10 Andrew russ laps of a circuit in 4 minates. Bryan runs laps of the same circuit in 3 minules. Chris can ran laps of the same circuit in 6 minutes. They all slart together on the starting line and run a race that goes for 36 minules. inat is the first time, after the sart, that they will all croos over the starting line together?

To find the first time after the start that Andrew, Bryan, and Chris will all cross the starting line together, we need to determine the time it takes for each of them to complete one lap and then find the least common multiple (LCM) of those times. ### Step 1: Determine the lap times for each runner - **Andrew** completes 10 laps in 4 minutes, so his lap time is: \[ A = \frac{4 \text{ minutes}}{10 \text{ laps}} = 0.4 \text{ minutes per lap} \] - **Bryan** completes 1 lap in 3 minutes, so his lap time is: \[ B = 3 \text{ minutes per lap} \] - **Chris** completes 1 lap in 6 minutes, so his lap time is: \[ C = 6 \text{ minutes per lap} \] ### Step 2: Find the least common multiple (LCM) of the lap times We need to find the LCM of \(0.4\), \(3\), and \(6\). To do this, we can convert \(0.4\) into a fraction: \[ 0.4 = \frac{2}{5} \] Now we have: - \(A = \frac{2}{5}\) - \(B = 3\) - \(C = 6\) Next, we will find the LCM of these values. To do this, we can express all numbers in terms of a common denominator. ### Step 3: Convert to a common denominator The least common multiple of the denominators \(5\), \(1\) (for \(3\)), and \(1\) (for \(6\)) is \(5\). We can express \(3\) and \(6\) in terms of fifths: - \(3 = \frac{15}{5}\) - \(6 = \frac{30}{5}\) Now we have: - \(A = \frac{2}{5}\) - \(B = \frac{15}{5}\) - \(C = \frac{30}{5}\) ### Step 4: Find the LCM The LCM of the numerators \(2\), \(15\), and \(30\) can be calculated. The LCM of these numbers is \(30\). Thus, the LCM of the lap times in terms of minutes is: \[ \text{LCM} = \frac{30}{5} = 6 \text{ minutes} \] ### Conclusion The first time after the start that Andrew, Bryan, and Chris will all cross the starting line together is at \(6\) minutes.