Distributing a Sum of Money in a Given Ratio with a Practical Example

Imagine you're faced with a problem where a sum of money needs to be distributed among four individuals, A, B, C, and D, in the ratio 2:4:3:5. Additionally, the condition provided is that C gets $100 less than D, and you are tasked with determining A's share. This article delves into the step-by-step approach to solving this problem, offering both a traditional algebraic method and a simpler, intuitive approach.

Problem Description

The problem involves distributing a sum of money among A, B, C, and D in the ratio of 2:4:3:5. We are also given that C's share is $100 less than D's share.

Conventional Algebraic Approach

To begin, let's represent the shares of A, B, C, and D with variables based on the given ratio. We'll use (x) as a common factor:

As share (2x) Bs share (4x) Cs share (3x) Ds share (5x)

According to the problem, C's share is $100 less than D's share. This can be translated into the following equation:

[3x 5x - 100]

Let's solve for (x):

Rearrange the equation: [3x - 5x -100] [-2x -100] [2x 100] [x 50]

Now that we have (x 50), we can find the shares of A, B, C, and D:

As share: (2x 2 times 50 100) Bs share: (4x 4 times 50 200) Cs share: (3x 3 times 50 150) Ds share: (5x 5 times 50 250)

Finally, let's confirm the condition that C gets $100 less than D:

[D - C 250 - 150 100]

This condition is satisfied. Therefore, A's share is $100.

A More Intuitive Approach

Here is an alternative method to simplify the problem. Instead of working with the specific values, we can use a simpler trick based on the sum of the ratio parts.

Let's assume the total amount to be distributed is 14 units of money. The ratio 2:4:3:5 can be represented by a sum of these parts:

[2 4 3 5 14]

Based on the ratio, we can distribute the parts as follows:

As share 2 units Bs share 4 units Cs share 3 units Ds share 5 units

When the difference between C's and D's shares is 2 units (5 - 3 2), A's share is also 2 units. So, let's denote A's share by (x), where (x 2) units. Now, if the difference between C's and D's shares is $100, we can use the same logic to find A's share:

[text{A's share} frac{text{x}}{2} times 100 frac{2}{2} times 100 100]

Thus, A's share is $100.

Conclusion

By using the algebraic approach, we calculated the shares and ensured the condition is met. In the simpler intuitive approach, we used the sum of the ratio parts to directly find the share of A. Both methods lead to the same result: A's share is $100.

This problem demonstrates the practical application of algebraic equations and the importance of understanding the underlying principles when solving share distribution problems.