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 →

If a motorist

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 159
Joined: Sun Jul 27, 2008 8:41 pm
Thanked: 3 times
by ankit1383 » Fri Dec 19, 2008 1:43 am
If a motorist has driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
A. 100
B. 120
C. 140
D. 150
E. 160
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Mon Dec 08, 2008 9:55 pm
Thanked: 4 times

by 720dreaming » Fri Dec 19, 2008 1:50 am
D. 70 more miles at 5 miles per hour faster means his real rate was 65. Since we know he only drove 1 additional hour.

At ten miles per hour faster, his new rate is 75. 75*2=150.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Tue Dec 16, 2008 2:54 pm

by gmatman1 » Fri Dec 19, 2008 5:42 am
In my opinion the question is not formulated very clearly. I interpretted the question differently, which results in an unsolvable problem.

I assumed that the motorist drove the whole day at the new speed, instead of the last extra hour(s). This means there will be 3 variables. So there is one variable too much to solve the two equations shown below.

s = speed
t = time
d= distance

1. st = d
2. (s + 5)(t + 1) = d + 70
Last edited by gmatman1 on Fri Dec 19, 2008 10:29 am, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Sun Nov 30, 2008 8:41 pm
Thanked: 4 times

by mrsmarthi » Fri Dec 19, 2008 7:46 am
gmatman1 wrote:In my opinion the question is not formulated very clearly. I interpretted the question differently, which results in an unsolvable problem.

I assumed that the motorist drived the whole day at the new speed, instead of the last extra hour(s). This means there will be 3 variables. So there is one variable too much to solve the two equations shown below.

s = speed
t = time
d= distance

1. st = d
2. (s + 5)(t + 1) = d + 70
Let s be the speed, t be the time that he travelled actually. So distance travelled is

st = d -----> Eq1

Hypothetical speed when speed is 5 more and time is 1 hr additional then distance travelled is

(s+5)(t+1) = d + 70. ==> st + s + 5t + 5 = d+70

d + s + 5t = d + 65 (Substituting st = d from Eq 1)

==> s + 5t = 65 --> Eq 2.

Hupothetical situation 2 where speed is 10 miles greater and time is 2 hrs more then

(s + 10)(t + 2) = ?

==> st + 2s + 10t + 20 = ?

==> d + 130 + 20

==> d + 150.

So additional distance travelled would be 150 Miles.

IMO D.
[/b]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Thu Nov 06, 2008 7:46 am

by trnguyen » Fri Dec 19, 2008 8:02 am
(r+5)(t+1) = 70+d

rt+r+5t+5 = 70 +d

rt+r+5t-d=65

d + r + 5t – d = 65

r + 5t = 65

(r+10)(t+2) = x+d

rt+2r+10t+20 = x +d

rt + 2r + 10t – d = x -20

d + 2r + 10t – d = x – 20

2r + 10t = x - 20

Multiply the first equation by 2: 2r + 10t = 130

130 = x – 20

X = 150
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 259
Joined: Thu Jan 18, 2007 8:30 pm
Thanked: 16 times

by amitabhprasad » Fri Dec 19, 2008 9:29 am
Diff in velocity = +5
total time = t+1
Distance covered = 70
==> 70 = 5(t+1) ==> t = 13
if relative velocity = +10
extra time = +2
then distance = (13+2)*10 = 150
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 62
Joined: Thu Oct 23, 2008 9:20 am

by tritrantran » Fri Dec 19, 2008 10:22 am
amitabhprasad wrote:Diff in velocity = +5
total time = t+1
Distance covered = 70
==> 70 = 5(t+1) ==> t = 13
if relative velocity = +10
extra time = +2
then distance = (13+2)*10 = 150
I like this method. Clear and efficient!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Tue Dec 16, 2008 2:54 pm

by gmatman1 » Fri Dec 19, 2008 10:28 am
I think 720dreaming uses an approach that is only valid if you assume that the extra distance (and change in speed) is made in the last hour(s).

The approach by mrsmarthi is based on the assumption that the motorist drives the whole day at the new speed, instead of the last extra hour(s).

How can it be then that both approaches result in the same answer? Or am I just confused?

By the way, I think mrsmarthi's method is the best.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Tue Dec 16, 2008 2:54 pm

by gmatman1 » Tue Dec 23, 2008 8:37 am
Nobody? I would really appreciate some further explanation. :D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1