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

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

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New post Updated on: 08 Jan 2013, 03:25
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A
B
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E

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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

Originally posted by samsikka23 on 08 Jan 2013, 03:14.
Last edited by Bunuel on 08 Jan 2013, 03:25, edited 1 time in total.
Renamed the topic.
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Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

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New post 08 Jan 2013, 13:12
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2
Attachment:
Rate Table Algebraic.jpg
Rate Table Algebraic.jpg [ 45.04 KiB | Viewed 5472 times ]

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) = \(\frac{ab}{(a+b)}\) * \((\frac{c}{b}-\frac{c}{a})\) = \(\frac{ab}{(a+b)}\) * \(\frac{c(a-b)}{ab}\) = \(\frac{c(a-b)}{(a+b)}\)

Hence answer is (B).
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Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

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

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New post 08 Jan 2013, 13:17
1
1
Attachment:
Pick numbers.jpg
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)
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Re: Jim takes a seconds to swim c meters at a constant rate from  [#permalink]

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

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

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

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

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

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

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

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New post 25 Jan 2017, 10:21
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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)


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

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New post 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 \(=\frac{c}{b}*\frac{ab}{(a+b)}-\frac{c}{a}*\frac{ab}{(a+b)} =\frac{ca}{(a+b)}-\frac{cb}{(a+b)}=\frac{c(a-b)}{(a+b)}\)
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