seekmba wrote:

\(t = s/v/1.5 = 1.5s/v = 1.5t\) (sorry if the equation looks awkward but I dont know how to put the division sign)

\(1.5t = t + 15\)

We can solve the equation and find the normal time.

You can see that \(time = \frac{distance}{speed}\) ---> This is the fundamental equation used to Calculate Bill's normal time.

Since the question states that Bill drove 1.5 times slower, this means that New Speed = \(\frac{Normal Speed}{1.5}\)

Now if we take the slower speed (V/1.5) and plug it in the fundamental formula \(time = \frac{distance}{speed}\) we get:

\(time =\frac{distance}{speed/1.5}\)

\(time = \frac{1.5 * distance}{speed}\)

The above equation also equals to:

\(1.5 * t2 = \frac{distance}{speed}\) where t2 is the new time and t1 is the original time.

Therefore, the equation boils down to: 1.5 * New Time = Original Time + 15

which equals to 1.5 t2 = t1 + 15 ---> Delta T = 1.5t - t = 15 ---> 0.5 t = 15 therefore the new time it took for him to drive to school with a speed 1.5 times slower is 30 minutes. Therefore (A) is SUFFICIENT.

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