Ratio's

A

ave13co

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Sep 16, 2019
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Hello and thanks in advance. I have a question about ratio's. ...... If I had a container with 7 quarts of liquid that was the wrong liquid and wanted to replace the wrong liquid with the right liquid. But could only remove and replace 3 quarts at a time creating a mix of right and wrong liquids, what would be the procedure to calculate proportions? I have been able to calculate after the first 3 quarts are replaced. Divide 100 by 7 and multiply that number to give proportions of 57.14 wrong and 42.85 right. This is where I get stuck. Since the remaining liquid is a mix, calculations are not so easy. Any help on how to do this will be appreciated.
 
Every time you are removing 3/7 of the remaining wrong liquid.
 
The remaining liquid is a mix. So I am removing 57.14 percent wrong liquid and 42.85 percent right liquid. Not 100 percent wrong liquid.
 
But after that step, wouldn't the proportions change? There would be less wrong liquid and more right liquid. I am getting a headache thinking about this.
 
Yes, every time the ratio changes. How? You are removing 3/7 of the mix. But since it's a mix, you are removing 3/7 of the wrong and 3/7 of the right liquid. So it's easy to calculate the remaining amount of the wrong liquid at each step.
 
BUT...Upon removing the second 3 quarts, the 3 quarts of right liquid isn't there any more. Only a portion of it. Granted, you are removing an equal amount of wrong liquid but the original 3 quarts of right liquid will become less than before. So the proportions would change. Confusing huh?
 
What is your goal? Your original question does say what you are trying to do.

Do you want to find the amount of "bad liquid" left after n replacements? Or something else?
 
ave13co said:
BUT...Upon removing the second 3 quarts, the 3 quarts of right liquid isn't there any more. Only a portion of it. Granted, you are removing an equal amount of wrong liquid but the original 3 quarts of right liquid will become less than before. So the proportions would change. Confusing huh?
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Let's see.
Initially, the amount of wrong liquid is 7.
At each step the amount of wrong liquid, W, becomes W - (3/7)*W = (4/7)*W.
After first step: Wrong liquid amount is (4/7)*7 = 4.
After 2nd step: (4/7)*(4/7)*7 = 16/7
...
After nth step: (4/7)n*7

Right liquid after nth step: 7 - (4/7)n*7 (whatever is not wrong is right)
Ratio of wrong to right: (4/7)n*7 : 7 - (4/7)n*7
 
I just did this problem recently. I have 7 quarts of transmission fluid in my car but could only remove just under a gallon at a time. I wanted to know how many times to do this to get at least 90% of new fluid in my transmission. Algebra came to the rescue! It really works in real life. And my old Jaguar is driving much better.
 
Dr.Peterson said:
What is your goal? Your original question does say what you are trying to do.

Do you want to find the amount of "bad liquid" left after n replacements? Or something else?
Click to expand...
Want to replace all bad liquid with good liquid.
 
Jomo said:
I just did this problem recently. I have 7 quarts of transmission fluid in my car but could only remove just under a gallon at a time. I wanted to know how many times to do this to get at least 90% of new fluid in my transmission. Algebra came to the rescue! It really works in real life. And my old Jaguar is driving much better.
Click to expand...
Ditto. Only my auto is a Ford Focus. Trouble is, I am not as algebra literate as you. I do think that after the third step, the amount of bad liquid would be insignificant. Also comes to the point of just waisting perfectly good trans fluid. But still much less expensive than taking the car to the Ford garage at $100.00 per hour.
 
Did you do the calculation and find what the fraction is after 3 steps?

What do you define as insignificant for your purposes? That (rather than "all bad liquid with good liquid") should be your goal.
 
lev888 said:
I am afraid (4/7)n*7 = 0 does not have a solution.
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Thank you for your reply. And I am sure you are right. Problem is that I have no idea what your message means. I took algebra 50 years ago and was able to get by my classes and graduate but haven't had much need for it since then. Thanks again.
 
ave13co said:
Thank you for your reply. And I am sure you are right. Problem is that I have no idea what your message means. I took algebra 50 years ago and was able to get by my classes and graduate but haven't had much need for it since then. Thanks again.
Click to expand...
After step n the amount of wrong liquid is (4/7)n*7. The question is: when does it become 0? So, we equate the expression to 0 and try to solve for n. But the equation does not have a solution - no matter have many steps you perform it will not be zero.
 
Hello ave13co. You didn't reply with your tolerance. After 13 swaps, you'll have a bit less than a teaspoon of baaad fluid left. After three swaps, it would be one quart plus about 1¼ cups left.

We've assumed that the two fluids are thoroughly mixed, before draining three quarts.

?
 
ave13co said:
Want to replace ALL bad liquid with good liquid.
Click to expand...
Can't be done, sorry. But you can get extremely close to 100% new fluid but never 100%.
What you should do is he following, but this is not exactly math. Change the fluid three times with driving a few miles in between. Then every time you change your engine oil also pump out one quart of transmission fluid.
 
Jomo said:
Can't be done, sorry. But you can get extremely close to 100% new fluid but never 100%.
What you should do is he following, but this is not exactly math. Change the fluid three times with driving a few miles in between. Then every time you change your engine oil also pump out one quart of transmission fluid.
Click to expand...
Problem solved. The auto in question was made to use a certain trans fluid. I put Valvoline MaxLife ATF in. The jug has a great big Mercon LV on it. The auto parts guy said it was the right stuff. After checking my owner's manual, I see that it says Mercon V. Also after checking Google, I find that LV and V are not compatible. Made me really worried and wanted to remove the bad trans fluid. This morning, I went to the Valvoline website to read about this MaxLife ATF stuff. And I found that it is compatible with both LV and V. Crazy huh? So I will be leaving all the trans fluid in place. But I guess there was a math problem that was at least interesting. Thanks all for your help.
 
ave13co said:
.… The auto parts guy said it was the right stuff …
Click to expand...
A common lament (although, not in this case). I just returned wiper blades two days ago; the guys pulled blades with two different arm attachments.

Glad to hear your own research paid off, and, if you're up for another interesting math problem …

You have a single mug of coffee and a small pitcher of cream. Each container has 10 ounces of liquid. You have a spoon which holds one ounce of liquid. Transfer one full spoonful of cream to the coffee, mix thoroughly and then transfer one full spoonful of the mixture back to the cream. Is there now more cream in the coffee or more coffee in the cream?

:devilish:
 
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