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

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. ðŸ™‚ 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.

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. ã€€ 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.

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

This site uses Akismet to reduce spam. Learn how your comment data is processed.