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March 3rd, 2012, 09:29 PM  #1 
Newbie Joined: Mar 2012 Posts: 6 Thanks: 0  I am a little Confused.
Hello, I have a problem that I don't know how to solve. Could someone please explain to me what I am suppose to do? "A car leaves a town traveling at 40 miles per hour. Two hours later a second car leaves the town traveling the same route and overtakes the first car in 5 hours and 20 minutes. How fast was the second car traveling?" 
March 3rd, 2012, 09:43 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: I am a little Confused.
Both cars travel the same distance, while the first car travels for 2 hr more. Let t be the time for the second car then t + 2 is the time for the first car. Let v be the speed of the second car which we assume is constant. Using distance equals rate times time, we may state: We are told giving us: 
March 4th, 2012, 06:47 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,622 Thanks: 2075 
The second car made up 80 miles in hours, so its relative velocity was 15 mph. Hence the answer is 55 mph.

March 8th, 2012, 02:39 PM  #4  
Newbie Joined: Mar 2012 Posts: 6 Thanks: 0  Re: I am a little Confused. Quote:
 
March 8th, 2012, 03:17 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: I am a little Confused.
We begin with distance equals rate times time. We let v be the rate of the second car, since we don't know the rate and this is what we're asked to find. If the second car traveled for 5 1/3 hr and the first car left 2 hours before the second then the second car traveled for 7 1/3 hr at 40 mph. Since they both traveled the same distance d we may state: Now we solve for v: 
March 9th, 2012, 05:52 PM  #6 
Newbie Joined: Mar 2012 Posts: 6 Thanks: 0  Re: I am a little Confused.
Thanks for the explanation. I have another one as well. How much water needs to be removed from 20 gallons of 30% salt solution to change it to a 40% salt solution? 
March 9th, 2012, 06:20 PM  #7 
Senior Member Joined: Jan 2012 Posts: 131 Thanks: 0  Re: I am a little Confused.
30%(Salt) of 20 gallons of solution= 6 gallons 40%(Salt) of x gallons solution= 6 gallons x=15 gallons. H20 to be removed=5 gallons. 
March 9th, 2012, 06:24 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: I am a little Confused.
I would state 30% of 20 gallons is 40% of how many (x) gallons? Writing this mathematically where 20 and x are in gallons: Since 20  15 = 5, we see 5 gallons of water must be removed. 
March 10th, 2012, 12:26 PM  #9 
Newbie Joined: Mar 2012 Posts: 6 Thanks: 0  Re: I am a little Confused.
Okay, I have another one. "Suppose that the perimeter of a square equals the perimeter of a rectangle. The width of the rectangle is 9 inches less than twice the side of the square, and the length of the rectangle is 3 inches less then twice the side of the square. Find the dimensions of the square and the rectangle." 
March 10th, 2012, 01:29 PM  #10 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond  Re: I am a little Confused.
Hi, Siedas. In future please start a new topic for a different (unrelated) question. Let W be the width of the rectangle. Let L be the length of the rectangle. Let s be the side of the square. From the given information W = 2s  9 and L = 2s  3. Since perimeter equals 2 times length plus two times width we have 2W + 2L = 4s 2(2s  9) + 2(2s  3) = 4s 4s  18 + 4s  6 = 4s 4s  24 = 0 4s = 24 s = 6, so the side of the square is 6, the width of the rectangle is 2(6)  9 = 3 and the length of the rectangle is 2(6)  3 = 9. 

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