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IIFT 2019 Question Paper | Quants

IIFT Previous Year Paper | IIFT Quants Questions | Question 16

The best way to boost your IIFT prep is to practice the actual IIFT Question Papers. 2IIM offers you exactly that, in a student friendly format to take value from this. In the 2019 IIFT, quants were a mixed bag of questions of varying difficulty, with some routine questions and the others were very demanding. Some beautiful questions that laid emphasis on Learning ideas from basics and being able to comprehend more than remembering gazillion formulae and shortcuts.

Question 16: A man standing on the line joining the two poles finds that the top of the poles make an angle of elevation of 60° and 45° respectively. After walking for sometime towards the other pole, the angles change to 30° and 60° respectively. The ratio of the height of the poles is :

  1. \\frac{\sqrt{3}-1}{2})
  2. \\frac{\sqrt{3}+1}{3})
  3. \\frac{\sqrt{3}-1}{4})
  4. \\frac{\sqrt{3}+1}{4})

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Explanatory Answer

Lets say the heights of the poles is h 1 and h2
After moving the angle to the first pole becomes 60 degree and to the second pole becomes 60 degree.
Now by Tan of the given angles, we can find the distances on the ground as indicated in the diagram below.

poles
We see that
h1 + \\frac{h_1}{\sqrt{3}}) = \\frac{h_1}{\sqrt{3}}) + √3 h2
h1 (1-\\frac{1}{\sqrt{3}}) ) = h2 \\sqrt{3}) - \\frac{1}{\sqrt{3}}))
\\frac{h_1(\sqrt{3}-1)}{\sqrt{3}}) = h2 × \\frac{2}{\sqrt{3}})
So, the ratio of the heights is,
\\frac{\sqrt{3}-1}{2})


The question is "The ratio of the height of the poles is :"

Hence, the answer is, "\\frac{\sqrt{3}-1}{2})"

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