Question

A fire engine starts pumping water at 9:20 am at the rate of 800 gallons per minute. Another fire engine, pumping at the rate of 1000 gallons per minute, starts at 9:30 am. At what time will the two engines have pumped the same number of gallon

Answer 1

Answer:

At 10:10 am

Step-by-step explanation:

Hi to answer this question we have to write a system of equations:

• Fire engine 1: 800 m  

Where m: pumping time of the engine

• Fire engine 2: 1000 (m-10)

Because it starts 10 minute later  

So, putting together both equations:

800m = 1000(m-10)

800m = 1000m - 10000

10000 = 1000m-800m

10000= 200m

10000/200=m

50 =m (50 minutes after the first engine starts)

So, 9:20 am + 50 minutes : 10:10 am .

Answer 2

Answer:

10:10 AM

Step-by-step explanation:

The first thing is to use an identical time variable for both cases, we will do it as follows:

Let t = number of minutes of pumping time of the first fire engine

Therefore, for the second fire truck it would be:

(t-10) = pumping time of the second fire engine, since it started 10 min after the first engine.

To find the value of t, we equalize the equations of the first engine and the second engine:

We know that the first one would be: 800 * t

And the second: 1000 * (t-10)

Thus

1000 * (t-10) = 800 * t

1000 * t - 10000 = 800t

1000 * t - 800 * t = 10000

200 * t = 10000

t = 10000/200

t = 50 minutes

In other words, 50 minutes after the first engine starts pumping, it equaled the second

To know the time they were matched it would be like this:

9:20 AM +: 50 = 10:10 AM

Therefore, at 10:10 AM both engines were matched.

To check the above we have to:

50 * 800 = 40000

40 * 1000 = 40000

Therefore, in that time, they were equalized.