Question

8.) A water container is 50 cm long, 70cm wide. What should be its height so that it can hold 6.3 litre of
water?​

Answer 1

Answer:

Here, l = 1.5 m, b = 1.25 m

∵ It is open from the top.

∵ Its surface area = [Lateral surface area] + [Base area]

= [2(l + b)h] + [l × b]

= [2(1.50 + 1.25)0.65 m2] + [1.50 × 1.25 m2]

= [2 × 2.75 × 0.65 m2] + [1.875 m2]

= 3.575 m2 + 1.875 m2 = 5.45 m2

∵ The total surface area of the box = 5.45 m2

∴ Area of the sheet required for making the box = 5.45 m2

(ii) Rate of sheet = Rs. 20 per m2

Cost of 5.45 m2 = Rs. 20 × 5.45

⇒ Cost of the required sheet = Rs.109

Q2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate off 7.50 per m2.

Sol: Length of the room (l) = 5m

Breadth of the room (b) = 4 m

Height of the room (h) = 3 m

The room is like a cuboid whose four walls (lateral surface) and ceiling are to be white washed.

∴ Area for white washing = [Lateal surface area] + [Area of the ceiling]

= [2(1 + b)h] + [1 × b]

= [2(5 + 4) } 3 m2] + [5 × 4 m2]

= [54m2] + [20m2] = 74m2

Cost of white washing:

Cost of white washing for 1 m2 = Rs. 7.50

∴ Cost of white washing for 74 m2 = Rs. 7.50 × 74

The required cost of white washing is Z555.

Q3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of 10 per m2 is Rs. 15000, find the height of the hall.

Sol: Note: Area of four walls = Lateral surface area.

A rectangular hall means a cubiod.

Let the length and breadth of the hall be ‘l’ and ‘b’ respectively.

∵[Perimeter of the floor] = 2(l + b)

⇒ [2(l + b)] = 250 m.

∵Area of four walls = lateral surface area

⇒ [2(l + b)] × h [where ’h’ is the height of hall.]

Cost of painting the four walls = Rs. 10 × 250 h = Rs. 2500 h

⇒ Rs. 2500 h = Rs. 15000

Thus, the required height of the hall 6 m

Q4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

Sol: Total area that can be painted = 9.375 m2, since a brick is like a cuboid

∴ Total surface area of a brick = 2[lb + bh + hl]

= 2[(22.5 × 10) + (10 × 7.5) + (7.5 × 22.5)] cm2

= 2[(225) + (75) + (168.75)] cm2

Thus, the required number of bricks = 100

Answer 2

Answer:

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