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).

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