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 November 8th, 2016, 01:57 PM #1 Senior Member   Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10 One car travels 140 miles in the same amount of time it takes a second car traveling One car travels 140 miles in the same amount of time it takes a second car traveling 6 miles per hour slower than the first to go 116 miles. What are the speeds of the cars? sorry if I asked this before could not find it. Thanks November 8th, 2016, 02:04 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521 $v \cdot t = 140$ $(v-6)t = 116$ solve the system for $v$, the speed of the faster car, then subtract 6 to get the speed of the slower car Last edited by skipjack; November 8th, 2016 at 04:49 PM. November 8th, 2016, 04:16 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Notice that you can eliminate "t" by dividing one equation by the other! November 8th, 2016, 05:08 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,641 Thanks: 2083 As the first car travels 24 miles further when its speed is 6 mph more than the second car's, their journey time is 4 hours each. Hence the first car travels at 140/4 miles per hour (which is 35 mph), and the second car travels at 116/4 miles per hour (which is 29 mph). November 10th, 2016, 01:32 AM #5 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $v⋅t=140....(1)$ $(v−6)t=116.......(2)$ divide equation(1) by (2) $\dfrac{v\not t}{(v-6)\not t}=\dfrac{140}{116}$ $116v=140v-840$ $-24v=-840$ $v=35\ (\text{faster car})$ $v=29\ (\text{slower car})$ Last edited by skipjack; November 10th, 2016 at 01:35 AM. November 11th, 2016, 09:59 AM #6 Member   Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 I first approached the problem with a testing method: If both cars drove for 1 hour. The first car would have to drive 140 mph to cover 140 miles. The slower car would have to drive 116 mph. However, 140-116 is 24mph difference; not the stated 6 mph difference. Then made each car drive for 2 hours. The first car would have to drive 70mph (140 divided by 2), while the second car would have to drive 58 mph (116 divided by 2). 70-68 is a 12 mph difference; not a 6 mph difference. I noticed when I doubled the time, the speed different cut in half from 24 down to 12 mph. Therefore I knew the next doubling of time to 4 hours would show a 6 mph difference in speed. And it does. At 4 hours of driving, the faster car travels 35 mph (140 divided by 4) and the slower car travels 29mph (116 divided by 4). 35-29 =6 mph. __________________________________________________ ___________ To solve this problem another way, I used Distance = Rate x Time; D=R x T; D=RT and solved it for Time first: 140=RT note: R is the speed for the faster car 116=(R-6)T (R-6) is the speed of the slower car Next, I subtracted those two equations above 140-116 = 24 RT - (R-6)T factor out T T[R-(R-6)] simplify T[-(-6)] T =6T 24 = 6T T= 4 Finally, I solved for rate D = R x T 140= R x 4 R= 35 mph Faster car 35-6 = 29 mph Slower car December 19th, 2016, 10:58 AM #7 Member   Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 I first approached the problem with a testing method: if both cars drove for 1 hour. In this case, the first car would have to drive 140 mph to cover 140 miles. The slower car would have to drive 116 mph. However, 140-116 is 24mph difference; not the stated 6 mph difference. Then I made each car drive for 2 hours. The first car would have to drive 70mph (140 divided by 2), while the second car would have to drive 58 mph (116 divided by 2). 70-68 is a 12 mph difference; not a 6 mph difference. I noticed when I doubled the time, the speed difference cut in half from 24 down to 12 mph. Therefore I knew the next doubling of time to 4 hours would show a 6 mph difference in speed. And it does. At 4 hours of driving, the faster car travels 35 mph (140 divided by 4) and the slower car travels 29mph (116 divided by 4). 35-29 =6 mph. __________________________________________________ ___________ To solve this problem another way, I used Distance = Rate x Time; D=R x T; D=RT and solved it for Time first: 140=RT note: R is the speed for the faster car 116=(R-6)T (R-6) is the speed of the slower car Next, I subtracted those two equations above 140-116 = 24 RT - (R-6)T factor out T T[R-(R-6)] simplify T[-(-6)] T =6T 24 = 6T T= 4 Finally, I solved for rate D = R x T 140= R x 4 Tags 140, amount, car, miles, takes, time, traveling, travels Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post AKA Differential Equations 2 January 25th, 2016 04:44 AM gju7 Differential Equations 5 August 18th, 2015 03:22 AM caters Math 5 March 12th, 2015 08:21 AM Aloysius Algebra 2 March 27th, 2013 07:01 AM fc_groningen Advanced Statistics 0 February 4th, 2011 07:54 AM

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