two climbers are hikingup a 3000 ft mountain. each day they climb 1000 ft and then come down 250 feet to make camp. how many days willit take them to reach the top
2006-09-12 17:15:33 · 6 answers · asked by hereo 1 in Science & Mathematics ➔ Mathematics
each day climbs 1000ft
returns 250 ft so, net climbing (1000-250)= 750ft
first 3 days total climbing= 750*3 = 2250 ft
last day need to be climbed = 3000-2250= 750ft
as they will not come down after reaching top, so
last 750 ft will be climbed in 750/1000 days= 0.75 days
so total time taken is 3+0.75=3.75 days
2006-09-12 18:51:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
first day = 0 + 1000 - 250 = 750
2nd day = 750 + 1000 -250 = 1500
3rd day = 1500 + 1000 -250 = 2250
4th day = 2250 + 1000 (reach top)
2006-09-12 17:30:06 · answer #2 · answered by Anonymous · 0⤊ 0⤋
If they climb 1,000 feet and come down 250 feet each day, then they're only climbing 750 feet per day. Divide 3,000 by 750 and you get 4 days.
2006-09-12 17:25:24 · answer #3 · answered by Bob P 1 · 0⤊ 0⤋
4 days
3000/750
3000=total
750=total per day
2006-09-12 17:18:11 · answer #4 · answered by rmr2990 2 · 0⤊ 0⤋
(1000 - 250)x = 3000
750x = 3000
x = 4
ANS : 4 days
2006-09-12 18:40:06 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
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2016-10-14 22:56:52 · answer #6 · answered by ? 4 · 0⤊ 0⤋