Aptitude - Volume and Surface Area - Discussion
Discussion Forum : Volume and Surface Area - General Questions (Q.No. 9)
9.
A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:
Answer: Option
Explanation:
| Area of the wet surface | = [2(lb + bh + lh) - lb] |
| = 2(bh + lh) + lb | |
| = [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] m2 | |
| = 49 m2. |
Discussion:
40 comments Page 2 of 4.
Kailash singhania said:
8 years ago
@Sandya.
1m and 25 cm is actually converted into meter which will be 1.25 m.
1m and 25 cm is actually converted into meter which will be 1.25 m.
Samim Ali said:
9 years ago
Thanks for the explanation.
Sandya said:
9 years ago
Can anyone explain how did you get 1.25?
Narinder said:
9 years ago
@Vyom:
Question is just about the wet area. If there is any question regarding total height then it Total length or something should be mentioned accordingly.
Question is just about the wet area. If there is any question regarding total height then it Total length or something should be mentioned accordingly.
Vyom sharma said:
9 years ago
The side walls are also not fully wet, as everybody said that the tank is not fully filled with water.
We don't know the total height of cistern, we just know how much depth of water is cistern containing.
Then, the area of side walls must NOT be fully included as a wet area.
Then How we conclude this?
We don't know the total height of cistern, we just know how much depth of water is cistern containing.
Then, the area of side walls must NOT be fully included as a wet area.
Then How we conclude this?
(1)
Chirag Barad said:
10 years ago
Because there is ask only about wet surface, not total surface.
Piyush said:
10 years ago
How do we know upper part not containing height?
(1)
Rakhshanda Khan said:
1 decade ago
1.25 is height and 6 m is length of the tank. So we must take whole 6 m length while we are calculating it. Because whole lower most part of the tank is wet. That's why we are taking whole 6 m.
Naveen said:
1 decade ago
Water is only at 1.25 m depth. Then why I have to take whole 6 m to calculate the wet surface?
Mounica said:
1 decade ago
What about the height of this cuboid?
It is possible to answer this question by assuming the height difference between water level and cuboid top surface as a smaller value, now the area of top surface can be deducted.
Considering each plane of cuboid total area is lb+bh+lh+lb+bh+lh and top surface is area is lb, by subtracting this lb the total wet surface = 2(6x4+4X1.25+1.25x6) - (6x4) = 49 sq units.
It is possible to answer this question by assuming the height difference between water level and cuboid top surface as a smaller value, now the area of top surface can be deducted.
Considering each plane of cuboid total area is lb+bh+lh+lb+bh+lh and top surface is area is lb, by subtracting this lb the total wet surface = 2(6x4+4X1.25+1.25x6) - (6x4) = 49 sq units.
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