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July 4th, 2017, 12:09 PM  #1 
Newbie Joined: Jul 2017 From: USA Posts: 2 Thanks: 0 Math Focus: Basic Arthmetics :D  A Ratio Problem
The ratio of red counters to blue counters is 9:11b There are y blue counters. Express the number of red counters in terms of y. I learned that the answer is 9y:11, but why is that? 
July 4th, 2017, 12:39 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,250 Thanks: 1439 
If the numbers of red and blue counters are precisely in the ratio 9:11, the number of blue counters must be an (exact) integer multiple of 11, namely 11n, where n is a natural number, and the corresponding number of red counters must be 9n. If you are effectively told that 11n = y, multiplying both sides of that equation by 9/11 gives you the equation 9n = 9y/11, which means that the number of red counters is 9y/11. 
July 5th, 2017, 02:29 AM  #3  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,081 Thanks: 698 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
Hint: you can only change ratios using multiplication and division. Starting ratio: 9 red : 11 blue Divide both sides by 11 gives: $\displaystyle \frac{9}{11}$ red : 1 blue Multiply both sides by $\displaystyle y$: $\displaystyle \frac{9y}{11}$ red : $\displaystyle y$ blue Therefore, for every $\displaystyle y$ blue counters, there are $\displaystyle \frac{9y}{11}$ red counters. Then you can do SkipJack's trick if you want (i.e. substitute $\displaystyle y = 11n$ where n is a whole number) because you know that you have a whole number of red counters and blue counters (no bits of counter allowed).  

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