Try this “rates” question on your own!
A 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) 2 mph
(B) 3 mph
(C) 4 mph
(D) 5 mph
(E) 6 mph
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First, let’s consider Day 1 and Day 2’s hours.
If x = hours on Day 1, then x + 2 = hours on Day 2. The question said he walked 18 hours total, so we can set up a simple equation:
x + (x + 2) = 18
2x + 2 = 18
2x = 16
x = 8
Therefore he walked 8 hours on Day 1 and 10 hours on Day 2.
We are told he went 1mph FASTER on Day 2. So if Day 1’s mph is y, then Day 2’s mph is y + 1.
Let’s look at the D = R x T formula.
D1 = R1 x T1
D2 = R2 x T2
If we plug in what we know:
D1 = (y) x 8 hrs
D2 = (y + 1) x 10 hrs
We know that D1 + D2 must equal 64, so we can sum the two equations and set them equal to 64.
(y) x 8hrs + (y + 1) x 10hrs = 64
Simplifying…
64 = 8y + 10y + 10
64 = 18y + 10
54 = 18y
3 = y
The answer is (B).