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February 24th, 2017, 01:06 AM  #1 
Senior Member Joined: Apr 2008 Posts: 166 Thanks: 3  another speed problem
Al and Bob, who live in North Vancouver, are Seattle Mariners fans. They regularly drive the 264 km from their home to the ballpark in Seattle. On one particular day, Bob drove to the game. On the return journey, Al was able to increase their average speed by 10% and save 18 minutes on the travelling time. Calculate the average speed when Bob drove to the game. This is how I have attempted it. Let x be the time Bob took to drive to the game and let x18 be the time Al took to drive back home. 264/x=(0.1)(264)/(xeighteen) Solving the equation gives x=20. But the answer key says x=80. I don't get it. Please explain how to do it. Thanks. Last edited by davedave; February 24th, 2017 at 01:10 AM. 
February 24th, 2017, 02:27 AM  #2 
Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30 
Your attempt is wrong. Have a look at the following steps. Let the speed at which Bob drive = $x \;km/p$ Speed of Al will be = $x+10\%$ of $x$ $=x+0.1x\implies1.1x$ Time saved by Al = $18\;min=\dfrac{18}{60}=0.3h$ Time is taken by Bob  Time is taken by Al = $0.3$ $=\dfrac{264}{x}\dfrac{264}{1.1x}=0.3$ $1.1(264)264=0.33x$ $26.4=0.33x$ $x=80$ 
February 24th, 2017, 07:02 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,752 Thanks: 604 
b = Bob's speed 264/b  264/(1.10b) = 18/60 Carry on... 
February 24th, 2017, 11:14 PM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 968 Thanks: 344 Math Focus: Yet to find out.  
February 25th, 2017, 05:27 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,752 Thanks: 604  
February 26th, 2017, 12:02 AM  #6 
Senior Member Joined: Apr 2008 Posts: 166 Thanks: 3 
Thanks a lot, deesuwalka.


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