Jim takes a seconds to swim c meters at a constant rate : GMAT Problem Solving (PS)
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 03 Dec 2016, 17:57

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Jim takes a seconds to swim c meters at a constant rate

Author Message
TAGS:

### Hide Tags

Manager
Joined: 12 Oct 2012
Posts: 112
Followers: 1

Kudos [?]: 42 [1] , given: 194

Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

07 Dec 2012, 19:12
1
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

60% (05:31) correct 40% (02:58) wrong based on 174 sessions

### HideShow timer Statistics

Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

A) c(b-a)/ a+b

B) c(a-b)/a+b

C) c(a+b)/a-b

D) ab(a-b)/a+b

E) ab(b-a)/a+b

The above question can be done by picking numbers. Is there an alternative way to solve this.
[Reveal] Spoiler: OA
VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1420
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Followers: 174

Kudos [?]: 1299 [2] , given: 62

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

07 Dec 2012, 19:43
2
KUDOS
2
This post was
BOOKMARKED
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

A) c(b-a)/ a+b

B) c(a-b)/a+b

C) c(a+b)/a-b

D) ab(a-b)/a+b

E) ab(b-a)/a+b

The above question can be done by picking numbers. Is there an alternative way to solve this.

Draw a RTD chart.
Jim's speed will come out to be as $$c/a$$ and that of Roger will come out to be as $$c/b$$

Note that this is a "kissing" problem where the two trains are literally approaching each other. In such problems, the easiest approach conceptually is that when the two trains will meet, they must have travlled for same amount of time, no matter if one has travelled greater distance than that of the other.

So let the two trains be travelling for, let us assume, t hours.

So $$(c/a)*t + (c/b)*t=C$$, where C is the total distance
i.e. $$(c/a)*t + (c/b)*t= c$$

On solving, $$t=(ab)/(a+b)$$

To find the difference between the distance travelled by Roger and Jim,
Speed of Roger * t - Speed of Jim * t i.e.

$$(c/b)*(ab)/(a+b) - (c/a)*(ab)/(a+b)$$

OR

$$c(a-b)/(a+b)$$.

Hope that helps.
_________________
Manager
Joined: 12 Oct 2012
Posts: 112
Followers: 1

Kudos [?]: 42 [0], given: 194

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

07 Dec 2012, 20:18
Marcab wrote:
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

A) c(b-a)/ a+b

B) c(a-b)/a+b

C) c(a+b)/a-b

D) ab(a-b)/a+b

E) ab(b-a)/a+b

The above question can be done by picking numbers. Is there an alternative way to solve this.

Draw a RTD chart.
Jim's speed will come out to be as $$c/a$$ and that of Roger will come out to be as $$c/b$$

Note that this is a "kissing" problem where the two trains are literally approaching each other. In such problems, the easiest approach conceptually is that when the two trains will meet, they must have travlled for same amount of time, no matter if one has travelled greater distance than that of the other.

So let the two trains be travelling for, let us assume, t hours.

So $$(c/a)*t + (c/b)*t=C$$, where C is the total distance
i.e. $$(c/a)*t + (c/b)*t= c$$

On solving, $$t=(ab)/(a+b)$$

To find the difference between the distance travelled by Roger and Jim,
Speed of Roger * t - Speed of Jim * t i.e.

$$(c/b)*(ab)/(a+b) - (c/a)*(ab)/(a+b)$$

OR

$$c(a-b)/(a+b)$$.

Hope that helps.

Thanx marcab.

I was trying to solve by this method. Let me know where I am going wrong.

Time Travelled = Distance travelled/ Relative speed of jim & roger
Hence, T= c/ (c/b-c/a)

which results in T = ab/a-b

Hence Jim must have travelled in T time = c/a * ab/a-b >>> cb/a-b.
VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1420
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Followers: 174

Kudos [?]: 1299 [0], given: 62

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

07 Dec 2012, 21:17
Use the concept of relative speed when the two trains are travelling in same direction.
_________________
Senior Manager
Joined: 22 Dec 2011
Posts: 298
Followers: 3

Kudos [?]: 224 [0], given: 32

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

07 Dec 2012, 23:01
Marcab wrote:
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

A) c(b-a)/ a+b

B) c(a-b)/a+b

C) c(a+b)/a-b

D) ab(a-b)/a+b

E) ab(b-a)/a+b

The above question can be done by picking numbers. Is there an alternative way to solve this.

Draw a RTD chart.
Jim's speed will come out to be as $$c/a$$ and that of Roger will come out to be as $$c/b$$

Note that this is a "kissing" problem where the two trains are literally approaching each other. In such problems, the easiest approach conceptually is that when the two trains will meet, they must have travlled for same amount of time, no matter if one has travelled greater distance than that of the other.

So let the two trains be travelling for, let us assume, t hours.

So $$(c/a)*t + (c/b)*t=C$$, where C is the total distance
i.e. $$(c/a)*t + (c/b)*t= c$$

On solving, $$t=(ab)/(a+b)$$

To find the difference between the distance travelled by Roger and Jim,
Speed of Roger * t - Speed of Jim * t i.e.

$$(c/b)*(ab)/(a+b) - (c/a)*(ab)/(a+b)$$

OR

$$c(a-b)/(a+b)$$.

Hope that helps.

Thanx marcab.

I was trying to solve by this method. Let me know where I am going wrong.

Time Travelled = Distance travelled/ Relative speed of jim & roger
Hence, T= c/ (c/b-c/a)

which results in T = ab/a-b

Hence Jim must have travelled in T time = c/a * ab/a-b >>> cb/a-b.

You have done everything right, but there is a penultimate error (next to last) in your sol

As Marcab has advised we can do this by relative speed concepts.

Jim speed =$$c/a$$
R speed = $$c/b$$

Time at which they pass each other = Gap / sum of speeds

Gap = C
Sum of speeds = $$c/a + c/b$$= $$c (b+a) /ab$$
Therefore time at which they pass will be $$ab / a+b$$

question is asking how many fewer miles Jim would have traveled when they pass each other.. (you didn't solve this step)

Distance traveled by R MINUS Distance traveled by Jim (as Roger is faster)
$$c / b * ab / a+b$$ - $$c/a * ab / a+b$$
On solving gives $$c(a-b)/a+b$$

@marcab long time no see what's new dude?

Cheers
VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1420
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Followers: 174

Kudos [?]: 1299 [0], given: 62

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

07 Dec 2012, 23:19
Hii JP.
Since there are few new verbal questions coming, so started focussing on official questions. You say how is prep going? When have you decided to take the test?
_________________
Senior Manager
Joined: 22 Dec 2011
Posts: 298
Followers: 3

Kudos [?]: 224 [0], given: 32

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

08 Dec 2012, 00:19
Marcab wrote:
Hii JP.
Since there are few new verbal questions coming, so started focussing on official questions. You say how is prep going? When have you decided to take the test?

Planning to take it by Jan Mid. I wasn't able to study at all last 3 weeks because of work. When you are planning to give?

Cheers
VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1420
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Followers: 174

Kudos [?]: 1299 [0], given: 62

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

08 Dec 2012, 01:37
Hii JP.
Planning to take by mid Feb, though haven't taken the date yet.
_________________
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1858
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Followers: 43

Kudos [?]: 1829 [1] , given: 193

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

14 Jun 2013, 01:01
1
KUDOS
Both Jim & Roger are travelling at constant speed & in opposite direction:
So, speed of Jim = c/a & speed of Roger = c/b
Let say Jim travelled distance x from P where it met Roger, it means that Roger travelled (c-x) from point Q
[x would be less than (c-x) as Jim is travelling slow]
From above, time taken by Jim to travel x = xa/c....................... (1)
Also, time taken by Roger to travel (c-x) = (c-x)b/c.....................(2)
Time taken by both Jim & Roger is same, so (1) = (2)
xa/c = (c-x)b/c,
Solving further, x = bc/(a+b).................... (3)
We require to find how many fewer meters will Jim have swum i.e
additional distance travelled by Roger = (c - x) - x
= c-2x
Substituting value of x from (3) & solving the equation further, we get Answer = c(a-b)/a+b
_________________

Kindly press "+1 Kudos" to appreciate

Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 631
Followers: 78

Kudos [?]: 1086 [1] , given: 136

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

14 Jun 2013, 03:38
1
KUDOS
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

A) c(b-a)/ a+b

B) c(a-b)/a+b

C) c(a+b)/a-b

D) ab(a-b)/a+b

E) ab(b-a)/a+b

The above question can be done by picking numbers. Is there an alternative way to solve this.

We know from the question stem that a>b, as because for the same distance, Roger will take lesser time than Jim.Eliminate options A and E(as they will lead to negative answer).

Also, as the answer is representing distance, we can straightaway eliminate D, which is representing $$time^2$$

Out of the 2 options remaining , i.e. B and C, we know that the answer has to be less than c(The difference between the distance covered by Jim and Roger can-not be more than c) For option C, the expression (a+b)/(a-b) WILL always be greater than 1. Thus, by process of elimination, the answer is B.

B.
_________________
Senior Manager
Joined: 13 May 2013
Posts: 472
Followers: 3

Kudos [?]: 156 [0], given: 134

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

08 Aug 2013, 12:42
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

Rate (Jim): c/a

Rate (Roger): c/b

Rate (Roger) > Rate (Jim)

Plugging in numbers: c=100, a=10, b=20

rate (Jim) = 100/10 minutes
rate = 10 meters/minute

rate (Roger) = 100/5 minutes
rate = 20 meters/minute

time = distance/rate
time = 100/(10+20)
time = 3.33 minutes until they meet one another.

distance = rate*time
distance (Jim) = 10*3.33
distance (Jim) = 33.3 meters.

He swims 100-33.3 = 66.7m less than Roger

100(10-20)/(10+20)
-1000/30
= -66.7
i.e. he swam 66.7 m less.

Is that correct?
Director
Joined: 14 Dec 2012
Posts: 841
Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
GPA: 3.6
Followers: 57

Kudos [?]: 1243 [1] , given: 197

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

08 Aug 2013, 12:49
1
KUDOS
WholeLottaLove wrote:
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

Rate (Jim): c/a

Rate (Roger): c/b

Rate (Roger) > Rate (Jim)

Plugging in numbers: c=100, a=10, b=20

rate (Jim) = 100/10 minutes
rate = 10 meters/minute

rate (Roger) = 100/5 minutes
rate = 20 meters/minute

time = distance/rate
time = 100/(10+20)
time = 3.33 minutes until they meet one another.

distance = rate*time
distance (Jim) = 10*3.33
distance (Jim) = 33.3 meters.

He swims 100-33.3 = 66.7m less than Roger

100(10-20)/(10+20)
-1000/30
= -66.7
i.e. he swam 66.7 m less.

Is that correct?

yes IMO All fine in your approach only thing in place minutes it should be seconds as it is already stated in question than a and b are in seconds.
moreover since ROGER is faster he is going to take less time thats why b<a
hence assume b=10 and a=20...then you will not get negative value.
_________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

GIVE VALUE TO OFFICIAL QUESTIONS...

learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment

Senior Manager
Joined: 13 May 2013
Posts: 472
Followers: 3

Kudos [?]: 156 [0], given: 134

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

08 Aug 2013, 15:49
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

We are given jim's rate and rogers rate in addition to the size of the pool. Also, we know that Roger is faster than Jim.

The portion in blue implies that Roger can swim the same distance in less time. However, when Jim and Roger cross paths in the pool, the time they will have traveled will be the same. Because the time that they spend swimming is the same we can say time = t.

If t is the same for both swimmers than why can't we just multiply their respective rates by t? Why do we have to solve out for t? It seems that for other problems, where we have two objects leaving from point A and B respectively and meeting somewhere in the middle, we do not need to solve out for 'T'.

We know the rate of both swimmers. We also know that the time they spend swimming is the same. We also know the length of the pool. We can do a combined distance formula where:

(rate Jim)*(t) + (rate Roger)*(t) = c

(c/a)*t + (c/b)*t = c
ct/a + at/b = c

Last edited by WholeLottaLove on 15 Aug 2013, 08:27, edited 1 time in total.
Director
Joined: 14 Dec 2012
Posts: 841
Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
GPA: 3.6
Followers: 57

Kudos [?]: 1243 [0], given: 197

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

08 Aug 2013, 18:36
WholeLottaLove wrote:
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

We are given jim's rate and rogers rate in addition to the size of the pool. Also, we know that Roger is faster than Jim.

The portion in blue implies that Roger can swim the same distance in less time. However, when Jim and Roger cross paths in the pool, the time they will have traveled will be the same. Because the time that they spend swimming is the same we can say time = t.

If t is the same for both swimmers than why can't we just multiply their respective rates by t? Why do we have to solve out for t?

We know the rate of both swimmers. We also know that the time they spend swimming is the same. We also know the length of the pool. We can do a combined distance formula where:

(rate Jim)*(t) + (rate Roger)*(t) = c

(c/a)*t + (c/b)*t = c
ct/a + at/b = c

this thinking is also correct.
you have:
(c/a)*t + (c/b)*t = c
solving for t
cancel c from both sides
on solving for t you will get t= ab/a+b
now we have too calculate difference in distance...
distance by jim = rate*time(t) = $$(\frac{c}{a})*(\frac{(ab)}{(a+b)})$$
distance by roger = rate*time(t) =$$( \frac{c}{b})*(\frac{(ab)}{(a+b)})$$

now suntract these two distance and you will get
$$\frac{(c(a-b))}{(a+b)}$$
hope this helps
_________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

GIVE VALUE TO OFFICIAL QUESTIONS...

learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment

Senior Manager
Joined: 10 Jul 2013
Posts: 335
Followers: 3

Kudos [?]: 305 [0], given: 102

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

08 Aug 2013, 21:26
Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in b seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?

A) c(b-a)/ a+b

B) c(a-b)/a+b

C) c(a+b)/a-b

D) ab(a-b)/a+b

E) ab(b-a)/a+b

The above question can be done by picking numbers. Is there an alternative way to solve this.

....
Jim = c/a
roger=c/b
let jim x distance in t time. so x = c/a t
so, roger c-x distance in t time, c-x = c/b t
....................................................................
(+) , or, t = ab/a+b
Now, more distance = c/b t - c/a t = c(a-b)/(a+b) Answer
_________________

Asif vai.....

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12855
Followers: 559

Kudos [?]: 157 [0], given: 0

Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

15 Nov 2014, 08:56
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Senior Manager
Joined: 07 Dec 2014
Posts: 453
Followers: 3

Kudos [?]: 76 [0], given: 2

Jim takes a seconds to swim c meters at a constant rate [#permalink]

### Show Tags

18 Nov 2015, 15:45
let r=roger's distance at passing
c-r=jim's distance at passing
r/(c-r)=a/b
r=ac/(a+b)
r-(c-r)=2r-c
substituting, 2r-c=2ac/(a+b)-c➡
c(a-b)/(a+b)
Jim takes a seconds to swim c meters at a constant rate   [#permalink] 18 Nov 2015, 15:45
Similar topics Replies Last post
Similar
Topics:
2 At a constant Rate of flow, it takes 20 minutes to fill a swimming poo 4 23 Dec 2015, 12:12
86 Running at their respective constant rates, Machine X takes 20 18 Mar 2014, 00:39
18 Rebecca runs at a constant rate on the treadmill and it take 12 14 Jun 2013, 10:06
19 Jim takes a seconds to swim c meters at a constant rate from 13 08 Jan 2013, 03:14
141 Running at their respective constant rates, machine X takes 41 05 Aug 2010, 01:01
Display posts from previous: Sort by