My Math Forum A Ratio Problem

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 July 4th, 2017, 11:09 AM #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, 11:39 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 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, 01:29 AM   #3
Senior Member

Joined: Apr 2014
From: Glasgow

Posts: 2,164
Thanks: 736

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
 Originally Posted by magicmetal03 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).

 Tags problem, ratio

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post WarmCustard Elementary Math 9 May 12th, 2016 07:15 AM marcus2015 Elementary Math 1 April 23rd, 2015 10:37 AM jon619 Algebra 2 October 1st, 2013 01:12 PM homeschooling Elementary Math 2 June 5th, 2013 03:34 PM piggysmile Algebra 4 May 31st, 2010 03:41 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top