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Given:

Radius of tank, r = 6.3 m

Speed of water through pipe = 14 l/s

Formula used:

Volume of tank = (4/3) × πr³ 

1 m³ = 1000 litres

Calculation:

Volume of tank = (4/3) × πr3 

⇒ Volume of tank = (4/3) × (22/7) × (6.3)³

⇒ 1047.816 m³

Volume of petrol in 1 second = 14 litres

⇒ Volume of petrol in 1 second = (14/1000) m³ 

Time taken by pipe to empty the tank = (Volume of tank)/(Volume of petrol in 1 second)

⇒ (1047.816)/(14/1000)

⇒ 74844 seconds

⇒ (74844)/60 minutes

⇒ 1247.4 minutes

∴ The time taken by pipe to empty the tank is 1247.4 minutes

Given:

The radius of the tank, r = 6.3 m

Speed of water through pipe = 14 litres/sec

Formula used:

Volume of tank = (4/3) × πr³ 

1 m³ = 1000 litres

Calculation:

Volume of tank = (4/3) × πr3 

⇒ The volume of tank = (4/3) × (22/7) × (6.3)³

⇒ 1047.816 m³

The volume of pipe in 1 second = 14 litres

⇒ The volume of pipe in 1 second = (14/1000) m³ 

Time is taken by pipe to empty the tank = (Volume of the tank)/(Volume of pipe in 1 second)

⇒ (1047.816)/(14/1000)

⇒ 74844 seconds

⇒ (74844)/60 minutes

⇒ 1247.4 minutes

∴ The time taken by pipe to empty the tank is 1247.4 minutes

Given:

Side of cube = 14 cm

Formula used:

Volume of sphere = (4/3)π × r³

Volume of cube = (side)³

Calculation:

Side of cube = Diameter of sphere

⇒ Diameter of sphere, d = 14 cm

Radius of sphere, r = d/2 = 7 cm 

Volume of cube = (14)³

⇒ Volume of cube = 2744 cm³

Volume of sphere = (4/3)π × r³

⇒ Volume of sphere = (4/3)(22/7) × (7)³

⇒ (4 × 22 × 343)/21

⇒ 1437.33 cm³

Volume of remaining cube = (Volume of cube ) – (Volume of sphere)

⇒ Volume of remaining cube = 2744 – 1437.34

⇒ 1306.67 cm³

∴ Volume of remaining cube is 1306.67 cm3.

Given:

The angle formed by bisector of any two angles of triangle is zero

Formula Used:

In triangle ABC if OB and OC are the angle bisector

∴ ∠BOC = ∠A/2 + 90

Calculation:

Let ABC is a triangle and OB and OC are the interior angle bisector of triangle ABC also it is given that the angle formed by bisector of any two angles of triangle is zero

∴ ∠BOC = 0

⇒ ∠A/2 + 90 = 0

⇒ ∠A = - 180°

This is not possible 

Given:

Three angles (2x – 30)°, 2x° and (3x - 40)° are given

Formula Used:

If a₁, a₂ and a₃ are the terms in arithmetic progression, then

a₂ – a₁ = a₃ – a₂

Calculation:

It is given in the question that the angles are in arithmetic progression,

2x – (2x – 30)° = (3x – 40)° - 2x

⇒ 2x – 2x + 30 = x – 40

⇒ 30 = x – 40

⇒ x = 70°

Angles = 110°, 140° and 170° 

∴ The largest angle is 170°

Given:

The side of the equilateral triangle = 10 cm

Formula used:

Area of equilateral triangle = (√3/4) × (side)² 

Calculation:

Area of the triangle = (√3/4) × 10² 

⇒ Area of the equilateral triangle = 100√3/4 = 25√3 cm² 

∴ Area of the equilateral triangle is 25√3 cm² 

Given 

length of rectangle  = X₁

Breadth of rectangle = X₂

Formula Used 

Area of rectangle = Length × Breadth 

Calculation 

⇒ X₁ + X₂ = 40 

⇒ We know that, all the rectangle, a square has largest area 

⇒ For given rectangle to be square X₁ = X₂

⇒ X₁ = X₂ = 20 (for maximum area) 

⇒ so Maximum Area = 20 × 20 = 400 sq m

Given:

Length of a side of an equilateral triangle = 6 cm

Formula used:

Height (h) = (a × √3)/2

Calculation:

Height = (6 × √3)/2

\(\therefore Height= 3\sqrt{3} \ cm\)

Given:

Side length = 6 meter

Formula used:

Area = side²

Calculation:

Area of square = (6 m)²

⇒ 36 m²

∴ The area of square is 36 m²

Given:

Length of the water tank = 15 m 

The breadth of the water tank = 10 m 

Depth of the water tank = 3 m

The cost of repair = Rs 24 per m²

Formula used:

The total surface area of the cuboid = 2(lb + bh + lh)

Calculation:

The upper part of the tank is not included, so

The area of the five walls = 2(lb + bh + lh) – lb

⇒ 2(bh + lh) + lb

⇒ 2(10 × 3 + 15 × 3) + 15 × 10

⇒ 2(30 + 45) + 150

⇒ 150 + 150 = 300 m²

Total cost of repair walls = 300 × 24 = 7200

∴ The total cost of repair walls of the water tank is Rs 7200

Calculation:

A cube is a three dimensional solid object bounded by six square faces with three meeting at each vertex.

It has 6 faces, 12 edges and 8 vertices

∴ A cube has 6 surfaces

Given:

Length of tank = 5 meter

Width of tank = 4 meter

Depth of tank = 3 meter

Formula used:

1 m³ = 1000 litres

Calculation:

Capacity of tank = (5 × 4 × 3) × 1000 litres

⇒ 60 × 1000 litres

⇒ 60000 litres

∴ The capacity of tank is 60000 litres

Given:

4(side of rhombus)² = 900 m²

Concept:

The side of a rhombus is equal.

Calculation:

Let the side of the rhombus 'a'

∴ Sum of side of Rhombus = 4a

∵ 4a² = 900 m²

⇒ a² = 900/4 m²

⇒ a = √(900/4) m

⇒ a = 30/2 m

⇒ a = 15 m

Given:

Length of medians is 9 cm, 10 cm, and 11 cm

Formula used/Concept Used:

The formula used to calculate the area of triangles when length of medians is given

s = (u + v + w)/2

Area of triangle = (4/3) × √s × (s - u) × (s - v) × (s - w)

Where u, v and w are the length of medians

Calculation:

Length of median is 9 cm, 10 cm, and 11 cm

∴ s = (9 + 10 + 11)/2 = 15 cm

Now, Area of triangle = (4/3) × √15 × (15 - 9) × (15 - 10) × (15 - 11) = (4/3) × 30√2 = 40√2

∴ The area of triangle is 40√2 square cm

Given:

The length of the diagonal is 7.5√2 cm

Formula used:

Area of square = Side × Side

Diagonal of square = √2 × side

Calculation:

The length of diagonal is 7.5√2 cm

∴ 7.5√2 = √2 × side

⇒ Side = 7.5 cm

Now, Area of square = 7.5 × 7.5 = 56.25 square cm

According to the given question

Given:

Perimeter of equilateral triangle is 25√3 cm

Formula used/Concept Used:

Height of equilateral triangle = a√3/2

Where ‘a’ is the side of equilateral triangle

Perimeter of equilateral triangle = 3 × side

Calculation:

The perimeter of equilateral triangle is 25√3 cm

⇒ Side = 25√3/3 cm

⇒ a = 25√3/3 cm

Now, the height of equilateral triangle = (25√3/3) × (√3/2) = 12.5 cm

∴ The height of equilateral triangle is 12.5 cm

Given:

The width of the hall(rectangular) = 3/4 times of its length

Area of the hall = 300 m²

Formula used:
Area of a rectangle = lb

l = length of the rectangle; b = width of the rectangle

Calculation:

b : l = 3 : 4

Let the width be 3x and length be 4x

According to the question:

lb = 300

⇒ 12x² = 300

⇒ x² = 25

⇒ x = 5

∴ Difference between length and width of the hall = 4x - 3x = 5 m

Given:
Length of two sides of the triangle is 12 cm and 14 cm

Perimeter of the triangle = 36 cm

Formula used:

Perimeter of a triangle = Sum of all three sides

Calculation:

According to the question:

36 = 12 + 14 + Length of the third side

∴ Length of the third side = 10 cm

Calculations :

When we construct a triangle taking diameter as one of the side, then 

The triangle will always be a right triangle

So, 

The angle in a semi-circle is always a right angle.

∴ The correct answer is option 4.

Given:

Circumference of the tank = 396 cm

Height of the tank = 19 cm

Formula used:

Circumference of the base = 2 π r

The volume of the cylinder = π r2 h

Here, h and r is the height and radius respectively

Calculation:

Circumference of the base = 2 π r

⇒ 2 × 22/7 × r = 396

⇒ r = 63 cm

Now, Quantity of water it can hold = π r2 h

⇒ 22/7 × 63 × 63 × 19

⇒ 237006

∴ The quantity of water tank can hold is 237006 cm3

Given:
Volume(V) of the cylinder = 1.54 m³

Diameter of the base of the cylinder = 140 cm

Formula used:

V of a cylinder = πr²h

r = radius of the base, h = height of the cylinder

Calculation:

r = 140/2 = 70 cm = 0.7 m

According to the question:

1.54 = (22/7) × 0.7 × 0.7 × h

⇒ h = 1 

∴ Height of the cylinder = 1 m

GIVEN:

Three statements.

CONCEPT:

Mensuration

FORMULA USED:

Volume of cuboid = LBH

Volume of cube = S²

CALCULATION:

A:

Maximum number of small cubes that can be put at bottom most layer = Base area of large cuboid / Base area of one small cube

= (32 × 24) / 8²

= 4 × 3

= 12

B:

Maximum number of small cubes that can be put at second layer = Base area of large cuboid / Base area of one small cube

48 = (32 × 24) / a²

⇒ a² = 16

⇒ a = 4

Side of each small cube = a = 4 cm

C:

Maximum number of small cubes that can be put at third layer = Base area of large cuboid / Base area of one small cube

= (32 × 24) / 1.6²

= 20 × 15

= 300

Given:

Diameter of each wheel of truck = 210 cm

Rotation of wheel = 200 times/minute

Concept:

Total distance covered by the truck is n times of circumference

Formula used:

Circumference of the circle = 2πr

Calculation:

Circumference of wheel = 2πr

⇒ Circumference of wheel = 2 × (22/7) × 105

⇒ Circumference of wheel = 660 cm

Total distance covered by wheel of the truck = (660 × 200)/(1000 × 100) = (66 × 2)/100 km

Now, Speed of the truck (in km/h) = (66 × 2 × 60)/100 = 79.2 km/h

∴ Speed of truck is 79.2 km/hr

13200 cm²

Given:

Radius of conical cap = 7 cm

Height of conical cap = 24 cm

Formula used:

Curved surface area of cone = πrl

l² = r² + h²

Where, r = radius of cone

l = slant height of cone

h = height of cone

Calculation:

l² = 7² + 24²

⇒ l² = 49 + 576 = 625

⇒ l = √625 = 25 cm

Curved surface area of cone = π × 7 × 25 = 550 cm²

Area of sheet for 2 dozen caps = 550 × 2 × 12 = 13200 cm²

∴ The area of the sheet required to make 2 dozen caps is 13200 cm²

Given:

Radius of conical cap = 12 cm

Height of conical cap = 16 cm

Formula used:

Curved surface area of cone = πrl

l² = r² + h²

where, r = radius of cone

l = slant height of cone

h = height of cone

Calculation:

l² = 12² + 16²

⇒ l² = 144 + 256 = 400

⇒ l = √400 = 20 cm

Curved surface area of cone = 22/7 × 12 × 20 = 5280/7 cm²

Cost of painting the surface of the cap = 5280/7 × 1.40 = 1056 Rs.

∴ The cost of painting the surface of the cap at the rate of 1.40 Rs per sq cm is Rs. 1056

Given:

The radius of the hemisphere is decreased by 37.5%.

Formula used:

The volume of a hemisphere = (2/3)πr3

Where r → The radius of the sphere

Calculations:

Let r₁ be the old radius and r₂ be the new radius.

As, 37.5% = 37.5/100 = 3/8

According to the question,

The ratio of the old radius to the new one = 8 ∶ (8 - 3) = 8 ∶ 5

So, The ratio of the volumes = {(2/3)π(r₁)³} ∶ {(2/3)π(r₂)³}

⇒ (r₁)³ ∶ (r₂)³ = 512 ∶ 125

Decrease = 512 - 125 = 387

% Decrease = (387/512) × 100 ≈ 75.6%

∴ The volume of the hemisphere will decrease by 75.6% approximately

Given:

Height of the well (H) = 40 m

The radius of the well (r) = 7 m

Length of the field (l) = 56 m

The breadth of the field (b) = 11 m

Formula used:

The volume of a cylinder (V) = πr2H

The volume of a cuboid (V') = l × b × h

Where l → length

b → breadth

h → height of the cuboid

r → radius 

H → height of the cylinder

Calculations:

Let the required height be h.

The volume of earth = (area of the field –  area of the well) × h

So, (22/7) × 7 × 7 × 40 = {56 × 11 - (22/7) × 7 × 7} × h

⇒ h = (154 × 40)/(616 - 154)

⇒ h = 13.33 m

∴  The increase in the level of the field is 13.33 m.

Given:

The radii of two circles are 4 cm and 3 cm.

Formula Used:

Area of circle = πr²

Here, r = radius of circle

Calculation:

The radii of two circles are 4 cm and 3 cm.

The area of circle having radius 4 cm = π(4)² = 16π 

The area of circle having radius 3 cm = π(3)2 = 9π 

The sum of area of circles = 16π + 9π  = 25π 

According to the question;

⇒ πr² = 4 × 25π 

⇒ r² = 100

⇒ r = 10 cm

Diameter = 2r = 2 × 10 = 20 cm

∴ The diameter of the circle is 20 cm.

Given:

Length of rectangular field = 32 cm

Breadth of rectangular field = 20 cm

Length of tile = 4 cm

Formula Used:

Area of square = side × side

Area of rectangle = length × breadth

Calculations:

Length of rectangular field = 32 cm

Breadth of rectangular field = 20 cm

Length of tile = 4 cm

⇒ Area of rectangular field = (32 × 20) = 640 cm²

Area of each tile = (4 × 4) = 16 cm²

⇒ Number of tiles = (640/16) = 40

∴ The number of square tiles used to cover the rectangular field is 40.

Given:

The length and breadth of a rectangular floor are 14.35 m and 11.55 m

Formula used:

Area of Rectangle = Length × breadth

Area of square = (Side)²

Required number = Area of the floor/Area of each tile

Concept used:

We require minimum number of square tiles, the size of the tile is given as the H. C.F of two sides of the room.

Calculation:

Length of rectangular floor = 14.35 × 100 = 1435 cm

Breadth of rectangular floor = 11.55 × 100 = 1155 cm

According to the question:

1155 = 3 × 5 × 11 × 7

1453 = 5 × 7 × 41

H. C. F = 5 × 7 = 35

Now,

Required number = Area of the floor/Area of each tile

⇒ (Length × breadth)/(Side)2

⇒ (1435 × 1155)/(35 × 35)

⇒ 41 × 33 = 1353

∴ The minimum number of square tiles are 1353.

Given:

Perimeter of equilateral triangle = 36 cm

Concepts used:

Perimeter of equilateral triangle = 3 × side

Area of equilateral triangle = √3/4 × (side)² 

Calculation:

Perimeter of equilateral triangle = 3 × side

⇒ 3 × side = 36 cm

⇒ Side = 36/3 cm

⇒ Side = 12 cm

Area of equilateral triangle = √3/4 × (side)² 

⇒ √3/4 × (12)² cm²

⇒ √3/4 × 144 cm² 

⇒ (1.732 × 144)/4 cm²

⇒ 62.352 cm². 

∴ The area of equilateral triangle is 62.352 cm².

Given:

Perimeter of circle = Area of circle

Formula used:

Perimeter of circle = 2πr

Area of circle = πr²

Calculation:

2πr = πr²

⇒ 2r = r²

⇒ 2 = r

∴ The radius of circle is 2 unit

Given:

4 (A) = 6 (B)

h = 12 cm

Where,

A = Area of CSA of cylinder

B = Sum of the area of its base

h = height of cylinder

Formula used:

Curved surface area of cylinder = 2π × radius × height

Volume of cylinder = π × (radius)2 × height

Area of base = π × (radius)2

Calculation:

4 (A) = 6 (B)

⇒4 (2 π r h) = 6 (2 π r²)

⇒ 8 h = 12 r

⇒ 8 × 12 = 12 r (given h= 12 cm)

⇒ r = 8 cm

Volume of cylinder = π × 8² × 12

= 768 π 

Given:

Diameter = 7 cm

Formula used:

Area of the circle = π r²

Calculation:

Area of the circle = π r2

⇒ (22/7) × (7/2) × (7/2)

⇒ 38.5 cm2

∴ 4 times area of circle = 38.5 × 4 = 154 cm²

Given:

The diameter of sphere is 21 cm

Formula used:

Volume of sphere = (4/3)π r3

Radius (r) = diameter/2

Calculation:

Diameter of sphere = 21 cm

Radius (r) = diameter/2

⇒ (21/2) cm

Volume of sphere = (4/3)π r³

⇒ (4/3) × (22/7) × (21/2) × (21/2) × (21/2)

⇒ 4851 cm³

∴ The volume of Sphere is 4851 cm³.

Given:

Diagonal of cube = 10√3 cm

Formula used:

Diagonal of cube = √3x

Volume of cube = x³

Where, x = Side of cube

Calculation:

According to the question,

√3x = 10√3

⇒ x = 10 cm

Volume of cube = (10)³ cm³

⇒ 1000 cm³

∴ The volume of cube is 1000 cm³

Given:

Diameter of sphere = 21 cm

⇒ Radius (r) = 21/2 cm

Formula used:

The volume of the sphere = (4/3) π r³, where r will be the radius of the sphere

Calculation:

The volume of the sphere = (4/3) π r3

⇒ (4/3) × (22/7) × (21/2) × (21/2) × (21/2)

⇒ 4851 cm³

∴ The volume of the sphere is 4851 cm3

Given:

Diameter of sphere = 21 cm

Diameter of lead balls = 3 cm

Formula used:

Volume of sphere = 4/3(πr³)

Calculation:

Let ‘n’ be the number of lead balls that can be made from a sphere

Radius of sphere = 21/2 cm

Radius of lead balls = 3/2 cm

⇒ Volume of sphere = 4/3(πr³)

⇒ Volume of sphere = 4/3[π(21/2)³]

⇒ Volume of lead ball = 4/3[π(3/2)³]

According to the question,

Volume of sphere = n × Volume of lead balls

⇒ 4/3[π(21/2)³] = n × 4/3[π(3/2)³]

⇒ (21/2)³ = n × (3/2)³

⇒ 9261/8 = n × 27/8

⇒ n = 9261/27

⇒ n = 343 

∴ The number of lead balls, each with a diameter of 3 cm, that can be made from a sphere of diameter 21 cm is 343.

Measurement: It is defined as the description of data in terms of numbers. More precisely, measurement is defined as the assignment of numerals to objects or events according to rules.

3-dimensional body: ​A body that has length, breadth, and height. For example, glass, ball, and cylinder are three-dimensional objects as it has length, breadth, and height.

• Weigh is the measurement that signifies how heavy a body is. To be more specific, it also signifies the bigness or the smallness of the pull of the earth on 3-D bodies.
• A 3-dimensional body occupies space and hence, its volume can also be measured.
• Distance is measured between two points.

Hence, we conclude that the above statement is about weight-measure.