Container A and container B have leaks. Container A has 800 ml of water,and is leaking 6 ml per minute. Container B has 1000ml,and is leaking 10 ml per minute. How many minutes,m,will it take for the two containers to have the same amount of water.
800 - 6 m = 1000 - 10 m
so 4 m = 200
50
To determine the number of minutes it will take for Container A and Container B to have the same amount of water, we need to set up an equation to represent the situation.
Let's denote the number of minutes as "m", and let "x" represent the amount of water (in milliliters) that both containers will have at the end of those m minutes.
For Container A, the amount of water remaining after m minutes can be represented as: 800 - 6m.
For Container B, the amount of water remaining after m minutes can be represented as: 1000 - 10m.
Since we want both containers to have the same amount of water (x), we can set up the following equation:
800 - 6m = 1000 - 10m
Now, let's solve the equation to find the value of m:
800 - 6m = 1000 - 10m
Rearranging the equation:
-6m + 10m = 1000 - 800
4m = 200
Dividing both sides by 4:
m = 200 / 4
m = 50
Therefore, it will take 50 minutes for Container A and Container B to have the same amount of water.