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Nicole cycles at a constant rate of 20 kilometers per hour, and is pas

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Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 06 Sep 2015, 00:27
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Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

a) x minutes
b) x/2 minutes
c) 2x/3 minutes
d) 3x/2 minutes
e) 2x minutes
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Re: Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 08 Jan 2017, 00:19
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Jessica rides for x minutes and in this time let's say she covers distance d.

Nicole would have covered 2d/3 (since their speeds are in the ratio 2:3)

Thus he has d/3 left, which is half of 2d/3. Hence he would take half the time taken for 2d/3 which is x/2.
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Re: Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 06 Sep 2015, 02:05
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chandeep01 wrote:
Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

a) x minutes
b) x/2 minutes
c) 2x/3 minutes
d) 3x/2 minutes
e) 2x minutes


Speed of nicole = 20km/h or 20/60 km/min = 1/3 km/min.

Once jessica passed the nicole, the distance between the nicole and jessica will increase at the rate of (30-20) = 10km/h or 1/6 km/min

now jessica is cycling for x minutes after passing the nicole, so in those x minutes distance between jessica and nicole would be (1/6)*x = x/6 km.

So, the time taken by nicole to travel x/6 km = (x/6)/(1/3) = x/2. hence answer should be B.
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Re: Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 06 Sep 2015, 05:55
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chandeep01 wrote:
Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

a) x minutes
b) x/2 minutes
c) 2x/3 minutes
d) 3x/2 minutes
e) 2x minutes


Let, x = 60

Relative Speed = 30-20 = 10 Km/Hr

Jessica passes Nicole and runs for x(=60)mins i.e. makes a distance of 10*1 = 10 Km ahead of Nicole

Nicole has speed of 20 Km/Hr so a distance of 10 km will be covered by her in = 10/20 = 1/2 hour = 30 mins = x/2 mins

Answer: Option B
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Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 06 Sep 2015, 05:56
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.



Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hours. If Jessica cycles at her constant rates for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to waits for Nicole to catch up to her.
a) x minutes
b) x/2 minutes
c) 2x/3 minutes
d) 3x/2 minutes
e) 2x minutes


After Jessica passed Nicole, since the relative speed of Nicole and Jessica is 10 kilometers per hour, the distance between two during x/60 hours(=x minutes) is (x/60) * (10) kilometers. Because Jessica stops after x/60 hours(=x minutes) Nicole should catch up (x/60) * (10) kilometers with her 20 kilometers per hour's speed. If it takes t minutes(=t/60 hours) then t should satisfy (t/60)*(20)=(x/60) * (10). So t= x/2. That is Jessica have to waits x/2minutes for Nicole to catch up to her.The answer is b).
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Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 07 Sep 2015, 13:05
N loses 1/6 k/m to J
In x minutes N is x/6 k behind J
N cycles at 1/3 k/m
x/6 / 1/3 = x/2 minutes to catch up
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Re: Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 07 Jan 2017, 21:42
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chandeep01 wrote:
Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

a) x minutes
b) x/2 minutes
c) 2x/3 minutes
d) 3x/2 minutes
e) 2x minutes



after meeting each other
distance travelled by jessica in x minutes=30xkm
for x minutes nicole travelled =20x km

distance between them =30x-20x=10x
this distance (10x) will be covered by nicole in 10x/20=x/2 minutes
thus jessica have to wait for x/2 minutes

Ans B
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Re: Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 07 Jan 2017, 22:57
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Hi Chandeep01,

Here is my 2 cents for this question,

Best way to solve this problem is Plugging in for x minutes.

“x” we can take anything, just make sure here we take a multiple of 30 and 20 because its easy for calculation.

Let say x = 120 minutes, that is 2 hours.

Then for two hours Jessica would have travelled,

D = 2 * 30 = 60 kms,

And Nicole would have travelled,

D = 2 * 20 = 40kms,

So Jessica is ahead of Nicole by 20kms and wait for Nicole to join.

So Nicole has to cover 20kms, so it would take an hour for Nicole to cover 20kms.

That’s it, since the answer choices are minutes, one hour = 60 mins.

So we should get the answer as 60 while substituting x = 120.

So the answer is B.

Hope this helps
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Re: Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 08 Jan 2017, 01:59
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CrackVerbalGMAT wrote:
Hi Chandeep01,

Here is my 2 cents for this question,

Best way to solve this problem is Plugging in for x minutes.

“x” we can take anything, just make sure here we take a multiple of 30 and 20 because its easy for calculation.

Let say x = 120 minutes, that is 2 hours.

Then for two hours Jessica would have travelled,

D = 2 * 30 = 60 kms,

And Nicole would have travelled,

D = 2 * 20 = 40kms,

So Jessica is ahead of Nicole by 20kms and wait for Nicole to join.

So Nicole has to cover 20kms, so it would take an hour for Nicole to cover 20kms.

That’s it, since the answer choices are minutes, one hour = 60 mins.

So we should get the answer as 60 while substituting x = 120.

So the answer is B.

Hope this helps


Dear CrackVerbal,

When I tried the rule that you mentioned in your post below. it did not work properly.

cars-p-and-q-started-simultaneously-from-opposite-ends-of-a-159355.html#p1782046

After Nicole passed Jessica, Jessica cycled for certain time the stopped till Jessica cached her up. So it mean that both cycled for same DISTANCE. So I applied the following rule as you mentioned

S1T1 = S2T2...........However It gave me wrong answer D.

Where did I go wrong?

thanks in advance for you help
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Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 08 Jan 2017, 15:53
chandeep01 wrote:
Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

a) x minutes
b) x/2 minutes
c) 2x/3 minutes
d) 3x/2 minutes
e) 2x minutes


J cycles 1/2 kpm
N cycles 1/3 kpm
after passing N, J cycles x*1/2=x/2 k
in same time, N cycles x*1/3=x/3 k
(x/2)-(x/3)=x/6 k remaining for N to catch up to J
x/6 k/1/3 kpm=x/2 minutes
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Re: Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 09 Jan 2017, 01:48
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Mo2men wrote:
CrackVerbalGMAT wrote:
Hi Chandeep01,

Here is my 2 cents for this question,

Best way to solve this problem is Plugging in for x minutes.

“x” we can take anything, just make sure here we take a multiple of 30 and 20 because its easy for calculation.

Let say x = 120 minutes, that is 2 hours.

Then for two hours Jessica would have travelled,

D = 2 * 30 = 60 kms,

And Nicole would have travelled,

D = 2 * 20 = 40kms,

So Jessica is ahead of Nicole by 20kms and wait for Nicole to join.

So Nicole has to cover 20kms, so it would take an hour for Nicole to cover 20kms.

That’s it, since the answer choices are minutes, one hour = 60 mins.

So we should get the answer as 60 while substituting x = 120.

So the answer is B.

Hope this helps


Dear CrackVerbal,

When I tried the rule that you mentioned in your post below. it did not work properly.

cars-p-and-q-started-simultaneously-from-opposite-ends-of-a-159355.html#p1782046

After Nicole passed Jessica, Jessica cycled for certain time the stopped till Jessica cached her up. So it mean that both cycled for same DISTANCE. So I applied the following rule as you mentioned

S1T1 = S2T2...........However It gave me wrong answer D.

Where did I go wrong?

thanks in advance for you help


Hi Mo2men,

You have done the first part right in identifying that the distance traveled by both Nicole and Jessica is a constant.

Let SN and TN be the speed and time of Nicole and SJ and TJ be the speed and time of Jessica.

Since the distance is constant SN*TN = SJ * TJ

Now the time taken by Jessica after she passes Nicole is 'x' minutes. After travelling for 'x' minutes Jessica stops and waits for Nicole. So the total time taken by Jessica to travel the constant distance is 'x' minutes.

Jessica travels for the initial 'x' minutes and an additional 'y' minutes when Nicole is at rest (waiting for her to catch up). So the total time taken by Jessica to travel the constant distance is 'x + y' minutes.

Here we need to find the value of y with respect to x

So substituting the speeds and times in SN*TN = SJ * TJ we get

20(x + y)/60 = 30(x)/60 ; Dividing by 60 since the time is in minutes

Simplifying we get x = 2y -----> y = x/2

Answer : B

Hope this helps!

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Re: Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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New post 11 Jan 2017, 08:01
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chandeep01 wrote:
Nicole cycles at a constant rate of 20 kilometers per hour, and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. If Jessica cycles at her constant rate for x minutes after passing Nicole, then stops to wait for her, how many minutes will Jessica have to wait for Nicole to catch up to her?

a) x minutes
b) x/2 minutes
c) 2x/3 minutes
d) 3x/2 minutes
e) 2x minutes


We can classify this problem as a “catch-up” rate problem, for which we use the fact that distance travelled by Nicole = distance travelled by Jessica.

We are given that Nicole cycles at a constant rate of 20 kilometers per hour and is passed by Jessica, who cycles at a constant rate of 30 kilometers per hour. We are also given that Jessica cycles at her constant rate for x minutes after passing Nicole, and then stops to wait for her. We can let t be the waiting time in minutes; that is, Nicole has to cycle x + t minutes to catch up to Jessica, who has to cycle x minutes only.

Since distance = rate x time, we can calculate each person’s distance in terms of t and x. However, the rate is in km per hour, but the time is in minutes. Thus we need to convert the time from minutes to hours. Recall that there are 60 minutes in an hour. Thus, x minutes = x/60 hours and x + t minutes = (x + t)/60 hours.

Nicole’s distance = 20[(x + t)/60] = (x + t)/3

Jessica’s distance = 30(x/60) = x/2

We can equate the two distances and determine t in terms of x.

(x + t)/3 = x/2

2(x + t) = 3x

2x + 2t = 3x

2t = x

t = x/2

Answer: B
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Re: Nicole cycles at a constant rate of 20 kilometers per hour, and is pas  [#permalink]

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