
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
August 28th, 2017, 05:36 PM  #1 
Member Joined: May 2015 From: U.S.A. Posts: 44 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, 07:16 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,413 Thanks: 717 
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, 08:19 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386  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. 

Tags 
adding, changing, percent, proportion 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Lost on proportion........  Opposite  Algebra  9  November 22nd, 2014 02:28 AM 
Ratio and Proportion  sachinrajsharma  Algebra  2  March 13th, 2013 10:22 AM 
Direct proportion  MathematicallyObtuse  Algebra  4  January 15th, 2011 07:26 PM 
Proportion  haftakhan  Algebra  2  July 5th, 2010 01:49 PM 
What proportion  Mahonroy  Advanced Statistics  2  September 8th, 2009 08:39 AM 