A hiker walked for two days. On the second day the hiker walked 2 hours longer

[BTGModeratorVI]

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

[Brent@GMATPrepNow]

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

If we let x = the hiker's walking speed on day 1, then x+1 = the hiker's walking speed on day 2 (since the hiker walked 1 mph faster on the second day)
So, the hiker's AVERAGE speed will be BETWEEN x and x+1 mph

The hiker walked 64 miles and 18 hours.
Average speed = distance/time = 64/18 ≈ 3.5 mph

So, according to the above property, it must be the case that x = 3 mph, and x+1 = 4 mph

Answer: B

Cheers,
Brent

[Brent@GMATPrepNow]

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

[swerve]

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

Let \(x\) be the number of miles in first day; \(64-x =\) number of miles in second day
given,
\(t+ t+2 =18 \rightarrow t=10\)
Speed in first day \(=\) speed in second day \(- 1\)

Therefore, \(\dfrac{64−x}{10}-\dfrac{x}{8}= 1\)

\(x = 24\)

Therefore, speed \(= \dfrac{d}{t} = \dfrac{24}{8} = 3\)