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…


  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.


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