BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Boater

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 21
Joined: Sun Feb 08, 2009 12:56 pm
Location: Midwest
Thanked: 1 times
by charlie33 » Sat Jun 27, 2009 9:03 am
A boater travels 21 miles upstream against a 2 mi/h current, then returns downstream to the starting point. If the entire trip took 10 hours, what is the rate of the boat in still water?

A. 4 mi/h
B. 7 mi/h
C. 5 mi/h
D. 6 mi/h
E. 3 mi/h
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 43
Joined: Tue Feb 24, 2009 3:12 am
Thanked: 3 times
GMAT Score:730

Answer

by chintanjadwani » Sat Jun 27, 2009 9:10 am
IMO C[/spoiler]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 316
Joined: Mon Sep 22, 2008 12:04 amFollowed by: 1 member
Thanked: 36 times

by Morgoth » Sat Jun 27, 2009 9:35 am
Let speed of the boat in still water = x

Upstream

Since, the boat is going upstream, more speed is required. The speed would be x+2

D/S = T

T= 21/(x+2)


Downstream

Since, the boat is going downstream, less speed is required. The speed would be x-2

D/S = T

T= 21/(x-2)


Total time = 21/(x+2) + 21/(x-2) = 10

Solve the equation, you'll get

5x^2 -21x -20 = 0

(5x+4)(x-5) = 0

discard 5x+4, coz speed can never be in negative

x-5 = 0

x = 5 mph

Hence C.


Hope this helps.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 392
Joined: Thu Jan 15, 2009 12:52 pm
Location: New Jersey
Thanked: 76 times

by truplayer256 » Sat Jun 27, 2009 9:39 am
Answer should be C. I had the same explanation as Morgoth.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1161
Joined: Mon May 12, 2008 2:52 amFollowed by: 1 member
Location: Sydney
Thanked: 23 times

by mehravikas » Tue Jun 30, 2009 9:37 pm
An easy solution would be to plug in the answer choices. Pick C

U = 3
D = 7

downstream travel time - 21 = T * 7 -> T = 3

upstream travel time - 21 = 3 (10 - T)
21 = 3 * (10-3) -> 21 = 21

Hence answer C.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1