5 thoughts on “XAT Scan – 8

    • That’s a lot more complicated, I suspect 🙂 A possible approach might be to eliminate all cases whee c^2 > a^2 + b^2, but even that needs to be handled carefully. Shall think about it and reply again if I find a reasonable answer, but from the point of view of aptitude tests, you may safely assume it is out of scope. If you’re writing an olympiad, however…

      regards
      J

  1. The graphical method is as always very elegant.

    I tried doing it in a different way and landed on 1/4. Can you please let me know if it is legit?

    Let the pieces be A,B and C. Let us choose A first. Clearly it cannot be greater than 1/2 of the total length. So probability of choosing an appropriate A which can lead to a triangle is 1/2.

    Now for B.
    Let the length of A be x. So 1-x is left. Suppose we take half of it for B. So length of B is (1-x)/2. Then length of C will be 1-(x+(1-x)/2) = 1/2 + x/2. Sum of lengths of A and B = x + (1-x)/2 = 1/2+x/2, which is same as that of C and so violates the triangle rule. So length of B has to be less than half of what is left after A. So we have 1/2 probability of choosing B which leads to a triangle. So total probability = 1/2 * 1/2 = 1/4.

    I know it looks rather clumsy, but do you think it is correct? Please let me know

    • No, I think you have made an error in this statement “Then length of C will be 1-(x+(1-x)/2) = 1/2 + x/2”. It will in fact be (1-x)/2 only. You can use this kind of approach too, but to prove its correctness requires integration.

      regards
      J

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s