23 June 2017

A logic puzzle after a long time…

Found an interesting one from my friend Prodipta Chatterjee. Took me a few minutes till I realized that the trick is to not think of it as other problems of similar descriptions… Try it…

From a bag that has 26 bills (of three denominations – $1, $2 and $5), you pick out 20 of them blindfolded. You are guaranteed (probability = 1) that you will have at least one $1 bill AND two $2 bills and five $5 bills.

You have to solve for the following question: How much total money is there in the bag?



Posted June 23, 2017 by Rajib Roy in category "Puzzles

8 COMMENTS :

  1. By Rajib Roy on

    Now for the answer. My mistake initially was to think about it as a probability problem. As Pradeep pointed out – that is my favorite hammer. 🙂 Actually, if you think about it, it is a much simpler problem….

    If you are guaranteed that there will be one $1 in any 20 bills you pick from 26. there has to be seven $1 bills. The worst case scenario is that you picked all the non-$1 bills with the first 19 and then you were left with no option but to pick a $1 bill because everything else was $1 bill. So 26-19 = 7 is the number of $1 bills.

    Similarly, you can calculate that there are eight $2 bills and eleven $5 bills.

    Making the total as $78. The logic is simple but it fooled me for some time as I was thinking of modeling it as a probability problem.

    Reply
  2. By rajibroy (Post author) on

    Now for the answer. My mistake initially was to think about it as a probability problem. As Pradeep pointed out – that is my favorite hammer.   Actually, if you think about it, it is a much simpler problem….

    If you are guaranteed that there will be one $1 in any 20 bills you pick from 26. there has to be seven $1 bills. The worst case scenario is that you picked all the non-$1 bills with the first 19 and then you were left with no option but to pick a $1 bill because everything else was $1 bill. So 26-19 = 7 is the number of $1 bills.

    Similarly, you can calculate that there are eight $2 bills and eleven $5 bills.

    Making the total as $78. The logic is simple but it fooled me for some time as I was thinking of modeling it as a probability problem.

    Reply
  3. By Manajit Sengupta on

    Or x +y = 19
    y+z = 18
    X+z = 15
    Where the number of bills are repesented by x for $2, y for $5 and z ifor $1

    X = 8
    Y = 11
    Z = 7

    Reply

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