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.

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- Jul 31st 2007, 08:28 PMAHDDMWord 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.

- Jul 31st 2007, 08:51 PMAHDDM
Lol I feel so stupid lol

- Aug 3rd 2007, 10:19 PMticbol
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. - Aug 3rd 2007, 10:56 PMCaptainBlack
We remove $\displaystyle x$ litres, so the concentration of the diluted juice is $\displaystyle \frac{10-x}{10}$, and then we remove another $\displaystyle x$ litres of the mixture which contains $\displaystyle x\frac{10-x}{10}$ litres of pure juice, so in total we have removed:

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

litres of pure juice. So the concentration is now:

$\displaystyle

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

$

Now multiply through by $\displaystyle 100$ and simplify:

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

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

RonL