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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

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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

by M7MBA » Tue Mar 23, 2021 7:19 am

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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
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Re: A hiker walked for two days. On the second day, the hiker walked 2 hours longer and at an average speed 1 mile per h

by Brent@GMATPrepNow » Wed Mar 24, 2021 6:05 am
M7MBA wrote: ↑
Tue Mar 23, 2021 7:19 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
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 Hanneson - Creator of GMATPrepNow.com
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