In a competition, 36 medals were awarded in dance, 12 in dramatics and 18 in music. If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, how many received medals in exactly two of these categories?
Answer: C
Let D, R, and M be the sets of people who got medals in Dance, Dramatics, and Music respectively.
Given: n(D)=36, n(R)=12, n(M)=18, n(D ∪ R ∪ M)=45, n(D ∩ R ∩ M)=4.
We use the formula:
n(D∪R∪M) = n(D)+n(R)+n(M) - [n(D∩R) + n(R∩M) + n(M∩D)] + n(D∩R∩M)
45 = (36 + 12 + 18) - [Sum of intersections of pairs] + 4
45 = 66 - [Sum of pairs] + 4
45 = 70 - [Sum of pairs]
Sum of pairs = n(D∩R) + n(R∩M) + n(M∩D) = 25.
The number of people who received medals in exactly two categories is given by:
[n(D∩R) + n(R∩M) + n(M∩D)] - 3 × n(D∩R∩M)
= 25 - 3 × 4 = 25 - 12 = 13.
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The Idempotent Law for set union states that:
Answer: A
The Idempotent Law states that when an operation is applied to an element with itself, the result is the same element.
For set theory, this means:
1. A ∪ A = A (Idempotent Law of Union)
2. A ∩ A = A (Idempotent Law of Intersection)
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If n(A) = 115, n(B) = 326, n(A-B) = 47, then what is n(A ∪ B) equal to?
Answer: A
We know that n(A) = n(A-B) + n(A ∩ B).
115 = 47 + n(A ∩ B)
n(A ∩ B) = 115 - 47 = 68.
Now we find the union:
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
n(A ∪ B) = 115 + 326 - 68 = 441 - 68 = 373.
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Which of the following is a finite set?
Answer: C
A finite set is a set that has a countable number of elements.
The set of all days in a week is {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}, which has 7 elements. This is a finite set.
The other options represent infinite sets.
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In a survey of 400 students, 100 were listed as drinking apple juice, 150 as drinking orange juice and 75 were listed as drinking both. How many students were drinking neither apple juice nor orange juice?
Answer: C
Let A be the set for apple juice and O for orange juice.
n(A) = 100, n(O) = 150, n(A ∩ O) = 75.
Number of students drinking at least one juice = n(A ∪ O) = n(A) + n(O) - n(A ∩ O) = 100 + 150 - 75 = 175.
Total students = 400.
Number of students drinking neither = Total - n(A ∪ O) = 400 - 175 = 225.
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If A = {1, 2, {3, 4}}, which of the following is a true statement?
Answer: B
The elements of set A are 1, 2, and the set {3, 4}.
Option A is false because 3 is not an element of A; it's an element of the element {3, 4}.
Option B is true because 1 and 2 are elements of A, so the set containing them, {1, 2}, is a subset of A.
Option C is false because {3, 4} is an element of A, not a subset. For it to be a subset, 3 and 4 would have to be elements of A themselves.
Option D is false for the same reason as A.
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In a class, 72% students like Physics and 44% students like Chemistry. If each student likes at least one subject and there are 40 students who like both, what is the total number of students in the class?
Answer: C
Let the total number of students be T.
n(P) = 0.72T, n(C) = 0.44T.
Since each student likes at least one subject, n(P ∪ C) = T.
n(P ∩ C) = 40.
Using the formula n(P ∪ C) = n(P) + n(C) - n(P ∩ C):
T = 0.72T + 0.44T - 40
T = 1.16T - 40
40 = 1.16T - T
40 = 0.16T
T = 40 / 0.16 = 4000 / 16 = 250.
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Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey; 24 played all the three games. The number of boys who did not play any game is:
Answer: D
We use the formula for the union of three sets:
n(C ∪ H ∪ B) = n(C) + n(H) + n(B) - n(C ∩ H) - n(H ∩ B) - n(C ∩ B) + n(C ∩ H ∩ B)
n(C ∪ H ∪ B) = 224 + 240 + 336 - 40 - 64 - 80 + 24
n(C ∪ H ∪ B) = 800 - 184 + 24 = 640.
This is the number of boys who played at least one game.
Number of boys who did not play any game = Total boys - n(C ∪ H ∪ B) = 800 - 640 = 160.
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In a competition, a school awarded medals in different categories. 36 medals in dance, 12 in dramatics and 18 in music. If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, how many received medals in exactly two of these categories?
Answer: C
Let D, R, M be the sets for Dance, Dramatics, Music. n(D)=36, n(R)=12, n(M)=18. n(D∪R∪M)=45. n(D∩R∩M)=4. Let x be the number of people who received medals in exactly two categories.
We know n(D∪R∪M) = (Sum of exactly one) + (Sum of exactly two) + (Sum of exactly three). Also, n(D∪R∪M) = n(D)+n(R)+n(M) - (Sum of pairs) + n(D∩R∩M). 45 = 36+12+18 - (Sum of pairs) + 4 => 45 = 70 - (Sum of pairs). So Sum of pairs = 25.
We know (Sum of pairs) = (Exactly two) + 3 * (Exactly three). 25 = x + 3 * 4 => 25 = x + 12 => x = 13. Wait, let me re-check. Yes, Sum of n(A∩B) = Sum of exactly two + 3 * n(A∩B∩C). That gives x=13. Why is the answer 11? Let me try another formula. (Sum of exactly one) = (n(A)+..) - 2(Sum of pairs) + 3(All three). Let's use the basic one. 45 = (n(D)+n(R)+n(M)) - (n(D∩R)+..) + n(D∩R∩M) => 45 = 66 - (n(D∩R)+..) + 4 => Sum of intersections = 25. This is correct. Let's use the formula for total people = (only A + only B + only C) + (exactly two) + (exactly three). Let 'Exactly two' be x. Then 45 = (Only D+..+Only M) + x + 4. Only D = 36 - n(D∩R) - n(D∩M) + n(D∩R∩M). This becomes circular. Let's re-verify: sum of pairs = 25. Exactly two = 25 - 3*4 = 13. I am consistently getting 13. I will modify the answer to be 13.
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In a group of 52 people, 16 drink tea but not coffee and 33 drink tea. How many drink coffee but not tea?
Answer: B
Let T be for tea and C for coffee. n(T only) = 16, n(T) = 33. Total = 52.
We can find the number who drink both: n(T ∩ C) = n(T) - n(T only) = 33 - 16 = 17.
The union of the sets (people who drink tea or coffee or both) is n(T ∪ C) = n(T only) + n(C only) + n(both) = 16 + n(C only) + 17.
Since everyone is assumed to drink something, total = n(T ∪ C). So, 52 = 16 + n(C only) + 17 = 33 + n(C only).
n(C only) = 52 - 33 = 19.
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