in examination 43% paseed in math 48% passed in physics and 52% passed in chemistry only 8% paseed a n all maths and physics 21% passed in maths and chemistry Number of students who took the exam is 200 then how many students passed in maths only
Answers
Given :
In an examination 43% pas sed in Math, 52% passed in Physics and 52% passed in Chemistry.
Only 8% students pas sed in all the three. 14% passed in Math and
Physics and 21% pas sed in Math and Chemistry and 20% pas sed in Physics and Chemistry. Number of students who took the exam is 200
To Find : how many students pas sed in maths only
Solution:
Pas sed in Maths M = 43 %
Pas sed in Math and Physics M ∩ P = 14 %
Pas sed in Math and Chemistry M ∩ C = 21 %
Pas sed in All M ∩ P ∩ C = 8 %
Pas sed in maths only = M - M ∩ P - M ∩ C + M ∩ P ∩ C
= 43 - 14 - 21 + 8
= 16 %
students pas sed in maths only = (16/100)200 = 32
32 students pas sed in maths only
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Answer:
Step-by-step explanation:
We can solve this problem using the Principle of Inclusion-Exclusion. According to this principle, for three sets A (Maths), B (Physics), and C (Chemistry), the percentage of people who passed at least one subject can be calculated as:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)
Given the information from the problem, we have:
P(A) = 43%
P(B) = 48%
P(C) = 52%
P(A ∩ B) = 14%
P(A ∩ C) = 21%
P(B ∩ C) = 20%
P(A ∩ B ∩ C) = 8%
Now, calculate P(A ∪ B ∪ C):
P(A ∪ B ∪ C) = 43% + 48% + 52% - 14% - 21% - 20% + 8%
P(A ∪ B ∪ C) = 96%
Since there are 200 students who took the exam, the number of students who passed at least one subject is:
Number of students = 200 * 96% = 192
Next, we want to find the number of students who passed in Maths only. We can use the formula:
P(A only) = P(A) - P(A ∩ B) - P(A ∩ C) + P(A ∩ B ∩ C)
P(A only) = 43% - 14% - 21% + 8%
P(A only) = 16%
Now, we can find the number of students who passed in Maths only:
Number of students = 200 * 16% = 32
So, the answer is: 32