Solving for the Value of a Positive Number x

The problem provides a relationship between two positive numbers, x and y, where x is greater than y. We are given that the difference between these two numbers is equal to 25% of x. We are also given the specific value of y, which is 15. Our goal is to find the value of x.

Setting up the Equation

Let the two positive numbers be x and y. We are told that x > y. The difference between x and y is \(x - y\).

We are given that this difference is 25% of x. In mathematical terms, 25% of x can be written as \(0.25x\) or \(\frac{25}{100}x\).

So, the relationship can be written as an equation:

\(x - y = 25\% \text{ of } x\)

\(x - y = 0.25x\)

Substituting the Given Value and Solving

We are given that the value of y is 15. We can substitute this value into the equation:

\(x - 15 = 0.25x\)

Now, we need to solve this linear equation for x. To do this, we gather all the terms involving x on one side of the equation and the constant terms on the other side.

Subtract \(0.25x\) from both sides of the equation:

\(x - 0.25x - 15 = 0.25x - 0.25x\)

\(0.75x - 15 = 0\)

Add 15 to both sides of the equation:

\(0.75x - 15 + 15 = 0 + 15\)

\(0.75x = 15\)

To isolate x, we can divide both sides by 0.75. Remember that \(0.75 = \frac{75}{100} = \frac{3}{4}\).

\(x = \frac{15}{0.75}\)

\(x = \frac{15}{\frac{3}{4}}\)

Dividing by a fraction is the same as multiplying by its reciprocal:

\(x = 15 \times \frac{4}{3}\)

Now, perform the multiplication:

\(x = \frac{15 \times 4}{3}\)

\(x = \frac{60}{3}\)

\(x = 20\)

Verification

Let's check if the value x = 20 satisfies the original condition with y = 15.

The difference between x and y is \(x - y = 20 - 15 = 5\).

25% of x is \(0.25 \times 20 = \frac{1}{4} \times 20 = 5\).

Since the difference (5) is equal to 25% of x (5), and x (20) > y (15), the value x = 20 is correct.

Conclusion

Based on the given information and our calculations, the value of x is 20.

Revision Table: Number Difference and Percentage

Additional Information: Percentage Calculations and Equations

Understanding percentages and how to work with algebraic equations is fundamental in solving problems like this. A percentage can always be converted into a decimal or a fraction to make calculations easier. For example, 25% is equivalent to \(\frac{25}{100}\), which simplifies to \(\frac{1}{4}\), or as a decimal, 0.25.

When solving an algebraic equation, the goal is to isolate the variable (in this case, x). This is done by performing the same operation on both sides of the equation to maintain equality. Operations include addition, subtraction, multiplication, and division.

In our solution, we used the fact that subtracting 0.25x from x (\(1x - 0.25x\)) results in \(0.75x\). Then, to get x by itself from \(0.75x = 15\), we divided both sides by 0.75.

This problem combines concepts of number properties, percentages, and basic algebra, which are common in many quantitative aptitude questions.