Question

Among 50 students in a class 26 got an A in the first exam and 21got A in second exam. If 17 students did not get an A in either exam,
how many students got an A in both exams?
ii) If number of students getting A in first exam is equal to number of
students getting A in second exam, if total number of students who got
A in exactly one exam is 40 and 4 students did not get an A in either
exam, then find number of students:
1. getting A in the first exam only
2. who got an A in second exam only
3. who got an A in both exams​

Answer 1

Step-by-step explanation:

i)

let the number of students who got A in first exam be denoted as A and those who got be in second exam as B

then

A=26

B=21

If 17 students did not get an A in either exam, then the number that got A in either exams (ie A union B) is

50-17=33

From laws of addition;

(A union B)=A+B - (A intersect B)

33=26+21 - (A intersect B)

(A intersect B)=26+21-33

(A intersect B)= 14

Therefore, the number of students who got an A in both exams is 14.

ii)

If number of students getting A in first exam is equal to number of

students getting A in second exam,

then A=B

if total number of students who got A in exactly one exam is 40,

then

(A intersect B') union (B intersect A')=40

If 4 students did not get an A in either exam,

then the number of those who got A in either exam is 50-4= 46

(A union B)=46

(A union B)= (A intersect B') union (B intersect A') + (A intersect B)

46=40 + (A intersect B)

(A intersect B) =46-40=6

Therefore the number of students who got A in both exams is 6.

Since A=B, then the number of students who got A in the first exam only is 20 and in the second exam only is 20

Answer 2

I don't know if it was going for anything but