Question 619474: I was given a practice test and I missed a section in class. I know the answer to the word problems below.. but I am not sure how to set up the equations to get the correct answers. There are 2 questions below:
1-A lab has a 10% acid solution and a 25% acid solution. How many liters of each are required to get 300L of a 15% acid solution?
Answer: 200 at 10%
100 at 25%
BUT HOW?!?!
2-Two trains leave Chicago at the same time and in opposite directions. One is going 120MPH and one is going 130MPH. How long before they are 750 miles apart?
Answer: 6
once again.. HOW?!
THANK YOU!!!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! I'll do the first one to get you started. Post the other question if you still need help with it.
# 1
Let x = amount of 10% solution and y = amount of 25% solution
We want our final solution to be 300 L, so x+y = 300. Solve this for y to get y = 300 - x
Since you have x liters of the 10% solution, you have exactly 0.1x liters of acid. Similarly, since you have y liters of the 25% solution, you have exactly 0.25y liters of acid
In total, you have 0.1x+0.25y liters of pure acid. We want our final solution to be 15% acid, so there must be 300*0.15 = 45 liters of pure acid in the final solution (mixed with water and other chemicals)
So we can say 0.1x+0.25y = 45
0.1x+0.25y = 45
0.1x+0.25(300 - x) = 45
0.1x+0.25(300) - 0.25x = 45
0.1x + 75 - 0.25x = 45
-0.15x + 75 = 45
-0.15x = 45 - 75
-0.15x = -30
x = -30/(-0.15)
x = 200
So you need 200 liters of the 10% solution
Now plug x = 200 into y = 300 - x to get
y = 300 - x
y = 300 - 200
y = 100
and you need 100 liters of the 25% solution
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