GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Oct 2019, 04:58

### 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

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

# Alan and Peter are cycling at different constant rates on a

Author Message
TAGS:

### Hide Tags

Manager
Joined: 06 Jun 2012
Posts: 119
Alan and Peter are cycling at different constant rates on a  [#permalink]

### Show Tags

Updated on: 11 May 2015, 03:34
2
7
00:00

Difficulty:

25% (medium)

Question Stats:

75% (01:24) correct 25% (01:43) wrong based on 332 sessions

### HideShow timer Statistics

Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

(1) One hour ago, Alan was 7 miles ahead of Peter.
(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.

_________________
Please give Kudos if you like the post

Originally posted by summer101 on 05 Aug 2013, 04:54.
Last edited by Bunuel on 11 May 2015, 03:34, edited 2 times in total.
Edited the question.
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1020
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

### Show Tags

05 Aug 2013, 04:58
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
Sufficient

II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
$$Speed*time=space$$.
$$P*t=A*t+4$$ (4=3 miles behind + 1 mile ahead) where $$P=18$$ and $$A=14$$.
$$18*t=14*t+4$$, $$4*t=4$$ and $$t=1$$ hour or 60 minutes.
Sufficient

For the theory refer here: distance-speed-time-word-problems-made-easy-87481.html
or here: time-speed-and-distance-simplified-150163.html
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
Senior Manager
Joined: 10 Jul 2013
Posts: 289
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

### Show Tags

05 Aug 2013, 14:09
1
Zarrolou wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
Sufficient

II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
$$Speed*time=space$$.
$$P*t=A*t+4$$ (4=3 miles behind + 1 mile ahead) where $$P=18$$ and $$A=14$$.
$$18*t=14*t+4$$, $$4*t=4$$ and $$t=1$$ hour or 60 minutes.
Sufficient

For the theory refer here: distance-speed-time-word-problems-made-easy-87481.html
or here: time-speed-and-distance-simplified-150163.html

zarrolou :
One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
i cant understand the reasoning.
peter gained just four miles in an hour, how can we be sure about it?
its only possible if Alan was in a dormant state i mean not moving at all.
but both were moving.
suppose , Alan was 7 miles ahead of peter so peter was 7 miles behind of peter.
in one hour Alan went 6 miles and peter went 2 miles, still they are 3 miles apart.
we dont know still their velocity....
can you explain the subtle fact?
_________________
Asif vai.....
Senior Manager
Joined: 10 Jul 2013
Posts: 289
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

### Show Tags

06 Aug 2013, 14:10
Asifpirlo wrote:
Zarrolou wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
Sufficient

II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
$$Speed*time=space$$.
$$P*t=A*t+4$$ (4=3 miles behind + 1 mile ahead) where $$P=18$$ and $$A=14$$.
$$18*t=14*t+4$$, $$4*t=4$$ and $$t=1$$ hour or 60 minutes.
Sufficient

For the theory refer here: distance-speed-time-word-problems-made-easy-87481.html
or here: time-speed-and-distance-simplified-150163.html

zarrolou :
One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
i cant understand the reasoning.
peter gained just four miles in an hour, how can we be sure about it?
its only possible if Alan was in a dormant state i mean not moving at all.
but both were moving.
suppose , Alan was 7 miles ahead of peter so peter was 7 miles behind of peter.
in one hour Alan went 6 miles and peter went 2 miles, still they are 3 miles apart.
we dont know still their velocity....
can you explain the subtle fact?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour,
Sufficient
so he will take another hour (60 minutes) to be 1 mile ahead of Alan.

he gained 4 miles means? Was Alan stationary ? please explain..
_________________
Asif vai.....
Manager
Joined: 06 Jul 2013
Posts: 85
GMAT 1: 620 Q48 V28
GMAT 2: 700 Q50 V33
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

### Show Tags

07 Aug 2013, 09:02
1
Since both are travelling at constant speed, peter would gain another 4 miles in next 1 hr since he gained 4 miles in last hour.
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1020
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

### Show Tags

07 Aug 2013, 09:06
Asifpirlo wrote:
Asifpirlo wrote:
Zarrolou wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
Sufficient

II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
$$Speed*time=space$$.
$$P*t=A*t+4$$ (4=3 miles behind + 1 mile ahead) where $$P=18$$ and $$A=14$$.
$$18*t=14*t+4$$, $$4*t=4$$ and $$t=1$$ hour or 60 minutes.
Sufficient

For the theory refer here: distance-speed-time-word-problems-made-easy-87481.html
or here: time-speed-and-distance-simplified-150163.html

zarrolou :
One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour, so he will take another hour (60 minutes) to be 1 mile ahead of Alan.
i cant understand the reasoning.
peter gained just four miles in an hour, how can we be sure about it?
its only possible if Alan was in a dormant state i mean not moving at all.
but both were moving.
suppose , Alan was 7 miles ahead of peter so peter was 7 miles behind of peter.
in one hour Alan went 6 miles and peter went 2 miles, still they are 3 miles apart.
we dont know still their velocity....
can you explain the subtle fact?

I.One hour ago, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour,
Sufficient
so he will take another hour (60 minutes) to be 1 mile ahead of Alan.

he gained 4 miles means? Was Alan stationary ? please explain..

Hi Asifpirlo,

sorry for the late reply. Here is my reasoning:

Alan and Peter are cycling at different constant rates on a straight track.Now Alan is now 3 miles ahead of Peter, and one hour ago, Alan was 7 miles ahead of Peter.

So I think that you would agree to the fact that in an hour Peter gained 4 miles on Alan: before was 7 miles behind, now is only 3.
Since A and B are are cycling at different constant rates, this pattern : in 1 hour Peter gains 4 miles on Alan will repeat.

So, as I said above, Peter will take another hour (60 minutes) to be 1 mile ahead of Alan.

Hope I've explained myself well
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
Senior Manager
Joined: 10 Jul 2013
Posts: 289
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

### Show Tags

07 Aug 2013, 09:26
summer101 wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

I.One hour ago, Alan was 7 miles ahead of Peter.
II.Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.

P.S Can some please provide a link or explain how to solve these difference in speed problem?

Zarrolou, yes now what are you saying is true indeed.
1st you said, Alan was 7 miles ahead of Peter.
And now he is 3 miles ahead, so Peter gained 4 miles in an hour
now you told, in an hour Peter gained 4 miles on Alan .

this two sentences confer two completely disparate meaning, i mean gaining 4 miles and gaining 4 miles on someone......

however nice work done again by you.... thanks for replying.....
_________________
Asif vai.....
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4000
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

### Show Tags

23 Nov 2018, 09:30
Top Contributor
1
summer101 wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

(1) One hour ago, Alan was 7 miles ahead of Peter.
(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.

Given: Alan is NOW 3 miles ahead of Peter

Target question: How many minutes from now will Peter be 1 mile ahead of Alan?

Statement 1: ONE HOUR AGO, Alan was 7 miles ahead of Peter.
Alan is NOW 3 miles ahead of Peter
So, in one hour, the GAP between Alan and Peter decreased by 4 miles
Another way to put it: Peter's speed is 4 mph greater than Alan's speed.
Another way to put it: in one hour, Peter traveled 4 miles more than Alan
So, in ONE HOUR FROM NOW, Peter will travel another 4 miles more than Alan. This means the Peter will not only close the 3-mile gap BUT ALSO travel 1 mile further than Alan travels.
So, the answer to the target question is in 60 MINUTES, Peter will be 1 mile ahead of Alan
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.
Another way to put it: Peter's speed is 4 mph greater than Alan's speed.
Another way to put it: in one hour, Peter travels 4 miles more than Alan
At this point, we can apply the same logic we applied to statement 1 to conclude that statement 2 is SUFFICIENT

RELATED VIDEO FROM OUR COURSE

_________________
Test confidently with gmatprepnow.com
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: Alan and Peter are cycling at different constant rates on a  [#permalink]

### Show Tags

23 Nov 2018, 16:21
summer101 wrote:
Alan and Peter are cycling at different constant rates on a straight track. If Alan is now 3 miles ahead of Peter, how many minutes from now will Peter be 1 mile ahead of Alan?

(1) One hour ago, Alan was 7 miles ahead of Peter.
(2) Alan is cycling at 14 miles per hour and Peter is cycling at 18 miles per hour.

Excellent opportunity for the relative velocity (speed) technique:

$$?\,\,\,:\,\,\,\min \,\,{\rm{for}}\,\,4\,\,{\rm{miles - }}\underline {{\rm{gaining}}} \,\,\,{\rm{Peter}}/{\rm{Alan}}\,\,$$

$$\left( 1 \right)\,\,{\text{RelativeSpee}}{{\text{d}}_{{\text{Peter/Alan}}}}\,\,\, = \,\,\,\,\,\frac{{4\,\,{\text{miles}}}}{{\boxed{1\,\,{\text{hour}}}}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \boxed{1\,{\text{h}}}\,\,\left( { = 60\min } \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.$$

$$\left( 2 \right)\,{\text{RelativeSpee}}{{\text{d}}_{{\text{Peter/Alan}}}}\,\,\,\, = \,\,\,\,\frac{{4\,\,{\text{miles}}}}{{\boxed{1\,\,{\text{hour}}}}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = \boxed{1\,{\text{h}}}\,\,\left( { = 60\min } \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Re: Alan and Peter are cycling at different constant rates on a   [#permalink] 23 Nov 2018, 16:21
Display posts from previous: Sort by