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A closed wooden rectangular box made of 1 cm thick wood has the following outer dimensions: length 22 cm, breadth 17 cm, and height 12 cm. It is filled with cement. What is the volume of the cement in the box?

A. 1488 cu. cm

B. 3000 cu. Cm

C. 4488 cu. Cm

D. 2880 cu. cm

Option 4 : B

**Given:**

Thickness of wood = 1 cm

Length of box = 22 cm

Breadth of box = 17 cm

Height of box = 12 cm

**Calculation:**

Inner length of the box = (22 − 2) = 20 cm

Inner breadth of the box = (17 − 2) = 15 cm

Inner height of the box = (12 − 2) = 10 cm

Inner volume of the box = (20 × 15 × 10) = 3000 cu. Cm

**∴ Volume of cement in the box is 3000 cu. cm**

Option 4 : 50,000 m^{3}

**Given:**

Length of pipe = 2 m

Breadth of pipe = 10 m

Speed of water = 10 km/hr

Time = 15 minutes

**Calculation:**

Speed = 10 km/hr = 10 × (5/18) m/sec = 25/9 m/s

Time = 15 minutes = (15 × 60) sec = 900 sec

Length (H) = (25/9) × 900 = 2500

Volume = L × B × H

= 2 × 10 × 2500

= 50,000 m^{3}

**∴ Volume of water collected in 15 minutes is 50,000 m ^{3}.**

Option 2 : 16 cm

**Concept used:**

Volume of cuboidal box = length × breadth × height

**Calculation:**

Suppose the length of the box is 4x, then, its breadth will be 3x and height will be 2x.

Now, Volume = 4x × 3x × 2x = 24x^{3}

⇒ 24x^{3} = 1536

⇒ x^{3} = 64

⇒ x = 4

Length of the box is 4x = 16 cm

∴ The the length of the box is 16 cm.

__Alternate Method__

Let breadth of the box be x.

(4x/3) × x × (4x/6) =1536

⇒ 16x3/18 = 1536

⇒ x3 = 1536 × 18/16

⇒ x3 = 1728

⇒ x = 12

length = 4x/3

⇒ 4 × 12/3

⇒ 16

**∴ The the length of the box is 16 cm.**

Option 2 : 250 litres

**Given:**

Length of the tank = 1 m

Breadth of the tank = (1/2) m

Height of the tank = (1/2) m

**Concept used:**

Volume of a tank = Length × Breadth × Height

1000 litre = 1 m^{3}

**Calculation:**

Volume of the tank = (1 × 1/2 × 1/2) m^{3}

= 1/4 m^{3}

Now, as we know that:

1m^{3} = 1000 litre

⇒ 1/4 × 1000 = 250 litres

**∴ 250 litres can be stored in the tank**

Option 1 : 15 meters

**Given:**

Length of room = 10 m

Breadth of room = 10 m

Height of room = 5 m

**Concept used:**

The length of the longest rod is the diagonal of room

Diagonal of a cuboid = √[(l^{2}) + (b^{2}) + (h^{2})]

**Calculation:**

Diagonal of cuboid = √[(10)^{2} + (10)^{2} + (5)^{2}] m

⇒ √[100 + 100 + 25] m

⇒ √[225] m

⇒ 15 m

**∴ The maximum length of rod that can be placed is 15 m**

A metallic solid cuboid of sides 44 cm, 32 cm and 36 cm melted and converted into some number of spheres of radius 12 cm. How many such sphere can be made with the metal (π = 22/7)?

Option 3 : 7

**Given:**

The sides of the cuboid are 44 cm, 32 cm, and 36 cm

The radius of the sphere is 12 cm

**Concept Used:**

The volume of a cuboid of sides l, b and h = l × b × h

The volume of the sphere of radius r = (4/3)πr^{3}

**Calculation:**

The volume of the metallic cuboid is (44 × 32 × 36) cm^{3}

The volume of the sphere is (4/3) × π × 12^{3}

Let, the total number of such sphere is n

Accordingly,

44 × 32 × 36 = n × (4/3) × π × 12^{3}

⇒ 44 × 32 × 36 = n × (4/3) × (22/7) × 12 × 12 × 12

⇒ n = 44 × 32 × 36 × (3/4) × (7/22) × (1/12) × (1/12) × (1/12)

⇒ n = 7

**∴**** Such 7 spheres can be made by given metallic cuboid.**

Option 4 : 20 days

⇒ Total water consumption of Kazipet of 1 day = 4000 × 9 = 36000 litres

⇒ Volume of cuboidal tank = 720 m^{3} = 720 × 1000 litres = 720000 litres

Option 4 : No change

**Given:**

Length = 15 cm, Breadth = 12 cm, Height = 11 cm

**Formula used:**

The total area of the four side faces of cuboid = 2h(l + b)

**Calculation:**

The total area of the four side faces of cuboid = 22 × 27 = 594 cm^{2}

Now, converting percentage to fraction to make calculation easy

\({\rm{}}6\frac{2}{3}\% = \frac{1}{{15}}\;,\;8\frac{1}{3}\% {\rm{}} = {\rm{}}\frac{1}{{12}}\;\)

Length after decrease = 15 × 14/15 = 14 cm

Breadth after increase = 12 × 13/12 = 13 cm

Height (no change) = 11 cm

⇒ The total area of the four side faces after change in length and breadth = 2 × 11(14 + 13)

⇒ The total area of the four side faces after change in length and breadth = 594 cm^{2}

⇒ Change in total area = 594 – 594 = 0

Option 1 : 160 m^{3}

**Given:**

The surface area of three faces = 25 m2, 32 m2 and 32 m2

**Concept used:**

The surface area of one face

1.) Length × Breadth

2.) Breadth × Height

3.) Height × Length

Volume of cuboid = Length × Breadth × Height

**Calculations:**

We have,

⇒ Length × Breadth = 25 m^{2}

⇒ Breadth × Height = 32 m^{2}

⇒ Height × Length = 32 m^{2}

Multiplying the above three equations, we get,

⇒ (Length × Breadth × Height)^{2} = 25 × 32 × 32

Taking square root on both sides,

⇒ (Length × breadth × Height) = 5 × 32

⇒ Length × breadth × Height = 160

⇒ Volume of cuboid = 160 m^{3 }

**∴ The volume of cuboid is 160 m ^{3}**

The length of a room is (3x + 10) m and the breadth of the room is (2x + 5) m. The area of four walls of the room is (60x + 180) m^{2}. What is the height of the room?

Option 2 : 6 m

**Given:**

The length of the room is (3x + 10) m and breadth of the room is (2x + 5) m. The area of four walls of the room is (60x + 180) m^{2}. We have to find the height of the room.

**Concept Used:**

If the length, breadth and height of a room be l, b and h then the area of four walls of the room is [2 × (l + b) × h]

**Calculation:**

Let, the height of the room be h

Accordingly,

2 × {(3x + 10) + (2x + 5)} × h = (60x + 180)

⇒ (5x + 15)h = 30x + 90

⇒ (5x + 15)h = 6(5x + 15)

⇒ h = 6

**∴ The height of the room is 6 m.**

Option 1 : \(3\sqrt {21\;} cm\)

Let the height of this cuboid be h.

⇒ Total surface area = 2[l × b + b × h + h × l]

⇒ 340 = 2[10 × 8 + 8 × h + h × 10]

⇒ 170 = 80 + 8h + 10h

⇒ 90 = 18h

⇒ h = 5 cm

⇒ The length of the longest stick = diagonal of cuboid

⇒ The length of the longest stick = √(l^{2} + b^{2} + h^{2}) = √(100 + 64 + 25) = √189

⇒ The length of the longest stick = 3√21 cm

∴ The length of the longest stick that can be fitted inside the cuboid is 3√21 cm.Option 1 : 100 cm^{2}

Volume of a cube = side^{3} = 15^{3} = 3375 cm^{3}

Volume of a cuboid = Volume of a cube – 175

Area of base × h = 3375 – 175

Area of base = 3200/32 = 100 cm^{2}

Option 3 : 7 cm

**Given:**

The volume of cuboid = 756 cm^{3}

The area of the base = 108 cm^{2}

**Formula used:**

The volume of a cuboid = area of base × height

**Calculation:**

Let the height of the cuboid be ‘h’.

Height of cuboid

⇒ 756 = 108 × h

⇒ h = 756/108

⇒ h = 7 cm

Option 3 : 18

Side of the square = 7 cm

Area of each small square which cut out from the plate = 0.25

Side of the each smaller square which cut out from the plate = √0.25 = √(0.5 × 0.5) = 0.5 cm

If remaining plate folded along the cuts to form a cuboid, then

Length of the cuboid = 6 cm

Breath of the cuboid = 6 cm

Height of the cuboid = 0.5 cm

As we know, volume of the cuboid = lbh

Volume of the cuboid = 6 × 6 × 0.5 = 18 cmOption 3 : 25 metres

**Given:**

Length = 16 meters

Breadth = 12 meters

Height = 15 meters

**Concept: **

Length of the largest rod that can be put in a room is equal to the length of the diagonal of that

room.

**Formula used:**

Diagonal of a cuboid = √(l^{2} + b^{2} + h^{2})

Where l, b, and h are respectively the length, breadth, and height of a room

**Calculation:**

Length of longest rod = √(16^{2} + 12^{2} + 15^{2})

⇒ Length of longest rod = √(256 + 144 + 225) = √625

∴ Length of longest rod = 25 cm

Option 2 : 516

**Given:**

Length = 15 cm

Breadth = 8 cm

Height = 6 cm

**Concept used:**

Total Surface area = 2 × (lb + bh + hl)

l → length, b → breadth , h → height

**Calculation:**

Total surface area = 2 × (15 × 8 + 8 × 6 + 6 × 15)

Total surface area = 2 × 258 = 516

**∴ The surface area is 516 cm ^{2}.**

Option 4 : 104

Side of each cube = 2 cm

Six cubes are joined end to end, then length of newly formed cuboid = 2 × 6 = 12 cm

Breadth of newly formed cuboid = 2 cm

Height of newly formed cuboid = 2 cm

As we know,

Total surface area of cuboid = 2 (lb + bh + hl)

⇒ 2 (12 × 2 + 2 × 2 + 2 × 12)

⇒ 2 (24 + 4 + 24)

⇒ 2 × 52

⇒ 104 cm^{2}

Option 2 : 4√138

Given:

Volume of cuboid = 17920 cm^{3}

Formula used:

Volume of cuboid = l × b × h

where, l is Length of cuboid

b is Breadth of cuboid

h is Height of cuboid

Diagonal of cuboid = √(l^{2} + b^{2} + h^{2})

Solution:

Let the length, breadth and height of the cuboid be 8x, 5x and 7x respectively

8x × 5x × 7x = 17920

⇒ x^{3} = 17920/280 = 64

⇒ x = 4

Length of cuboid = 8 × 4 = 32 cm

Breadth of cuboid = 5 × 4 = 20 cm

Height of cuboid = 7 × 4 = 28 cm

**∴ Diagonal of cuboid = √(32 ^{2} + 20^{2} + 28^{2}) = 4√138 **

Option 3 : 75200

**Formula used:**

Total surface area of cuboid = 2(lb + bh + hl)

Here l is length, b is breadth and h is height

**Calculation:**

The cuboid is cut from the middle of each face to get 8 identical cuboids

Total surface area of each small cuboid = 2(50 × 40 + 40 × 30 + 30 × 50)

⇒ Total surface area of each small cuboid = 2× (4700) = 9400

Total surface area of 8 small cuboids = 8 × 9400 = 75200

Option 1 : 10

**Given :**

Dimensions of the room are 10 cm, 8 cm and 6 cm respectively

**Formula used :**

Length of the longest diagonal (rod) = √[(length)^{2} + (breadth)^{2} + (height)^{2}]

**Calculations :**

According to the the question

x√2 = √[(10)^{2} + (8)^{2} + (6)^{2}]

x√2 = √(100 + 64 + 36)

⇒ x√2 = √200

⇒ x = 10

**∴ the value of x will be 10 **