# Bag of marbles

1. Jun 7, 2008

### LocationX

In bag K, 3/8 of the marbles are red. In bag M, 1/2 of the marbles are red. Bag K has four times as many red marbles as bag M. If the two bags were combined, what is the percentage of red marbles in the bag?

My work:

Say we had 6 red marbles in bag M. This means that bag K must have 24 red marbles. If this is the case, then we can say that 3/8=24/x

solving for x, we get 64 marbles total for bag K

if both bags were combined, then we would get 24+6 red marbles and 64+12 total marbles.

the ratio of red to total marbles is 30/76=39%

did I do this problem correctly?

2. Jun 7, 2008

### konthelion

Yes, the answer is correct. I guess instead of plugging in numbers, you could've set up an equation, it would've been easier.

3. Jun 8, 2008

### tiny-tim

Hi LocationX!

You got the right answer, but no, you didn't do the problem correctly.

As konthelion indicates, you should use algebra … that is, call something "x", and get an equation in x.

4. Jun 8, 2008

### LocationX

could someone show me how to set this up? Or at least the beginning steps?

5. Jun 8, 2008

### tiny-tim

ok … we'll start with what you did, and just adjust it slightly … you started with:

Say we had A marbles in bag M. This means that bag M must have A/2 red marbles. This means that bag K must have …

6. Jun 8, 2008

### LocationX

2A red marbles? Where do I go from here?

7. Jun 8, 2008

### tiny-tim

The next step is exactly the same as in your original proof … find the total number of marbles in bag K (and do it the same way )

8. Jun 8, 2008

### HallsofIvy

Staff Emeritus
I see nothing wrong with the way it was done initially! Certainly, in order that the problem have a specific solution, it must be true that the solution not depend upon the specific number of marbles in the bags.

If you want to show that, in fact, that is true, LocationX, just do exactly what you did but use a variable, say "x", in place of your "6 red marbles":

Suppose there were "x" red marbles in bag M. Then, since 1/2 of the marbles in M are red, there are a total of 2x marbles in bag M. Also, since bag K has four times as many red marbles as M, there are 4x red marbles in bag K. Since 3/8 of the marbles in K are red and (3/8)?= 4x gives ?= (8/3)(4x) there must be 32x/3 total marbles in K.

Now put all that together. There are x red marbles in M and 4x red marbles in K that means that there must be x+ 4x= 5x red marbles altogether. There are 2x marbles in M and 32x/3 marbles in K so there must be 2x+ 32x/3= (6x+ 32x)/3= 38x/3 total marbles. That means that the ratio of red marbles to total marbles is (5x)/(38x/3)= 15/38= 39%.

Notice that the fact that the total number of marbles in K, 32x/3, is a whole number means that x must be a multiple of 3, just as your original "6" was. Also, since you used 2*3= 6 instead of just 6, you got 2*15/2*38= 30/76 which is still 39%.

But I, at least, have no problem with your recognizing that, as I said, in order that the problem be solvable the result must NOT depend on the exact number of marbles in the bags and so you are free to choose any convenient number of red marbles in M- such as your "6 red marbles" ("3 red marbles in M" would have been even easier).

Here's a much much harder version of the same idea:

A cylindrical hole, of length 6 cm, is drilled all the way through a wooden sphere. What volume of wood is left in the sphere?

Notice that you are not told either the radius of the cylindrical hole or the radius of the sphere. The hard way to solve that problem is this: Assume the cylindrical hole has radius r. You can then calculate the volume of the cylinder itself. You can alos use the Pythagorean theorem to find the radius of the sphere, R, as a function of r, so that a cylinder of that radius and length will go all the way through the sphere. You can use that information (and an integral) to find the volume of the two "caps" on either end of the cylindrical hole that will be cut away. Finally, subtract those three volumes from the volume of the sphere and see that "r" magically cancels!

The easy way is this: Assume that since you were not given either the radius of the cylindrical hole or the radius of the sphere, the answer is independent of those. Thus, you are free to choose one of those to be any convenient number. A radius for the cylinder of "0 cm" is very convenient! The "cylindrical hole" becomes a single line which removes no wood from the sphere. In order that the single line of length 6 cm go all the way through the sphere, the sphere must have diameter 6 cm, so radius 3 cm and volume $(4/3)\pi (3^3)= 36\pi$ cubic cm. Since this "imaginary" hole removed no wood, the "volume of wood left in the sphere" is also $36\pi$ cubic cm!