Container A has 1600 ml of water and is leaking 12 ml per minute. Container B has 2000 ml and is leaking 20 ml per minute. Let m represest the number of minutes the contains are leaking. How many minutes, m, will it take for the two containers to have the same amount of water

Given:

Container A has 1600 ml of water and is leaking 12 ml per minute. 

Container B has 2000 ml and is leaking 20 ml per minute.

Representation:

Let m represent the number of minutes.

Container A = 1600 - 12m

Container B = 2000 - 20m

Determine the time it would take for the two containers to have the same amount of water.

Solution:

Equate the two given representation of the containers in one solution and then solve for m.

1600 - 12m = 2000 - 20m

Transpose -20 and 1600 on their opposite sides, note that when transposing, the sign of the number changes.

-12 m + 20m = 2000 - 1600

8m = 400

Divide both sides by 8 in order to find the value of m.

m = 50 minutes

Check:

Substitute the value of m in the equation for each container and see if they have the same amount of water.

Container A = 1600 - 12(50)

Container A = 1000 ml 

Container B = 2000 - 20(50)

Container B = 1000 ml

Final answer:

m= 50 or 50 minutes