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November 8th, 2016, 02:57 PM   #1
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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.

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November 8th, 2016, 03:04 PM   #2
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$v \cdot t = 140$

$(v-6)t = 116$

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.
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November 8th, 2016, 05:16 PM   #3
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Notice that you can eliminate "t" by dividing one equation by the other!
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November 8th, 2016, 06:08 PM   #4
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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).
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November 10th, 2016, 02:32 AM   #5
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$v⋅t=140....(1)$

$(v−6)t=116.......(2)$

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.
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November 11th, 2016, 10:59 AM   #6
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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
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December 19th, 2016, 11:58 AM   #7
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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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