Math question

Discus: SAT/ACT Tests and Test Preparation: October 2003 Archive: Math question
 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

biump

 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
100 70
x 2
cross multiply
70x=200
x=20\7=2.851

hope this helps
rohan

 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

why x/2?

 By Akaflex (Akaflex) on Wednesday, October 22, 2003 - 04:10 pm: Edit

AHHHHHHHHHHHHHHHHHHHHHHHHHHH HELPPPPPPPPPPPPP

 By Rohan_Gokhale (Rohan_Gokhale) on Wednesday, October 22, 2003 - 04:44 pm: Edit

dude
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
therefore x=200\70

rohan

 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
(.3x+1-y)/(x+1-x)=.5
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