A hiker walked for two days. On the second day the hiker walked 2 hours longer and at an average speed 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?

general knowledge math & puzzles

  1. 1

  2. 4

  3. 3

  4. 2

Show answer
Correct Option: C

AI Explanation

To solve this problem, let's denote the average speed on the first day as "x" mph.

On the first day, the hiker walks for a certain number of hours, and on the second day, the hiker walks for 2 hours longer. Therefore, let's denote the number of hours walked on the first day as "t" hours.

According to the problem, the hiker walked a total of 64 miles in 18 hours. We can create the following equation based on the given information:

$t \cdot x + (t + 2) \cdot (x + 1) = 64$

Simplifying this equation, we get:

$tx + (t + 2)(x + 1) = 64$

Expanding the equation, we get:

$tx + tx + 2x + t + 2 = 64$

Combining like terms, we get:

$2tx + 2x + t + 2 = 64$

Simplifying further, we get:

$2tx + 2x + t = 62$

Now, we need to use the fact that the total time spent walking is 18 hours. This gives us the equation:

$t + (t + 2) = 18$

Simplifying this equation, we get:

$2t + 2 = 18$

$2t = 16$

$t = 8$

Now that we have the value of "t", we can substitute it back into the equation:

$2tx + 2x + t = 62$

$2(8)x + 2x + 8 = 62$

$16x + 2x + 8 = 62$

$18x + 8 = 62$

$18x = 54$

$x = 3$

Therefore, the hiker's average speed on the first day was 3 mph.

Therefore, the correct answer is option C) 3.

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