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Jim takes a seconds to swim c meters at a constant rate [#permalink]
 07 Dec 2012, 20:12
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51% (13:38) correct
48% (02:33) wrong based on 31 sessions
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.
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Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]
07 Dec 2012, 20:43
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aditi2013 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/bNote 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= cOn 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.
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Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]
07 Dec 2012, 21:18
Marcab wrote: aditi2013 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/bNote 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= cOn 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.
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Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]
07 Dec 2012, 22:17
Use the concept of relative speed when the two trains are travelling in same direction.
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Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]
08 Dec 2012, 00:01
aditi2013 wrote: Marcab wrote: aditi2013 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/bNote 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= cOn 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/aR speed = c/bTime at which they pass each other = Gap / sum of speeds Gap = C Sum of speeds = c/a + c/b= c (b+a) /abTherefore time at which they pass will be ab / a+bquestion 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+bOn solving gives c(a-b)/a+b@marcab long time no see what's new dude? Cheers
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Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]
08 Dec 2012, 00:19
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?
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Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]
08 Dec 2012, 01:19
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
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Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]
08 Dec 2012, 02:37
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Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]
14 Jun 2013, 02:01
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
Answer = (B)
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Re: Jim takes a seconds to swim c meters at a constant rate [#permalink]
14 Jun 2013, 04:38
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aditi2013 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. 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^2Out 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.
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Re: Jim takes a seconds to swim c meters at a constant rate
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14 Jun 2013, 04:38
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