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

A student asked me to explain why statement 2 is sufficient. So,......

Target question: What is the volume of gasoline in the tank?

Statement 2: The top surface of the gasoline forms a rectangle that has an area of 24 square feet.
We know that the tank is 6 ft long
Let x = the width of the surface
We have the following:



Area of rectangle = (base)(height)
So, we can write: 24 = (6)(x)
Solve get: x = 4
So, the diagram looks like this:


IMPORTANT: For geometry Data Sufficiency questions, we’re typically checking to see whether the statements "lock" a particular angle, length, or shape into having just one possible measurement. This concept is discussed in much greater detail in the following video: https://www.gmatprepnow.com/module/gmat ... /video/884

So we want to determine whether there is exactly one tank size such that when the gasoline is 2 feet deep, the surface of the gasoline is 4 feet wide.

So let's start with some random tank like this...

... and fill it with gasoline to a depth of 2 feet

When we measure the width of the surface, we get 3 feet

So, this particular tank is too small.

Let's try slightly bigger tank. Here's one.

Still too small.

As we can see, if we gradually make the tank bigger and bigger and bigger, there will be EXACTLY ONE tank such that the width of the surface is 4 feet.


Since there is EXACTLY ONE tank that meets the given conditions, we know that statement 2 is SUFFICIENT

Answer: D

ASIDE: Some people might be wondering, "Sure, but what is the actual volume of the gasoline in the tank?"
Fortunately, we don't need to find the actual volume. We need to only demonstrate that we COULD find the volume.
One way to do this would be to use geometric properties and formulas. But we can also find the volume in other ways.
For example, if I had very precise measuring equipment and a way to construct cylindrical tanks of any shape, then I COULD keep testing actual tanks until I find one that meets the given conditions. Then, I'd fill that tank with gasoline to a depth of 2 feet. Then I'd pour that gasoline into a very precise measuring cup. Done!!

Cheers,
Brent

Solution:

Question Stem Analysis:

We need to determine the volume of gasoline in the tank, given that 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, and the tank is filled with gasoline to a depth of exactly 2 feet.

Notice that when the tank is resting vertically with the circular base on the ground (like an upright soda can), the height of the tank is 6 feet.

Statement One Alone:

Since the diameter of the tank is 4 feet, which is exactly twice the depth of 2 feet currently of the gasoline filled inside the tank, we see that the tank is actually half full. Since the diameter of the tank is 4 feet, its radius is 2 feet. Since in the question stem analysis, we determined the height of the tank is 6 feet, we can determine the full capacity of the tank. Since the gasoline is filled to half capacity, dividing the full capacity of the tank by 2 will yield the volume of the gasoline in the tank. Statement one alone is sufficient.

Statement Two Alone:

We see that the rectangle (i.e, the top surface of the gasoline) has a width of 4 feet since the area is 24 square feet and its length is 6 feet. The width of the rectangle is actually a chord of the circular base of the tank. Let the distance between the center of the circle and this chord be x. Recall that we are told that the depth of the gasoline is 2 feet, therefore, the radius of the circular base is x + 2 if the center is above the gasoline level and 2 - x if the center is below the gasoline level. In the former case, a right triangle is formed where the legs have length 2 and x, and where the hypotenuse (which is the radius) has length x + 2. Using the Pythagorean theorem, we obtain:

(x + 2)^2 = x^2 + 2^2

x^2 + 4x + 4 = x^2 + 4

4x = 4

x = 0

In this case, we see that the distance between the center and the gasoline level is 0, which means that the tank is actually half full and the diameter of the circular ends is 4.

Assuming the level of the gasoline is above the center, we can proceed as above and solve (2 - x)^2 = x^2 + 2^2, which also yields x = 0. Once again, the tank is half full and the diameter is 4.

We see that in either case, we have enough information to determine the volume of gasoline in the tank.

Answer: D

Answer: Option D

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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 \(2^2+(2\sqrt{3}-r)^2=r^2\)

.

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

Hi reto
here it is

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.

Wait, I don't understand how you need any more information at all. Given,that the length of the cylinder is 6ft and height of the water is 2, won't the water take the shape of half of a cylinder with radius 2 ? After that you don't need any more information regarding the volume of water, right ?

You need to know the radius of the cylinder and how the radius stacks up against the depth of the water in the cylinder. Without this information, you will not know if the statements are sufficient or not.

Take a look at: a-tank-is-filled-with-gasoline-to-a-depth-of-exactly-2-feet-the-tank-202262.html#p1553082

Your statement is correct if a statement tells you the radius = height of the water but if radius \(\neq\) height of the water, you will not be able to apply any volume formulae.

Hope this helps.

Anytime you have the height of a liquid and the chord that it's surface makes, we can find the radius. In this case, if you understand that statement two gives us a chord, you can confirm it's the radius by using this formula:

(1/2*chord)(1/2*chord) = (height of liquid)(distance from surface to top of cylinder)

In this case 2*2=2*(distance from surface to the top of the cylinder) and we see that the distance from the surface of the gasoline to the top of the cylinder is 2, making the surface of the gasoline right in the middle of the cylinder.

So if the chord were length of 4 and the height of the liquid were 1, you could still use that formula to find the radius of 2.5. I'm not sure what you would do with it, but it's easy to get at should you find it necessary on test day.

Hello everyone

Could someone explain to me why statement II is sufficient? If we look at the cut that the rectangle is doing on the cylinder it could happen either under the centre or above it. In both cases it'll respect the premises but have completely different volumes of gasoline in it....

I'm really scratching my head with this!

No. It will be at the centre only. Note that the height is 2 and diameter is 4 which means the radius is 2. Hence it will be a semi circle only.

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.

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.

Hi Brent, thank you for the excellent explanation. BrentGMATPrepNow

One question: if we keep the dimensions of the top layer (6 x 4) the same and gradually make the can larger, the level would drop. Would this change the volume?
For instance, if we look at a can with a diameter of 100, the amount of gas would be very low within that can, but the top layer would still be 6 x 4 with a depth of 2.