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Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveli

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Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveli  [#permalink]

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New post 22 Jul 2018, 20:48
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Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveling toward Cincinnati, and another car leaves Cincinnati at the same time, traveling toward Toledo. The car leaving Toledo averages 15 miles per hour faster than the other, and they meet after 1 hour and 36 minutes. What is the speed of the faster car in miles per hour?

A. 55
B. 66
C. 70
D. 81
E. 110

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Re: Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveli  [#permalink]

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New post 22 Jul 2018, 20:56
Bunuel wrote:
Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveling toward Cincinnati, and another car leaves Cincinnati at the same time, traveling toward Toledo. The car leaving Toledo averages 15 miles per hour faster than the other, and they meet after 1 hour and 36 minutes. What is the speed of the faster car in miles per hour?

A. 55
B. 66
C. 70
D. 81
E. 110


Let the speed be X for car leaving Toledo. Therefore, the speed of the car leaving Cincinnati is (X-15).
Now from question: X * 96/60 + (X-15)*96/60 = 200
=> After simplifying, (2X-15)*8/5 = 200
=> 2X-15 = 200*5/8= 125
=>2X= 140
=> X= 70

C is the Answer
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Re: Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveli  [#permalink]

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New post 22 Jul 2018, 20:59
Bunuel wrote:
Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveling toward Cincinnati, and another car leaves Cincinnati at the same time, traveling toward Toledo. The car leaving Toledo averages 15 miles per hour faster than the other, and they meet after 1 hour and 36 minutes. What is the speed of the faster car in miles per hour?

A. 55
B. 66
C. 70
D. 81
E. 110


Lets assume B travels from C to T and avg speed is x mph & A travels from T to C and avg speed is (x+15) mph.
Now 1 hour and 36 minutes = 1.6 hrs
Hence by the given condition :
1.6x+1.6(x+15)= 200
or 2*1.6x+1.6*15=200
or 3.2x+24=200
or 3.2x=176
or x=55 .......Hence the speed of the faster car = 55+15= 70 mph...............Answer C
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Re: Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveli  [#permalink]

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New post 22 Jul 2018, 21:13
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OA: C
Speed of car(Cincinnati --> Toledo) = \(x\) miles per hour
Speed of car(Toledo --> Cincinnati) =\(x+15\) miles per hour
Time from start to when they meet = \(1\)hour\(36\) minutes = \(1 +\) \(\frac{36}{60}\) =\(1.6\)hours
Distance between Toledo and Cincinnati = \(200\) miles
Realtive Speed of car(Toledo --> Cincinnati) with the respect to car moving(Cincinnati --> Toledo) \(= x+15-(-x) =x+15+x =2x+15\)
\(\frac{{200}}{{2x+15}}= 1.6\)
\(2x+15 = \frac{200}{1.6}\)
\(2x+15=125\)
\(2x=110\)
\(x=55\)
Speed of car (Toledo --> Cincinnati) = \(x+15\) miles per hour = \(55+15\) miles per hour = \(70\) miles per hour
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Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveli  [#permalink]

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New post 23 Jul 2018, 08:27
Bunuel wrote:
Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveling toward Cincinnati, and another car leaves Cincinnati at the same time, traveling toward Toledo. The car leaving Toledo averages 15 miles per hour faster than the other, and they meet after 1 hour and 36 minutes. What is the speed of the faster car in miles per hour?

A. 55
B. 66
C. 70
D. 81
E. 110

"Closing the gap" approach
We need the speed of the faster car, "Car A." Slower car = "Car B"

Distance between Car A and Car B = 200 miles = gap distance

Time taken to close the distance gap: 1 hour 36 minutes
\(=1\frac{36}{60}hrs=1\frac{3}{5}hrs=\frac{8}{5}\) hours

Speeds/rates*. Car A travels 15 miles per hour faster than Car B
Let \(r\) = speed of faster Car A
So \((r-15)\) = speed of slower Car B

Relative / Combined speed
Cars travel in opposite directions, so ADD speeds to find the relative speed at which the distance gap is closed
Relative speed: \((A_{r}+B_{r})=r+ (r-15)\)

Solve with the standard RT=D formula
\(r_{relative}*t=D\)
\((r+r-15)*\frac{8}{3}=200\)
\((2r-15)(8)=1,000\)
\((16r-120)=1,000\)
\(16r=1,120\)
\(r=70\)

Faster Car A travels at 70 mph

Answer C

* Or use \(r\) = slower car and \(r+15\) = faster car. In that case, we are solving for the rate of the slower car, which is trap answer A. We need to add 15 to the rate of the slower car to get the correct answer.
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Re: Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveli  [#permalink]

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New post 25 Jul 2018, 17:33
Bunuel wrote:
Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveling toward Cincinnati, and another car leaves Cincinnati at the same time, traveling toward Toledo. The car leaving Toledo averages 15 miles per hour faster than the other, and they meet after 1 hour and 36 minutes. What is the speed of the faster car in miles per hour?

A. 55
B. 66
C. 70
D. 81
E. 110


Let’s let r = the rate of the slower car, and so (r + 15) is the rate of the faster car. Note that the time traveled for both cars at the moment when they meet is 1 hour and 36 minutes, which we express as (1 + 36/60) hours. We can use the equation distance 1 + distance 2 = total distance:

r(1 + 36/60) + (r + 15)(1 + 36/60) = 200

r(1 + 3/5) + (r + 15)(1 + 3/5) = 200

r(8/5) + (r + 15)(8/5) = 200

8r + 8(r + 15) = 1000

8r + 8r + 120 = 1000

16r = 880

r = 55

The speed of the faster car is 55 + 15 = 70 mph.

Alternate Solution:

Let’s let r = the rate of the slower car, and so (r + 15) is the rate of the faster car. Since the cars are moving towards each other, the distance between the two cars decrease at a rate of r + r + 15 = 2r + 15 miles per hour. We know the distance between the two cars is 200 miles and they meet after 1 hour 36 minutes = 8/5 hours; thus:

200/(2r + 15) = 8/5

1000 = 16r + 120

880 = 16r

r = 55

The speed of the faster car is 55 + 15 = 70 mph.

Answer: C
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Re: Toledo and Cincinnati are 200 miles apart. A car leaves Toledo traveli &nbs [#permalink] 25 Jul 2018, 17:33
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