Question

The amount of water used when taking a shower varies directly with the number of minutes the shower is run. If a 4 -minute shower uses 7 gallons of water, how much water is used in a 9-minute shower?

Short answer

15.75 gallons

Step-by-step solution

Understanding the Problem

Determine what is given and what needs to be found. The problem states that water usage varies directly with the number of minutes. Given a 4-minute shower uses 7 gallons, find the water usage for a 9-minute shower.

Set Up a Direct Variation Equation

Since water usage varies directly with the number of minutes, we can use the direct variation formula: \[ W = kM \]where - \( W \) is the water used in gallons, - \( M \) is the number of minutes, and - \( k \) is the constant of variation.

Find the Constant of Variation

Use the given values to find the constant \( k \). Substitute the values for a 4-minute shower and 7 gallons of water: \[ 7 = k \times 4 \]Solve for \( k \): \[ k = \frac{7}{4} = 1.75 \]

Use the Constant to Find Water for 9 Minutes

Now that \( k \) is known, use it to find the water used in a 9-minute shower. Substitute \( k \) and \( M = 9 \): \[ W = 1.75 \times 9 \]Calculate the multiplication: \[ W = 15.75 \]

Conclude the Solution

So, the amount of water used for a 9-minute shower is 15.75 gallons.

Key concepts

Understanding Water Usage Calculation

The amount of water used during a shower can be calculated by understanding the concept of direct variation. In this case, water usage depends on the duration of the shower. The longer the shower, the more water is consumed. This relationship helps us predict water usage based on the number of minutes the shower runs.

For example, if we know a 4-minute shower uses 7 gallons of water, we can set up an equation to determine water usage for any other duration. This forms the basis of water usage calculation, especially when planning to conserve water or monitor household usage.

Setting Up a Direct Variation Equation

Direct variation describes a relationship where one quantity increases or decreases in direct proportion to another. In mathematical terms, this can be expressed as: \ \[ W = kM \ \] where:

• W = the amount of water used (in gallons),
• M = the number of minutes of the shower,
• k = the constant of variation, which remains the same for all values of M and W.

Let's set up this equation using our example. If we know that a 4-minute shower uses 7 gallons of water, we can substitute these values into the equation to find the constant of variation:
\[ 7 = k \times 4 \ \] This lets us solve for k, helping us understand the rate at which water is used per minute of showering.

Calculating the Constant of Variation

To find the constant of variation, we use the values we have. Plugging into the direct variation equation: \ \[ 7 = k \times 4 \ \] we simplify to solve for k: \ \[ k = \frac{7}{4} = 1.75 \ \]
With k known, we can now calculate water usage for any shower duration. For a 9-minute shower, replace M with 9 in the equation: \ \[ W = 1.75 \times 9 \ \]
After performing the multiplication, we find: \[ W = 15.75 \] Therefore, a 9-minute shower uses 15.75 gallons of water. The constant allows for easy calculations, helping predict usage in various scenarios. Understanding and calculating this constant can be very useful in practical situations, like saving water or budgeting utility costs.