If a car had traveled 20 kmh faster than it actually did, the trip

Using answer choices, it is much easier if you use 30 minutes = \(\frac{1}{2}\) hour

Start with Answer C to get a benchmark
If original speed was 50, then time, t = D/r:
Time at 50 mph: \(\frac{60}{50}=\frac{6}{5}\) hours
20 mph faster? (50 + 20) = 70 mph. t = D/r
Time at 70 mph: \(\frac{60}{70}=\frac{6}{7}\) hours

That hour fraction in minutes will not yield an integer (30 minutes), but find the difference in times to get a benchmark
\((\frac{6}{5} - \frac{6}{7})=\frac{(42-30)}{35}=\frac{12}{35}\) hours
\(\frac{12}{35}\approx{\frac{1}{3}}\) hour
\(\frac{1}{3} < \frac{1}{2}\) hr

We need time to be longer. That means speed must be slower (decreased speed = increased travel time). Eliminate answers D and E.

Try B) 40
Time at 40 mph: \(\frac{60}{40} = \frac{3}{2}\) hour
20 mph faster: (40 + 20) = 60 mph
Time at 60 mph: \(\frac{60}{60}=1\) hour
Time difference? \(\frac{3}{2} - 1 = \frac{1}{2}\) hour = 30 minutes

Answer B