Mathematical Puzzles and Their Solutions: A Sphere Problem Analysis

Mathematics is a language that allows us to describe and solve complex problems in a systematic and precise way. Below is an analysis of a sphere problem that involves a box containing a total of 30 red, white, and black spheres distributed in a specific pattern. We will break down the problem and provide a solution using algebraic equations.

The initial setup of the problem is as follows:

There are 30 red, white, and black spheres in a box. White spheres are 3 less than black spheres and 12 less than red spheres. Determine the number of spheres for each color.

Algebraic Representation

To represent this problem algebraically, we introduce the following variables:

R: Red spheres W: White spheres B: Black spheres

The total number of spheres is 30. Thus, we can write the following equation:

R W B 30

We are also given two additional pieces of information:

W (white spheres) are 3 less than B (black spheres): B W 3 W (white spheres) are 12 less than R (red spheres): R W 12

Solving the Equations

Let's start substituting the equations to solve for W, B, and R.

Step 1: Substitute B in the total equation

Using the equation B W 3, we can substitute B in the total equation:

R W (W 3) 30

Simplifying, we get:

R 2W 3 30

R 2W 27

Step 2: Substitute R in the simplified equation

Using the equation R W 12, we substitute R in the simplified equation:

(W 12) 2W 27

Combining like terms, we get:

3W 12 27

Subtracting 12 from both sides:

3W 15

Dividing by 3:

W 5

Step 3: Determine B and R

Now that we have W, we can find B and R using the substitution equations:

B W 3 5 3 8 R W 12 5 12 17

Therefore, the number of spheres for each color is:

Red spheres (R): 17 White spheres (W): 5 Black spheres (B): 8

Verification

Let's verify the solution by substituting the values back into the original total equation:

R W B 30

17 5 8 30

Thus, the solution is correct.

Bonus Interpretation

To further explore the solution, let's consider an alternative initial condition where we have only one white sphere. If there were one white sphere, we would have four black and thirteen red ones, as follows:

White (W): 1 Black (B): 4 (3 more than white) Red (R): 13 (12 more than white)

The total would then be 18:

1 4 13 18

The remaining 12 spheres (30 - 18 12) can be distributed equally among the three colors. If we distribute these 12 spheres equally, each color would get four additional spheres:

White (W): 1 4 5 Black (B): 4 4 8 (3 more than white) Red (R): 13 4 17 (12 more than white)

This solution matches the previous one and can be considered an alternative distribution scenario.