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

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08 Jan 2013, 04:14

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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

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

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08 Jan 2013, 04:50

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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)

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) _________________

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

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09 Jan 2013, 16:20

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

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05 Mar 2014, 06:33

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

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01 May 2014, 10:48

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.

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

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31 Aug 2015, 03:08

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).

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

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22 Jan 2017, 16:49

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.
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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)

Answer: B
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