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

 It is currently 15 Dec 2018, 19:39

### 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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar

##### General Discussion
Manager
Joined: 12 Mar 2012
Posts: 80
Location: India
Concentration: Technology, Strategy
GMAT 1: 710 Q49 V36
GPA: 3.2
WE: Information Technology (Computer Software)
Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

08 Jan 2013, 03:50
1
Speed of Jim is c/a meter per second. Speed of Roger is c/b meter per second.
Let us suppose when they both meet or cross each other Jim has covered x distance and Roger has covered c-x distance. Since time taken is same, as they both started together we can have the equation:

x/(c/a) = (c-x)/(c/b)

ax/c = b(c-x)/c or ax = b(c-x)
Solving for x, x = cb/(a+b)
Jim travelled, x = cb/(a+b)
Roger travelled, c-x = ca/(a+b)
Ans: ca/(a+b) - cb/(a+b) = c(a-b)/(a+b)
Senior Manager
Joined: 27 Jun 2012
Posts: 374
Concentration: Strategy, Finance
Schools: Haas EWMBA '17
Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

08 Jan 2013, 13:17
1
1
Attachment:

Pick numbers.jpg [ 43.32 KiB | Viewed 5481 times ]

Pick Numbers Approach:
Pick a=15 (time taken by Jim)
b=10 (time taken by Roger)
c=30 (Total distance PQ)

Rate for Jim = 30/15 = 2
Rate for roger = 30/10 = 3
Mutual rate = 2+3 = 5
Total time taken by both to cross each other = 30/5 = 6

Roger's distance - Jim's distance= (Roger's rate * mutual time) - (Jim's rate * mutual time) = mutual time * (Roger's rate - Jim's rate) = 6 (3-2) = 6

Choice analysis with Plug-in the numbers a=15, b=10, c=30
A. c(b-a)/(a+b) = 30(-5)/25= -6 (this is negative)
B. c(a-b)/(a+b) = 30(5)/25= 6 (this is correct)!
C. c(a+b)/(a-b) = 30(25)/5 = 150 (does not match with 6)
D. ab(a-b)/(a+b) = 150(5)/25= 30 (does not match with 6)
E. ab(b-a)/(a+b) = 150(-5)/25= -30 (does not match with 6)
_________________

Thanks,
Prashant Ponde

Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
VOTE GMAT Practice Tests: Vote Here
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here
Finance your Student loan through SoFi and get $100 referral bonus : Click here Board of Directors Joined: 01 Sep 2010 Posts: 3293 Re: Jim takes a seconds to swim c meters at a constant rate from [#permalink] ### Show Tags 09 Jan 2013, 12:35 PraPon wrote: Attachment: Pick numbers.jpg Pick Numbers Approach: Pick a=15 (time taken by Jim) b=10 (time taken by Roger) c=30 (Total distance PQ) Rate for Jim = 30/15 = 2 Rate for roger = 30/10 = 3 Mutual rate = 2+3 = 5 Total time taken by both to cross each other = 30/5 = 6 Roger's distance - Jim's distance= (Roger's rate * mutual time) - (Jim's rate * mutual time) = mutual time * (Roger's rate - Jim's rate) = 6 (3-2) = 6 Choice analysis with Plug-in the numbers a=15, b=10, c=30 A. c(b-a)/(a+b) = 30(-5)/25= -6 (this is negative) B. c(a-b)/(a+b) = 30(5)/25= 6 (this is correct)! C. c(a+b)/(a-b) = 30(25)/5 = 150 (does not match with 6) D. ab(a-b)/(a+b) = 150(5)/25= 30 (does not match with 6) E. ab(b-a)/(a+b) = 150(-5)/25= -30 (does not match with 6) Like your explanation and pick numbers: 30 distance 3 seconds for J and 2 for R and move from here is the best approach algebra is quite painful _________________ Senior Manager Joined: 27 Jun 2012 Posts: 374 Concentration: Strategy, Finance Schools: Haas EWMBA '17 Re: Jim takes a seconds to swim c meters at a constant rate from [#permalink] ### Show Tags 09 Jan 2013, 13:50 carcass wrote: PraPon wrote: Attachment: Pick numbers.jpg Pick Numbers Approach: Pick a=15 (time taken by Jim) b=10 (time taken by Roger) c=30 (Total distance PQ) Rate for Jim = 30/15 = 2 Rate for roger = 30/10 = 3 Mutual rate = 2+3 = 5 Total time taken by both to cross each other = 30/5 = 6 Roger's distance - Jim's distance= (Roger's rate * mutual time) - (Jim's rate * mutual time) = mutual time * (Roger's rate - Jim's rate) = 6 (3-2) = 6 Choice analysis with Plug-in the numbers a=15, b=10, c=30 A. c(b-a)/(a+b) = 30(-5)/25= -6 (this is negative) B. c(a-b)/(a+b) = 30(5)/25= 6 (this is correct)! C. c(a+b)/(a-b) = 30(25)/5 = 150 (does not match with 6) D. ab(a-b)/(a+b) = 150(5)/25= 30 (does not match with 6) E. ab(b-a)/(a+b) = 150(-5)/25= -30 (does not match with 6) Like your explanation and pick numbers: 30 distance 3 seconds for J and 2 for R and move from here is the best approach algebra is quite painful Thanks carcass. Yes it got much simpler with pick numbers. Algebraic approach took longer - about 2.5-3 mins - to plan/solve. _________________ Thanks, Prashant Ponde Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7 Reading Comprehension notes: Click here VOTE GMAT Practice Tests: Vote Here PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here Finance your Student loan through SoFi and get$100 referral bonus : Click here

Manager
Joined: 27 Feb 2012
Posts: 119
Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

09 Jan 2013, 15:20
2
samsikka23 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)

Hi, I am having trouble with this question. Can you please help explain it algebraically as well as with picking numbers, thanks

OK. Unit of distance is mt. Check the unit of options only A B and C satisfies.
b<a so option would be negative.
Between B and C now. Educated guess is B because we have relative speed in denominator. But let me explain why.....

The relative speed will be b+a for objects travelling in opposite direction. Let both of them meet in t seconds. Distance traveled is c.
t = c / (c/a+c/b) = ab/(a+b).
Distance traveled by jim whose speed is c/a is c/a * ab/ (a+b) = cb/(a+b)
Distance traveled by roger... c/b*ab/(a+b) = ca/(a+b)
Difference c*(a-b)/(a+b)
_________________

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Please +1 KUDO if my post helps. Thank you.

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1825
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

06 Mar 2014, 21:54
1
Attachments

sp.jpg [ 35.5 KiB | Viewed 4553 times ]

_________________

Kindly press "+1 Kudos" to appreciate

Retired Moderator
Joined: 20 Dec 2013
Posts: 173
Location: United States (NY)
GMAT 1: 640 Q44 V34
GMAT 2: 710 Q48 V40
GMAT 3: 720 Q49 V40
GPA: 3.16
WE: Consulting (Venture Capital)
Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

06 Mar 2014, 22:19
Jim Rate = c/a
Roger Rate c/b
b<a

Combined Rate = c(a+b)/ab
Time to Complete Course = c/[c(a+b)/ab] = ab/(a+b)
Roger Distance - Jim Distance = [ab/(a+b)][c/b)-(c/a)] = c(a-b)/(a+b)

B
_________________
Manager
Joined: 28 Apr 2014
Posts: 217
Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

01 May 2014, 09:48
Abhii46 wrote:
Speed of Jim is c/a meter per second. Speed of Roger is c/b meter per second.
Let us suppose when they both meet or cross each other Jim has covered x distance and Roger has covered c-x distance. Since time taken is same, as they both started together we can have the equation:

x/(c/a) = (c-x)/(c/b)

ax/c = b(c-x)/c or ax = b(c-x)
Solving for x, x = cb/(a+b)
Jim travelled, x = cb/(a+b)
Roger travelled, c-x = ca/(a+b)
Ans: ca/(a+b) - cb/(a+b) = c(a-b)/(a+b)

How come Roger's distance be taken as c-x ? I dont see it being mentioned that distance between the two points is c.
Math Expert
Joined: 02 Sep 2009
Posts: 51218
Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

02 May 2014, 00:40
himanshujovi wrote:
Abhii46 wrote:
Speed of Jim is c/a meter per second. Speed of Roger is c/b meter per second.
Let us suppose when they both meet or cross each other Jim has covered x distance and Roger has covered c-x distance. Since time taken is same, as they both started together we can have the equation:

x/(c/a) = (c-x)/(c/b)

ax/c = b(c-x)/c or ax = b(c-x)
Solving for x, x = cb/(a+b)
Jim travelled, x = cb/(a+b)
Roger travelled, c-x = ca/(a+b)
Ans: ca/(a+b) - cb/(a+b) = c(a-b)/(a+b)

How come Roger's distance be taken as c-x ? I dont see it being mentioned that distance between the two points is c.

Jim takes a seconds to swim c meters at a constant rate from point P to point Q in a pool.
_________________
VP
Joined: 07 Dec 2014
Posts: 1128
Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

31 Aug 2015, 18:56
time to passing = c/(c/a+c/b) ➔ ab/a+b
time x rate difference = (ab/a+b)(c/b-c/a) ➔ c(a-b)/a+b
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

25 Jan 2017, 10:21
1
samsikka23 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)

We are given that Jim takes a seconds to swim c meters. Since rate = distance/time, Jim’s rate is c/a. We are also given that Roger can swim c meters in b seconds. Roger’s rate = c/b.

Since Jim leaves point P at the same time Roger leaves point Q, we can let t = the time it takes them to pass each other, and we can use the following formula:

distance of Jim + distance of Roger = total distance

(c/a)t + (c/b)t = c

Let’s multiply both sides of the equation by ab:

bct + act = abc

Now divide both sides of the equation by c and solve for t:

bt + at = ab

t(b + a) = ab

t = ab/(a +b)

In t seconds (when they pass each other), Roger has swum (c/b)[ab/(a +b)] = ac/(a + b) meters and Jim has swum (c/a)[ab/(a +b)] = bc/(a + b) meters. Therefore, the difference between the distances swum is:

ac/(a + b) - bc/(a + b) = (ac - bc)/(a + b) = c(a - b)/(a + b)

_________________

Jeffery Miller

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

Senior Manager
Joined: 18 Jun 2018
Posts: 253
Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

### Show Tags

10 Aug 2018, 08:04
samsikka23 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)

Hi, I am having trouble with this question. Can you please help explain it algebraically as well as with picking numbers, thanks

OA:B
Average speed of Jim $$Av_{Jim}:\frac{c}{a}$$
Average speed of Roger $$Av_{Roger}:\frac{c}{b}$$
Relative speed of Jim with respect to Roger :$$Av_{Jim}+Av_{Roger}$$(as they are swimming towards each other) $$= \frac{c}{a} +\frac{c}{b}=\frac{c(a+b)}{ab}$$
Time taken from when they start and when they meet in pool$$= \frac{c}{Relative speed} =\frac{c}{\frac{c(a+b)}{ab}}=\frac{ab}{(a+b)}$$
Distance covered by Jim in this time $$= \frac{c}{a}*\frac{ab}{(a+b)}$$
Distance covered by Roger in this time $$= \frac{c}{b}*\frac{ab}{(a+b)}$$
Difference $$=\frac{c}{b}*\frac{ab}{(a+b)}-\frac{c}{a}*\frac{ab}{(a+b)} =\frac{ca}{(a+b)}-\frac{cb}{(a+b)}=\frac{c(a-b)}{(a+b)}$$
Re: Jim takes a seconds to swim c meters at a constant rate from &nbs [#permalink] 10 Aug 2018, 08:04
Display posts from previous: Sort by