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 →

rate prob gmat

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 96
Joined: Fri Apr 23, 2010 1:14 amFollowed by: 1 member
Thanked: 1 times
by quantskillsgmat » Tue Dec 27, 2011 10:28 pm
The reduction in speed of a railway engine is directly proportional to square root of number of compartment attached to it. with no compartments attached speed is 42m/hr. with 9 compartments speed is 24m/hr.what are maximum number of compartments that can be attached to the engine.
a)49
b)48
c)46
d)47
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members
Thanked: 230 times

by shankar.ashwin » Wed Dec 28, 2011 2:54 am
Original speed = 42

Reduction in speed = k * Sqrt(No. of compartments)

Given speed = 24 for 9 compartments.

(42-24) = k * Sqrt(9)

18 = 3k

k = 6

The train will stop when 'Reduction in speed = 42'

So, 42 = 6 * Sqrt(No of compartments)

Sqrt(No of compartments) = 7

No. of compartments = 49.

Since the train will be stationary when 49 compartments are attached, 48 is the maximum that can be attached to keep it moving. B
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1