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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: 2,771 Thanks: 1426 
$v \cdot t = 140$ $(v6)t = 116$ 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: 3,261 Thanks: 894 
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: 19,733 Thanks: 1810 
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) $\dfrac{v\not t}{(v6)\not t}=\dfrac{140}{116}$ $116v=140v840$ $24v=840$ $v=35\ (\text{faster car})$ $v=29\ (\text{slower car})$ 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: 34 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, 140116 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). 7068 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). 3529 =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=(R6)T (R6) is the speed of the slower car Next, I subtracted those two equations above 140116 = 24 RT  (R6)T factor out T T[R(R6)] 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 356 = 29 mph Slower car 
December 19th, 2016, 10:58 AM  #7 
Member Joined: Nov 2016 From: USA Posts: 34 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, 140116 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). 7068 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). 3529 =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=(R6)T (R6) is the speed of the slower car Next, I subtracted those two equations above 140116 = 24 RT  (R6)T factor out T T[R(R6)] 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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140, amount, car, miles, takes, time, traveling, travels 
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