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Translate the word problem below into an equation; then solve.



Do-Nothing #1 can count 5 more ceiling dots per minute than Do-Nothing #2. If they work together for 30 minutes straight, they can count a combined total of 3,450 dots. How many dots can Do-Nothing #2 count per minute?

Verified Answer :

  • Define the rates of the two machines as \(r_1 = r_2 + 5\) and their combined work as \(30(r_1 + r_2) = 3450\).
  • Substitute the rate expression into the total work equation: \(30(2r_2 + 5) = 3450\).
  • Divide by 30 and solve for \(r_2\): \(2r_2 = 110 \implies r_2 = 55\).
  • The count rate for Do-Nothing #2 is \(\boxed{55}\) dots per minute.

Explanation

  1. Defining Variables and Establishing Equations

Let \(r_1\) be the number of dots that Do-Nothing #1 can count per minute, and let \(r_2\) be the number of dots that Do-Nothing #2 can count per minute. We are given that Do-Nothing #1 counts 5 more dots per minute than Do-Nothing #2, which gives us the linear relationship:
\[r_1 = r_2 + 5\]
Furthermore, when working together, they count a total of 3,450 dots in 30 minutes. The combined rate of the two is \(r_1 + r_2\). Thus, their combined production over 30 minutes is:
\[30(r_1 + r_2) = 3450\]

  1. Solving the System of Equations

To find the value of \(r_2\), we substitute the expression for \(r_1\) into the combined work equation:
\[30((r_2 + 5) + r_2) = 3450\]
First, we can simplify the equation by dividing both sides by 30:
\[(r_2 + 5) + r_2 = \frac{3450}{30}\]
\[2r_2 + 5 = 115\]
Now, we subtract 5 from both sides:
\[2r_2 = 110\]
Finally, dividing by 2 yields:
\[r_2 = 55\]
So, Do-Nothing #2 can count 55 dots per minute.

Examples

Understanding how to set up and solve systems of equations involving rates is essential for optimizing work processes. For example, if you are managing two automated printing machines with different speeds, you can use these equations to predict how many pages they will produce in a set amount of time or to determine the capacity needed to finish a project by a specific deadline. This type of algebraic modeling helps in logistics, manufacturing, and resource management to ensure efficiency and meet production goals.

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