A girl travels along a straight line, from point A to B at a constant

Question

A girl travels along a straight line, from point A to B at a constant speed, meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of meters/sec for another T seconds. BC makes an angle 105° with AB. If CA makes an angle 30° with BC, how much time will she take to travel back from point C to A at a constant speed of meters/sec, if she travels along a straight line from C to A?

A \(0.1(\sqrt{3} +1)T\)
B \(0.2(\sqrt{3} +1)T\)
C \(0.5(\sqrt{3} +1)T\)
D \(0.6(\sqrt{3} +1)T\)
E \((\sqrt{3} +1)T\)

A  0.1(\sqrt{3} +1)T 
B  0.2(\sqrt{3} +1)T 
C  0.5(\sqrt{3} +1)T 
D  0.6(\sqrt{3} +1)T 
E  (\sqrt{3} +1)T

TestPrepUnlimited · Answer

We need one of the speeds to answer this question.

The graph for this question is fixed though. It is shown below and most importantly we need to draw BD perpendicular for AC to solve this question.

The two triangles would be 45-45-90 and 30-60-90 so every length can be solved with basic trig.

We can still figure out that AB and BC have a ratio of  \sqrt{2}:2 , hence only one speed is needed.

yashikaaggarwal · Answer

Is there any easy way to solve this? I couldn't even comprehend the question.

A girl travels along a straight line, from point A to B at a constant

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GMAT Problem Solving (PS) Questions

A girl travels along a straight line, from point A to B at a constant

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