1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine
(I) The area of the sheet required for making the box.
(ii) The cost of the sheet for it, if a sheet measuring 1m2 costs Rs. 20.
Given: length (l) of box = 1.5m
Breadth (b) of box = 1.25 m
Depth (h) of box = 0.65m
(i) Box is to be open at the top.
Area of sheet required.
= 2lh+2bh+lb
= [2×1.5×0.65+2×1.25×0.65+1.5×1.25]m2
= (1.95+1.625+1.875) m2 = 5.45 m2
(ii) Cost of sheet per m2 area = Rs.20
Cost of sheet of 5.45 m2 area = Rs (5.45×20)
= Rs.109.
2. The length, breadth and height of a room are 5 m, 4 m and 3 m, respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs 7.50 per m2.
Ans:
Length (l) of room = 5m
Breadth (b) of room = 4m
Height (h) of room = 3m
It can be observed that four walls and the ceiling of the room are to be whitewashed.
Total area to be whitewashed = Area of walls + Area of the ceiling of the room
= 2lh+2bh+lb
= [2×5×3+2×4×3+5×4]
= (30+24+20)
= 74
Area = 74 m2
Also,
Cost of whitewash per m2 area = Rs.7.50 (Given)
Cost of whitewashing 74 m2 area = Rs. (74×7.50)
= Rs. 555
3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15,000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area.]
Ans:
Let the length, breadth, and height of the rectangular hall be l, b, and h, respectively.
Area of four walls = 2lh+2bh
= 2(l+b)h
Perimeter of the floor of hall = 2(l+b)
= 250 m.
Area of four walls = 2(l+b) h = 250h m2
Cost of painting per square metre area = Rs.10
Cost of painting 250h square metre area = Rs (250h×10) = Rs.2500h.
However, it is given that the cost of painting the walls is Rs. 15,000.
15000 = 2500h
Or h = 6
Hence, the height of the hall is 6 m.
4. 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?
Ans:
Total surface area of one brick = 2(lb +bh+lb)
= [2(22.5×10+10×7.5+22.5×7.5)] cm2
= 2(225+75+168.75) cm2
= (2×468.75) cm2
= 937.5 cm2
Let n bricks can be painted out by the paint of the container.
Area of n bricks = (n×937.5) cm2 = 937.5n cm2
As per the given instructions, the area that can be painted by the paint of the container = 9.375 m2 = 93750 cm2
So, we have 93750 = 937.5n
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.
4. 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?
Ans:
Total surface area of one brick = 2(lb +bh+lb)
= [2(22.5×10+10×7.5+22.5×7.5)] cm2
= 2(225+75+168.75) cm2
= (2×468.75) cm2
= 937.5 cm2
Let n bricks can be painted out by the paint of the container.
Area of n bricks = (n×937.5) cm2 = 937.5n cm2
As per the given instructions, the area that can be painted by the paint of the container = 9.375 m2 = 93750 cm2
So, we have 93750 = 937.5n
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.