Problem 59 - Prealgebra

Question

Find a fraction that satisfies all of the conditions below. Then write a sentence explaining why you think your fraction is or is not the only solution that satisfies the conditions. \(\cdot\) The fraction can be written as a percent greater than \(1 \%\) \(\cdot\) The fraction can be written as a percent less than \(50 \%\) \(\cdot\) The decimal equivalent of the fraction is a terminating decimal. \- The value of the denominator minus the value of the numerator is 3 .

Step-by-Step Solution

Answer

The possible fractions are \( \frac{1}{4} \) and \( \frac{2}{5} \); there could be multiple solutions due to different denominators.

Step 1: Understand the Conditions

First, let's interpret the conditions. We need a fraction that can be expressed as a percentage between 1% and 50%, the decimal equivalent must be terminating, and the difference between the denominator and numerator needs to be 3.

Step 2: Determine the Range for the Fraction

Since the fraction must be a percent between 1% and 50%, the fraction itself must be between 0.01 and 0.5. Therefore, we can express the fraction as \( \frac{n}{d} \), where \( \frac{n}{d} = 0.01 \) to \( 0.5 \).

Step 3: Consider Terminating Decimals

Since the decimal is terminating, the denominator \( d \) must be a product of only 2's and 5's, i.e., a factor of 10, 100, etc. Possible values for \( d \) include 2, 4, 5, 10, 20, 25, 50, 100, etc.

Step 4: Use the Condition for Denominator Minus Numerator

Given that \( d - n = 3 \), we have \( n = d - 3 \). Thus, the fraction becomes \( \frac{d-3}{d} \). Plug this into the range from Step 2 as \( 0.01 < \frac{d-3}{d} < 0.5 \).

Step 5: Solve for Values of d

Consider values of \( d \) that are products of 2's and/or 5's. For example, start with \( d = 4 \): - The fraction becomes \( \frac{1}{4} \). - \( \frac{1}{4} = 0.25 \), which is a terminating decimal. - 0.25 is 25%, which lies between 1% and 50%.

Step 6: Verify and Explore Other Possible Solutions

Check the next possible \( d \), such as \( d = 5 \) or \( d = 10 \), to see if they offer additional solutions: - With \( d = 5 \), \( n = 2 \), the fraction \( \frac{2}{5} = 0.4 \) or 40%. - Both satisfy all conditions, but have \( d = 10 \) results in \( \frac{7}{10} = 0.7 \) which exceeds 50%.Thus, potential solutions are \( \frac{1}{4} \) and \( \frac{2}{5} \).

Key Concepts

Understanding Fractions

Fractions are a way to express parts of a whole. They consist of a numerator and a denominator. The numerator is the top number, telling us how many parts we have. The denominator is the bottom number, telling us how many of those parts make up a whole.
For example, in the fraction \( \frac{1}{4} \), the numerator is 1 and the denominator is 4. This means one part out of four equal parts of a whole. When working with fractions, it's vital to understand how these two components interact, especially when solving problems like the given exercise. Here, the denominator minus the numerator needs to be 3, forming specific fractions that fit conditions.

Exploring Percentages

Percentages are another way to represent fractions, showing parts per hundred. To convert a fraction to a percentage, you multiply by 100. For instance, \( \frac{1}{4} \) as a percentage is calculated by:

• \( \frac{1}{4} \times 100 = 25\%\)

In our initial exercise, the less than 50% and greater than 1% constraints help us refine which fractions to use. While it might seem tricky, converting fractions to percentages is as simple as multiplying and comparing numbers. We want fractions that fall comfortably in this percentage range.

Demystifying Terminating Decimals

Terminating decimals are decimals that come to an end. They do not repeat endlessly. To understand if a fraction converts to a terminating decimal, examine its denominator. This must be solely a product of the factors 2 and 5.
For example, \( \frac{1}{4} = 0.25 \). Here, 4, the denominator, is a factor of 2, which allows for a clean termination. When the denominator has any primes other than 2 or 5, like 3 or 7, the decimal expansion would keep going. Understanding this helps pick the right fractions when solving problems with decimal constraints.

Numerator and Denominator Relationship

The relationship between numerator and denominator is crucial in forming meaningful fractions. In many exercises, such as the one above, specific conditions need to be met, like having the denominator minus the numerator equal a certain value. This affects the fraction's value significantly.
Consider the exercise condition where \( d - n = 3 \). When solving this, rearrange it to find that the numerator can be calculated by \( n = d - 3 \). By substituting possible values for the denominator (products of 2 and 5), you can find suitable fractions that meet all conditions. This process helps deepen understanding of how numerator-denominator relationships work.