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August 28th, 2017, 06:36 PM   #1
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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?
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August 28th, 2017, 08:16 PM   #2
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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
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August 28th, 2017, 09:19 PM   #3
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Quote:
Originally Posted by Rexan View Post
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.
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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 View Post
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 View Post
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 View Post
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"?
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November 17th, 2017, 01:49 PM   #5
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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.
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November 17th, 2017, 02:50 PM   #6
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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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