Answer to Question #99980 in Discrete Mathematics for Gabriel Kabore

Question

In a class, 20 students study agric, 21 study biology and those who study chemistry are 5 more than those who study agric. it is also known that 3 students study agric only and 4 students study only chemistry. Half as many students who study chemistry only study biology only and twice as many students who study exactly one of the subjects study 2 of the 3 subjects. If one third of the students who study biology study all there subjects and the number of students who study agric and chemistry only is equal to the number of students who do not study any of the 3 subjects. a) Illustrate the above information on a vane diagram b) How many students i) are in the class ii) study at least 2 subjects iii) study at most one subject iv) Do not study any of the subjects

Answer 1

Illustration of all available information

b) Now we will record all the information we have:

A=20 - the number of students who study agric;

B=21 - the number of students who study biology;

C=25 - the number of students who study chemistry;

a=3 - the number of students who study agric only;

c=4 - the number of students who study chemistry only;

b=4/2=2 - the number of students who study biology only;

Let,

(ab) - the number of students who study agric+biology;

(bc) - the number of students who study biology+chemistry;

(ac) - the number of students who study agric+chemistry;

(abc) - the number of students who study agric+biology+chemistry;

We know that

$\left\{\begin{array}{c} (ab)+(bc)+(ac)=2\cdot(a+b+c)\\ (abc)=B\div3\\ a+(ab)+(ac)+(abc)=20\\ b+(ab)+(bc)+(abc)=21 \end{array}\right.\longrightarrow\\[0.5cm] \left\{\begin{array}{c} (ab)+(bc)+(ac)=2\cdot(2+3+4)\\ (abc)=21\div3=7\\ 3+(ab)+(ac)+7=20\\ 2+(ab)+(bc)+7=21 \end{array}\right.\longrightarrow\\[0.5cm] \left\{\begin{array}{c} (abc)=7\\ (ab)+(bc)+(ac)=18\\ (ab)+(ac)=10\\ (ab)+(bc)=12 \end{array}\right.\longrightarrow \left\{\begin{array}{c} (abc)=7\\ (bc)=8\\ (ac)=6\\ (ab)=4 \end{array}\right.$

Then,

$n=3+4=7-do\,not\,study\,any\,of\,3\,subject.$

Conclusion,

$N=a+b+c+(ab)+(bc)+(ac)+(abc)+n=\\ =3+2+4+4+6+8+7+7=41\\ \boxed{N=41-total\,number\,of\,students\,in\,the\,class}$

$N(study\, at\,least\,2)=(ab)+(bc)+(ac)+(abc)=4+6+8+7\\ \boxed{N(study\, at\,least\,2)=25}$

$N(most\,one\,subject)=a+b+c+n=3+2+4+7=16\\ \boxed{N(most\,one\,subject)=16}$

ANSWER

$N=41-total\,number\,of\,students\,in\,the\,class\\[0.5cm] N(study\, at\,least\,2)=25\\[0.5cm] N(most\,one\,subject)=16\\[0.5cm] n=7-do\,not\,study\,any\,of\,3\,subject.$