Problem 37 - Algebra A Combined Function

Question

Perform the indicated operation. $$ \begin{array}{r} 16.003 \\ \times 5.31 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The result of the multiplication is 85,355.93.

Step 1: Set Up the Multiplication

Write the numbers 16.003 and 5.31 in standard vertical multiplication format, aligning the digits by their place values. We have:\[\begin{array}{r}16.003 \\times 5.31 \\hline\end{array}\]

Step 2: Ignore Decimal Points for Initial Calculation

Temporarily ignore the decimal points and treat 16.003 as 16003 and 5.31 as 531. Set up the multiplication as if these were whole numbers.

Step 3: Multiply as Whole Numbers

Multiply 16003 by 531. This can be broken down into smaller steps:1. Multiply 16003 by 1 to get 16003.2. Multiply 16003 by 3 to get 48009 (remember it represents 30 so place one zero to the right).3. Multiply 16003 by 5 to get 80015 (represents 500 so place two zeros to the right).Add these results:\[\begin{array}{r}16003 \+ 480090 \+ 8001500 \\hline\end{array}8535593\]

Step 4: Adjust for Decimal Points

Count the total number of decimal places in the original numbers. 16.003 has 3 decimal places and 5.31 has 2, totaling 5 decimal places. Therefore, place the decimal point in the product so that there are 5 digits counting from the right. This gives us:\[85355.93\]

Step 5: Verify the Result

Double-check both the multiplication steps and the placement of the decimal point to ensure accuracy.

Key Concepts

Place Value Alignment

When multiplying decimal numbers, aligning the digits according to their place value is crucial. Think of place value alignment as ensuring the numbers are properly lined up according to their positions in whole numbers.
For example, if you are multiplying two numbers like 16.003 and 5.31, write them down with their decimal points lined up vertically. This is similar to stacking them so each digit falls into its respective category: ones, tens, hundreds, etc.

• This helps in organizing the calculation.
• It ensures each digit is in the correct column, simplifying the later steps.

By keeping each column distinct, you simplify both the arithmetic and subsequent placement of the decimal point.

Decimal Point Adjustment

Adjusting the decimal point correctly is a key step in decimal multiplication. Initially, you might ignore the decimal points to simplify the arithmetic. But once you have finished multiplying the raw numbers, you must adjust the decimal point to reflect the precision of the original decimal numbers.
In our example, we treat the decimals 16.003 and 5.31 as whole numbers, 16003 and 531, allowing for straightforward multiplication. The decimal point is then reintroduced based on the number of decimal places in the original numbers.
This adjustment ensures the resulting number accurately depicts the value of the original multiplication with decimals.

Whole Number Multiplication

Once place value alignment and initial decimal point consideration are set aside, the bulk of the work in decimal multiplication involves treating the numbers as whole numbers.
This involves going through the process of long multiplication, which includes:

• Multiplying the whole number parts as if no decimal point existed.
• Breaking down the multiplication into simpler steps like multiplying with single digits and adding these intermediate products.

In our exercise, 16003 is multiplied by 531 resulting in a series of simpler multiplications that are then added to yield the final product of 8535593. Consistently applying this step ensures the accuracy of the numeric result before placing the decimal.

Decimal Places Counting

To correctly place the decimal in the final product of a decimal multiplication, count the total number of decimal places in the original numbers. This step is required right before inserting the decimal into the answer.
For instance:

• 16.003 has three decimal places.
• 5.31 has two decimal places.

Adding these gives a total of five decimal places. Therefore, when inserting the decimal back into the multiplication result, count five places from the right of the final product. This yields the accurate result of 85355.93.
Ensuring this count is accurate helps maintain the precision required for decimal numbers.