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Cyclist A leaves point X at 12 noon and travels at constant velocity

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Cyclist A leaves point X at 12 noon and travels at constant velocity [#permalink]

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New post 14 Jan 2018, 05:45
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Cyclist A leaves point X at 12 noon and travels at constant velocity in a straight path. Cyclist B leaves point X at 2 p.m. travels the same path at constant velocity, and overtakes cyclist A at 4p.m. At what speed was cyclist B traveling?

(1) Cyclist A traveled 15 miles in the first hour.
(2) The rate of cyclist B is twice that of cyclist A.
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity [#permalink]

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New post 14 Jan 2018, 13:28
Bunuel wrote:
Cyclist A leaves point X at 12 noon and travels at constant velocity in a straight path. Cyclist B leaves point X at 2 p.m. travels the same path at constant velocity, and overtakes cyclist A at 4p.m. At what speed was cyclist B traveling?

(1) Cyclist A traveled 15 miles in the first hour.
(2) The rate of cyclist B is twice that of cyclist A.


We know the total time of B so all we need is their total distance.
We'll look for an answer that gives us this information, a Logical approach.

(1) Since A travels at a constant speed then we can multiply 15 by 4 hours to get the distance A cycled which is also the distance B cycled.
Sufficient!

(2) This gives us no information on the distance and is not helpful.
(in fact, it only restates information we already know as B travels the same distance as A in half the time = twice the rate)
Insufficient.

(A) is our answer.
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity [#permalink]

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New post 14 Jan 2018, 23:00
Cyclist A leaves point X at 12 noon and travels at constant velocity in a straight path. Cyclist B leaves point X at 2 p.m. travels the same path at constant velocity, and overtakes cyclist A at 4p.m. At what speed was cyclist B traveling?

(1) Cyclist A traveled 15 miles in the first hour.
(2) The rate of cyclist B is twice that of cyclist A.

Hi,

Distance covered by B in 2 hrs = distance covered by A in 4 hrs.
Therefore 2B=4A
OR B=2A where A and B are the speed of A and B

1.A=15 hence b=30;
sufficient;
2. repeats the given premise. insufficient.
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Cyclist A leaves point X at 12 noon and travels at constant velocity [#permalink]

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New post 15 Jan 2018, 04:34
DavidTutorexamPAL wrote:
Bunuel wrote:
Cyclist A leaves point X at 12 noon and travels at constant velocity in a straight path. Cyclist B leaves point X at 2 p.m. travels the same path at constant velocity, and overtakes cyclist A at 4p.m. At what speed was cyclist B traveling?

(1) Cyclist A traveled 15 miles in the first hour.
(2) The rate of cyclist B is twice that of cyclist A.


We know the total time of B so all we need is their total distance.
We'll look for an answer that gives us this information, a Logical approach.

(1) Since A travels at a constant speed then we can multiply 15 by 4 hours to get the distance A cycled which is also the distance B cycled.
Sufficient!

(2) This gives us no information on the distance and is not helpful.
(in fact, it only restates information we already know as B travels the same distance as A in half the time = twice the rate)
Insufficient.

(A) is our answer.


Hi David,

In statement 1, Could the speed second and third hour be different than the speed in first hour? does not it affect the calculation?

Suppose this example 1:

speed of 1st hr = 15,
speed of 2nd hr = 20
speed of 3rd hr = 10
speed of 4th hr = 5

Total distance = 50

Suppose this example 2:

speed of 1st hr = 15,
speed of 2nd hr = 25
speed of 3rd hr = 15
speed of 4th hr = 25

Total distance = 80

So we in the same 4 hours there would different speeds leading to different values of distance.

Where did I go wrong?
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity [#permalink]

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New post 15 Jan 2018, 04:50
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Mo2men wrote:
DavidTutorexamPAL wrote:
Bunuel wrote:
Cyclist A leaves point X at 12 noon and travels at constant velocity in a straight path. Cyclist B leaves point X at 2 p.m. travels the same path at constant velocity, and overtakes cyclist A at 4p.m. At what speed was cyclist B traveling?

(1) Cyclist A traveled 15 miles in the first hour.
(2) The rate of cyclist B is twice that of cyclist A.


We know the total time of B so all we need is their total distance.
We'll look for an answer that gives us this information, a Logical approach.

(1) Since A travels at a constant speed then we can multiply 15 by 4 hours to get the distance A cycled which is also the distance B cycled.
Sufficient!

(2) This gives us no information on the distance and is not helpful.
(in fact, it only restates information we already know as B travels the same distance as A in half the time = twice the rate)
Insufficient.

(A) is our answer.


Hi David,

In statement 1, Could the speed second and third hour be different than the speed in first hour? does not it affect the calculation?

Suppose this example 1:

speed of 1st hr = 15,
speed of 2nd hr = 20
speed of 3rd hr = 10
speed of 4th hr = 5

Total distance = 50

Suppose this example 2:

speed of 1st hr = 15,
speed of 2nd hr = 25
speed of 3rd hr = 15
speed of 4th hr = 25

Total distance = 80

So we in the same 4 hours there would different speeds leading to different values of distance.

Where did I go wrong?


Hi Mo2men,

The question states "at constant velocity" for both cyclists.
So the speed is the same during all the hours... otherwise you would be correct!
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity [#permalink]

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New post 15 Jan 2018, 04:53
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DavidTutorexamPAL wrote:

Hi Mo2men,

The question states "at constant velocity" for both cyclists.
So the speed is the same during all the hours... otherwise you would be correct!


Good point :-)

thanks for your fast reply
Re: Cyclist A leaves point X at 12 noon and travels at constant velocity   [#permalink] 15 Jan 2018, 04:53
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