Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

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

Expert replies
by BTGModeratorVI » Mon Sep 07, 2020 6:55 am
Timer
00:00
Your answer
Difficulty–

right wrong

Pick an answer to see how members did

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
Join the discussion
Source: — Problem Solving |

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
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
Image
Join the discussion

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
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

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
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 \(= \dfrac{d}{t} = \dfrac{24}{8} = 3\)
Join the discussion