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A girl travels along a straight line, from point A to B at a constant speed, V1 meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of V2 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 Vj meters/sec, if she travels along a straight line from C to A?
  • a)
    0.53(√3- 1)T
  • b)
    T
  • c)
    0.5(√3+ 1)T
  • d)
    √3
  • e)
    None of the above
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A girl travels along a straight line, from point A to B at a constant ...
We draw BD perpendicular to AC.
In right angled triangle BDC, BD / BC = sin 30°
or, BD = (V2 * T)/2 ……. (i)
In right angled triangle BDA, BD / BA = sin 45°
Or, BD = (V1 * T)/a/2 …… (ii)
From (i) and (ii),we get
V2/V1 = √2
Total distance to be travelled from C to A = CD + DA = √3 BD + BD
= BD(1 +√3 )
Replacing BD = (V2 * T)/2 in the above equation,
CA = (V2 * T) /2(1 +√3 )
Time taken at speed V2 = 0.5(√3 + 1)T
Hence, option C is the correctanswer.
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Most Upvoted Answer
A girl travels along a straight line, from point A to B at a constant ...
To find the distance traveled from point A to point B, we can use the formula:

Distance (AB) = Speed (V1) * Time (T)

To find the distance traveled from point B to point C, we can use the same formula:

Distance (BC) = Speed (V2) * Time (T)

The angle between AB and BC is given as 105 degrees. Since the girl is traveling in a straight line, the angle between AB and BC is equal to the angle between BC and the extension of AB. Therefore, the angle between BC and the extension of AB is also 105 degrees.

Using the Law of Cosines, we can find the length of BC:

BC² = AB² + AC² - 2 * AB * AC * cos(105)

Since AB is equal to the distance traveled in the first leg of the journey (AB = V1 * T), we can substitute it into the equation:

BC² = (V1 * T)² + AC² - 2 * V1 * T * AC * cos(105)

Similarly, AC is equal to the distance traveled in the second leg of the journey (AC = V2 * T), so we can substitute it into the equation:

BC² = (V1 * T)² + (V2 * T)² - 2 * V1 * T * V2 * T * cos(105)

Simplifying further:

BC² = V1² * T² + V2² * T² - 2 * V1 * V2 * T² * cos(105)

BC = √(V1² * T² + V2² * T² - 2 * V1 * V2 * T² * cos(105))

Therefore, the length of BC can be found using the given speeds (V1 and V2), time (T), and the angle between AB and BC (105 degrees).
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A girl travels along a straight line, from point A to B at a constant speed, V1 meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of V2 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 Vj meters/sec, if she travels along a straight line from C to A?a)0.53(√3- 1)Tb)Tc)0.5(√3+ 1)Td)√3e)None of the aboveCorrect answer is option 'C'. Can you explain this answer?
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A girl travels along a straight line, from point A to B at a constant speed, V1 meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of V2 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 Vj meters/sec, if she travels along a straight line from C to A?a)0.53(√3- 1)Tb)Tc)0.5(√3+ 1)Td)√3e)None of the aboveCorrect answer is option 'C'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about A girl travels along a straight line, from point A to B at a constant speed, V1 meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of V2 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 Vj meters/sec, if she travels along a straight line from C to A?a)0.53(√3- 1)Tb)Tc)0.5(√3+ 1)Td)√3e)None of the aboveCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A girl travels along a straight line, from point A to B at a constant speed, V1 meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of V2 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 Vj meters/sec, if she travels along a straight line from C to A?a)0.53(√3- 1)Tb)Tc)0.5(√3+ 1)Td)√3e)None of the aboveCorrect answer is option 'C'. Can you explain this answer?.
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