cuboidal beam = 8 m
One edge of beam = 0.5 m
Let third edge of beam = h m
No. of cubes of side 1 cm (.01 m) produced = 4000
So,
=
=
=
Length of third edge = 0.001 m
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?
Given,
Dimensions of metal block = 2.25m × 1.5m ×27 cm = 2.25m × 1.5m × .27m
Side of each cube formed = 45 cm = 0.45 m
So,
Number of cubes can formed = cubes
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.
Given,
Dimensions of solid rectangular piece = 6m × 6cm ×2 cm
Volume of rectangular iron =
Weight of 1 cm3 iron = 8 gm
∴
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 a wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm3. The height of the cuboid is …50…. 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.
(i)
(ii)
(iii)
(iv) Side of cube = 8 cm
Volume of cube =
(v) Volume of cuboid = 4000 cm3
Length = 10 cm , breadth = 8cm
Then,
Height =
(vi)
=
(vii)
(viii)
(ix)
(x)
1
Find the surface area of a cuboid whose
(i) length = 10 cm, breadth = 12 cm, height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2m, breadth = 4 m, height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.
(i) Given,
Length = 10 cm
Breadth = 12 cm
Height = 14 cm
So,
Surface area of cuboid = =
=
(ii) Given,
Length = 6 dm
Breadth = 8 dm
Height = 10 dm
So,
Surface area of cuboid =
=
(iii) Given,
Length = 2m
Breadth = 4m
Height = 5m
So,
Surface area of cuboid =
=
(iv) Given,
length = 3.2 m= 32 dm
breadth = 30 dm
height = 250 cm= 25 dm
so,
surface area of cuboid =
=
Find the surface area of a cube whose edge is
(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1 m
(i) We have,
Edge of cube = 1.2 m
Surface area of cube =
(ii) We have,
Edge of cube = 27 cm
Surface area of cube =
(iii) We have,
Edge of cube = 3 cm
Surface area of cube =
(iv) We have,
Edge of cube = 6 m
Surface area of cube =
(v) We have,
Edge of cube = 2.1 m
Surface area of cube =
A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area.
Given,
Dimensions of cuboidal box = 5cm × 5cm × 4cm
Surface area of cuboid =
Find the surface area of a cube whose volume is
(i) 343 m3
(ii) 216 dm3
(i) Given,
Volume of cube = 343 m3
Side of cube = a =
So,
Surface area of cube =
(ii) Given,
Volume of cube = 216 dm3
Side of cube =
So,
Surface area of cube =
Find the volume of a cube whose surface area is
(i) 96 cm2
(ii) 150 m2
(i) Given,
Surface area of cube = 96 cm2
=
=
=
So,
Volume of cube =
(ii) Given,
Surface area of cube = 150 m2
=
=
=
So,
Volume of cube = 53 = 125 m3
The dimensions of a cuboid are in the ratio 5 : 3 : 1 and its total surface area is 414 m2. Find the dimensions.
Given,
Ratio of dimensions of cuboid = 5:3:1
Total surface area of cuboid = 414 m2
Let dimensions are =
So,
=
=
=
=
=
So,
Dimensions are =
=
=
Find the area of the cardboard required to make a closed box of length 25 cm, 0.5 m and height 15 cm.
Given,
Dimensions of closed box = 25 cm×0.5 m× 15 cm = 25 cm ×50 cm × 15 cm
So,
Area of cardboard required =
=
Find the surface area of a wooden box whose shape is of a cube, and if the edge of the box is 12 cm.
Given,
Edge of a cubic wooden box = 12 cm
Surface area of cubic wooden box =
The dimensions of an oil tin are 26 cm× 26 cm× 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 tin sheet used for these 20 tins.
Given,
Dimensions of oil tin are = 26 cm × 26 cm × 45 cm
Then,
Area of tin sheet required for making one oil tin = total surface area of oil tin
=
=
Area of tin sheet required for 20 oil tins =
So,
Cost of 1 m2 tin sheet = Rs. 10
∴cost of 12.064 m2 tin sheet =
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.)
Given,
Dimensions of class room = 11m × 8 m ×5 m
Where,
Length = 11m , Breadth = 8m , Height = 5 m
Ten,
Area of floor =
Area of four walls ( including doors & windows) =
=
∴ Sum of areas of floor and four walls = area of floor + area of four walls
= 88 + 190 = 278 m2
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.
Given,
Dimensions of swimming pool are = 20m × 15m ×3m
Where,
Length = 20m , Breadth = 15m , Height = 3m
Then,
Area of floor & walls of swimming pool =
=
So,
Cost of repairing 1 m2 area = Rs.25
∴ Cost of repairing 510 m2 =
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.
Given,
Perimeter of floor = 30 m
Height of floor = 3 m
So,
=
=
Area of four walls of room =
Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume.
Given,
Let length of cuboid =
Let breadth of cuboid = b cm
Let height of cuboid = h cm
So,
Area of floor =
Product of areas of two adjacent walls =
Product of areas of floor and two adjacent walls =
=
=
The walls and ceiling of a room are to be plastered. The length, breadth nad height of the room are 4.5 m, 3 m and 350 cm, respectively. Find the cost of plastering at the rate of Rs. 8 per square metre.
Given,
Length of room = 4.5 m
Breadth of wall = 3 m
Height of wall = 350 cm =
So,
Area of ceiling + area of walls =
=
=
Cost of plastering 1 m2 area = Rs.8
∴ Cost of plastering 66 m2 area = 66 ×8 =Rs.528
A cuboid has total surface area of 50 m2 and lateral surface area os 30 m2. Find the area of its base.
Given,
Total surface area of cuboid = 50 m2
Lateral surface area of cuboid = 30 m2
So,
=
And,
=
=
=
=
So,
Area of base =
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.
Given,
Dimensions of class room = 7m × 6m ×3.5m
Where,
Length = 7m, Breadth = 6m , Height = 3.5 m
Area of four walls (including doors & windows) =
=
Then,
Area of walls without doors & windows =
= area including doors & windows – area occupied by doors & windows
Area of only walls = 91 – 17 = 74 m2
So,
Cost of white washing 1 m2 area of walls = Rs.1.50
∴ Total cost of white washing = 74 × 1.50 = Rs.111
The central hall of a school is 80 m long and 8 m high. It has 10 doors each of size 3 m ×1.5 m and 10 windows each of size 1.5 m× 1 m. If the cost od 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.
Given,
Dimensions of central hall of a school = Length = 80 m , height = 8m
Let breadth of hall = b metre
So,
Area of each door = 3m × 1.5 m = 4.5 m2
∴ Area of 10 doors = 10 × 4.5 = 45 m2
Area of each window = 1.5m × 1m = 1.5 m2
∴ Area of 10 windows = 10 ×1.5 = 15 m2
Area occupied by doors and windows = 45 + 15 = 60 m2
Area of the walls of the hall including doors and windows =
= 2(80×8+b×8) = 2(640+8b) m2
Then,
Area of only walls = (area of walls including doors & windows – area occupied by doors & windows)
=
Total cost of white washing = Rs.2385.60
Given, Rate of white washing = Rs.1.20 per m2
So,
=
=
=
=
=
Hence,
Breadth of hall = 48 m
Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.
Given,
Length of room = 12 m
Breadth = 9m
Height = 8m
So,
Length of longest rod that can be placed in room = diagonal of room (cuboid) =
If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that
Given,
V = volume of cuboid
S = surface area of cuboid
= a,b,c = dimensions of cuboid
So,
S =
V = abc
=
=
The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V, prove that V2 = xyz.
Given,
Areas of three faces of cuboid =
Let length of cuboid =
So,
=
=
=
Or we can write ,
=
If ‘V’ is volume of cuboid = V =
=
=
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.
Given,
Capacity of water reservoir = 105 m3
Length of base of reservoir = 12 m
Width of base = 3.5 m
Let depth of reservoir = h m
So,
=
=
= 2.5 m
Depth of reservoir = 2.5 m
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.
Given,
Edge length of cube A = 18 cm
Edge length of cube B = 24 cm
Edge length of cube C = 30 cm
So,
Volume of cube A = =
Volume of cube B = =
Volume of cube C = =
Total volume of three cubes = 5832 + 13824 + 27000 = 46656 cm3
Let ‘a’ be the length of edge of new cube formed .
=
=
So,
Edge of bigger cube = 36 cm
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.
Given,
Breadth of room is twice of its height =
Breadth is one half of length =
Volume of the room =
=
=
=
Hence,
Breadth of cube = b = 8 dm
Length of cube = 2b = 2×8 = 16 dm
Height of cube =
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.
Given,
Length of tank = 12 m
Width of tank = 9m
Depth of tank = 4m
So,
Area of sheet required = total surface area of tank
=
=
Let
=
=
=
Then,
Cost of iron sheet at rate Rs.5 per metre = 5×192 = Rs. 960
A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12m×8m×6m, find the cost of iron sheet at Rs. 17.50 per metre.
Given,
Dimensions of tank = 12m × 8m × 6m
Where length = 12m , breadth = 8m , height = 6m
Area of sheet required for making the tank = total surface area of tank with one top open
=
Let
=
=
=
So,
Cost of iron sheet at rate Rs.17.50 per metre = 17.50 ×84 = Rs.1470
Three equal cubes are placedadjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Given,
Let edge length of three equal cubes = a
Then,
Sum of surface area of 3 cubes =
When these cubes are placed in a row adjacently they form a cuboid.
Length of new cuboid formed = a+a+a = 3a
Breadth of cuboid = a
Height of cuboid = a
Total surface area of cuboid =
=
Hence,
=
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.
Given,
Dimensions of room = 12.5m ×9m × 7m
Dimensions of each door = 2.5m ×1.2m
Dimensions of each window = 1.5m ×1m
Area of four walls including doors and windows =
=
Area of 2 doors and 4 windows =
Area of only walls = 301 – 12 = 289 m2
Hence,
Cost of painting walls at rate Rs.3.50 per square metre = Rs.(3.50 ×289) = Rs.1011.50
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 is the level of field raised?
Given,
Length of field = 150m
Width of field = 100m
Area of field = 150m × 100m = 15000 m2
Length of plot = 50 m
Breadth = 30 m
Depth upto which it dug = 8 m
So volume of earth taken out from it = 50×30×8 = 12000 m3
let raise in earth level of field on which it spread = h metre
so,
=
=
The level of field raised by 0.8 metre
Two cubes, each of volume 512 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Given,
Volume of each cube = 512 cm3
Let length of edge of each cube = a cm
So,
=
= a
When these two cubes are joined end to end a cuboid is formed :
Length of cuboid = 8+8 = 16 cm
Breadth = 8 cm
Height = 8 cm
Surface area of resulting cuboid =
=
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.
Given,
Edge of three cubes are respectively = 3cm , 4 cm , 5 cm
So,
Sum of volume of these cubes =
After melted these cubes a new cube is formed.
Let edge length of this new cube = a cm
So,
=
=
Edge of new cube = 6 cm
Surface area of new cube =
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.
Given,
Length of room = 12 m
Let width of room = b metre
Let height of room = h metre
Now,
Area of floor = 12b m2
Cost of matting floor @rate 85 paise per square metre = Rs.91.80
=
=
Breadth of room = 9 m
Area of 4 walls =
Cost of preparing walls at rate Rs.1.35 per square metre = Rs.340.20
=
=
Height of room = 6 m
The length of a hall is 18 m and the 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.
Given,
Length of hall = 18 m
Width of hall = 12 m
Let height of hall = h metre
Then,
Sum of area of floor & flat roof =
Sum of area of 4 walls =
Now,
=
=
Height of hall = 7.2 metre
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.
Given,
Edge of metal cube = 12 cm
Edge of smaller two cubes = 6 cm , 8 cm
Let edge of third cube = a cm
So,
Volume of metal cube = sum of volume of three small cubes
=
=
=
So,
Edge of third smaller cube would be = 10 cm
The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the gall, if each person required 150 m3 of air?
Given,
Dimensions of cinema hall are = 100m ×50m × 18m
Where , length = 100m , breadth = 50m , height = 18 m
Each person air requirement = 150 m3
Now,
Volume of cinema hall =
So,
Number of person can sit in cinema hall =
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× 3 cm× 0.75 cm can be put in this box?
Given,
External dimensions of wooden box = 48cm×36cm×30cm
Dimensions of bricks = 6cm×3cm×0.75cm
Thickness of wood = 1.5 cm
So,
Internal dimensions of box =
= 45cm ×33 cm× 27 cm
Hence,
Number of bricks can be put in box = bricks
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. 1248. Find the dimensions of the box.
Given,
Ratio of dimensions of rectangular box = 2:3:4
Let length of box = 2x m
Let breadth = 3x m
Let height = 4x m
Area of sheet of paper required for covering it = total surface area of cuboid
=
Cost of covering it with sheet of paper at Rs.9.50 /m2=
Cost of covering it with sheet of paper at rate Rs.8/m2 =
=
=
=
=
So,
Length of box = 2x = 2×4 = 8 m
Breadth of box = 3x = 3×4 = 12 m
Height of box = 4x = 4×4 = 16 m