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April 30th, 2017, 08:10 PM  #1 
Member Joined: Jan 2017 From: US Posts: 86 Thanks: 5  Inverse Variation in Algebra The cost per person to ride a bus to a concert varies inversely with the number of people who ride the bus. The bus holds 54 people. If it costs $42 per person when 20 people ride the bus, solve for how much it would cost if the bus was filled to capacity. Be sure to define your variables, find your constant variable, construct your variation equation, and then solve. 
May 3rd, 2017, 07:17 PM  #2 
Member Joined: Jan 2017 From: US Posts: 86 Thanks: 5 
y=42 when x=20 y=kx 42=k*20 So, if x=54: 42=k*54 54/42= 1.3 (rounded to the nearest tenth) 1.3*54=\$70.20 So it would cost \$70.20 when the bus is filled to capacity. Did I do this right? Last edited by greg1313; May 3rd, 2017 at 07:42 PM. Reason: Added ' \' before dollar signs for correct rendering. 
May 3rd, 2017, 07:48 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,475 Thanks: 886 Math Focus: Elementary mathematics and beyond 
The cost per person is inversely proportional to the number of people riding the bus. The more people riding the bus the less cost per person. Start with 42 = k/20. 
May 3rd, 2017, 08:28 PM  #4 
Member Joined: Jan 2017 From: US Posts: 86 Thanks: 5 
Ok so So 42=k/20 42*20= 840 k=840 y=kx 42=840/54 42 = 15.5 So if 42 people ride, the cost would be $15.50..? 
May 3rd, 2017, 09:24 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,475 Thanks: 886 Math Focus: Elementary mathematics and beyond 
There are errors in your work though your result is almost correct. 42 does not equal 840/54. Write C = 840/54 where C is the cost per person. The result is \$15.56 per person to the nearest penny, as 840/54 = 15.5555... 
May 4th, 2017, 01:30 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 16,947 Thanks: 1255 
Does the driver of the bus "ride the bus"?

May 4th, 2017, 09:53 PM  #7 
Member Joined: Jan 2017 From: US Posts: 86 Thanks: 5  

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algebra, inverse, variation 
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