BTGModeratorVI wrote: ↑Mon Sep 07, 2020 6:55 am
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
Answer:
B
Source: Official guide
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
