There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics?
I ended up getting 110. Let me explain my thought process. 15 people aren't taking classes, so we're really only considering 360 – 15 = 345 people here. How do we get the 15 that take all 3? Using inclusion-exclusion, it's 345 – 2P – P – 180 + 75 + 75 + 30 = 15. Solving for P gets us 110. Is this correct?
Update: According to a commenter, there is an answer given here:
https://web2.0calc.com/questions/please-help-me-i-have-one-trie-lefft
There are 360 people and 345 people taking courses. 15 take all 3 and 60 only take Calculus because 75+30+15+60=180.
Next, to find the total number of people who are only taking physics or only taking chemistry, I subtracted 240 from 345 because 240 (69+75+75+30) is everyone taking courses except for the people only taking physics of chemistry. I found that 105 are either only taking phy. or only taking chem.
lets say z=only physics
y=only chemistry
105=y+z or y=105-z
Also, x=the total number of people taking physics (so 2x is the total number of people taking chemistry)
Now:
2x=75+75+15+y
x=45+y-z
75+45+z=45+y-z
75+2z=y and from before: y=105-z
75+2z=105-z
z=10 so 10 people only take physics
Finally, the number of people taking physics is 10 (only) + 75 (phy. and chem) +30 (and calc) + 15 (all three) = 130. 130 students are taking physics.
But they get a different answer – 130 compared to my 110. Who's correct here?
Best Answer
Your answer of $(110)$ is right. My alternative approach uses a truth table.
\begin{array}{| r | r | r | r| r |} \hline \text{Calculus} & \text{Physics} & \text{Chemistry} & \text{Variable} & \text{Value} \\ T & T & T & x_1 & 15 \\ \hline T & T & F & x_2 & 15 \\ \hline T & F & T & x_3 & 60 \\ \hline T & F & F & x_4 & 90 \\ \hline F & T & T & x_5 & 60 \\ \hline F & T & F & x_6 & \\ \hline F & F & T & x_7 & \\ \hline F & F & F & x_8 & 15 \\ \hline \end{array}
For the moment, ignore the posted values. Constraints:
$$x_1 + x_2 + x_3 + \cdots + x_8 = 360.\tag1 $$ $$x_1 = 15.\tag2 $$ $$x_8 = 15.\tag3 $$ $$x_1 + x_2 + x_3 + x_4 = 180.\tag4$$ $$x_1 + x_3 = 75.\tag5 $$ $$x_1 + x_2 = 30.\tag6 $$ $$x_1 + x_5 = 75.\tag7 $$ $$x_1 + x_3 + x_5 + x_7 = 2(x_1 + x_2 + x_5 + x_6).\tag8$$
$$x_1 + x_2 + x_5 + x_6 = ?.\tag9$$
Equations (2), (3), (5), (6), and (7) above lead immediately to the posted values for $x_1, x_8, x_2, x_3,$ and $x_5$.
With $x_1, x_2, x_3$ now determined, equation (4) solves for $x_4$.
It remains to solve for $x_6$.
With all variables but $x_6, x_7$ determined, equations (1) and (8) yield:
$$x_6 + x_7 = 105.$$
$$135 + x_7 = 2(90 + x_6) \implies $$
$$x_7 = 45 + 2x_6 = 105 - x_6 \implies $$ $$3x_6 = 60 \implies x_6 = 20.$$
Now, equation (9) can be solved:
$$x_1 + x_2 + x_5 + x_6 = 15 + 15 + 60 + 20 = 110.$$