Students ask me about this question frequently, and I think this problem is a great example of the fact that you often need to break a larger problem into several smaller steps.

A cylindrical bucket, with height 10 and radius r is 3/4 filled with water. A boy is dropping marbles with volume r/10 in the bucket at a rate of 12 a minute. How many seconds will it take before the water overflows the bucket ?

The problem is more easily thought about in pieces:
1) How much space is left in the bucket to fill ?
2) How many marbles will it take to fill that space ?
3) How much time will it take to drop that many marbles in ?

For (1), the overall capacity of a cylindrical bucket is (pi) r-squared * height. So for this bucket, the overall volume is 10*pi*r-squared. 1/4 of that volume remains empty, so the space left to be filled is (1/4)*10*pi*r-squared or (5/2)*pi*r-squared.

For (2), you want to divide the space you have (from part 1) by the size of each marble.

If you have difficulty seeing that this is the next step, think of a simpler problem. Suppose you had a flat rectangle that was 10 x 20 inches, and you wanted to know how many rectangles 5 x 1 inches it would take to fill that space. The concept is the same, but the shape is easier for people to visualize and to instinctively calculate. A 200 square inch rectangle being filled by 5 inch rectangles. You need 40 of the little rectangles. Of course, the number 40 has no relevance for this problem, but the way you approach the calculation is identical. You take the total space to be filled (200) and divide it by the size of the unit used to fill the space (5).


Here, this calculation is (5/2 pi r-squared) / (r/10)

This = 25*pi*r

3) So how long will it take to drop 25*pi*r marbles in the bucket ?
This is a work problem. The amount of work to get done is the number of marbles to be dropped. The rate is 12 marbles per minute. We solve for time.

Watch out for the units. The problem gives the rate is minutes but asks for the answer in seconds.

(12 marbles/min)* a conversion of (1 min/60 sec) =1 marble every 5 seconds
So the rate we will use is (1 marble/5 sec).
W = R*T
(25*pi*r) = (1 marble/5 sec) * T

T = 125*pi*r which is answer (C)

An alternative way to work on this problem, or any rate problem, is to focus on your units of measurement. With rate problems, students are often left wondering do I divide by a number, or multiply by a number. Viewing the rate problems as a ratio, and then focusing on the units of measurement makes it nearly impossible to screw up.

The final answer asks you for how many seconds. So start by writing down "sec=?".

Then write down what you know:
*Empty Space - 1/4 (Vol Cyl) = 1/4(10)pi*r^2=2.5*pi*r^2 SPACE
* r/10 SPACE/MARBLE
* 12 MARBLE/MINUTE

Since the problem does not give you a specific units of measurement for distance (for instance, inches, cm), you must create a consistent one. Since the only time I work with distance is with volume, I simply substitute the word SPACE.

Now line up your units of measurements.

1) Since your final answer is "sec" you must have time in the top. The means you must flip your "12 MAR/MIN" to "1/12 MIN/MAR".
2) MIN is not in your final answer, so convert it by inserting "60SEC/MIN"
3) MAR is not in your final answer, so you must multiply by MAR. However to cancel, MAR must be in the top, so flip that to "10/r MAR/SPACE"
4) SPACE is not in your final answer, so you must multiply by 2.5 pi* r^s SPACE. This is already set up to cancel.

You now have:

1/12 MIN/MAR * 60SEC/MIN * 10/r MAR/SPACE * 2.5 pi*r^2 SPACE

Notice that you now have SEC as your only unit of measurement left, so you know you are good. Now multiply. Notice that many things cancel. They do that on purpose to help you. First, you see that you have "pi*r" remaining which is in all of your final answers. Then the "60/12" which becomes "5" so you have "5*10*2.5" --> 125 pi * r

Focus on your units and your life will be easy peasy.