My Math Forum I am a little Confused.

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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: 520 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: $$$40\text{ mph}$$$$(t+2)\text{ hr}$$=$$v\text{ mph}$$$$t\text{ hr}$$$ $v=\frac{40(t+2)}{t}\:\text{mph}$ We are told $t=\frac{16}{3}\:\text{hr}$ giving us: $v=\frac{40$$\frac{16}{3}+2$$}{\frac{16}{3}}\:\text {mph}=55\text{ mph}$
 March 4th, 2012, 06:47 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,480 Thanks: 2039 The second car made up 80 miles in $\small5\frac13$ 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:
 Originally Posted by MarkFL 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: $$$40\text{ mph}$$$$(t+2)\text{ hr}$$=$$v\text{ mph}$$$$t\text{ hr}$$$ $v=\frac{40(t+2)}{t}\:\text{mph}$ We are told $t=\frac{16}{3}\:\text{hr}$ giving us: $v=\frac{40$$\frac{16}{3}+2$$}{\frac{16}{3}}\:\text {mph}=55\text{ mph}$
Could you explain to me how you got that answer?

 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: 520 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: $d=$$40\text{ mph}$$$$\frac{22}{3}\:\text{hr}$$=v$$\frac{16}{3}\ :\text{hr}$$$ Now we solve for v: $v=\frac{$$40\text{ mph}$$$$\frac{22}{3}\:\text{hr}$$}{$$\frac{16}{3}\ :\text{hr}$$}=\frac{40\cdot22}{16}\:\text{mph}=\fr ac{2^4\cdot5\cdot11}{2^4}\:\text{mph}=55\text{ mph}$
 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: 520 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: $30%\cdot20=40%\cdot x$ $x=\frac{3}{4}\cdot20=15$ 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: 7,932 Thanks: 1127 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.

 Tags confused

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post charlieboon Calculus 5 February 23rd, 2013 10:01 AM ccfoose Algebra 3 April 16th, 2012 09:55 PM katfisher1208 Algebra 1 April 20th, 2010 06:59 PM momoftwo New Users 4 August 19th, 2009 06:15 PM charlieboon Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top