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Given:
In a group of 15 students,
7 have studied Latin,
8 have studied Greek,
3 have not studied either.
To find:
The number of students who studied both Latin and Greek.
Solution:
In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.
Therefore,
n(A∪B) = 15 – 3
n(A∪B) = 12
7 have studied Latin,
n(A) = 7
8 have studied Greek,
n(B) = 8
n(A∩B) is the number of students who studied both Latin and Greek.
n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = 7 + 8 – 12
n(A∩B) = 15 – 12
n(A∩B) = 3
The number of students who studied both Latin and Greek is 3
Final answer:
3 of them studied both Latin and Greek.
Thus, the correct answer .3
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.
LNTKCHMF
south east
Second
X filling rate is 1/18 or 4/72 per hour
Y filling rate is 1/24 or 3/72 per hour
Combined as a pair it’s 7/72 in a complete 2 hour alternating sequence
72+ 7/72 =10 sequences = 70/72 on 20 hours, The outstanding balance is just 2/72
So at 20.5 hours with X turn filling is 100% complete with 72/72
Answer: It will take 20.5 hours to completely fill this tank.
Analysis of measured filling process, from X = 42/72 share & from Y = 30/72 share
D = 180m
S = 42 – 6 = 36 * 5/18 = 10
T = D/S = 180/10
T = 18s
p=5/6+5/16-5/6
both 5/6’s get cancelled.
so, p=5/16
p=0.3125, therefore (p-1)3 value is (0.3125-1)3= -0.6875*3
=-2.0625.
C
20&130