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Retired Moderator V
Joined: 22 Jun 2014
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Concentration: General Management, Technology
GMAT 1: 540 Q45 V20 GPA: 2.49
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An enclosed rectangular tank with dimensions 2 meters by 3 meters  [#permalink]

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1
2 00:00

Difficulty:   15% (low)

Question Stats: 83% (01:31) correct 17% (01:58) wrong based on 97 sessions

### HideShow timer Statistics An enclosed rectangular tank with dimensions 2 meters by 3 meters by 4 meters is filled with water to a depth of 1 meter as shown by the shaded region in the figure above. If the tank is turned so that it rests on one of its smallest faces, the depth, in meters, of the water will be

(A) $$\frac{2}{3}$$
(B) $$1$$
(C) $$\frac{4}{3}$$
(D) $$\frac{5}{3}$$
(E) $$2$$

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Attachment: Tank.JPG [ 14.93 KiB | Viewed 1088 times ]

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GMAT 1: 630 Q48 V27 An enclosed rectangular tank with dimensions 2 meters by 3 meters  [#permalink]

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2
total volume of water at present=L*B*H = 4*2*1 =8
When standing on one of shortest sides, volume=2*3*H = 6*H.
Now 6H=8
So, H = 4/3

Hence, C is the answer.
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An enclosed rectangular tank with dimensions 2 meters by 3 meters  [#permalink]

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anurag16589 wrote:
total volume of water at present=L*B*H = 4*2*1 =8
When standing on one of shortest sides, volume=2*3*H = 6*H.
Now 6H=8
So, H = 4/3

Hence, C is the answer.

hi anurag16589 what concept do you apply ? do you have any links ? I am bad at those volume problems may be you can give some advice ganand Senior PS Moderator V
Joined: 26 Feb 2016
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Location: India
GPA: 3.12
An enclosed rectangular tank with dimensions 2 meters by 3 meters  [#permalink]

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1 Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

The various faces on which the cuboid can be turned are 2*3(6) or 3*4(12) since the cuboid is
already on its third face which is 2*4(8). The smallest face towards which the tank is moved
has an area of 6 meter^2.

Therefore, the depth of water is meter in the cube can be measured as $$\frac{4}{3}$$ meter(Option C)
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An enclosed rectangular tank with dimensions 2 meters by 3 meters  [#permalink]

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pushpitkc wrote: Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

The various faces on which the cuboid can be turned are 2*3(6) or 3*4(12) since the cuboid is
already on its third face which is 2*4(8). The smallest face towards which the tank is moved
has an area of 6 meter^2.

Therefore, the depth of water is meter in the cube can be measured as $$\frac{4}{3}$$ meter(Option C)

pushpitkc I omitted this step / couldnt see it ---> Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

but i did calculate the are of large rectangle which is 8

and area of smallest rectangle which is 6

so al we have to do is to write ratio the areas of above mentioned rectangles ? if So

my question IS why you wrote 4/3 and not 3/4, how should i figure out this - to write 4/3 or 3/4 ?

i hope you can take time to reply to my post thank you Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3386
Location: India
GPA: 3.12
Re: An enclosed rectangular tank with dimensions 2 meters by 3 meters  [#permalink]

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1
Hey dave13

Once we have established the volume of water contained in the rectangular tank is 8 m^3
and the smallest face on which the rectangular tank is turned is 2*3 or 6 m^2, in order to
understand how high the water is the calculations will be as follows

6m^2 * height of water(in m) = 8m^3 (volume of water we need to fit in the vessel)

Now the height in meters is $$\frac{8}{6} = \frac{4}{3}$$

Hope that helps!
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Re: An enclosed rectangular tank with dimensions 2 meters by 3 meters  [#permalink]

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1
dave13 wrote:
pushpitkc wrote: Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

The various faces on which the cuboid can be turned are 2*3(6) or 3*4(12) since the cuboid is
already on its third face which is 2*4(8). The smallest face towards which the tank is moved
has an area of 6 meter^2.

Therefore, the depth of water is meter in the cube can be measured as $$\frac{4}{3}$$ meter(Option C)

pushpitkc I omitted this step / couldnt see it ---> Since the depth of water is 1 meter, the volume of water contained is 4*2*1 = 8 meter^3

but i did calculate the are of large rectangle which is 8

and area of smallest rectangle which is 6

so al we have to do is to write ratio the areas of above mentioned rectangles ? if So

my question IS why you wrote 4/3 and not 3/4, how should i figure out this - to write 4/3 or 3/4 ?

i hope you can take time to reply to my post thank you Hi dave13,

I think you are confused with the application of Area and Volume.

In the above problem, the quantity(volume ) of water is going to be constant. Hence we equate the volume.

In the Original Diagram, we have the following dimensions:

Height = 3 cm but the height of the water is only 1 cm, hence consider height as 1 cm. Length = 2 cm and Width = 4 cm.

Volume = 2*4*1 = 8 $$cm^3$$ ---- (1)

After rotation, we have the following diagram. Please note that the water level is not depicted in the diagram.

Attachment: rotated tank1.png [ 17.19 KiB | Viewed 731 times ]

After rotation we have the following dimensions:

Height = 4 cm but the height of water level is unknown, hence assume as $$h$$ cm. Length = 3 cm, and Width = 2 cm.

The volume of water =$$h*3*2 = 6h$$ --- (2)

By equating (1) and (2) we have $$h = \frac{4}{3}$$cm.

Hope this helps.

Please go through the following article:

https://study.com/academy/lesson/differ ... olume.html

Thank you.
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Re: An enclosed rectangular tank with dimensions 2 meters by 3 meters  [#permalink]

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ganand and pushpitkc thanks a lot  Re: An enclosed rectangular tank with dimensions 2 meters by 3 meters   [#permalink] 29 Nov 2018, 01:03
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# An enclosed rectangular tank with dimensions 2 meters by 3 meters

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