[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?

Guest 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

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

KeyLimePi Jul 23, 2019