Problem 29 - Prealgebra

Question

One driver can circle a one-mile track in 30 seconds. Another driver takes 20 seconds. If they both start at the same time, in how many seconds will they be together again at the starting line?

Step-by-Step Solution

Answer

They will be together again at the starting line in 60 seconds.

Step 1: Determine the Speed of Each Driver

Let's calculate the speed of each driver in terms of laps per second. The first driver completes one lap in 30 seconds, so their speed is \( \frac{1}{30} \) laps per second. The second driver completes one lap in 20 seconds, so their speed is \( \frac{1}{20} \) laps per second.

Step 2: Set Up the Condition for Meeting

Both drivers need to complete an integer number of laps and meet at the starting line. Let \( t \) be the time in seconds after which they meet again. Both drivers would have completed a whole number of laps in this time. We set up the equations for the number of laps: 1. For the first driver: \( \frac{t}{30} \) is an integer. 2. For the second driver: \( \frac{t}{20} \) is an integer.

Step 3: Find the Least Common Multiple (LCM)

To find the time \( t \) when both conditions are met simultaneously, we need to find the least common multiple (LCM) of 30 and 20. The LCM of two numbers is the smallest number that both numbers can divide evenly. Let's calculate it:- The prime factorization of 30 is \( 2 \times 3 \times 5 \).- The prime factorization of 20 is \( 2^2 \times 5 \).To find the LCM, take the highest power of each prime number that appears:\[ \text{LCM}(30, 20) = 2^2 \times 3 \times 5 = 60 \]

Step 4: Conclusion

The minimum time after which both drivers will be together again at the starting line is 60 seconds. Therefore, \( t = 60 \) seconds is the time at which both complete integer numbers of laps and meet at the starting line.

Key Concepts

Speed Calculation

Speed calculation involves determining how fast an object is moving. In this exercise, speed is measured in terms of laps per second. If a driver circles a track in a fixed amount of time, we can find their speed by dividing one lap by the time it takes to complete it.
This gives us a measure of their consistency and pace per unit time.To calculate the drivers' speeds in our example:

• The first driver completes one lap in 30 seconds, so their speed is \( \frac{1}{30} \) laps per second.
• The second driver completes one lap in 20 seconds, resulting in a speed of \( \frac{1}{20} \) laps per second.

Understanding these speeds helps us predict how soon they'll meet back at the start, as each driver progresses at these rates around the track.

Integer Solutions

When calculating the time at which both drivers meet again at the starting line, it's critical that both have completed a whole number—or integer—of laps. This concept is central to understanding when two periodic events converge.To achieve integer solutions in this scenario:

• We set up a condition for each driver. For the first driver, let \( \frac{t}{30} \) be an integer.
• For the second driver, let \( \frac{t}{20} \) be an integer.

This means the time \( t \) in which they meet again must be a multiple of both 30 and 20. Only a time that satisfies both conditions will result in each driver having completed an exact integer number of laps. This is crucial in ensuring they both arrive at the start line simultaneously.

Prime Factorization

To solve the problem of determining how many seconds it will take both drivers to meet again at the starting line, prime factorization is used to find the least common multiple (LCM) of two numbers—in our case, 30 and 20.Prime factorization breaks down a number into its basic building blocks—prime numbers. Here's how we do it:

• For 30: The prime factorization is \( 2 \times 3 \times 5 \).
• For 20: The prime factorization is \( 2^2 \times 5 \).

By taking the highest power of all primes involved, we find the LCM:

• The highest power of 2 is \( 2^2 \).
• The highest power of 3 is 3.
• The highest power of 5 is 5.

Thus, the LCM(30, 20) = \( 2^2 \times 3 \times 5 = 60 \). This indicates that after 60 seconds, both drivers will have completed integer numbers of laps and will be back together at the starting line.