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?
Sushil solved it already.
As did Steve
Which suggests that this one was way to easy.
I have sent it to my boyfriend’s son to solve ! ๐
Dhananjay, Pradeep and Manivannan all got the answers too.
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. ๐ 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.
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. ใ 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.
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