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

Redeem

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.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
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
GMATOnDemand Start 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

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

500 PS section2 #12

Expert replies
by dunkin77 » Sun Apr 01, 2007 10:40 am
Hi,

My answer was A, but the correct answer is C) (so, obviously my calculation was wrong) Can you help?? Thanks!


A bus trip of 450 miles would have taken 1 hour less if the average speed S for the trip had been greater by 5 miles per hour. What was the average speed S, in miles per hour, for the trip?
(A) 10
(B) 40
(C) 45
(D) 50
(E) 55
Join the discussion
Source: — Problem Solving |

by jayhawk2001 » Sun Apr 01, 2007 12:05 pm
The difference in time (1 hr) can be represented by following equation

450/S - 450/(S+5) = 1

Solving for S, we get 45, -50. Hence C

It might be easier to substitute values from the answer to arrive at
the answer quickly. Choose the middle answer for starters.
We have 450/45 - 450/50 = 1. Which is what we are looking for !
Join the discussion

by dunkin77 » Sun Apr 01, 2007 12:50 pm
Thanks Jay- it's very clear and good idea to try from C). :)
Join the discussion

by BTGmoderatorRO » Sun Oct 15, 2017 7:36 am
The total distance covered during the bus trip is 450 miles.
and, average speed, S, =Total distance covered/ total time taken
S=450/ total time taken
total time taken= 450/S .hour

from the question, the time taken could have been reduced by 1 hour if the average speed S have been increased by 5miles/hr
i.e if Avg. speed (S+5) miles/hr
$$then\ time=\left(\frac{450}{S}d-\frac{1}{ }\right)\ hr$$
$$Therefore,S+5=\frac{total\ dis\tan ce}{\frac{450}{S}-1}$$
Since distance is still the same,
$$Therefore,S+5=\frac{450}{\frac{450}{S}-1}$$
$$\left(S+5\right)\left(\frac{450}{S}-1\right)=450$$
By expanding the expression on the left, we have,
$$450-S+\frac{2250}{S}-5=450$$
$$S+5-\frac{2250}{S}=0$$
$$S^2+5S-2250=0$$
solving this quadratically, we obtain
(S+50)(S-45)=0
S=-50 or S=45 miles/hr

since average speed cannot be negative, we will discard the negative value obtained.
Average speed, S= 45 miles/hr
Join the discussion

by [email protected] » Sun Oct 15, 2017 2:08 pm
Hi All,

While this is an old post, the math concepts in this prompt will be ones that you will see on Test Day. This particular prompt has a built-in shortcut that you can take advantage of. Since the total Distance is 450 miles - and the DIFFERENCE in TIMES is exactly 1 hour, we will almost certainly need two different speeds that BOTH divide evenly into 450... and that DIFFER by 5 miles/hour and would lead to a 1 hour difference in time-traveled. Looking at the answer choices, the only possibility that stands out would be if the speeds were 45 and 50 (since both of those values divide evenly into 450 and differ by 5). If we 'TEST' those speeds against that distance, we find....

450 mi = (45 mi/hour)(10 hours)
450 mi = (50 mi/hour)(9 hours)

This is an exact match for what we were told. We're asked for the slower speed...

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Join the discussion

by Scott@TargetTestPrep » Sun Nov 24, 2019 5:21 pm
dunkin77 wrote:Hi,

My answer was A, but the correct answer is C) (so, obviously my calculation was wrong) Can you help?? Thanks!


A bus trip of 450 miles would have taken 1 hour less if the average speed S for the trip had been greater by 5 miles per hour. What was the average speed S, in miles per hour, for the trip?
(A) 10
(B) 40
(C) 45
(D) 50
(E) 55
Let t = the time to complete the trip of 450 miles when the average speed is S. Thus, we have:

St = 450

and

(S + 5)(t - 1) = 450

Isolating t in the first equation, we have: t = 450/S. Substituting this in the second equation, we have:

(S + 5)(450/S - 1) = 450

450 - S + 2250/S - 5 = 450

-S + 2250/S - 5 = 0

S + 5 - 2250/S = 0

S^2 + 5S - 2250 = 0

(S + 50)(S - 45) = 0

S = -50 or S = 45

Since S can't be negative, S = 45.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Join the discussion

by Brent@GMATPrepNow » Mon Nov 25, 2019 10:38 am
dunkin77 wrote: A bus trip of 450 miles would have taken 1 hour less if the average speed S for the trip had been greater by 5 miles per hour. What was the average speed S, in miles per hour, for the trip?
(A) 10
(B) 40
(C) 45
(D) 50
(E) 55
Let's start with a word equation:
travel time at actual speed = travel time at faster speed + 1 hour
In other words: travel time at S mph = travel time at (S + 5) mph + 1 hour

travel time = distance/speed
So, we get: 450/S= 450/(S + 5) + 1
Multiply both sides by S to get: 450 = 450S/(S+5) + S
Multiply both sides by S+5 to get: 450(S + 5) = 450S + S(S+5)
Expand: 450S + 2250 = 450S + S² + 5S
Subtract 450S from both sides: 2250 = S² + 5S
Rewrite as: S² + 5S - 2250 = 0
Factor: (S + 50)(S - 45) = 0
So, EITHER S = 50, OR S = 45
Since the speed can't be negative, the correct answer must be S = 45

Answer: C

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