It is currently 20 Oct 2017, 13:22

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Adam began driving from home on a trip averaging 30 miles per hour. Ho

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Director
Director
User avatar
P
Joined: 05 Mar 2015
Posts: 964

Kudos [?]: 287 [0], given: 41

Adam began driving from home on a trip averaging 30 miles per hour. Ho [#permalink]

Show Tags

New post 26 Dec 2016, 12:01
3
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

83% (01:29) correct 17% (01:50) wrong based on 86 sessions

HideShow timer Statistics

Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60
[Reveal] Spoiler: OA

Kudos [?]: 287 [0], given: 41

1 KUDOS received
Math Forum Moderator
User avatar
P
Joined: 13 Apr 2015
Posts: 1503

Kudos [?]: 1112 [1], given: 884

Location: India
Concentration: Strategy, General Management
WE: Information Technology (Consulting)
GMAT ToolKit User Premium Member CAT Tests
Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho [#permalink]

Show Tags

New post 26 Dec 2016, 12:26
1
This post received
KUDOS
Distance traveled by Adam = 30 * 3.5 = 105

Jimmy must complete 105 miles in 3 hours --> Speed = 105/3 = 35

Answer: C

Kudos [?]: 1112 [1], given: 884

Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 220

Kudos [?]: 77 [0], given: 151

Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho [#permalink]

Show Tags

New post 23 Aug 2017, 22:33
Somehow I am not arriving at the right answering using Shrinking/Expanding Distance approach. Any help?

Kudos [?]: 77 [0], given: 151

1 KUDOS received
Manager
Manager
User avatar
S
Status: Math Tutor
Joined: 12 Aug 2017
Posts: 72

Kudos [?]: 29 [1], given: 13

WE: Education (Education)
Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho [#permalink]

Show Tags

New post 23 Aug 2017, 23:19
1
This post received
KUDOS
Jimmy starts after half an hour. Adam will have traveled 15 miles by now.
So Jimmy will have to cover 15 miles in 3 hours. (Because jimmy starts half an hour after Adam)
So the relative difference between there speeds when travelling in same direction \(s_1\)-\(s_2\)= \(\frac{15}{3}\) = 5
Adam's speed is 30
so Jimmy's speed is 30+5 =35
Option C
_________________

Abhishek Parikh
Math Tutor
Whatsapp- +919983944321
Mobile- +971529685639
Website: http://www.holamaven.com

Kudos [?]: 29 [1], given: 13

1 KUDOS received
Director
Director
avatar
P
Joined: 22 May 2016
Posts: 819

Kudos [?]: 266 [1], given: 552

Adam began driving from home on a trip averaging 30 miles per hour. Ho [#permalink]

Show Tags

New post 24 Aug 2017, 10:36
1
This post received
KUDOS
1
This post was
BOOKMARKED
rohit8865 wrote:
Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60

1. Gap = Distance
When Adam begins driving before Jimmy, Adam creates a distance "gap" between them:

\(D = rt\), so \(D_{gap} = 30 mph *
\frac{1}{2}\) hour = 15 miles

2. Relative speeds / rates

Jimmy must be driving faster than Adam to catch Adam.

Use relative speeds where, if two travelers move in same direction ("chase"), subtract slower rate from faster one.

Relative speed = \(Rate_{Jimmy} - Rate_{Adam}\)

Call them \(R_2\) and \(R_1\)

3. Equation and answer

(\(R_2 - R_1) * t = D\) -->

(relative) \(r * t = D\), so

relative \(r = \frac{D}{t}\)

Time is 3 hours (to close the gap when both are driving)

Jimmy's rate is \(R_2\)
Adam's rate is 30

relative \(r = \frac{D}{t}\), where RHS = \(\frac{15}{3}\)

\(R_2\) - 30 = \(\frac{15}{3}\)

\(R_2\) - 30 = 5

\(R_2\) = 35

Answer C

Kudos [?]: 266 [1], given: 552

Manager
Manager
User avatar
G
Joined: 30 May 2012
Posts: 220

Kudos [?]: 77 [0], given: 151

Location: United States (TX)
Concentration: Finance, Marketing
GPA: 3.3
WE: Information Technology (Consulting)
Premium Member
Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho [#permalink]

Show Tags

New post 24 Aug 2017, 13:01
genxer123 wrote:
rohit8865 wrote:
Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60

1. Gap = Distance
When Adam begins driving before Jimmy, Adam creates a distance "gap" between them:

\(D = rt\), so \(D_{gap} = 30 mph *
\frac{1}{2}\) hour = 15 miles

2. Relative speeds / rates
.
.
.
.

Time is 3 hours (to close the gap when both are driving)




Thanks, Genxer but why is the time 3 hours and not 3.5? Is it because we have already accounted for 30 mins when Jimmy will be catching up with Adam?

Kudos [?]: 77 [0], given: 151

Director
Director
avatar
P
Joined: 22 May 2016
Posts: 819

Kudos [?]: 266 [0], given: 552

Adam began driving from home on a trip averaging 30 miles per hour. Ho [#permalink]

Show Tags

New post 24 Aug 2017, 13:47
Blackbox wrote:
genxer123 wrote:
rohit8865 wrote:
Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60

1. Gap = Distance
When Adam begins driving before Jimmy, Adam creates a distance "gap" between them:

\(D = rt\), so \(D_{gap} = 30 mph *
\frac{1}{2}\) hour = 15 miles

2. Relative speeds / rates
.
Time is 3 hours (to close the gap when both are driving)


Thanks, Genxer but why is the time 3 hours and not 3.5? Is it because we have already accounted for 30 mins when Jimmy will be catching up with Adam?

Blackbox - Hmm. I'm not sure exactly what you mean when you say "we have already accounted for [the first 30 minutes]."

If you mean "we ignore that extra 30 minutes because in this and similar gap problems it can't be used to find the time for catch-up or the rate of catch-up [and we've used it to find the distance of catch-up]," then yes, that is correct. We have used that 30-minute time block to find the distance of the gap between the drivers; it has nothing to do with the chase while the chase is on.

If you mean, "we ignore that extra 30 minutes because it is used only to find the distance/gap (and hence is accounted for)," then the answer is -- not all the time.

Here? Yes. That's not true in all problems. If the question asked you to find the ratio of miles driven by Jimmy to miles driven by Adam, you would have to go back and use that 30-minute time block again (to find Adam's total distance.) I'm just being cautious, because these problems can get confusing.

Generally, in gap problems, when you are asked about anything that pertains only to the time when both travelers are traveling, yes, you use the time the second traveler started. Generally.

Finally (wow am I trying to be careful!): In "chase" problems, if you are using relative speed and trying to find the rate of one person during the chase, or the time it takes to catch up, although you are using the distance created by the person who started first, you may only use the time that both are traveling at the same time -- which is, as you note, when the second person starts traveling.

It helped me to think about the time issue this way: When "solving for the gap" . . . . that gap only begins to close when the second driver starts (and gradually catches up to the slower one). So if the question has to do with the chase, you use the second traveler's beginning time. Does that make sense? :-)

Kudos [?]: 266 [0], given: 552

Expert Post
Target Test Prep Representative
User avatar
S
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 1648

Kudos [?]: 842 [0], given: 3

Location: United States (CA)
Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho [#permalink]

Show Tags

New post 29 Aug 2017, 16:14
rohit8865 wrote:
Adam began driving from home on a trip averaging 30 miles per hour. How many miles per hour must jimmy drive on average to catch up to him in exactly when Adam drives for 3.5 hours, if jimmy leaves 30 minutes after Adam?

A. 39
B. 55
C. 35
D. 40
E. 60


We can let Jimmy’s rate = r, Jimmy’s time = 3, and Adam’s time = 3.5; thus:

3r = 30(3.5)

3r = 105

r = 35

Answer: C
_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 842 [0], given: 3

Re: Adam began driving from home on a trip averaging 30 miles per hour. Ho   [#permalink] 29 Aug 2017, 16:14
Display posts from previous: Sort by

Adam began driving from home on a trip averaging 30 miles per hour. Ho

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.