My Math Forum speed of a car!

 Algebra Pre-Algebra and Basic Algebra Math Forum

July 19th, 2014, 05:28 AM   #1
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speed of a car!

question is attached!

i tried solving it like this:
Mack's time : 40/v
Jack's time : 40/v+40 -1/3

therefore,

40/v = 40/v+40 -1/3
i solved for v and ended up with:
v = 79.5 and 0.503

i obviously did something wrong because the answer at the back of the book is 52.11kmph
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 July 19th, 2014, 06:16 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond Let M be Mack's speed and t be time in hours. 40 = (M + 40)(t - 1/3) 40 = tM, t = 40/M Now solve 40 = (M + 40)(40/M - 1/3) for M. Thanks from wrightarya
 July 19th, 2014, 06:30 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond You have an error in your expression for Jack's time; it should simply be 40/(v + 40). I'll assume you intended 40/v = 40/(v + 40) + 1/3, which gives a correct result. Last edited by greg1313; July 19th, 2014 at 06:45 AM.
July 20th, 2014, 04:41 AM   #4
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From: Australia

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Quote:
 Originally Posted by greg1313 You have an error in your expression for Jack's time; it should simply be 40/(v + 40). I'll assume you intended 40/v = 40/(v + 40) + 1/3, which gives a correct result.
im still not sure why this question has 1/3 added onto the 40/(v+40)

40/v = 40/(v + 40) + 1/3

But from the other problem i posted (the 'trains problem') the 1/2 is subtracted from the 105/(x-10)

105/x = 105/(x-10) -1/2

why is the 1/2 subtracted? and why is the 1/3 added? arent the questions sort of the same?

 July 20th, 2014, 12:03 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 40/v > 40/(v + 40) 105/v < 105/(v -10)

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