harsh.jain wrote:If speed increases by 33.33%, what is the percent reduction in the time taken to travel the same
distance?
It's a challenging question for me, please some one can help me solve this.
Rate and time are RECIPROCALS.
1/3 faster = 4/3 of the regular speed.
4/3 of the regular speed implies 3/4 of the regular time.
Thus, the time decreases by 1/4 = 25%.
An alternate approach is to plug in.
Let the distance = 12 miles.
Let the regular rate = 3 miles per hour.
Regular time = d/r = 12/3 = 4 hours.
1/3 faster = 3 + (1/3)3 = 4 miles per hour.
Time required at the faster rate = d/r = 12/4 = 3 hours.
Time decrease = 4-3 = 1 hour.
Percent decrease in the time = (time difference)/(regular time) * 100 = 1/4 * 100 = 25.
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