Question

Biology Water makes up about \(\frac{2}{3}\) of a person's body weight. a. A student weighs 90 pounds. Determine how many pounds of the student's weight are attributed to water by finding a fraction equivalent to \(\frac{2}{3}\) with a denominator of 90 b. A student weighs 75 pounds. Determine how many pounds of the student's weight are attributed to water.

Short answer

A 90-pound student's weight attributed to water is 60 pounds. A 75-pound student's weight attributed to water is 50 pounds.

Step-by-step solution

- Convert the fraction

Convert the fraction \(\frac{2}{3}\) to have a denominator of 90. This can be done by multiplying both the numerator and the denominator by the same number.

- Determine the factor

The denominator is currently 3 and needs to be converted to 90. Divide the new denominator by the old one: \(\frac{90}{3} = 30\).

- Multiply the numerator

Multiply the numerator and denominator of \(\frac{2}{3}\) by 30: \(\frac{2 \times 30}{3 \times 30} = \frac{60}{90}\).

- Calculate the weight attributed to water for 90 pounds

Since \(\frac{2}{3}\) of a 90-pound student's weight is water and it's equivalent to \(\frac{60}{90}\), 60 pounds of the student's weight are attributed to water.

- Calculate the weight attributed to water for 75 pounds

Use the original fraction \(\frac{2}{3}\) to determine the water weight for 75 pounds: \(\frac{2}{3} \times 75\).

- Multiply the fraction

Calculate the product: \(\frac{2}{3} \times 75 = \frac{150}{3} = 50\). Therefore, 50 pounds of the 75-pound student's weight are attributed to water.

Key concepts

Fractions

Understanding fractions is crucial in both mathematics and biology. A fraction represents a part of a whole. It's written in the form \( \frac{a}{b} \), where ‘a’ is the numerator and ‘b’ is the denominator. For example, \( \frac{2}{3} \) means 2 parts out of a total of 3 parts.
In the given exercise, water makes up about \( \frac{2}{3} \) of a person's body weight. This fraction has to be equivalent to another fraction with a different denominator. This is often done to match the fraction with given values in practical problems.
Equivalent fractions are fractions that describe the same portion of a whole, even though they have different numerators and denominators. To make two fractions equivalent, multiply or divide both the numerator and the denominator by the same number. This helps make calculations easier and helps in accurate problem-solving.
For example:
\( \frac{2}{3} \) can be converted to \( \frac{60}{90} \) by multiplying the numerator (2) and denominator (3) by 30. The output fraction \( \frac{60}{90} \) still represents the same value as \( \frac{2}{3} \), just in a different form.
Understanding these basic principles makes solving bigger problems simpler.

Weight Calculation

Calculating weight using fractions is a common task in biology for understanding proportions of body compositions.
In the exercise, we need to calculate the weight attributed to water for students weighing different amounts. For example, a student's body weight is 90 pounds, and water makes up \( \frac{2}{3} \) of this weight.
To determine the exact pounds, convert \( \frac{2}{3} \) to a fraction with a denominator of 90. This is done because if the total weight is 90 pounds, the fraction representing the water weight should have 90 as the denominator.

Here are the steps broken down:

• Identify the current fraction representing the water proportion ( \( \frac{2}{3} \) ).
• Determine the factor to make 3 into 90, which is done by dividing 90 by 3. This results in 30.
• Multiply both the numerator (2) and the denominator (3) of \( \frac{2}{3} \) by 30.
• The new equivalent fraction is \( \frac{60}{90} \).

This means 60 pounds out of 90 pounds are attributed to water. The same process applies to other weights, like converting \( \frac{2}{3} \) to determine the water weight in a student weighing 75 pounds.
Using these steps makes calculations straightforward.

Algebra Applications

Algebra forms the foundation of solving equations and understanding relationships between variables. Applying algebraic principles can simplify the problem-solving process in biology and other sciences.
For the given exercise, multiplying fractions by whole numbers helps determine the weight attributed to water. Let's break down the application of algebra here:

• Given the fraction \( \frac{2}{3} \), find the water weight for a student weighing 75 pounds.
• Set up the equation \( \frac{2}{3} \) multiplied by 75 pounds: \( \frac{2}{3} \times 75 \).
• Perform the multiplication: \( \frac{2 \times 75}{3} \).
• Calculate the result: \( \frac{150}{3} = 50 \) pounds.

Therefore, for a student weighing 75 pounds, 50 pounds of that weight are attributed to water.
This use of basic algebraic operations ensures that students can break down complex problems into simpler, manageable calculations. It involves understanding multiplication and division of both fractions and whole numbers.