Word Problem: We have 5mL of 80% acid. We want to

sacb52man

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Jan 18, 2007
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We have 5mL of 80% acid. We want to dilute it with 2% acid to get 10% acid. How much 2% do we need?
 
This problem is identical to the other. Where's your trouble?.
 
Mixtures

Galactus: I cannot set up a formula for either mixture problem.

5ml of 80%acid dilute with 2% acid to get 10%acid. How much 2% is needed.

sol 5*80%+*2%=(5+X)*10% ??????????? This is the best I can do

2L 75% apple juice how much 10% apple juice is needed to mix to get 20% apple juice
sol I don't get it because,if you start out with two liters of 75% apple,you could mix an infinite amount of 10% apple juice into it and still be over 10% apple juice,so I thought there would be no answer. I was wrong because the answer is 11L I don't get it.

HELP
 
Re: Mixtures

sacb52man said:
Galactus: I cannot set up a formula for either mixture problem.

5ml of 80%acid dilute with 2% acid to get 10%acid. How much 2% is needed.

sol 5*80%+*2%=(5+X)*10% ??????????? This is the best I can do

2L 75% apple juice how much 10% apple juice is needed to mix to get 20% apple juice
sol I don't get it because,if you start out with two liters of 75% apple,you could mix an infinite amount of 10% apple juice into it and still be over 10% apple juice,so I thought there would be no answer. I was wrong because the answer is 11L I don't get it.

HELP
You appear to be in panic mode. Slow down. Take a deep breath. Relax. Think more clearly without the panic.

The goal is 20% Apple Juice. Why does it bother you that it iwll be greater than 10%? It's supposed to be. There is no cause for alarm.

One step at a time. Don't be tempted to jump to the end.

2 L of 75% Apple Juice = 1.5 L of Pure Apple Juicl
(Some) L of 10% Apple Juice = 0.10*(Some) L of Pure Apple Juice
(Some + 2) L of 20% Apple Juice = 0.20*(Some + 2) L of Pure Apple Juice

1.5 L + 0.10*(Some) L = 0.20*(Some+2) L

1.5 + 0.10*(Some) = 0.20*(Some+2)

1.5 + 0.10*(Some) = 0.20*(Some) + 0.2*(2)

1.5 + 0.10*(Some) = 0.20*(Some) + 0.4

1.5 - 0.4 + 0.10*(Some) = 0.20*(Some)

1.5 - 0.4 + 0.10*(Some) = 0.20*(Some)

1.1 + 0.10*(Some) = 0.20*(Some)

1.1 = 0.20*(Some) - 0.10*(Some)

1.1 = 0.10*(Some)

1.1/0.10 = (Some)

11 = (Some)

You check it and see if that is right.

Then, go back and work the problem by equating NONapple juice. Call it water, if you like.

How much water is in the 75% apple juice?

You should get the same answer.
 
Re: Mixtures

Galactus: I cannot set up a formula for either mixture problem.

5ml of 80%acid dilute with 2% acid to get 10%acid. How much 2% is needed.

sol 5*80%+*2%=(5+X)*10% ??????????? This is the best I can do


HELP


tkh done a fine job of explaining. Actually, your equation above isn't too far off. You forgot an x.

\(\displaystyle \L\\\underbrace{5(0.80)}_{\text{5ml of\\80% solution}}+\overbrace{0.02x}^{\text{x amount\\of 2% solution}}=\underbrace{0.10(5+x)}_{\text{total 10% solution}}\)

Now, try it with the portion which is not solution. You should get the same answer.

\(\displaystyle \L\\0.20(5)+0.98x=0.90(5+x)\)

See how that works?.
 
See if you can follow this "way":

General:
a @ x
b @ y
======
a+b @ z
(ax + by) / (a + b) = z

Your problem:
5 @ 80
x @ 2
======
5+x @ 10

(5(80) + x(2)) / (5 + x) = 10
(400 + 2x) / (5 + x) = 10
Solve for x
 
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