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FOR 15 = RS.20
FOR 20 = RS.18
FOR 25 = RS.16
FOR 30 = RS.14
FOR 35 = RS.12
FOR 40 = RS.10
The probability that no one opens the door would be
0.5×0.5×0.5×0.5= 0.0625
So, the probability that atleast one opens the door is 1-0.0625= 0.9375
answer can be 1,2,3,4 can’t be determined exact number
just explaining a case:
there are 10 people in the party.
name of people no. of people with they made handshake list of those people(this can vary but showing the possibility)
1 1 9
2 2 8, 9
3 3 7, 8, 9
4 4 6, 7, 8, 9
5 5 6, 7, 8, 9. 10
6 6 4, 5, 7, 8, 9, 10
7 7 3, 4, 5, 6, 8, 9, 10
8 8 2, 3, 4, 5, 6, 7, 9, 10
9 9 1, 2, 3, 4, 5, 6, 7, 8, 10
10 4 5, 6, 7, 8, 9
jack got 9 different answer so jack can be either 4th number or 10th number and jack’s wife know jack very well so she can’t have handshake with jack so if 4th is jack then she can’t be handshake with 6,7,8,9, in this case she can be 1,2,3, 5, 10 and now depending upon which no is jack’s wife she can have hand shake with- 1- 4 people, and if jack is number 10 then she can’t be 5,6,7,8,9 so again depending upon her number she can handshake with people in range of 1-4
8*1024+3*64+3
8192+192+3
8387
1000/0011/1000/0111
So ie 7. 1s here
Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:
1. Weigh ABC versus DEF.
Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.
2. Weigh G versus H.
Scenario a.i: If these (2) balance, the oddball is I.
Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.
Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.
2. Weigh A versus B.
Scenario b.i: If these (2) balance, the oddball is C.
Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.
answer is maximum of 2.
Answers as expected
Friends,
The Answer given by ‘Gaurav Sharma’ is correct and
the approach (bottom to top) suggested by ‘Shailesh’ is
good. But with minor correction we can arrive the solution
using this approach:
After 5th loot, No. of breads left = 3
after 4th loot, no. of breads left = (3+0.5)x2 = 7
after 3rd loot, no. of breads left = (7+0.5)x2 = 15
after 2nd loot, no. of breads left = (15+0.5)x2 = 31
after 1st loot, no. of breads left = (31+0.5)x2 = 63
So, before 1st loot, no. of breads left = (63+0.5)x2 = 127
Summation of 301 – summation of 99
=41550
Summation of n=((n*n+1)/2)