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# [Urgent] Distance, rate problem

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

Annnnnd CPhill beat me to it

Jul 23, 2019
#4
+2

No big deal, KLP.....we both got the correct answer.....I gave you a point, too !!!   CPhill  Jul 23, 2019