The hiker walked for two days. on the second day the hiker walked 2 hours longer and at an average speed 1 mile per hour faster than he walked on the first day . if during the two days he walked a total of 64 miles and spent a total of 18 hours walking , what was his average speed on the first day?
a)2mph
b)3mph
c)4mph
d)5mph
e)6mph
answer is 3 mph . Please explain it ?
Distance problem paper test code 14 section 5 question #16
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first dayifthyder wrote:The hiker walked for two days. on the second day the hiker walked 2 hours longer and at an average speed 1 mile per hour faster than he walked on the first day . if during the two days he walked a total of 64 miles and spent a total of 18 hours walking , what was his average speed on the first day?
a)2mph
b)3mph
c)4mph
d)5mph
e)6mph
answer is 3 mph . Please explain it ?
Let distance on the first day be x
D = S * T
x = S*T
Total Time = 18 hrs = T+T+2 = 2T + 2, T = 8
x = S*8----------I
Second day
D-x = (S+1)*(T+2)
64-x = (S+1) * 10 -----------II
Substitute x in the II equation
64-8S = (S+1)*10
S = 3
Hence B is the answer.
You can also try pluggin in the answer choices.
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can someone explain when its okay to add times (as done here) versus when it's not and you must use the inverse?
- neerajkumar1_1
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what u r confusing is a time and work problem with avg rate problem...missrochelle wrote:can someone explain when its okay to add times (as done here) versus when it's not and you must use the inverse?
try to practise a lot and identify what the question is asking...
and to further clear ur thoughts...
when we add inverse if time... its not time which we are adding... the inverse of time becomes "rate".... so what we add is rates of doing work by two people...
Hope this Helps!!!