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25 Oct 2016, 23:10
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
49% (01:53) correct
51%
(02:04)
wrong
based on 92
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While snail R starts to move from point A to C via AC directly, snail S starts to move from point B to C via BC directly. After the snails meet at point C, R moves to D with S's former speed, while S moves to E with Rs former speed. Do the snails reach their destinations simultaneously?
(1) S moves with the speed of 1 centimeter per minute.
(2) R moves twice as fast as S.
Attachment:
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
26 Oct 2016, 06:00
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We know the ratios(1:root 3 : 2 ) of lengths as per the diagram , if we can know at least the ratio of speeds or speeds of snails , We can answer the question as Y/N.
1. here , only the speed of one snail is given ---->not sufficient.
2. here , the ratios are given ----> Sufficient
Ans. B
Correct me if any thing wrong , if its helpful , give me KUDOS please
1. here , only the speed of one snail is given ---->not sufficient.
2. here , the ratios are given ----> Sufficient
Ans. B
Correct me if any thing wrong , if its helpful , give me KUDOS please
26 Oct 2016, 18:24
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BC = AC * Sin(30) = AC/2
DC = EC/2
snail R
AC = R * t
snail S
Ac/2 = S * t1
(1) S =1
Nothing is mentioned about the time when R or S reach point C. We cannot find a relation between the speeds.
Insufficient.
(2) R moves twice as fast as S.
We know the speeds, and distance values. Hence time can be calculated whether both reach simultaneously.
Sufficient.
Option B.
DC = EC/2
snail R
AC = R * t
snail S
Ac/2 = S * t1
(1) S =1
Nothing is mentioned about the time when R or S reach point C. We cannot find a relation between the speeds.
Insufficient.
(2) R moves twice as fast as S.
We know the speeds, and distance values. Hence time can be calculated whether both reach simultaneously.
Sufficient.
Option B.
