Problem 80 - Algebra for College Students

Question

The average monthly precipitation for Honolulu, HI, for October, November, and December is 3.11 in. If 2.98 in. falls in October and 3.05 in. falls in November, how many inches must fall in December so that the average monthly precipitation for these months exceeds 3.11 in.?

Step-by-Step Solution

Answer

December precipitation must exceed 3.30 inches.

Step 1: Understand the given information

The average monthly precipitation for October, November, and December should exceed 3.11 inches. Precipitation in October is 2.98 inches, and in November it is 3.05 inches. The task is to find how many inches must fall in December.

Step 2: Set up the formula for average precipitation

The formula for average precipitation is: \[\text{Average} = \frac{\text{Total Precipitation}}{\text{Number of Months}}\]In this case, the number of months is 3.

Step 3: Calculate the total precipitation needed

Let the precipitation in December be represented by \(D\). Given the average should exceed 3.11 inches, we set up the inequality:\[\frac{2.98 + 3.05 + D}{3} > 3.11\]

Step 4: Simplify the inequality

Multiply both sides by 3 to eliminate the denominator:\[2.98 + 3.05 + D > 3.11 \times 3\]\[2.98 + 3.05 + D > 9.33\]

Step 5: Solve for December precipitation

Combine the known amounts:\[6.03 + D > 9.33\]Subtract 6.03 from both sides:\[D > 3.30\]So, the precipitation in December must exceed 3.30 inches.

Key Concepts

inequality in algebra

Inequality in algebra helps us determine ranges of values rather than fixed numbers. In the given exercise, we use inequalities to find out how much precipitation is needed in December. An inequality shows the relationship between expressions that are not necessarily equal. It can use symbols like <, >, ≤, or ≥.
For example, the inequality we set up is: \(\frac{2.98 + 3.05 + D}{3} > 3.11 \)
In this setup:

• \(2.98 + 3.05 + D\) represents the total monthly precipitation for October, November, and December.
• \(>\text{3.11}\) shows that the average must be greater than 3.11 inches.

Understanding inequalities is crucial because it allows us to solve for a range where the solution works, rather than pinpointing a single value.

solving for a variable

Solving for a variable involves isolating that variable on one side of the equation or inequality. Take our problem: we’re solving for \(\text{D}\), the precipitation amount in December. First, we set up our inequality:
\(\frac{2.98 + 3.05 + D}{3} > 3.11\)
We need to isolate \(\text{D}\). Here’s how:

1. Multiply both sides by 3 to remove the fraction: \(\text{2.98 + 3.05 + D} > \text{9.33}\)
2. Combine known values: \(\text{6.03 + D} > \text{9.33}\)
3. Subtract 6.03 from both sides: \(\text{D} > \text{3.30}\)

This shows \(\text{D}\) must be greater than 3.30 inches. The goal with solving for a variable is to perform arithmetic operations that simplify the equation step-by-step until the variable is isolated and solved for.

arithmetic operations

Arithmetic operations are the basic building blocks of mathematics, and they are crucial for solving algebraic equations and inequalities. In this exercise, specific arithmetic operations help us manipulate and simplify the inequality.
Consider our steps:

• Addition: When we combine the precipitations for October and November: \(\text{2.98} + \text{3.05} = \text{6.03}\).
• Multiplication: To clear the fraction in the average precipitation formula, we multiply both sides by 3, which eliminates the denominator.
• Subtraction: Finally, to isolate our variable \(\text{D}\), we subtract 6.03 from both sides:
  \(\text{D} > \text{3.30}\).

These operations are simple yet powerful tools that enable us to break down complicated problems step-by-step and find clear solutions. By mastering these basic operations, you can confidently tackle a wide range of mathematical challenges.