MATH SOLVE

2 months ago

Q:
# On the second day of a hiking trip, the tourists covered 17% bigger distance than on the first day. By how many percent did their average speed change from first to second day, if on the second day they were hiking for 20% longer time?

Accepted Solution

A:

speed=distance/time

suppose the distance of the first day is d, and the time is t

distance of the second day: d+0.17d=1.17d

time of the second day: t+0.2t=1.2t

speed of the second day: 1.17d/1.2t=0.975(d/t)=(1-0.025)(d/t)

so the speed of the second day is 2.5% slower than the first day.Β

suppose the distance of the first day is d, and the time is t

distance of the second day: d+0.17d=1.17d

time of the second day: t+0.2t=1.2t

speed of the second day: 1.17d/1.2t=0.975(d/t)=(1-0.025)(d/t)

so the speed of the second day is 2.5% slower than the first day.Β