Problem Solving 2 - linear Equation

[paulmdrdo1]

here's the other problem.

1. An army of soldiers is marching down a road at 5 mi/hr. A messenger on horseback rides from the front to the rear and returns immediately, the total time taken being 10 minutes. Assuming that the messenger rides at the rate of 10mi/hr, determine the distance from the front to the rear.

I don't understand what's going in the problem. and what is it really requiring to solve. front of what and rear of what?

thanks!

 

[Evgeny.Makarov]

I don't understand what's going in the problem. and what is it really requiring to solve. front of what and rear of what?

Front and rear of the column of soldiers. You have to find the length of the column.

 

[paulmdrdo1]

how do I set-up the equation?

 

[paulmdrdo]

the messenger took two trips, front-rear and rear-front. Take the relative speed of the messenger and the column of soldiers then use it as the rate of each trip.

 

[CellCoree]

The problem is asking for the distance between the front and rear of the army of soldiers. The information given is that the army is marching at a speed of 5 miles per hour and the messenger on horseback can ride at a speed of 10 miles per hour. The messenger takes a total of 10 minutes to ride from the front to the rear and back.

To solve this problem, we can use the formula Distance = Speed x Time. Since the messenger rides at a speed of 10 miles per hour, we can set up the equation 10 miles = 10 miles/hour x 10 minutes. We need to convert the 10 minutes to hours, so we divide by 60 to get 10/60 = 1/6 hours. Plugging this into the equation, we get 10 miles = 10 miles/hour x 1/6 hours. This simplifies to 10 miles = 5/3 miles, or 6 miles.

Therefore, the distance between the front and rear of the army is 6 miles. This problem required us to use problem-solving skills and apply a linear equation to find the solution.