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Jim takes a seconds to swim c meters at a constant rate from [#permalink]
08 Jan 2013, 03:14

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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]
08 Jan 2013, 03:50

1

This post received KUDOS

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)

Algebraic Approach: As Jim and Roger as swimming in opposite direction we can add their rates Mutual speed = \frac{c}{a} +\frac{c}{b} = \frac{c(a+b)}{ab} Mutual Time (taken by both cross each other) = \frac{distance}{mutualspeed} = c/\frac{c(a+b)}{ab} = \frac{ab}{(a+b)} Roger's distance - Jim's distance = (Roger's rate * mutual time) - (Jim's rate * mutual time) = mutual time * (Roger's rate - Jim's rate) = \frac{ab}{(a+b)} * (\frac{c}{b}-\frac{c}{a}) = \frac{ab}{(a+b)} * \frac{c(a-b)}{ab} = \frac{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 _________________

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. _________________

Re: Jim takes a seconds to swim c meters at a constant rate from [#permalink]
09 Jan 2013, 15:20

2

This post received KUDOS

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) _________________

Re: Jim takes a seconds to swim c meters at a constant rate from [#permalink]
05 Mar 2014, 05:33

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

Re: Jim takes a seconds to swim c meters at a constant rate from [#permalink]
02 May 2014, 00:40

Expert's post

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. _________________

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