Two water tanks at a factory begin draining at the same time. The large tank contains

Question

1198
liters of water and drains at a rate of
19
liters each minute. The small tank holds
120
liters of water but is only
5
6
full. It drains at a much slower rate of only
1
liter each minute. How many liters of water are in each tank at the moment that they have the same amount of water?

Answer 1

Let's first find out how long it will take for both tanks to have the same amount of water remaining.

Let x be the number of minutes it takes for both tanks to have the same amount of water remaining.

For the large tank:
Remaining water in large tank = 1198 - 19x

For the small tank:
Remaining water in small tank = 120 * (5/6) - x

Setting the two expressions equal to each other:
1198 - 19x = 120 * (5/6) - x
1198 - 19x = 100 - x

Rearranging:
18x = 1098
x = 61

After 61 minutes, both tanks will have the same amount of water remaining.

Plugging x back into one of the expressions to find out how much water will be left in each tank:
Remaining water in large tank = 1198 - 19*61 = 1198 - 1159 = 39 liters
Remaining water in small tank = 120 * (5/6) - 61 = 100 - 61 = 39 liters

Therefore, after 61 minutes, both tanks will have 39 liters of water remaining.