An equation you can use to solve this problem would be y=9x+124 (where y represents the total amount of water and x equals the number of minutes that have passed). Using this equation, lets solve the problems.
How many liters of water are in Tank A after the following amounts of time have passed?
4 minutes --> y=9(4)+124 = 160
80 seconds --> 4/3 minutes --> y=9(4/3)+124 = 136
Your third time amount is cut off.
How many minutes have passed, , when Tank A contains the following amounts of water?
151 liters --> (151)=9x+124 [subtract 124 from both sides] --> 27=9x [divide both sides by 9] = 3=x, 3 hours
191.5 liters --> (191.5)=9x+124 [subtract 124 from both sides] --> 67.5=9x [divide both sides by 9] = 7.5=x, 7.5 hours or 7 hours 30 minutes
270.25 liters --> (270.25)=9x+124 [subtract 124 from both sides] --> 146.25=9x [divide both sides by 9] = 16.25=x, 16.25 hours or 16 hours 15 minutes
Tank A initially contained 124 liters of water. It is then filled with more water, at a constant rate of 9 liters per minute. How many liters of water are in Tank A after the following amounts of time have passed?
4 minutes
80 seconds
minutes
How many minutes have passed, , when Tank A contains the following amounts of water?
151 liters
191.5 liters
270.25 liters
liters
1 answer
2 answers