Adding Water to Car Radiator: 40% or 10% Antifreeze?

AI Thread Summary
To adjust the antifreeze concentration in a car radiator, calculations are necessary based on the current mixture. The radiator holds 10 gallons, with 5 gallons currently containing 60% antifreeze, equating to 3 gallons of antifreeze and 2 gallons of water. To achieve a 40% antifreeze concentration, additional water must be added, calculated through the formula (amount of antifreeze)/(total volume) x 100. For a 10% antifreeze concentration, even more water will need to be added. The total volume must remain within the radiator's 10-gallon capacity.
theoristo
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The radiator of a car can contain 10 gal of liquid. If it is half full with a
mixture having 60% antifreeze and 40% water, how much more water
must be added so that the resulting mixture has only
a) 40% antifreeze? b) 10% antifreeze?
Will it fit in the radiator?
 
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theoristo said:
The radiator of a car can contain 10 gal of liquid. If it is half full with a
mixture having 60% antifreeze and 40% water, how much more water
must be added so that the resulting mixture has only
a) 40% antifreeze? b) 10% antifreeze?
Will it fit in the radiator?

Show your work. At the very least, define some variables that must be found in order to solve your problem. Then try to write some equations that those variables must satisfy. You really do need to be able to do this on your own if you want to pass the course. We can help, but we can't do the problem for you.
 
Ray Vickson said:
Show your work. At the very least, define some variables that must be found in order to solve your problem. Then try to write some equations that those variables must satisfy. You really do need to be able to do this on your own if you want to pass the course. We can help, but we can't do the problem for you.
Sorry I didn't want the solution I just wanted to share it.
 
3/5 = % antifreeze
2/5 = % water
half-full = 5 gallons of mixture

If 5 gallons is in the radiator, and 3/5 of it is antifreeze, then there must be 3 gallons of antifreeze in the radiator. Similarly, the other 2/5 must be water.

a.) remember that (amount of antifreeze)/(total) x 100 = % antifreeze
Since you know the desired amount of antifreeze, we must add a certain amount of water to the total mixture in order to reduce the percentage of antifreeze. (3)/(5 + x gal of water) = (2/5)

b.) same as part a.) but more water must be added
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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