Page 21 - Pharmacy College Admission Test Review Book

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1.(b) These types of problems can be solved by Let A be the set of students drinking milk, then
using the SET THEORY. There are few terminologies n(A) = 400
and equations you should keep in mind to solve
these problems. Let B be the set of students drinking coffee, then
n(B) = 300
A. (U) = It is known as universal set. The universal
set depends on the context. The set of students drinking both milk and coffee is
A  B, therefore n(A  B) = 150
B. A  B = When every element of set A also
belongs to set B, then set A is said to be the subset Finally, let the set of students of the school be U,
of B. For example, therefore n(U) = 800

A = {1, 2, 3} and B = {0, 1, 2, 3, 4} The set of students drinking neither coffee nor milk
should be A’  B’ = (A  B)’
C. When A  B and B  A, then A = B or they are
called equal sets. n (A  B)’ = n(U) - n (A  B)
= n(U) - [n(A) + n(B) - n(A  B)
D. A  B = The set consisting of all elements which = 800 - 400 - 300+ 150
are in A or in B is called the union of A and B, and is = 250
denoted by A  B. For example,
2. (a) 8. This type of problem can also be solved by
A = {1, 2, 3} and B = {0, 1, 2, 3, 4} using a Venn diagram.
U
A  B = {0, 1, 2, 3, 4}
M B
E. A  B = The set consisting of all elements which y x z
are common in both A and B is called the
intersection of A and B, and is denoted by A  B.
For example,
Let x = Number of students taking mathematics as
A = {1, 2, 3} and B = {0, 1, 2, 3, 4} well as biology.
Let y = Number of students taking mathematics but
A  B = {1, 2, 3} not biology.

F. (A’) = The set consisting of all those elements of U From Venn diagram, we can say:
which are not in A is called the component of A and
is denoted by A’. For example, x + y = 24
x + 16= 24
U = {1, 2, 3, 4, 5} and A = {1, 2} then, x = 8
3. (b) This type of problem can be solved by using
A’ = {3, 4, 5} the Binomial Theorem. Let say a = 2x, b = y and n =
5
G. De Morgan’s Laws:
n
n
n
n
n
(a + b) = ( ) a + ( ) a n−1 + ⋯ ( ) a n−r
1. (A  B)’ = A’  B’ 0 1 r
2. (A  B)’ = A’  B’ (2x + y) =
5

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