Q32. - Algebra 1

Question

CHEERLEADING For the competition, the cheerleading team is preparing a dance routine that must last 4 minutes, with a variation of ±5 seconds.

a. Find the least and greatest possible times for the routine in minutes and seconds.

b. Find the least and greatest possible times in seconds.

Step-by-Step Solution

Answer

a. The least possible time for the routine will be 3 minutes 55 seconds and the greatest possible time for the routine will be 4 minutes 5 seconds.

b. The least possible time for the routine in seconds will be 235 seconds and the greatest possible time for the routine in seconds will be 245 seconds.

Part a. Step 1. Write down the given statement.

It is given that the dance routine must last 4 minutes.

The variation allowed is $\pm 5\text{ seconds}$.

Part a. Step 2. Evaluate the least and greatest possible time.

Let the time allowed for the dance routine be $x$.

Therefore, 

$$\begin{gathered} \text{x}=\text{4 minutes ± 5 seconds} \\ \text{x}=\left(\text{4 min - 5 sec}\right)\text{or x}=\left(\text{4 min + 5 sec}\right) \\ \text{x}=\text{3 minutes 55 seconds or x}=\text{4 minutes and 5 seconds} \end{gathered}$$

Part a. Step 3. The conclusion from the least and greatest possible time.

The least possible time for the routine will be 3 minutes 55 seconds.

The greatest possible time for the routine will be 4 minutes 5 seconds.

Part b. Step 1. Write down the given statement.

It is given that the dance routine must last 4 minutes.

The variation allowed is $\pm 5\text{ seconds}$.

Part b. Step 2. Evaluate the least and greatest possible time.

Let the time allowed for the dance routine be $x$. Therefore, 

 $$\begin{gathered} \text{x }=\text{ 4 minutes ± 5 seconds} \\ \text{x }=\left(\text{4 min - 5 sec}\right)\text{or x }=\left(\text{4 min + 5 sec}\right) \\ \text{x }=\text{ 3 minutes 55 seconds or x }=\text{ 4 minutes and 5 seconds} \\ \text{x }=\text{ 3}\left(60\right)+55\text{ seconds or x }=\text{ 4}\left(60\right)+5\text{ seconds} \\ \text{x }=\text{ 180+55 seconds or x }=\text{ 240+5 seconds} \\ \text{x }=\text{ 235 seconds or x }=\text{ 245 seconds} \end{gathered}$$

Part b. Step 3. The conclusion from the least and greatest possible time.

The least possible time for the routine in seconds will be 235 seconds.

The greatest possible time for the routine in seconds will be 245 seconds.