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

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

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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  [#permalink]

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07 Dec 2012, 20:43
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2
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  [#permalink]

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07 Dec 2012, 21:18
Marcab 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.

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]

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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]

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08 Dec 2012, 00:01
Marcab 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.

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 =$$c/a$$
R speed = $$c/b$$

Time at which they pass each other = Gap / sum of speeds

Gap = C
Sum of speeds = $$c/a + c/b$$= $$c (b+a) /ab$$
Therefore time at which they pass will be $$ab / a+b$$

question 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)
$$c / b * ab / a+b$$ - $$c/a * ab / a+b$$
On 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]

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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]

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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]

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08 Dec 2012, 02:37
Hii JP.
Planning to take by mid Feb, though haven't taken the date yet.
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Re: Jim takes a seconds to swim c meters at a constant rate  [#permalink]

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14 Jun 2013, 02:01
1
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
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Re: Jim takes a seconds to swim c meters at a constant rate  [#permalink]

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14 Jun 2013, 04:38
2
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  [#permalink]

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08 Aug 2013, 13:42
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?

Rate (Jim): c/a

Rate (Roger): c/b

Rate (Roger) > Rate (Jim)

Plugging in numbers: c=100, a=10, b=20

rate (Jim) = 100/10 minutes
rate = 10 meters/minute

rate (Roger) = 100/5 minutes
rate = 20 meters/minute

time = distance/rate
time = 100/(10+20)
time = 3.33 minutes until they meet one another.

distance = rate*time
distance (Jim) = 10*3.33
distance (Jim) = 33.3 meters.

He swims 100-33.3 = 66.7m less than Roger

100(10-20)/(10+20)
-1000/30
= -66.7
i.e. he swam 66.7 m less.

Is that correct?
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Re: Jim takes a seconds to swim c meters at a constant rate  [#permalink]

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08 Aug 2013, 13:49
1
WholeLottaLove 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?

Rate (Jim): c/a

Rate (Roger): c/b

Rate (Roger) > Rate (Jim)

Plugging in numbers: c=100, a=10, b=20

rate (Jim) = 100/10 minutes
rate = 10 meters/minute

rate (Roger) = 100/5 minutes
rate = 20 meters/minute

time = distance/rate
time = 100/(10+20)
time = 3.33 minutes until they meet one another.

distance = rate*time
distance (Jim) = 10*3.33
distance (Jim) = 33.3 meters.

He swims 100-33.3 = 66.7m less than Roger

100(10-20)/(10+20)
-1000/30
= -66.7
i.e. he swam 66.7 m less.

Is that correct?

yes IMO All fine in your approach only thing in place minutes it should be seconds as it is already stated in question than a and b are in seconds.
moreover since ROGER is faster he is going to take less time thats why b<a
hence assume b=10 and a=20...then you will not get negative value.
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Re: Jim takes a seconds to swim c meters at a constant rate  [#permalink]

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Updated on: 15 Aug 2013, 09:27
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?

We are given jim's rate and rogers rate in addition to the size of the pool. Also, we know that Roger is faster than Jim.

The portion in blue implies that Roger can swim the same distance in less time. However, when Jim and Roger cross paths in the pool, the time they will have traveled will be the same. Because the time that they spend swimming is the same we can say time = t.

If t is the same for both swimmers than why can't we just multiply their respective rates by t? Why do we have to solve out for t? It seems that for other problems, where we have two objects leaving from point A and B respectively and meeting somewhere in the middle, we do not need to solve out for 'T'.

We know the rate of both swimmers. We also know that the time they spend swimming is the same. We also know the length of the pool. We can do a combined distance formula where:

(rate Jim)*(t) + (rate Roger)*(t) = c

(c/a)*t + (c/b)*t = c
ct/a + at/b = c

Originally posted by WholeLottaLove on 08 Aug 2013, 16:49.
Last edited by WholeLottaLove on 15 Aug 2013, 09:27, edited 1 time in total.
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Re: Jim takes a seconds to swim c meters at a constant rate  [#permalink]

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08 Aug 2013, 19:36
WholeLottaLove 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?

We are given jim's rate and rogers rate in addition to the size of the pool. Also, we know that Roger is faster than Jim.

The portion in blue implies that Roger can swim the same distance in less time. However, when Jim and Roger cross paths in the pool, the time they will have traveled will be the same. Because the time that they spend swimming is the same we can say time = t.

If t is the same for both swimmers than why can't we just multiply their respective rates by t? Why do we have to solve out for t?

We know the rate of both swimmers. We also know that the time they spend swimming is the same. We also know the length of the pool. We can do a combined distance formula where:

(rate Jim)*(t) + (rate Roger)*(t) = c

(c/a)*t + (c/b)*t = c
ct/a + at/b = c

this thinking is also correct.
you have:
(c/a)*t + (c/b)*t = c
solving for t
cancel c from both sides
on solving for t you will get t= ab/a+b
now we have too calculate difference in distance...
distance by jim = rate*time(t) = $$(\frac{c}{a})*(\frac{(ab)}{(a+b)})$$
distance by roger = rate*time(t) =$$( \frac{c}{b})*(\frac{(ab)}{(a+b)})$$

now suntract these two distance and you will get
$$\frac{(c(a-b))}{(a+b)}$$
hope this helps
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Re: Jim takes a seconds to swim c meters at a constant rate  [#permalink]

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08 Aug 2013, 22:26
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.

....
Jim = c/a
roger=c/b
let jim x distance in t time. so x = c/a t
so, roger c-x distance in t time, c-x = c/b t
....................................................................
(+) , or, t = ab/a+b
Now, more distance = c/b t - c/a t = c(a-b)/(a+b) Answer
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Jim takes a seconds to swim c meters at a constant rate  [#permalink]

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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  [#permalink]

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12 Aug 2017, 23:47
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Re: Jim takes a seconds to swim c meters at a constant rate   [#permalink] 12 Aug 2017, 23:47
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