In a school, 400 students opt for Science, 250 students opt for Economics and 160 opt
for both Science and Economics (a) How many students have enrolled for Science only?
(b) How many have enrolled for Economics only?
Answers
Answer:
Science students = 560
Commerce student = 410
Step-by-step explanation:
Total science students = 400 + 160
= 560
Total economics students = 250 + 160
= 410
Concept:The intersection of the two sets is the set that includes every item shared by sets A and B.
The intersection of these two sets may be obtained as follows if B is the set of the first five multiples of 4 and A is the set of even numbers less than 10:
A = {2, 4, 6, 8}
B = {4, 8, 12, 16, 20}
The elements that both A and B share are 4 and 8.
Therefore, the collection of things at the point where A and B intersect is equal to 4, 8.
The intersection of the two sets A and B is the set of all items, whereas the union of the two sets A and B is the set of all elements that are either in A or in B. We may define the formulas for both union and intersection of the provided sets based on the cardinality of the sets, as seen below.
If A and B are finite sets such that A ∩ B = φ, then
n (A ∪ B) = n (A) + n (B).
If A ∩ B ≠ φ, then
n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
NOTE:
- ∪ represents Union
- ∩ represents Intersection
Given:
400 students opt for Science,
250 students opt for Economics, and
160 opt for both Science and Economics.
Find:
(a) How many students have enrolled for Science only?
(b) How many have enrolled for Economics only?
Solution:
Suppose, S = Students opting for Science
E = Students opting for Economics
Then, we have
n (S ∪ E) = n (S) + n (E) – n (S ∩ E)
and n (S ∩ E) = 160,
n (S) = 400, and
n (E) = 250.
(a) Number of students enrolling for Science only are n(S) – n(S ∩ E)
= 400 – 160
= 240
(b) Number of students enrolling for Economics only are n(E) – n(S ∩ E)
= 250 – 160
= 90
Hence,
(a) Number of students enrolling for Science only is 240
(b) Number of students enrolling for Economics only is 90
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