Aptitude - Area - Discussion

I wish to elaborate this further.

It is given that Tank is Long, Wide and Deep. It leads to assumption that Tank is Cuboid or Rectangular Prism.

It is not given that Cuboid is in shape of Rhombus or Parallelogram. Therefore, due to benefit of doubt, we assume it to be Rectangular Cuboid.

Following is diagram:-

__.
W |\__\ H.
\|__|.
L.

Length = 25 m.
Width = 12 m.
Height = 6 m.

There are total four walls. Area of four walls will not be equal, because Cuboid 's Length? Cuboid 's Width.

Therefore, 4*(Length*Height) is NOT possible.

Out of four walls, two walls lie opposite to each other with same dimensions. Therefore, its area will be equal.

Area of first two similar walls = 2*(Length*Height)....equation (1).

Remaining two walls will lie opposite to each other with same dimensions. Therefore, its area will be equal to each other, but different to equation (1).

Area of other two similar walls = 2*(Width*Height)....equation (2).

From equation (1) and equation (2),

Following will be true:-

Area of four walls = 2*(Length*Height)+2*(Width*Height).

= 2*Height*(Length+Width).
= 2*6*(25+12).
= 2*6*37.
= 444 sq.m.

Area of Bottom.

= Length*Width.
= 25 x 12.
= 300 sq.m.

Area of Walls and Bottom = 444+300 = 744 sq.m.

=> Cost of plastering = 744 sq.m. x Rs. 0.75 = Rs.558.

Forget the perimeter. the tank depth is 6m and it surrounded by 4 walls.
the two walls is 6x12 and another two walls is 6 x 25 in size.
So total wall area is= (2x6x12)+(2x6x25) = 444 sq m.
Now 4 walls completed plastering.
you have only the bottom side.

Area of bottom= 12x25 = 300 sq.m.
So Total area = 300 + 444 = 744 sq.m.
Now total cost is = 744 x 0.75 RS..

Draw a 3d sketch of a hollow rectangular pipe you will understand!

The tank is open at the top so it has 5 sides. We have to calculate the area of the 5 sides including the bottom.

We have length l, breadth b and height h.

Area of bottom part = l*b.

Now for those remaining 5 sides the area will include h. So area of the remaining four sides is:

(l*h)+(b*h)+(l*h)+(b*h).

= 2(l*h)+2(b*h).

= 2(l+b)*h.

So total area of 5 sides is 2(l+b)*h+l*b.

i agree with mahesh plastering is done to the tank perimeter not area.... this is not filling water , which occupies area it is similar to fencing the rectangular field..

so i think area to be plastered is the perimeter
2(l+b)*h...

Area of 4 walls of a room = 2 (Length + Breadth) x Height.(Standard Formula).

Here, 2(l+b)*h for four walls, and the bottom is extra.

So here we add the area of bottom.

[2(l+b)*h]*(l*b)...

I guess This will help!!

Formula for four walls is 2(l+b)*h so we have to multiply by h this is for four walls but here it is given as tank so v huv to cover the bottom area also so (l*b).

Hence to find we have to add both 2(l+b)h +(l*b).

Hi Ben,

In the question, it is 75 Paise per sq. m and the option are in Rupees. So if you divide by 100 to convert 75 Paise to Rs i.e 0.75 Rs.

Therefore the answer is 744 sq.m. x Rs. 0.75 = Rs.558.

Thanks

@All

The cost of plastering its 'walls' and 'bottom'.

Finding area:

For walls = [2(l + b) x h]

For bottom = (l x b)

So, total area to be plastered = [2(l + b) x h] + (l x b)

I think you got it.

@malar

2(l+b) is only the perimeter of the rectangle (2D) with only length and breadth and no thickness / depth.

Here it is the wall of the tank with depth of 6 m.

Hope you got the idea.

How do you know the tank is of rectangular shape, since it is not mentioned in the question. It can be of any other shape. What is the reason behind assuming it a rectangular shape.