When a cylindrical tank is filled with water at a rate of 22

Also, with the same logic, why is the capacity 22?

Is it perhaps that since we are not told about how long this rate will be filling in the tank, we can assume that with these numbers it will be totally full?

Hi pacifist85,

Imagine if we took a big cylinder and broke it into small "sections"; we'd just have a bunch of smaller cylinders that all had the same radius.

Here, we're given information about what happens in 1 HOUR: 22 cubic meters of water enter and the water level rises 0.7 meters.

From this data, the easiest thing to do is say that we have a "section" that is 0.7 meters high; in 1 hour that section will be FILLED by 22 cubic meters of water.

Next, we need the formula for Volume of a Cylinder:

V = pi(R^2)(H)

We have the Volume (22 cubic meters) and the Height (0.7 meters)...

22 = pi(R^2)(0.7)

At this point, the "math" would get pretty ugly, BUT the answer choices (and a little estimation) provide us with a way to avoid a lot of the math...

pi = about 3.14
.7(pi) = about 2.2

22 = 2.2(R^2)
10 = R^2
\sqrt{10} = R

Final Answer:

GMAT assassins aren't born, they're made,
Rich

use the units to solve.

22 (m^3/hr) / .7 (m/hr) = (slighty over) ~30 m^2 = pi x r^2 * surface area of the rising water AND area of the cylinder

r= sqrt(~30/pi) = sqrt ( 10) * assume slighty over 3 is pi and use the above area equation to solve for approximate value of radius.

We can simply inspect the tube after one hour to determine the radius since the radius of the tube won't change over time. We know that the amount of water in the tube is equal to 22, so we can set it equal to the formula for the volume of a cylidner (piR^2H). Since the height rises by 0.7m/hr, we can set the height of the tube equal to its height after an hour.

22 = pi*r^2*H
22 = pi*r^2(7/10)
10 = r^2
\sqrt{10} = r

Let's examine what occurs in a 1-hour period

The volume of water increases by 22 cubic meters.
The height of the water increases by 0.7 meters.

So, we need to find the radius of a 0.7 meter high cylinder that has a volume of 22 cubic meters.

Volume = (pi)(radius²)(height)
22 = (pi)(radius²)(0.7)

IMPORTANT: notice that (pi)(0.7) = approximately 2.2

So, we get: 22 ≈ (2.2)(radius²)
Divide both sides by 2.2: 10 ≈ radius²
Solve: radius ≈ √10

Answer: B

Cheers,
Brent

in 1 hr
vol= 22 m3
and h= 0.7
so
22=22/7*r^2*7/10
r=√10
IMO B

The filling rates are given... Which makes it little difficult to interpret the situation.
Instead taking account of the rates just take a fixed time (says one hour) to make situation easy to understand.
In 1 hours the volume filled is 22 cubic m
And the height of this filled volume is 0.7 m
And now all you need to do is calculate the radius of cylinder or the volume filled (both of which are same)
V = πR²H
22 = π*R²*0.7
(22/7)*10 =π *R²
R² = 10
R = √10
IMO B

Posted from my mobile device

We can let r = the radius of the tank, in meters. We know that the volume of the “slice” of the cylindrical tank that is filled in one hour is 22 cubic meters. We also know that the formula for the volume of that cylindrical “slice” is π x r^2 x .0.7, and we can create the equation:

π x r^2 x 0.7 = 22

Now let’s use 22/7 as the approximation for π:

22/7 x r^2 x 0.7 = 22

22 x r^2 x 0.1 = 22

r^2 x 0.1 = 1

r^2 = 10

r = √10

Answer: B

To make it easier, assume the height of the cylinder to be 0.7 m, and he volume to be 22, as stated in the question.

Plug these values in the volume of cylinder formula (\(πr^2h\)) to get the value of r has 10^(-1/2)

1. Let's assume the height and volume of the cylinder to be \(0.7\) \(m\) and \(22\) \(m^3\) respectively

2. With these attributes the cylinder will be filled to its full capacity in an hour

3. Substituting the above values in \(πr^2h = 22\) we get \(πr^2*0.7 = 22\)

4. Rearranging the elements \(r^2 = \frac{22 * 10}{7 * π}\)

5. Here, \(22\), \(7\) and \(π\) cancel out as \(7*π\) ~ \(22\) leaving us with \(10\)

6. \(r^2 = 10\) yields \(r\) \(=\) \(\sqrt{10}\)

Ans. B

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