Question 1101378
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The distance there is the same as the distance back.
That means the ratio of the speeds is the reciprocal of the ratio of the times.
The ratio of the times there and back is 6:5, so the ratio of the speeds there and back is 5:6.<br>
The problem asks for the percent increase in speed on the way back:
6/5 = 1.2 = 120%
The return speed is 120% of the speed going; it is 20% faster than the speed going.