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09 Sep 2008, 05:15
Kudos
Bookmarks
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 ?
a)2mph
b)3mph
c)4mph
d)5mph
e)6mph
answer is 3 mph . Please explain it ?
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09 Sep 2008, 05:46
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let time taken for day 1 = t and average speed = t
Then, time taken for day 2 = t + 2 and average speed = s + 1.
Now, t + t+2 = 18 => t = 8.
Also, ts + (t+2)(s+1) = 64
Or, s = 3.
Hence, answer is 3mph.
Then, time taken for day 2 = t + 2 and average speed = s + 1.
Now, t + t+2 = 18 => t = 8.
Also, ts + (t+2)(s+1) = 64
Or, s = 3.
Hence, answer is 3mph.
09 Sep 2008, 06:51
Kudos
Bookmarks
Hi..
I want to explain it very basics..
we have total 18 hours. and on second day hiker walks 2 hours more than yesterday.
let's first day he walked h hours, so second day is h + 2
h + ( h + 2 ) = 18
h = 8 hours.
let's assume he walked at x miles per hour on first day and x + 1 on second day.
And, hiker has walked 64 miles in last 2 days..
so, 8*x + 10*(x + 1) = 64
8x + 10x + 10 = 64
18x = 54
x = 3 mph. It is our answer.
I want to explain it very basics..
we have total 18 hours. and on second day hiker walks 2 hours more than yesterday.
let's first day he walked h hours, so second day is h + 2
h + ( h + 2 ) = 18
h = 8 hours.
let's assume he walked at x miles per hour on first day and x + 1 on second day.
And, hiker has walked 64 miles in last 2 days..
so, 8*x + 10*(x + 1) = 64
8x + 10x + 10 = 64
18x = 54
x = 3 mph. It is our answer.
