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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.

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

Hi Brent,

Thank you for the detailed images. However, I still need some help understanding why it is not possible for the width=4 to be below/above halfway mark, and still have depth=2. For instance in the below image, is it not possible?



Thank you!
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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BrentGMATPrepNow
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BrentGMATPrepNow
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

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.

Every time we change the diameter of the tank, the width of water (when the tank is 2 in deep) changes.
For example, if we take a tank with diameter 100 and fill it to a depth of 2 feet, the width of the water will be 28 ft (not 4 feet)

BrentGMATPrepNow - how did you calculate 28 feet in yellow ?
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BrentGMATPrepNow - how did you calculate 28 feet in yellow ?

Start here:


Apply Pythagorean Theorem to get x = 14
So, the other side has length 14 as well

14 + 14 = 28
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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

Hi BrentGMATPrepNow - Just wanted to confirm three quick points based on the above 'lock' mentioned in the yellow specifically

(i) if i understand, what you are saying is -- given the length of the cylinder is constant (=6 feet). Really the only change one can make is on the diameter of the cylinder.

The larger the diameter, for a given height of 2 feet of gasoline -- the depth of the gasoline is constantly going to change

So, a cylinder of 6 feet length, gasoline height of 2 feet -- if you want a depth of 4 feet.... there is only one diameter of a cylinder that will work (as one can see, the longer the diameter of the cylinder, for a given height of 2 feet ) --> that is 'lock' you are mentioning above

(ii) This 'lock' was only true as long as the cylinders length of 6 feet is constant, correct ?

What if the cylinders length of 6 feet WAS NOT constant ?

I dont think this 'lock' strategy would have worked.

Just confirming

(iii) How are you so sure from your 4 circles above -- the 'lock' was when the diameter of the cylinder was 4 feet specifically ?
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BrentGMATPrepNow
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

Hi BrentGMATPrepNow - Just wanted to confirm three quick points based on the above 'lock' mentioned in the yellow specifically

(i) if i understand, what you are saying is -- given the length of the cylinder is constant (=6 feet). Really the only change one can make is on the diameter of the cylinder.

The larger the diameter, for a given height of 2 feet of gasoline -- the depth of the gasoline is constantly going to change

So, a cylinder of 6 feet length, gasoline height of 2 feet -- if you want a depth of 4 feet.... there is only one diameter of a cylinder that will work (as one can see, the longer the diameter of the cylinder, for a given height of 2 feet ) --> that is 'lock' you are mentioning above

(ii) This 'lock' was only true as long as the cylinders length of 6 feet is constant, correct ?

What if the cylinders length of 6 feet WAS NOT constant ?

I dont think this 'lock' strategy would have worked.

Just confirming

(iii) How are you so sure from your 4 circles above -- the 'lock' was when the diameter of the cylinder was 4 feet specifically ?

My responses are....

i) in order for the top (rectangular) surface to have an area of 24 square feet, the width must be 4 feet (since the length is already 6 feet)

ii) Yes, the lock is only possible because the cylinder's length is locked in at 6 feet.
If the length of the cylinder is not locked in, the strategy still works. We just get a different answer.
If the length of the cylinder is not locked in, then neither statement is sufficient alone. In fact, the correct answer would be E.
In other words, if we don't know the height of a cylinder, there's no way we can ever find the volume of that cylinder (or, in this case, the volume of gasoline in that cylinder)

iii) I'm not actually saying the lock occurs when the diameter of the cylinder is 4 ft (Yes, it turns out that the lock does occur when the diameter is 4 ft, but that part plays no role in my solution). I'm just saying that there exists only one measurement of the diameter that satisfies statement 2. In other words, I know that one (and only 1) a lock occurs, which means statement 2 must be sufficient.
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KarishmaB while solving this question, I realized the following. Please correct me if I am wrong:

Diameter = 2 x radius so from (2) we get the depth as "2" and the width has "4" and in a circle, the diameter is the ONLY chord whose half-length can be drawn? In other words, the diameter is the ONLY chord from which we can draw a line from its center touching the circumference of the circle.


In other words, if I know that "2" and "4" are the measurements given I cannot have a chord other than the diameter that can be twice of "2"?

Also, I really liked what you and a few other experts did for (2). You guys just "fixed" the semi-circle that is formed from "4" and "2" thus realizing that the area of this semicircle can be found. Then we multiply it with length 6 and arrive at our answer :)
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KarishmaB while solving this question, I realized the following. Please correct me if I am wrong:

Diameter = 2 x radius so from (2) we get the depth as "2" and the width has "4" and in a circle, the diameter is the ONLY chord whose half-length can be drawn? In other words, the diameter is the ONLY chord from which we can draw a line from its center touching the circumference of the circle.


In other words, if I know that "2" and "4" are the measurements given I cannot have a chord other than the diameter that can be twice of "2"?

Also, I really liked what you and a few other experts did for (2). You guys just "fixed" the semi-circle that is formed from "4" and "2" thus realizing that the area of this semicircle can be found. Then we multiply it with length 6 and arrive at our answer :)

Hoozan - From what I understand, you are saying that you realised it is a semi circle and hence you realised that you can calculate the area and then the volume. In this question, it works.
But do note that even if it were not a semi circle, it would still be fully defined. Make a straight line of say 3 cm. Now at its centre, make a perpendicular of say 1 cm downwards. Now make an arc combining the three points. Is the area of this segment defined? or can you make it have variable area? There is only one way to make the arc around it and hence the area is defined. That is what we needed to know for our DS question.
Whether it is a semi circle or not doesn't have anything to do with it.
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Apparently half the width of the rectangle(here 2) will be equal to the depth(here 2) only when the liquid is half-filled i.e., the width of the rectangle is exactly in line with the diameter.
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KarishmaB ThatDudeKnows avigutman none of the explanations are providing me clarity on how we are sure that 4 has to be a diameter. It could simply be a chord too with a depth of two feet. Pls clarify intuitively
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KarishmaB ThatDudeKnows avigutman none of the explanations are providing me clarity on how we are sure that 4 has to be a diameter. It could simply be a chord too with a depth of two feet. Pls clarify intuitively

Draw a line of 4 cm (since rectangle's side length is found to be 4). From its centre, make a line down of 2 cm. This is given since it is the height. Now can you draw the arc around these 3 points such that you do not get a semi circle? Can you make a chord of length 4 such that its mid point is at a distance of 2 away from the circle but the chord is not a diameter? Try making this with a compass.
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KarishmaB ThatDudeKnows avigutman unfortunately I will not have a compass in the exam. Please tell another way to see intuitively. I would imagine just any circle can be formed with 4 as a chord and depth as 2 since the circle becomes thinner below the half mark and above the half mark
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KarishmaB ThatDudeKnows avigutman unfortunately I will not have a compass in the exam. Please tell another way to see intuitively. I would imagine just any circle can be formed with 4 as a chord and depth as 2 since the circle becomes thinner below the half mark and above the half mark

Elite097

Two ways to think about this in addition to those already discussed.

First, look at Figure 1. I'll assume that by this point in looking at the question, you understand where I got the horizontal line of length 4 and the vertical line of length 2. The vertical one bisects the horizontal one, dividing the horizontal one into two segments each of length 2. Can you draw a circle that connects points A, B, and C? Sure, it will have its center where the two lines intersect and the radius will be 2. Can you draw more than one circle? Nope. If you can only draw one circle, you have enough information to answer the question.

Second, look at figure 2. We have three points (A, B, and C) and need them to reside on the same circle. In order for any two points to reside on the same circle, they must be equidistant from the center of the circle. Look at points A and C. In order for them to be equidistant from a point, that point must reside on the red line. So, the center of the circle is somewhere on the red line. Now look at points B and C. In order for them to be equidistant from a point, that point must reside on the green line. So, the center of the circle is somewhere on the green line. (Note that we could have also drawn a line of a third color equidistant from A and B that would be vertical and pass through C, but we don't need more than two lines.) The center of the circle must therefore be where the red and green lines cross. That is the only possible center and we will be able to find the radius by measuring the distance to any of the three points. If you can only draw one circle, you have enough information to answer the question.
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KarishmaB ThatDudeKnows avigutman unfortunately I will not have a compass in the exam. Please tell another way to see intuitively. I would imagine just any circle can be formed with 4 as a chord and depth as 2 since the circle becomes thinner below the half mark and above the half mark


See if this helps you.

Draw the chord BC with center of BC at A. The height will be from A, and say the height to be AD.

Now, AB, AC and AD are all 2, and AD will be perpendicular to BC as AD is the height.

Take triangle ACD
\(\angle ACD= \angle ADC=45\)

Similarly, from triangle ABD
Take triangle ACD
\(\angle ABD= \angle ADB=45\)

Thus, angle BDC =45+45=90.

We know that a point on the circumference will make a 90 angle with the diameter only.
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avigutman ThatDudeKnows Hello. Could you please tell me if i Would i have been able to calculate the volume of the gasoline if the Tank had a Diameter more than4 feet. Say 6 ft/8ft or less than 4 ft, say 3ft?
Or what if the height of the gasoline was above the center of the Tank?
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Just another perspective:

From Statement II, it is clear that the horizontal line that cuts the circular ends is 4 ft. Now, this horizontal line can be either the diameter or any other chord. If the latter is true, then the diameter has to be greater than 4 ft, irrespective of the position of the chord. Now this clearly contradicts Statement I. But in DS, the two statements will never contradict. Thus, the only other possibility that the horizontal line is the diameter is true.

Statement II, hence is sufficient.

Please note that here only the fact that the two statements do not contradict is used. Statement I itself is not used. So. the correct answer is NOT C.
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