Out of 136 students in a school, 60 take french, 100 take chemistry and 48 take physics. if 28 take french and chemistry, 44 take chemistry and physics and 20 take french and physics, how many take all three subjects?

20

Step-by-step explanation:

Let us see the formula to find intersection of 3 sets.

n(A∪B∪C)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(C∩A)+n(A∩B∩C)

Here we are given three sets as the number of students in class of French , physics and Chemistry . Let these sets be denoted by F,C, P. Hence the formula for them will be

n(F∪C∪P)=n(F)+n(C)+n(P)-n(F∩C)-n(C∩P)-n(P∩F)+n(F∩C∩P)

Also we are given that

n(F∪C∪P)=136

n(F)=60

n(C)=100

n(P)=48

n(F∩C)=28

n(C∩P)=44

n(P∩F)=20

n(F∩C∩P)=x

Hence put these values in the above equation and solve for x

136=60+100+48-28-44-20+x

136=208-92+x

136=116+x

20=x

Hence , there are 20 students who took all the three subjects.

x + 4/7

step-by-step explanation:

use cymath

step-by-step explanation:

the answer is 128