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A hiker walked for two days. One the second day the hiker walked 2 hours longer and 1 mph 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?



I am clueless on this one

Typo on second line....It should read as follows:


A hiker walked for two days. On the second day the hiker walked 2 hours longer and 1 mph 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?

Day 1 = 8 hours x 3mph = 24 miles
Day 2 = 10 hours x4mph = 40 miles

I worked this out too but only by trial and error working backwards!

Let x = speed on day 1

8x + 10(x + 1) = 64

8x + 10x + 10 = 64

18x = 64 - 10 = 54

x = 54/18 = 3

Thanks Stevie Gee.

Let x = no. of hours walked on 1st day

x + 2 = no. of hours walked on 2nd day

x + (x +2) = 18
2x + 2 = 18
2x = 18 -2
2x= 16
x = 8

I divided the 18 in half (9) and added 1 to one and took 1 from the other.
Of course, the numbers were very simple!

That was how I got the answer 1st time but knew there would be a mathematical solution (having to think back a long time).

It's probably even longer ago for me!