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 August 28th, 2017, 06:36 PM #1 Member   Joined: May 2015 From: U.S.A. Posts: 45 Thanks: 0 Changing a proportion by adding more The question is: "A manufacturer of soft drinks advertises their OJ as 'naturally flavored' although it only contains 5% OJ. A new regulation calls for 10% juice. How much pure OJ must be added to 900 gal. of OJ to meet the regulation?" I tried a couple ideas: $\displaystyle 900 + x = .1$ because you need to add an unknown amount to get a 10% juice. Definitely not it, though. Then I tried $\displaystyle 900 + x = 900(1.1)$ because you need to add an unknown amount to get a 10% juice, and I tried the 900(1.1) because the final amount would have to be greater than 900, due to adding the juice to an already established amount. But the last one gives $\displaystyle x = 90$, and that's 10% of the initial amount, which isn't what the problem is looking for. It seems to be a problem of perpetually increasing dilution that I can't figure out how to account for: if you add 90 gallons of pure juice, you now have 990 gallons of OJ that needs to be at 10% juice, in which case you need 99 gallons of pure juice, but if you add 9 more gallons, you're at 999 gallons.... and so on. There seems to be something incredibly basic I'm missing. Any insights?
 August 28th, 2017, 08:16 PM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,607 Thanks: 819 we add $j$ gal of 100% juice and end up with $900+j$ gallons of total liquid that is 10% juice. Let's look at the amount of pure juice on each side. $900(0.05) + j = (900+j)(0.1)$ $45 + j = 90+ 0.1 j$ $0.9j = 45$ $j = 50$ so we must add 50 gallons of pure juice to obtain 950 gallons of 10% juice
August 28th, 2017, 09:19 PM   #3
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Quote:
 Originally Posted by Rexan I tried . . . $\displaystyle 900 + x = .1$
That equation should be $0.1(900 + x) = 0.05(900) + x$, which implies $900 + x = 450 + 10x$.
Hence $x = (900 - 450)/(10 - 1) = 50$.

Alternatively, use the method shown below.

There are currently 45 gallons of pure OJ with a remaining 855 gallons.
As 10/(100 - 10) = 1/9, the amount of pure OJ needs to be 1/9 of 855 gallons, i.e. 95 gallons,
so 50 gallons of pure OJ must be added.

November 17th, 2017, 01:38 PM   #4
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Profuse apologies for resurrecting this old thread. I've realized that I still don't understand how to work on this problem.

Quote:
 Originally Posted by romsek we add $j$ gal of 100% juice and end up with $900+j$ gallons of total liquid that is 10% juice. Let's look at the amount of pure juice on each side. $900(0.05) + j = (900+j)(0.1)$
Quote:
 Originally Posted by skipjack That equation should be $0.1(900 + x) = 0.05(900) + x$, which implies $900 + x = 450 + 10x$.
So, here, both of you came up with the same equation. What's the reasoning behind where to place the parentheses? I can see where the numbers come from (.05 for 5% juice, .1 for the 10% juice we're aiming for), but, besides trial and error, how would you know it's $\displaystyle 0.05(900) + x$ and not $\displaystyle 0.05(900 + x)$ since that's the format while working with the .1?

Quote:
 Originally Posted by skipjack Alternatively, use the method shown below. There are currently 45 gallons of pure OJ with a remaining 855 gallons. As 10/(100 - 10) = 1/9, the amount of pure OJ needs to be 1/9 of 855 gallons, i.e. 95 gallons, so 50 gallons of pure OJ must be added.
How did you create the original fraction of 10/(100-10)? The 10 seems to be from the "10%" part of the problem, but what do the numerator and denominator stand for? It doesn't seem to be "desired value"/"present value" since the present value was 95% (or 9.5). Is it "desired value"/"the whole" - "desired value"?

 November 17th, 2017, 01:49 PM #5 Senior Member   Joined: Jun 2015 From: England Posts: 697 Thanks: 199 This whole question is a subdivision of alligation, which is about mixing two strengths together to obtain a desired intermediate strength, perhaps also with a specific amount in the end result. Google offers insights into various tried and tested methods. https://www.google.co.uk/search?q=alligation+method Last edited by skipjack; November 17th, 2017 at 02:28 PM.
 November 17th, 2017, 02:50 PM #6 Global Moderator   Joined: Dec 2006 Posts: 18,166 Thanks: 1424 Originally, pure OJ was 5% of the juice, giving a 5:95 ratio of pure OJ to diluent. The desired ratio is 10:90. The original volume of diluent was 95% of 900 gallons, which is 855 gallons. The desired volume of pure OJ is therefore 10/90 times 855 gallons. which is 95 gallons. As the original volume of pure OJ was 5% of 900 gallons, which is 45 gallons, the volume of pure OJ that must be added is 50 gallons.

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