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Problem of The Day 2.07.09

with 10 comments


Two triangles are considered distinct if they cannot be superimposed on each other by rotation. How many distinct triangles, with integer sides,  exist, such that there perimeter is 30?

A) 19    B) 57    C) 114    d) 38    E) 36

Written by Implex

July 2, 2009 at 6:35 am

Posted in Geometry, Problem of the week

Tagged with , ,

10 Responses

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  1. option B????

    anup

    July 2, 2009 at 7:16 am

  2. Kindly post your process, i will know if it is right or not!

    Rahul

    July 2, 2009 at 7:29 am

    • suppose d given triangles are equilateral..den dere side=20..so 18 distinct triangles can be formed…den findout d area of circum circle..proceeding in dat way i found its near 60…m not sure..but help me out

      anup

      July 2, 2009 at 7:43 am

      • You need to think in terms of triangle inequalities

        Rahul

        July 2, 2009 at 7:44 am

  3. is the answer a) 19 ?
    i.e. how i approached this question
    since all(any two) the sides are less than half the perimeter.
    14 ,14 ,2
    14 ,13 ,3
    14 ,12 ,4
    . .
    8 8 = total 7 possibilities.
    similary
    13 ,13 ,4
    13 ,12 ,5
    13 ,11 ,6
    likewise we can do for 12 , 11 and 10.
    giving a total of 19.

    correct me if i m wrong.

    yodha

    July 4, 2009 at 3:32 pm

  4. One sincere question..

    Could we have a facility of subscribing to the posts of this blog. This blog is a boon to CAT aspirants.🙂

    And could you provide solutions at the end of 48 hours or a specific period?

    If it’s not asking too much🙂

    Anish

    July 4, 2009 at 8:03 pm

  5. https://quantologic.wordpress.com/feed/

    there is a feed button out there!🙂

    Yeah I generally provide solutions.
    I was away traveling and little busy tonight. I hope to satisfy all your needs by tomorrow or day after.

    Good Luck

    Rahul

    July 6, 2009 at 3:37 pm

  6. is the answer a) 19 ?
    i.e. how i approached this question
    since all(any two) the sides are less than half the perimeter.
    14 ,14 ,2
    14 ,13 ,3
    14 ,12 ,4
    . .
    8 8 = total 7 possibilities.
    similary
    13 ,13 ,4
    13 ,12 ,5
    13 ,11 ,6
    likewise we can do for 12 , 11 and 10.
    giving a total of 19.

    correct me if i m wrong

    yodha

    July 16, 2009 at 5:12 am

  7. what is the answer to this question ????

    yodha

    July 16, 2009 at 5:16 am

  8. it’s 19 i guess

    Rahul

    July 16, 2009 at 5:18 am


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