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 November 22nd, 2016, 01:13 PM #1 Senior Member   Joined: Apr 2008 Posts: 175 Thanks: 3 How do you approach this word problem? (ex) The Fine Arts High School Piano club has 30 students, some from Grade 11 and the remainder from Grade 12. Each pair of students from the club must play a piano duet together over the course of the year. When two grade 11 students play together, they need 2 hours of practice time. When a grade 11 and a grade 12 student play together, they need 3 hours of practice time. When two grade 12 students play together, they need 4 hours of practice time. In total, the students need 1392 hours of practice time. How many grade 11 students are there in total? my attempt let x be the number of grade 11 students let y be the number of grade 12 students let a be the number of hours of practice by two grade 11 students let b be the number of hours of practice by two grade 12 students let c be the number of hours of practice by one grade 11 and one grade 12 student Then, I set up the equations (1) x + y = 30 (2) a + b + c = 1392 Since two grade 11 students need 2 hours of practice, x=a. That means, if a=8 hours, x=8 (8 grade 11 students play for 8 hours altogether). Similarly, y=b/2. That means, if one grade 11 and one grade 12 have 8 hours of practice, y=4 (4 grade 12 students play for 8 hours). Next, I let x+y=2c/3. If one grade 11 and one grade 12 student practice for six hours, x+y=4. That means, two grade 11 and two grade 12 students play for six hours in total. Then, I rewrite (2) in terms of x and y, combine like terms, and get (3) x + y = 2784/7. Finding the difference of (1) and (3) gives 0 = 337.71, which does not make sense. Can someone explain how to do it? Thanks a lot.
 November 22nd, 2016, 01:44 PM #2 Global Moderator   Joined: May 2007 Posts: 6,275 Thanks: 516 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: 740 Thanks: 298 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: 175 Thanks: 3 Thanks, JeffM1!

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