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

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Math Expert
Joined: 02 Sep 2009
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Cyclist A leaves point X at 12 noon and travels at constant velocity  [#permalink]

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11 Apr 2017, 01:08
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Difficulty:

5% (low)

Question Stats:

94% (01:07) correct 6% (00:57) wrong based on 155 sessions

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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 A at 4p.m. What was the speed of B?

(1) Cyclist A weighs twice as much as cyclist B.
(2) When Cyclist B overtook cyclist A, both had traveled 75 miles.

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

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11 Apr 2017, 01:23
we need to find either of following parameters :
1. distance travelled
2. speed of A

weight is irrelevent

only statement 2 is sufficient.
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity  [#permalink]

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11 Apr 2017, 05:30
Statement 2 is sufficient since the distance travelled is provided and we know the time hence, one can calculate using the formula S=D/T

hence, statement 2 is sufficient.
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity  [#permalink]

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11 Apr 2017, 08:01
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 A at 4p.m. What was the speed of B?

(1) Cyclist A weighs twice as much as cyclist B.
(2) When Cyclist B overtook cyclist A, both had traveled 75 miles.

if combined distance travelled by them is 75 miles

if individual distance is 75 miles then the answer should be B
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity  [#permalink]

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11 Apr 2017, 09:04
B) 75/2hr is B's speed

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

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17 Jul 2017, 07:57
(2) When Cyclist B overtook cyclist A, both had traveled 75 miles.

B+A= 75
OR
B=75 , A=75

?
Confusing statement
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity  [#permalink]

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18 Jul 2017, 10:15
abhisheknandy08 wrote:
(2) When Cyclist B overtook cyclist A, both had traveled 75 miles.

B+A= 75
OR
B=75 , A=75

?
Confusing statement

It's not confusing at all. They each went 75miles. You don't even need to know about A

Posted from my mobile device
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity  [#permalink]

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07 Aug 2017, 23:46
But here are we not assuming that the total distance to be traveled is just 75 km??VeritasPrepKarishma Mam can you please explain this to me?? Thank you
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity  [#permalink]

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08 Aug 2017, 01:05
1
longhaul123 wrote:
But here are we not assuming that the total distance to be traveled is just 75 km??VeritasPrepKarishma Mam can you please explain this to me?? Thank you

You know that both A and B start from point X. The point where B "overtakes" A, they both are at the same point.
Stmnt 2 tells us that
"When Cyclist B overtook cyclist A, both had traveled 75 miles."

So they both would have been 75 miles away from point X. Perhaps to clarify, it would be a good idea to write "both had travelled 75 miles each". If their total distance would have been 75 miles, the stmnt would have said "both had travelled 75 miles together".

Anyway, here it doesn't matter. It is a DS question. Even if they had travelled 75 miles together, you would know that each had travelled 37.5 miles because they had travelled equal distances. Still information would be sufficient to answer the question using stmnt 2 alone.
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Re: Cyclist A leaves point X at 12 noon and travels at constant velocity  [#permalink]

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08 Aug 2017, 23:34
A starts at 12 noon
B starts at 2 noon

(1) Cyclist A weighs twice as much as cyclist B.
(2) When Cyclist B overtook cyclist A, both had traveled 75 miles.

Statement 1 is irrelevant

As per statement 2 B overtakes A at 4p.m so total time elapsed is

A - 4 hours
B- 2 hours

D = s x t
DA = sa X 4 ---- A
DB = sb x 2 ---- B

Statement 2 alone is sufficient.
Re: Cyclist A leaves point X at 12 noon and travels at constant velocity   [#permalink] 08 Aug 2017, 23:34
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