1. ## Word Problem

You have 10L pure juice you take out x liters of juice and replace with water. Then you take out x liters again and add water again. The final concentration is 69% juice. Find X.

2. Lol I feel so stupid lol

3. Originally Posted by AHDDM
You have 10L pure juice you take out x liters of juice and replace with water. Then you take out x liters again and add water again. The final concentration is 69% juice. Find X.
Umm, I miss Math. I am rusty now---no practice. Let me start practicing again with your question.

You take away x liters of pure juice then you put in x liters of water (no juice in it) to make it 10 liters again. So the "juiciness" of the4 new mixture based on the quantity of pure juice in the new mixture is:
(10-x)(100% pure juice) +x(0% pure juice)
= (10-x)(1.00) +x(0)
= 0.1(10-x) <------"juiciness", as if acidity.

Then you take away x liters of the new mixture then replace that with x liters of water again to go back to 10 liters again. The resulting newer mixture is 69% pure juice. So, basing on the quantity of pure juice again in the newer mixture,
(10-x)[0.1(10-x)] +x(0) = 10(0.69)
(10-x)^2 = 69
10-x = sqrt(69)
x = 10 -sqrt(69) = 1.6934 liters -----------answer.

4. Originally Posted by AHDDM
You have 10L pure juice you take out x liters of juice and replace with water. Then you take out x liters again and add water again. The final concentration is 69% juice. Find X.
We remove $x$ litres, so the concentration of the diluted juice is $\frac{10-x}{10}$, and then we remove another $x$ litres of the mixture which contains $x\frac{10-x}{10}$ litres of pure juice, so in total we have removed:

$x+x\frac{10-x}{10}$

litres of pure juice. So the concentration is now:

$
\frac{10-\left( x+x\frac{10-x}{10}\right)}{10}=0.69
$

Now multiply through by $100$ and simplify:

$x^2-20x+31=0$

which has roots $x\approx18.307\ \wedge\ x\approx 1.693$, the first of these is clearly non-physical, so the second is the solution we seek.

RonL