|By Akaflex (Akaflex) on Wednesday, October 22, 2003 - 01:57 pm: Edit|
A car radiator contains 10 liters of a 30 percent antifreeze solution. How many liters will have to be replaced with pure antifreeze if the resulting solution is to be 50 percent antifreeze?
can sum1 plz explain
|By Akaflex (Akaflex) on Wednesday, October 22, 2003 - 03:45 pm: Edit|
|By Rohan_Gokhale (Rohan_Gokhale) on Wednesday, October 22, 2003 - 03:48 pm: Edit|
rite now, the radiator has 30% of 10 = 3 liters
to make it 50 %, you have to replace 2 litres of water by antifreeze. however, the anitfreeze is mixed with water.so to remove 2 liters of water, you have to take into account the fact that 70% of the stuff is water. set up a rule of 3 equation
BAM that's your answer
hope this helps
|By Akaflex (Akaflex) on Wednesday, October 22, 2003 - 04:06 pm: Edit|
wait u lost me where did u get the 100?
|By Akaflex (Akaflex) on Wednesday, October 22, 2003 - 04:09 pm: Edit|
|By Akaflex (Akaflex) on Wednesday, October 22, 2003 - 04:10 pm: Edit|
|By Rohan_Gokhale (Rohan_Gokhale) on Wednesday, October 22, 2003 - 04:44 pm: Edit|
remember that in a percentage problem , the best number to put in for the "total quantity" of stuff is always 100.
my reasoning is : since 70% is water, every 100 liters of the stuff contains 70 liters of water.
so you get the relation
100 : 70
x : 2
simply stated , it means
"to get 70 liters of water you need 100 liters of the mixture. so how many liters of the mixture do you need to get 2 liters?"
when you use the relation you got (it's called direct proportion)
you get the equation
70x=200 by cross multiplying
|By Drusba (Drusba) on Wednesday, October 22, 2003 - 05:22 pm: Edit|
Maybe this will help:
The 10 liters are 30% antifreeze, 70% water. Thus there are 3 liters of antifreeze and 7 liters of water in there. You want to take out 2 liters of water and put in 2 liters of antifreeze to get the 50/50 mixture. Each liter that you would remove would be made up of 70% water and 30% anitfreeze and thus you can't just remove 2 liters and put back 2. But the math becomes simple. How many liters that are .7 water do you have to remove to get 2 liters of water? Let x equal the total number of liters you need to remove to get 2 liters of water. The equation becomes .7x=2. x=2.86 liters, the number you have to take out and replace with pure antifreeze to get a 50/50 mixture.
|By Volleygenius (Volleygenius) on Wednesday, October 22, 2003 - 05:25 pm: Edit|
i solved this another way that might be easier to follow since it seems like some of you were confused by rohan's way. i set up 2 equations.
(.3x+y)/(x+y)=.5 and x+y=1
sub in x for y and u get
simplify to get -.7x=-.5 and x=5/7
this means that 5/7 of the final product is the original and 2/7 is the added antifreeze. so multiply by 10 to find liters and u get 20/7
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