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A tank is filled with gasoline to a depth of exactly 2 feet. The tank

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A tank is filled with gasoline to a depth of exactly 2 feet. The tank  [#permalink]

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New post 08 Apr 2018, 02:45
ENGRTOMBA2018 wrote:
reto wrote:
iamschnaider wrote:
A tank is filled with gasoline to a depth of exactly 2 feet. The tank is a cylinder resting horizontally on its side, with its circular ends oriented vertically. The inside of the tank is exactly 6 feet long. What is the volume of gasoline in the tank?

(1) The inside of the tank is exactly 4 feet in diameter.

(2) The top surface of the gasoline forms a rectangle that has an area of 24 square feet.

From OG 2016 (question 86 DS)


I get why 1 works, but the main qualm I have about 2 is how we can be sure that the 4 feet resulting from the square is a diameter of the circle ie. how we know that it fills the tank exactly up to halfway.


Could someone draw this for Statement 2? I don't get it... :(


reto, see the attached picture.

The trick with statement 2 is that H = 0 when you calculate the value of H from the 2 equations:

\(H^2 + 2^2 = R^2\)

and

R = 2+H , you get H =0

This means that the depth of the gasoline in the cylinder = radius of the cylinder.

Thus, it is sufficient to answer the question.

FYI, we need to know about the radius as without the depth of gasoline = radius of the cylinder, it will be difficult to calculate the volume of the gasoline in the tank.

Hope this helps.


Hi Bunuel VeritasPrepKarishma

My way of thinking is if liquid level is below the centre of the circle then the H= (r-2) ;(Radius from the centre of the circle less 2 feet ) then the equation becomes
(r-2)^2 +2^2 = r^2

and if the liquid is filled till above the centre of the circle then H=(2-r), equation becomes
(2-r)^2+2^2 = r^2.

Though using either of the equation i am getting r=2

Plz correct if i am going wrong somewhere with my understanding. Thanks
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Re: A tank is filled with gasoline to a depth of exactly 2 feet. The tank  [#permalink]

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New post 09 Apr 2018, 02:08
cruiseav wrote:
Hi Bunuel VeritasPrepKarishma

My way of thinking is if liquid level is below the centre of the circle then the H= (r-2) ;(Radius from the centre of the circle less 2 feet ) then the equation becomes
(r-2)^2 +2^2 = r^2

and if the liquid is filled till above the centre of the circle then H=(2-r), equation becomes
(2-r)^2+2^2 = r^2.

Though using either of the equation i am getting r=2

Plz correct if i am going wrong somewhere with my understanding. Thanks


Yes, your logic is correct. Though, as I mentioned in my solution, this calculation is not required.
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New post 16 Apr 2018, 11:29
I'm not sure if this is the correct place to ask, but if the depth was let's say 3ft instead of 2. Would statement 2 still be sufficient?
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New post 17 Apr 2018, 00:26
1
ahawkins wrote:
I'm not sure if this is the correct place to ask, but if the depth was let's say 3ft instead of 2. Would statement 2 still be sufficient?


Yes, it would be sufficient. 3 points uniquely define a circle i.e. you can make only one circle with given 3 distinct points. It is the circumcircle of the triangle drawn by connecting the 3 points.
So no matter what the dimensions given to you, the dimensions are uniquely defined.
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New post 16 May 2018, 02:43
As mentioned in the question stem, the cylinder is lying horizontally and depth or height of the is 2 ft( in horizontal position) , actual height of the cylinder is 6 ft.
so, volume of the tank is 3.14* R^2* 6

S1) diameter is 4ft .
so, R= 2ft and as mentioned depth or height of the is 2 ft( in horizontal position) so volume of the fluid can be derived from here. so its sufficient.

S2) upper surface of fluid forms a rectangle of area 24.
as , h = 6 so A*6= 24, A= 4;
A/2 =2 which is depth or height of the is 2 ft( in horizontal position) so volume of the fluid can be derived from here. so its sufficient.

So, answer id D.
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New post 21 Mar 2019, 06:31
Let´s suppose the water filet is on the center of the cylinder.
Than the water´s height (2 feet) would be the radius of the supposed cylinder, consequently the diameter would be be 4 feet.
Which by the way is the only possible value for water filet´s width.
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New post 12 May 2019, 11:30
chetan2u wrote:
iamschnaider wrote:
A tank is filled with gasoline to a depth of exactly 2 feet. The tank is a cylinder resting horizontally on its side, with its circular ends oriented vertically. The inside of the tank is exactly 6 feet long. What is the volume of gasoline in the tank?

1) The inside of the tank is exactly 4 feet in diameter.

2) The top surface of the gasoline forms a rectangle that has an area of 24 square feet.

From OG 2016 (question 86 DS)


I get why 1 works, but the main qualm I have about 2 is how we can be sure that the 4 feet resulting from the square is a diameter of the circle ie. how we know that it fills the tank exactly up to halfway.


Hi,
i agree with you however we still can find the answer ..
we can find the chord/diameter of the circular end as 4..
now we have depth as 2, length as 6 and width as 4 but the volume will vary depending on what the dia would be..
there could be two scenarios..
1) where the center is below the chord... that is greater than 2\(\sqrt{3}\)..
eq becomes \(2^2+(2\sqrt{3}+r)^2=r^2\)
r comes out negative so not possible.
2) where the center is above the chord... that is greater than 2\(\sqrt{3}\).. r can be found so suff..
eq becomes \(2^2+(2\sqrt{3}-r)^2=r^2\)

.


Hi chetan2u

Though i understood the logic i am confused at the first place as how u got 2√3 in the beginning..

Seems like a silly doubt but i am not able to catch it

Thanks..
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A tank is filled with gasoline to a depth of exactly 2 feet. The tank   [#permalink] 12 May 2019, 11:30

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