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[tex] \large\frak{ \purple{ Question}}[/tex]
In a certain class, one-third of the students were absent. Half of the total strength attended the Maths test and one-fourth of the total strength attended the Physics test. If 6 students attended both the tests and every student who was present attended at least one of the two tests, then how many students were absent on that day?
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Sagot :

The number of students that were absent is; 24

How to find the Probability?

Let the total number of students be x .

The number of students which were absent = ¹/₃x.

Number of students which were present = ²/₃x

Thus;

Number of students who attended math's test = ¹/₂x

Number of students who attended physics test = ¹/₄x

No. of students who attended both tests = 6

A denotes number of students who attended math's test

B denotes number of students who attended physics test

Thus;

n(A) = ¹/₂x

n(B) = ¹/₄x

n(A ∩ B) = 6

n(A∪B) = number of students which were present

Thus;

n(A ∪ B) = n(A) + n(B) − n(A∩B)

²/₃x = ¹/₂x + ¹/₄x − 6

²/₃x = ³/₄x - 6

³/₄x - ²/₃x = 6

x = 72

Thus, number of students that were absent = ¹/₃ * 72

Number of students that were absent = 24

Read more about Probability at; https://brainly.com/question/251701

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