Question

Last year at Townsburg High School, 31% of the graduating seniors took the ACT exam, 41% of the graduating seniors took the SAT exam, and 19% of graduating seniors took both exams. Find the proportion of students that took neither of the two exams.

Answer 1

The proportion of students that took neither of the two exams = 47%

Step-by-step explanation

Let the proportion of students that took neither exams be n(A' u S') = x%

The total proportion of students = n(U) = 100%

The proportion of those that wrote the ACT exam consists of those that wote only the ACT exam and thse that wrote both the ACT and SAT exams

Proportion of those that wrote the ACT exam = n(A) = 31%

n(A) = n(A n S') + n(A n S)

31% = n(A n S') + 19%

n(A n S') = 31% - 19% = 12%

The proportion of those that wrote the SAT exam consists of those that wote only the SAT exam and thse that wrote both the ACT and SAT exams

Proportion of those that wrote the SAT exam = n(S) = 41%

n(S) = n(A' n S) + n(A n S)

41% = n(A' n S) + 19%

n(A' n S) = 41% - 19% = 22%

The venn diagram for this is presented below

So, we can sum all of these up to obtain the proportion who did not write either of the exams.

n(U) = n(A n S') + n(A' n S) + n(A n S) + n(A' u S')

100% = 12% + 22% + 19% + x

x = 100% - (12% + 22% + 19%)

x = 100% - 53%

x = 47%

Hope this Helps!!!