Solving for Total Money: A Fractional Distribution Problem

In this article, we address a mathematical problem involving the distribution of money among four boys. The problem presents a scenario where each boy receives a specific fraction of the total amount. Through detailed analysis and step-by-step calculations, we will determine the total sum of money that was distributed.

Introduction to the Problem

The problem states that four boys shared a certain amount of money. The first boy received 1/6 of the total, the second boy received 1/8, the third boy received 1/2, and the fourth boy received 55 units. We aim to calculate the total amount of money that was originally distributed among the four boys.

Step-by-Step Solution

Let the total amount of money be (textit{x}).

Step 1: Expressing the Fractions

We start by expressing the amounts received by each boy in terms of (textit{x}):

The first boy received (frac{1}{6} cdot textit{x}). The second boy received (frac{1}{8} cdot textit{x}). The third boy received (frac{1}{2} cdot textit{x}). The fourth boy received 55 units.

Step 2: Combining the Fractions

We are given that the sum of the amounts received by all four boys is equal to the total amount of money. Therefore, we can write:

[frac{1}{6} cdot textit{x} frac{1}{8} cdot textit{x} frac{1}{2} cdot textit{x} 55 textit{x}]

To combine the fractions, we find the least common multiple (LCM) of the denominators, which is 24. We rewrite each fraction with the common denominator:

[frac{4}{24} cdot textit{x} frac{3}{24} cdot textit{x} frac{12}{24} cdot textit{x} 55 textit{x}]

Next, we combine the fractions:

[frac{4 3 12}{24} cdot textit{x} 55 textit{x}] [frac{19}{24} cdot textit{x} 55 textit{x}]

Step 3: Isolating (textit{x})

Now, we isolate (textit{x}) by subtracting (frac{19}{24} cdot textit{x}) from both sides:

[55 textit{x} - frac{19}{24} cdot textit{x}] [55 frac{24}{24} cdot textit{x} - frac{19}{24} cdot textit{x}] [55 frac{5}{24} cdot textit{x}]

Multiplying both sides by (frac{24}{5}) to solve for (textit{x}), we get:

[textit{x} 55 cdot frac{24}{5}] [textit{x} 55 cdot 4.8] [textit{x} 264]

Therefore, the total amount of money that was distributed is 264 units.

Alternative Methods for Solving the Problem

There are several alternative methods that can be used to solve this problem. Let's explore a few of them:

Method 1: Using the Least Common Multiple (LCM)

Note that the LCM of 6, 4, and 2 is 12. Let the total money be 12(textit{x}).

The first boy received (frac{1}{6} cdot 12textit{x} 2x). The second boy received (frac{1}{4} cdot 12textit{x} 3x). The third boy received (frac{1}{2} cdot 12textit{x} 6x). The fourth boy received 5x.

Combining these amounts, we have:

[2x 3x 6x 5x 12textit{x}] [16x 12x 55] [4x 55] [12textit{x} 60]

Thus, the total money is 60 units.

Method 2: Simplifying Fractions

Using the LCM of 6, 8, and 2, which is 24, we rewrite the fractions with 24 as the common denominator:

[frac{1}{6} cdot 24textit{x} 4x, quad frac{1}{8} cdot 24textit{x} 3x, quad frac{1}{2} cdot 24textit{x} 12x]

The remaining fraction is:

[24x - (4x 3x 12x) 5x]

Therefore, we have:

[5x 55] [24x 264]

Again, the total money is 264 units.

Conclusion

The total amount of money that was distributed among the four boys is 264 units. This solution is verified using various methods, including the least common multiple and simplification of fractions. The problem showcases the importance of understanding fractions and their application in real-world scenarios.