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[Urgent] Distance, rate problem 

if a car travels from location a to b at 100km/h the driver will arrive 15 minutes late. if the car travels 120 km/h the driver will arrive 40 minutes early. what is the distance between location a and b?

 Jul 23, 2019
 #1
avatar+111393 
+3

Remember that Rate * Time = Distance

Let the normal driving time  be, T  (in hours)

Let 15 min  = (1/4)hr

Let 40 min  = (2/3)hr

 

So....equating distance betwen a and b, we have that

 

100 * ( T + 1/4)   =   120 * ( T - 2/3)      simpify

 

100T + 25  =  120T  - 80       add 80 to both sides , subtract 100T from both sides

 

105  =  20T         divide both sides by 20

 

105 / 20  = T   =  5.25 hrs

 

So....the total distance between a and b is  

 

100 * ( 5.25 + 1/4)   =   100 (5.25 + .25 )  = 100 (5.50)  =  550 km

 

 

cool cool cool

 Jul 23, 2019
edited by CPhill  Jul 23, 2019
 #2
avatar+142 
+3

Let \(d\) be the distance to location b in km. Since you drove \(100\) km/h, then the time that it took you to get there is \(\frac{d}{100}\). If you drove \(120\) km/h then the time would be \(\frac{d}{120}\). Since you drove arrived \(55 \) minutes later, and \(\frac{55}{60}\), or \(\frac{11}{12}\). Therefore we can create the equation, \(\frac{d}{120}=\frac{d}{100}-\frac{11}{12}\). Multiplying both sides by \(600\). The LCM. We get \(5d=6d-550\). Subtract \(6d\) from both sides. \(-d=-550\). Divide by \(-1\)\(d=550\). The distance between location \(a\) and location \(b\) is \(550\) km.

 

 

-\(\pi\) KeyLimePi

 Jul 23, 2019
 #3
avatar+142 
+3

Annnnnd CPhill beat me to it

 Jul 23, 2019
 #4
avatar+111393 
+2

No big deal, KLP.....we both got the correct answer.....I gave you a point, too !!!

 

 

cool cool cool

CPhill  Jul 23, 2019

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