Another approach:
On the second day the hiker walked 2 hours longer than he walked on the first day. During the two days he spent a total of 18 hours walking
Let h = # of hours walked on first day
So, h + 2 = # of hours walked on second day
We can write: h + (h + 2) = 18
Simplify: 2h + 2 = 18
Solve: h = 8
So, the hiker traveled for 8 hours on the first day and for 10 hours on the second day

On the second day the hiker walked at an average speed 1 mile per hour faster than he walked on the first day. During the two days he walked a total of 64 miles.
Let x = the hiker's speed (in kilometres per hour) on the first day
So, x + 1 = the hiker's speed (in kilometres per hour) on the second day

distance = (time)(speed)
Let's start with the following word equation: (distance traveled on the first day) + (distance traveled on the second day) = 64
Substitute to get: (8)(x) + (10)(x + 1) = 64
Simplify: 8x + 10x + 10 = 64
Simplify: 18x + 10 = 64
Subtract 10 from both sides: 18x = 54
Divide both sides by 18 to get: x = 3

So, the hiker's speed on day one was 3 kilometres per hour

Answer: B

Cheers,
Brent

In general, I spend some time traveling A mph, and the rest of the time traveling A mph, then my average speed will be BETWEEN A mph and B mph.
For example, if I spend some time traveling 5 mph, and the rest of the time traveling 10 mph, then my average speed will be BETWEEN 5 and 10 mph.

We know the hiker had two speeds: x mph and x+1 mph
Since the average speed was 3.5 mph, x must be 3, since this would mean m+1 equals 4, which case 3.5 mph is BETWEEN 3 mph and 4 mph.

BTW, I added a second, more traditional, solution above.

Cheers,
Brent

Thank you so much!

Day 1:
Time = T
Speed = X Mph
Distance = XT Miles

Day 2:
Time = (T+2)
Speed = (X+1) Mph
Distance = (X+1)(T+2) Miles

Also, Total time = 18 = T + (T+2)
So, T = 8 hours

Now, XT + (X+1)(T+2) = 64 Miles
Solve for X = 3 Mph

Hence, option B

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