My Math Forum One car travels 140 miles in the same amount of time it takes a second car traveling

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 November 8th, 2016, 02:57 PM #1 Senior Member   Joined: Feb 2016 From: seattle Posts: 354 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, 03:04 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,546 Thanks: 1259 $v \cdot t = 140$ $(v-6)t = 116$ 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 05:49 PM.
 November 8th, 2016, 05:16 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,479 Thanks: 627 Notice that you can eliminate "t" by dividing one equation by the other!
 November 8th, 2016, 06:08 PM #4 Global Moderator   Joined: Dec 2006 Posts: 17,139 Thanks: 1281 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, 02:32 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 02:35 AM.
 November 11th, 2016, 10:59 AM #6 Newbie   Joined: Nov 2016 From: USA Posts: 21 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[6] =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, 11:58 AM #7 Newbie   Joined: Nov 2016 From: USA Posts: 21 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[6] =6T 24 = 6T T= 4 Finally, I solved for rate D = R x T 140= R x 4

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