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

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

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New post 07 Dec 2012, 20:12
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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

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 from point P  [#permalink]

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New post 08 Jan 2013, 14:12
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Attachment:
Rate Table Algebraic.jpg
Rate Table Algebraic.jpg [ 45.04 KiB | Viewed 6638 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 point P  [#permalink]

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New post 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 \(c/b\)

Note 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. \((c/a)*t + (c/b)*t= c\)

On 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 from point P  [#permalink]

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

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New post 08 Jan 2013, 14:17
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Attachment:
Pick numbers.jpg
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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 point P  [#permalink]

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New post 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 point P  [#permalink]

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New post 14 Jun 2013, 02:01
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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 from point P  [#permalink]

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New post 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 \(time^2\)

Out 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 from point P  [#permalink]

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

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

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New post 02 May 2014, 01:40
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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Re: Jim takes a seconds to swim c meters at a constant rate from point P  [#permalink]

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New post Updated on: 13 Aug 2017, 18:23
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

let d=roger's distance at passing
c-d=jim's distance at passing
d/(c-d)=a/b
d=ac/(a+b)
d-(c-d)=2d-c
substituting, 2d-c=2*ac/(a+b)-c➡
c(a-b)/(a+b)
B

Originally posted by gracie on 18 Nov 2015, 16:45.
Last edited by gracie on 13 Aug 2017, 18:23, edited 1 time in total.
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Re: Jim takes a seconds to swim c meters at a constant rate from point P  [#permalink]

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New post 25 Jan 2017, 11:21
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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 point P  [#permalink]

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New post 16 Sep 2019, 00:38
(1) Time taken for Jim and Roger to meet: c/(a+b)... (Since they are swimming towards each other their combined speeds will come into play). So Jim and Roger each swim for c/(a+b) secs before they meet.
(2) Jim swims a*[c/(a+b)] meters.
(3) Roger swims b*[c(a+b)] meters.
(4) Difference: bc/(a+b) - ac/(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 point P   [#permalink] 16 Sep 2019, 00:38
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