A cuboidal water tank is 6 m long, -Maths 9th

1 Answer

Answer :

Volume of cuboidal tank = 1 x b x h = 6 m x 5 m x 4.5 m = 135 m3  = 135 x 1000 L = 135000 L

Related questions

Description : The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m -Maths 9th

Last Answer : Length (l) and depth (h) of tank is 2.5 m and 10 m respectively. To find: The value of breadth, say b. Formula to find the volume of a tank = l b h = (2.5 b 10) m3= 25b m3 Capacity ... of water (Given) Therefore, 25000 b = 50000 This implies, b = 2 Therefore, the breadth of the tank is 2 m.

Description : A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? -Maths 9th

Last Answer : Dimensions of a cuboidal water tank are: l = 6 m and b = 5 m and h = 4.5 m Formula to find volume of tank, V = l b h Put the values, we get V = (6 5 4.5) = 135 ... water, 135 m3volume hold = (135 1000) litres = 135000 litres Therefore, given cuboidal water tank can hold up to135000 litres of wate

Description : Rinku has built a cuboidal water tank in his house. The top of the water tank is covered with an iron lid. -Maths 9th

Last Answer : Total inner surface area of the water tank including the base without top = 2(l + b) h + l b = 2(180 + 120) 60 + 180 120 = 36000 + 21600 = 57600 cm2 Area of each title = 10 ... 57600 / 80 = 720 Total amount required for 720 tiles at the rate of 480 per dozen = 480 / 12 720 = 28800

Description : Rinku has built a cuboidal water tank in his house. The top of the water tank is covered with an iron lid. -Maths 9th

Last Answer : Total inner surface area of the water tank including the base without top = 2(l + b) h + l b = 2(180 + 120) 60 + 180 120 = 36000 + 21600 = 57600 cm2 Area of each title = 10 ... 57600 / 80 = 720 Total amount required for 720 tiles at the rate of 480 per dozen = 480 / 12 720 = 28800

Description : Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Last Answer : The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m. Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula) Required Volume is 144 m3 Now, Cost of digging per m3 volume = Rs 30 Cost of digging 144 m3 volume = Rs (144×30) = Rs 4320

Description : A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high -Maths 9th

Last Answer : From the question statement, we have Edge of a cube = 10cm Length, l = 12.5 cm Breadth, b = 10cm Height, h = 8 cm (i) Find the lateral surface area for both the figures Lateral surface ... . Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2

Description : A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th

Last Answer : 92239223% Volume of cone = 1313πr2h = 13×227×1×7=22313×227×1×7=223 cu. cm Volume of cubical block = (10 × 5 × 2) cm3 = 100 cm3 ∴ Wastage of wood = (100−227)100×100(100−227)100×100 = 27832783% = 92239223%

Description : Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Last Answer : Convert all to metres: 5 km = 5000 m 14 cm = 0.14 m 7 cm = 0.07 m Find the radius: Radius = Diameter 2 Radius = 0.14 2 = 0.07 m Find the amount of water that flowed out in an hour: Volume ... hours needed: Number of hours = 154 77 = 2 hours It takes 2 hours to fill up the tank to rise by 7 cm

Description : 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 meter , sheet being 2 m wide. -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : Water flows in a tank 150 m × 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm, at a speed of 15 km per hour. -Maths 9th

Last Answer : Volume of water discharged through the pipe = Volume increase of the tank First, consider the pipe. Volume discharge through pipe = length breadth speed time Let the time taken to fill the tank to 3 m depth be t. ... the tank V=150 100 3 cu. m ⇒ V=450 100 Therefore, 450t=450 100 ⇒ t=100 hours

Description : Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. -Maths 9th

Last Answer : Height of cylindrical tank, h = 4.5m Radius of the circular end , r = (4.2/2)m = 2.1m (i) the lateral or curved surface area of cylindrical tank is 2πrh = 2 (22/7) 2.1 4.5 m2 = (44 0.3 ... ) = 87.12 m2 This implies, S = 95.04 m2 Therefore, 95.04m2 steel was used in actual while making such a tank.

Description : A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Last Answer : Answer We have, r=0.7m, h=8m ∴ Total surface area = 2πr2+2πrh=2πr(r+h)=2×722​×0.7×8.7m2 Required cost = Rs. {2×722​×0.7×8.7×10}=Rs.382.80

Description : It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th

Last Answer : Radius of the closed cylindrical tank = 140/2 cm = 70 cm = 0.7 m Height of the closed cylindrical tank = 1 m Area of metal sheet required = 2πr(r + h) = 2 x 22/7 x 0.7 (1 + 0.7) = 7.48 m2

Description : Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. -Maths 9th

Last Answer : Edge of cubical tank = 1.5 m ∴ Area of 4 walls = 4 (side)² = 4(1.5)² m² = 4 x 225 = 9 m² Area of floor = (1.5)² = 2.25 m² ∴ Total surface area = 9 + 2.25 = 11.25 m² Edge of square tile = 25 m = 0.25 m² ∴ Area of 1 tile = (0.25)2 = .0625 m²

Description : A storage tank is in the form of a cube. When it is full of water, the volume of water is 15.625 m3. -Maths 9th

Last Answer : When it is full of water, the volume of water is 15.625m3. If the present depth of water is 1.3m, then, find the volume of water alredy used fom the tank. Hence, the volume of water ... 7500 L. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Description : A storage tank is in the form of a cube. When it is full of water, the volume of water is 15.625 m3. -Maths 9th

Last Answer : Let side of a cube be = x m ∴ Volume of cubical tank = 15.625 m3 [given] ⇒ x3 = 15.625 m3 ⇒ x = 2.5 m and present depth of water in cubical tank = 1.3 m ∴ Height of water used =2.5 - 1.3 ... 7. 5 x 1000 = 7500 L [∴ 1 m3 = 1000 L] Hence, the volume of water already used from the tank is 7500 L.

Description : Shahid has built a cubical water tank with lid for his house, -Maths 9th

Last Answer : Edge of tank = 2 m = 2 x 100 cm = 200 cm Area of five faces of the tank = 5a2 = 5 (200 cm)2 = 2,00,000 cm2 Area of a square tile = 25 cm x 25 cm = 625 cm2 Number of tiles required = Area of ... /12 do zen Cost of one dozen of tiles = ₹480 ∴ Cost of 320/12 dozen tiles = ₹480 x 320/12 = ₹12, 800

Description : A tank internally measuring 150 cm × 120 cm × 100 cm has 1281600 cm^3 of water in it. Porous bricks are placed in the water until the tank -Maths 9th

Last Answer : Volume of a brick = 20 cm 6 cm 4 cm = 480 cm3. Water absorbed by one brick = (110 480)(110 480) cm3 = 48 cm3. Let x bricks be placed in the water. Then, x bricks absorb 48x cm3 of water. ... tank ⇒ 1281600 + 480xx - 48xx = 150 x 120 x 100 ⇒ 432xx = 1800000 - 1281600 = 518400 ⇒ xx = 1200.

Description : It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? -Maths 9th

Last Answer : Let h be the height and r be the radius of a cylindrical tank. Height of cylindrical tank, h = 1m Radius = half of diameter = (140/2) cm = 70cm = 0.7m Area of sheet required = Total surface are of tank = 2πr( ... [2 (22/7) 0.7(0.7+1)] = 7.48 Therefore, 7.48 square meters of the sheet are required.

Description : A storage tank is in the form of a cube. -Maths 9th

Last Answer : Volume of water in cubical storage tank = 15.625 m3 = (2.5 m)3 = a3 ⇒ a = 2.5 m Volume of water in cubical tank when depth of water is 1.3 m = 2.5 X 2.5 x 1.3 = 8.125 m3 Volume of water already used from the tank = 15.625 - 8.125 = 7.5 m3

Description : A hemispherical tank is made up of an iron -Maths 9th

Last Answer : Inner radius of the hemispherical tank (r) = 1 m Outer radius of the hemispherical tank (R) = 1 + 0.01 = 1.01 m Volume of iron used to make the hemispherical tank = 2/3 πR3 - 2/3πr3 = 2/3π(R3 - r3) = 2/3 x 22/7 [(1.01)3 - 13] = 44/21(1.0303 - 1) = 44/21 x 0.0303 = 0.06349 m3

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

Description : How many angles of obliquity are there for a cuboidal body under complex stresses? a. 6 b. 8 c. 4 d. None

Last Answer : a. 6

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

Description : The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Last Answer : Let sides of △ are=6x,7x and 8x Perimeter=6x+7x+8x=21x 21x=410 x=20 Sides are 120,140 and 160 m Area =S(S−A)(S−B)(S−C)​ [Heron's Formula] S=2120+140+160​=210 m A=210(210−120)(210−140)(210−160)​=210015​ sq. m

Description : The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Last Answer : Let sides of △ are=6x,7x and 8x Perimeter=6x+7x+8x=21x 21x=410 x=20 Sides are 120,140 and 160 m Area =S(S−A)(S−B)(S−C)​ [Heron's Formula] S=2120+140+160​=210 m A=210(210−120)(210−140)(210−160)​=210015​ sq. m

Description : In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

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Description : In Fig. 6.7, if l||m, then find the value of x. -Maths 9th

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Description : In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

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Description : Determine the ratio in which the point P(m, 6) divides the join of A(– 4, 3) and B(2, 8). Also find the value of m. -Maths 9th

Last Answer : (b) 1 : 2Any point on the x-axis is (a, 0).Let the point (a, 0) divide the join of A(2, -3) and B(5, 6) in the ratio k : 1. Then the co-ordinates of the point of division are \(\bigg(rac{5k+2}{k+1},rac{6k-3}{k+1}\ ... 6k - 3 = 0 ⇒ k = \(rac{1}{2}\)Required ratio is k : 1 ⇒ \(rac{1}{2}\) : 1 = 1 : 2.

Description : The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it level in 5 revolutions? -Maths 9th

Last Answer : Radius of a garden roller (r) = 1.4 / 2 = 0.7 m and length of the garden roller (h) = 2 m ∴ Curved surface area of garden roller = 2πrh . So, the area levelled of a garden roller in one revolution is 8.8 m2. Now, area levelled of a garden in 5 revolutions = 8.8 x 5= 44m2

Description : The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it level in 5 revolutions? -Maths 9th

Last Answer : Radius of a garden roller (r) = 1.4 / 2 = 0.7 m and length of the garden roller (h) = 2 m ∴ Curved surface area of garden roller = 2πrh . So, the area levelled of a garden roller in one revolution is 8.8 m2. Now, area levelled of a garden in 5 revolutions = 8.8 x 5= 44m2

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Last Answer : S(△ABC)=60+80+402=90S(△ABC)=60+80+402=90 ar△ABDar△ABD =90(90−80)(90−60)(90−40)−−−−−−−−−−−−−−−−−−−−−−−√=90(90−80)(90−60)(90−40) =90×10×30×50−−−−−−−−−−−−−−√=90×10×30×50 =30015−−√m2=30015m2 ar□ABCE=2×ar△ABDar◻ABCE=2×ar△ABD =60015−−√m2

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2

Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Last Answer : Length of the plank=4m=400cm Breadth=50cm Height=20cm Volume of the plank=L*B*H =400*50*20 =400000cm^3 Length of the pit=16m=1600cm Breadth=12m=1200cm Height=4m=400cm Volume of the pit= L ... *1200*400 =768000000cm^3 Number of planks that can be fitted= 768000000/400000 =1920 planks is the answer.

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

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Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2.

Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

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Description : How many planks each of which is 2 m long, -Maths 9th

Last Answer : 2 m = 2 x 100 cm = 200 cm, 6 m = 6 x 100 = 600 cm Number of planks = Volume of wooden block/ Volume of each plank = 600 x 18 x 44/200 x 3 x 4 = 198.

Description : A field is 70 m long and 40 m broad. -Maths 9th

Last Answer : Area of the field on which earth taken out is to be spread = 70 X 40 m2 - 10 X 8 m2 = 2800 m2 - 80 m2 = 2720 m2 Volume of the earth dug out = 10 X 8 X 5 m3 = 400 m3 Rise in level of the field = ... earth dugout/ Area on which earth taken out is to be spread = 400/ 2720 = 0.147 m = 14.7 cm

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