If n(A Δ B) = 50, n(A) = 40, find n(B) if A and B are disjoint.
Answer: A
The symmetric difference A Δ B = (A ∪ B) - (A ∩ B).
If A and B are disjoint, their intersection is empty, so n(A ∩ B) = 0.
In this case, n(A Δ B) = n(A ∪ B) - 0 = n(A ∪ B).
Also for disjoint sets, n(A ∪ B) = n(A) + n(B).
Therefore, n(A Δ B) = n(A) + n(B).
50 = 40 + n(B) \(\Rightarrow\) n(B) = 10.
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In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each student likes to play at least one of the two games. How many students like to play both cricket and football?
Answer: A
n(C) = 24, n(F) = 16, n(C ∪ F) = 35.
We need to find n(C ∩ F).
n(C ∪ F) = n(C) + n(F) - n(C ∩ F)
35 = 24 + 16 - n(C ∩ F)
35 = 40 - n(C ∩ F)
n(C ∩ F) = 40 - 35 = 5.
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If A, B, C are three sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C, then:
Answer: B
This is a standard property of sets. Consider an element x ∈ B. Then x ∈ A ∪ B. Since A ∪ B = A ∪ C, we have x ∈ A ∪ C. So, x ∈ A or x ∈ C.
Case 1: x ∈ A. Then x ∈ A and x ∈ B, so x ∈ A ∩ B. Since A ∩ B = A ∩ C, we have x ∈ A ∩ C, which implies x ∈ C.
Case 2: x ∉ A. Since x ∈ A ∪ C, we must have x ∈ C.
In both cases, if x ∈ B, then x ∈ C. So, B ⊂ C. Similarly, we can prove that C ⊂ B. Therefore, B = C.
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The set {x | x is a real number and x² = -1} is a:
Answer: D
No real number squares to a negative value, since \( x^{2} \geq 0 \) for every real \( x \).
The set therefore has no members and is the empty set. A singleton would need exactly one element - which it would have over the complex numbers, but the condition here restricts to reals.
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In a school, there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teach both physics and mathematics. How many teach physics?
Answer: B
Let M be the set for Math teachers and P for Physics teachers.
n(M ∪ P) = 20, n(M) = 12, n(M ∩ P) = 4.
We use the formula n(M ∪ P) = n(M) + n(P) - n(M ∩ P).
20 = 12 + n(P) - 4
20 = 8 + n(P)
n(P) = 12.
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Let A = {1, 2, 3}. The number of proper subsets of A is:
Answer: B
A set with 'n' elements has \(2^n\) subsets in total.
Here, n=3, so the total number of subsets is \(2^3 = 8\).
A proper subset is any subset except the set itself. So, we subtract the set A from the total number of subsets.
Number of proper subsets = \(2^3 - 1 = 8 - 1 = 7\).
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In a town, 40% of the people have brown hair, 25% have brown eyes, and 10% have both brown hair and brown eyes. What percentage of people in the town have neither brown hair nor brown eyes?
Answer: B
Let H be the set for brown hair and E for brown eyes.
n(H) = 40%, n(E) = 25%, n(H ∩ E) = 10%.
Percentage of people with at least one trait = n(H ∪ E) = n(H) + n(E) - n(H ∩ E) = 40% + 25% - 10% = 55%.
Percentage of people with neither trait = 100% - n(H ∪ E) = 100% - 55% = 45%.
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In a class of 100 students, 55 students passed in Mathematics and 67 students passed in Physics. If the number of students who passed in Mathematics only is 22, how many students passed in Physics only?
Answer: D
Let M be the set of students who passed in Mathematics and P be the set of students who passed in Physics.
Given: n(M) = 55, n(P) = 67.
Number of students who passed in Mathematics only is n(M) - n(M ∩ P) = 22.
From this, we can find the number of students who passed in both subjects:
55 - n(M ∩ P) = 22 \(\Rightarrow\) n(M ∩ P) = 55 - 22 = 33.
Now, we can find the number of students who passed in Physics only:
Physics only = n(P) - n(M ∩ P) = 67 - 33 = 34.
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If A is the set of all even numbers and B is the set of all odd numbers, what is A ∩ B?
Answer: C
An even number is an integer that is exactly divisible by 2. An odd number is an integer that is not.
By definition, a number cannot be both even and odd at the same time.
Therefore, the two sets have no elements in common, and their intersection is the empty set (∅).
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In a group of 40 players, 25 play Cricket, 20 play Tennis, and 8 play both. How many players play only Tennis?
Answer: A
Let T be the set of players who play Tennis.
Let C be the set of players who play Cricket.
Given n(T) = 20 and n(T ∩ C) = 8.
The number of players who play only Tennis is given by n(T) - n(T ∩ C).
Number of players who play only Tennis = 20 - 8 = 12.
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