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July 4th, 2017, 11:09 AM   #1
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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?
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July 4th, 2017, 11:39 AM   #2
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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.
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July 5th, 2017, 01:29 AM   #3
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Quote:
Originally Posted by magicmetal03 View Post
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?
Here's an alternative method: let's make the ratio have $\displaystyle y$ blue counters and see what we end up with on the red counters side...

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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