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

 November 22nd, 2016, 01:44 PM #2 Global Moderator   Joined: May 2007 Posts: 6,558 Thanks: 602 Something is wrong with the way you stated it. The most practice would be if all students were grade 12. In that case there would be 15 pairs each practicing 4 hours. This gives total practice time of 60 hours. 1392 just doesn't fit.
 November 22nd, 2016, 02:24 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 Your identification of unknowns is fine. Your problem is in calculating hours of practicing time. How many distinct pairs of 11th grade students are there? $\dbinom{x}{2} = \dfrac{x!}{2! * (x - 2)!} = \dfrac{x(x - 1)}{2} \implies a = 2 * \dfrac{x(x - 1)}{2} = x^2 - x.$ How many distinct pairs of 12th grade students are there? $\dbinom{y}{2} = \dfrac{y!}{2! * (y - 2)!} = \dfrac{y(y - 1)}{2} \implies b = 4 * \dfrac{y(y - 1)}{2} = 2y^2 - 2y.$ And it should obvious that the number of distinct pairs containing one 11th grader and one 12th grader is $xy \implies c = 3xy.$ With me so far? Now you have $x + y = 30\ and\ x^2 - x + 2y^2 - 2y + 3xy = 1392 \implies$ $x = 30 - y \implies (30 - y)^2 - (30 - y) + 2y^2 - 2y + 3y(30 - y) = 1392.$ Follow that? Now what? Try it yourself before scrolling down for answer. $900 - 60y + y^2 + 900 - 30 + y + 2y^2 - 2y + 90y - 3y^2 = 1392 \implies$ $29y = 1392 - 870 = 522 \implies y = 18 \implies x = 12.$ So the number of distinct pairs of 11th graders is 66 and a = 132. The number of distinct pairs of 12th graders is 153 and b = 612. The number of distinct 11th / 12th grade pairs is 216 and c = 648. And 648 + 612 + 132 = 1392. It checks. Thanks from davedave and romsek Last edited by JeffM1; November 22nd, 2016 at 02:27 PM.
 November 25th, 2016, 01:33 AM #4 Senior Member   Joined: Apr 2008 Posts: 193 Thanks: 3 Thanks, JeffM1!

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