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

Question

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.

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.

PrashantPonde · Accepted Answer

Rate Table Algebraic.jpg 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).

Marcab · Answer

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.

Abhii46 · Answer

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)

PrashantPonde · Answer

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)

BangOn · Answer

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)

PareshGmat · Answer

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
 
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)

mau5 · Answer

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.

PareshGmat · Answer

Please refer screenshot below:

himanshujovi · Answer

How come Roger's distance be taken as c-x ? I dont see it being mentioned that distance between the two points is c.

Bunuel · Answer

Jim takes a seconds to swim  c  meters at a constant rate  from point P to point Q in a pool .

gracie · Answer

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.

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

JeffTargetTestPrep · Answer

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)  = ac/(a + b) meters and Jim has swum (c/a)  = 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

effatara · Answer

(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*  meters.
(3) Roger swims b*  meters.
(4) Difference: bc/(a+b) - ac/(a+b)...> c(a-b)/(a+b)

danthegmatmannyc · Answer

Check out my video solution here:

https://www.youtube.com/watch?v=adcVyZLrPvY&ab_channel=DantheGMATMan

KarishmaB · Answer

If you do not want to pick numbers, take ratios. Over same distance, time taken by Jim:Rogers = a:b Speed of Jim:Rogers = b:a (Inverse of time ratio since distance is constant) Distance covered in same time by Jim:Rogers = b:a So diff in distance covered on ratio scale = a - b (because a > b) Total distance of ratio scale = a+b Total actual distance = c Multiplier = c/(a+b) So diff in distance covered = (a-b)*c/(a+b) Answer (B) Check this post for a discussion on ratios: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... of-ratios/ https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... os-in-tsd/

Adit_ · Answer

I got to the answer but isnt the gap between them actually 2d?
I got d as c(a-b)/a+b)

Can you help me out  Bunuel

Bunuel · Answer

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.

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)

Jim’s speed  = \frac{c}{a} 
Roger’s speed  = \frac{c}{b}

Let t be the time (in seconds) until they meet.

Since they swim toward each other and the total distance between P and Q is c:

(\frac{c}{a})t + (\frac{c}{b})t = c

t(\frac{1}{a} + \frac{1}{b}) = 1

t = \frac{ab}{a + b}

Distances swum when they meet:

Jim = speed * time  = \frac{c}{a} * \frac{ab}{a + b} = \frac{bc}{a + b}

Roger = speed * time  = \frac{c}{b} * \frac{ab}{a + b} = \frac{ac}{a + b}

Difference (Roger - Jim):

\frac{ac}{a + b} - \frac{bc}{a + b} = \frac{c(a - b)}{a + b}

Answer: B.

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

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GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards