A certain league has five divisions. The respective divisions had 6, 7, 8, 9 and 11 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?
Options:
(A)  81
(B)  83
(C)  85
(D)  86
(E)  89

Overall, there can be 81 minuses. Tus, it is So the answer is: (A).

If P(A) = 2/15, P(B) = 4/15, and P(A âˆª B) = 12/15 Find P(B|A)
Options:
(A) 1/2
(B) 2/3
(C) 3/2
(D) 4/5
(E) 6

P(B|A) = P(A âˆª B)/P(A)
P(B|A) = (12/15)/(2/15) = 6. So the answer is: (E).

There are 3 teams each with 3 basket players. How many combinations are there if we were to pick 2 players from the 3 teams such that no team was picked twice?
Options:
(A) 50
(B) 23
(C) 27
(D) 28
(E) 45

There are altogether three arrangements, which are we can select two members each from team A and B, each from team A and C, and each from team B and C. For each arrangement, there are altogether 9 possibilities (3*3) since each team has 3 players. So there are a total of 27 possibilities (9*3). I hope this method is easier for you to understand. So the answer is: (C).

In what time a sum of money double itself at 3% per annum simple interest?
Options:
(A) 33 1/9%
(B) 33 1/3%
(C) 33 7/3%
(D) 33 2/3%
(E) 33 8/3%


P = (P*3*R)/100
R = 33 1/3% So the answer is: (B).
A certain league has five divisions. The respective divisions had 6, 7, 8, 9 and 11 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?
Options:
(A)  81
(B)  83
(C)  85
(D)  86
(E)  89

Overall, there can be 81 minuses. Tus, it is So the answer is: (A).

If P(A) = 2/15, P(B) = 4/15, and P(A âˆª B) = 12/15 Find P(B|A)
Options:
(A) 1/2
(B) 2/3
(C) 3/2
(D) 4/5
(E) 6

P(B|A) = P(A âˆª B)/P(A)
P(B|A) = (12/15)/(2/15) = 6. So the answer is: (E).

There are 3 teams each with 3 basket players. How many combinations are there if we were to pick 2 players from the 3 teams such that no team was picked twice?
Options:
(A) 50
(B) 23
(C) 27
(D) 28
(E) 45

There are altogether three arrangements, which are we can select two members each from team A and B, each from team A and C, and each from team B and C. For each arrangement, there are altogether 9 possibilities (3*3) since each team has 3 players. So there are a total of 27 possibilities (9*3). I hope this method is easier for you to understand. So the answer is: (C).

In what time a sum of money double itself at 3% per annum simple interest?
Options:
(A) 33 1/9%
(B) 33 1/3%
(C) 33 7/3%
(D) 33 2/3%
(E) 33 8/3%


user
P = (P*3*R)/100
R = 33 1/3% So the answer is: (B).
assistant
A certain league has five divisions. The respective divisions had 6, 7, 8, 9 and 11 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?
Options:
(A)  81
(B)  83
(C)  85
(D)  86
(E)  89

Overall, there can be 81 minuses. Tus, it is So the answer is: (A).

If P(A) = 2/15, P(B) = 4/15, and P(A âˆª B) = 12/15 Find P(B|A)
Options:
(A) 1/2
(B) 2/3
(C) 3/2
(D) 4/5
(E) 6

P(B|A) = P(A âˆª B)/P(A)
P(B|A) = (12/15)/(2/15) = 6. So the answer is: (E).

There are 3 teams each with 3 basket players. How many combinations are there if we were to pick 2 players from the 3 teams such that no team was picked twice?
Options:
(A) 50
(B) 23
(C) 27
(D) 28
(E) 45

There are altogether three arrangements, which are we can select two members each from team A and B, each from team A and C, and each from team B and C. For each arrangement, there are altogether 9 possibilities (3*3) since each team has 3 players. So there are a total of 27 possibilities (9*3). I hope this method is easier for you to understand. So the answer is: (C).

In what time a sum of money double itself at 3% per annum simple interest?
Options:
(A) 33 1/9%
(B) 33 1/3%
(C) 33 7/3%
(D) 33 2/3%
(E) 33 8/3%


P = (P*3*R)/100
R = 33 1/3% So the answer is: (B).