Rate Problem

Let the distance to Bill's workplace be $D$ miles and the time work start after Bill starts driving be $T$ minutes.

Then,

$\begin {cases} \dfrac {D}{36}\times 60 = T + 15 &\Rightarrow \dfrac {5}{3}D = T + 15 &...(1)\\ \dfrac {D}{60}\times 60 = T - 15 &\Rightarrow D = T - 15 &...(2) \end {cases}$

Eq.1 - Eq. 2: $\quad \Rightarrow \left( \dfrac {5}{3}-1 \right)D = 30\quad \Rightarrow \dfrac {2}{3}D = 30 \quad \Rightarrow D = 45$

Substiting $D = 45$ in Eq. 2: $\quad \Rightarrow 45 = T - 15\quad \Rightarrow T = 60$ minutes or $1$ hour.

The required speed for Bill to arrive at work on time $= \dfrac {D}{T} = \dfrac {45}{1} = \boxed {45}$ mph.

If D = Distance, T = Desired travel time, delta = deviation (=1/4 hour) Then, $\frac{D}{36}=T+\delta$ $\frac{D}{60}=T-\delta$ Adding the equations $\frac{D}{36} + \frac{D}{60}= \frac{2D}{45}= 2T$ $T = \frac{D}{45}$ Hence the ideal speed is 45 mph

let "t" be the original time taken by Bill to go to office ..... and "d" be the dist. frm home to office

using "Dist. = Speed * Time" , d = 36 (t + 0.25) [15 min = 0.25 hrs ]

and d = 60 (t - 0.25)

equating these 2 eqns, v get 36t +9 = 60t - 15 ... that means 24t = 24; therefore t = 1

using t=1 in one of the eqns, v get "d = 45 miles"

so Speed he should travel at = 45/1 = 45mph