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Distance problem paper test code 14 section 5 question #16

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by ifthyder » Tue Sep 09, 2008 4:11 am
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 ?
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Re: Distance problem paper test code 14 section 5 question

by parallel_chase » Tue Sep 09, 2008 5:23 am
ifthyder 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 ?
first day

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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by missrochelle » Mon Oct 18, 2010 6:46 pm
can someone explain when its okay to add times (as done here) versus when it's not and you must use the inverse?
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by neerajkumar1_1 » Mon Oct 18, 2010 7:13 pm
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?
what u r confusing is a time and work problem with avg rate problem...
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!!!
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