The radiator of a car contains 20 liters of water. Five liters are drained off and replaced by antifreeze. Then 5 liters of the mixture are drained off and replaced by antifreeze, and so on. This process is continued until six drain-offs and replacements have been made. How much antifreeze is the final mixture?
3 answers
16.44 L of antifreeze
Could you please work it out for me so I can see how you got that answer.
at each drainage even we are taking out 1/4 of the current mixture and replacing it with 5 L of pure antifreeze, leaving 3/4 of the previous mixture
after 1st drainage:
amount of antifreeze =5 L
after 2nd drainage:
amount of antifreeze = (3/4)(5) + 5 = 8.75 L
after 3rd drainage:
amount of antifreeze = (3/4)(8.75) + 5 = 11.5625 L
after 4th drainage:
amount of antifreeze = (3/5)(11.5625) + 5 = 13.671875 L
after 5th drainage:
amount of antifreeze = (3/4)(13.6718.. = 15.2539..
after 6th drainage = (3/4(15.2539..)+5 = appr 16.44 L
after 1st drainage:
amount of antifreeze =5 L
after 2nd drainage:
amount of antifreeze = (3/4)(5) + 5 = 8.75 L
after 3rd drainage:
amount of antifreeze = (3/4)(8.75) + 5 = 11.5625 L
after 4th drainage:
amount of antifreeze = (3/5)(11.5625) + 5 = 13.671875 L
after 5th drainage:
amount of antifreeze = (3/4)(13.6718.. = 15.2539..
after 6th drainage = (3/4(15.2539..)+5 = appr 16.44 L
2 answers