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I am trying exercises of Quantitative Aptitude and I am unable to work out how this problem can be solved:

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As a tetrahedron has 6 edges, I thought 5 cuts should be required. But that's wrong.

Answer is:

C ( 1 )

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    $\begingroup$ This is equivalent to the number of times that one needs to lift the pen from the paper to draw the figure if it's not allowed to re-draw a line $\endgroup$ – legrojan Jun 23 '20 at 6:53
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There needs to be

One cut

The tetrahedron has 4 vertices, and three wires meet at each – an odd number.
So at each vertex there must be one bend (2 wires), and one end (1 wire).
So at each vertex, there must be one free end.
So there are 4 ends needed, and so two wires are needed, and so one cut is made.
Each wire has two bends.

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