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.
Matches played: 60.
Matches won: 30% of 60 => (60*(30/100)) = 18 matches.
Iterative approach:
On adding 1 to matches played and matches won, on every iteration until the win percentage gets to 50. So
19 / 61 = 0.3114754098360656
20 / 62 = 0.3225806451612903
21 / 63 = 0.3333333333333333
22 / 64 = 0.34375
…
…
…
…
Similarly,
41 / 83 = 0.4939759036144578
42 / 84 = 0.5
So, after 60th match 24 more matches has to be played and won to get 50% average winning rate.
20000
1, 3, 10, 21, 64, 129, 356, 777
a
previous ans wrong again..!!!!
first find x its 2 hours
total journey 6 hrs 3km up and also down
jill takes one hr to come down and jack takes 3 hrs so x
is 2 hours….
And while going up jack goes at 1.5km/hr and jill goes at
0.75 km hr so to travel 3 km they take 2 hrs and 4 hrs
respectively…!!! x again is 2 hrs…
so jack totally takes 5 hrs to travel 6 km
therfore speed is 6/5 km per hour
thats is 1.2 km per hour is the correct answer
habeas corpus
11 14
12 15
13 16
=======
14 17
the answers is 14
Length of train = 125Mtrs
Speed of man = 5Km/Hr
=(5*1000)/(1*3600)
=1.4M/s
Time taken for train to pass = 10s
Spd = D/T
Total Distance (Man distance+ Train Length)
#Distance covered by Man in 10s
D=Spd*T
= (1.4M/s*10s)
= 14Mtrs
#Train length = 125Mtrs
Total D = 14Mtrs + 125Mtrs
= 139Mtrs
Time taken = 10s
Sp = D/t
=139M/10s
=13.9M/s
data inadequate
A. Rs. 410
Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125