[Urgent] Distance, rate problem

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

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

Annnnnd CPhill beat me to it