Question 172019: solve the problems A and B
A) how many liters of 4% brine solution must be added to 40 liters of 12% brine solution to dilute it to an 8% solution?
B) A student has a test scores of 85%,78%, and 73% in beginning algebra class. what must she score on the last exam to earn a B?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! A)
Letx=amount of 4% brine solution needed
Now we know that the amount of pure salt in the 4% solution that is added(0.04x) plus the amount of pure salt in the 12% solution(0.12*40) has to equal the amount of pure salt in the final mixture(0.08(40+x)), so our equation to solve is:
0.04x+0.12*40=0.08(40+x) get rid of parens (distributive)
0.04x+4.8=3.2+0.08x subtract 3.2 and also 0.04x from each side
0.04x-0.04x+4.8-3.2=3.2-3.2-0.04x+0.08x collect like terms
1.6=0.04x divide each side by 0.04
x=40 liters--------------------amount of 4% brine solution needed
CK
0.04*40+0.12*40=0.08*80
1.6+4.8=6.4
6.4=6.4
B)
I assume an average score of 80 is a B??????
Let x=score that she must make on the last test
Then (85+78+73+x)/4=80 multiply each side by 4
85+78+73+x=320 or
236+x=320 subtract 236 from each side
x=320-236=84----------------------------this would give her an average of 80
CK
(85+78+73+84)/4=320/4=80
Hope this helps---ptaylor
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