1
$\begingroup$

Here is my question.

A math examination has three questions. Twenty-six students took the examination, and every student answered at least one question. Six students did not answer the first question; twelve did not answer the second question; and five did not answer the third question. If eight students answered all three questions, how many students answered exactly one question?

The answer and the Venn diagram for the exercise is:

enter image description here

From the regions in the Venn diagram, we have

$a + b + c + d + e + f + g = 26$ : the total number of students

$g = 8$ : all three questions answered

$b + f + c = 6$ : Question 1 is not answered

$a + e + c = 12$ : Question 2 is not answered

$a + d + b = 5$ : Question 3 is not answered

My confusion starts here: Why do I want to find $a + b + c$, as stated below?

We want to find $a + b + c$, and we know that $a + b + c > 3$. Adding the last three equations, we have

Why do I do this next step? $2(a + b + c) + e + d + f = 23$

and

Why do I do this next step? $a + b + c + 23 - 2(a + b + c) + 8 = 26$:

How do I get this next step? Why is a + b + c = 5? So we conclude that $a + b + c = 5$.

Thanks for any help you can provide. I have been at this section on Venn diagrams all week and I just can't seem to get it.

$\endgroup$
1
$\begingroup$

We also know $g=8$, or equivalently $$a+b+c+d+e+f=26-8=18.$$

By summing $$\color{blue}{b}+f+\color{red}{c}=6,$$ $$\color{green}{a}+e+\color{red}{c}=12,$$ $$\color{green}{a}+d+\color{blue}{b}=5$$ we're essentially double-counting $a$, $b$ and $c$ (those who answered one question) whereas we're single-counting $d$, $e$ and $f$ (those who answered two questions). This imbalance between $a+b+c$ and $d+e+f$ can be exploited: it enables us to separate $a+b+c$ out from the first equation.

The rest is just arithmetic. Let $X=a+b+c$, then the final equation implies $X+23-2X+8=26$, and we solve for $X$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Hi Rebecca, thanks for you explanation. This was very useful. $\endgroup$ – user92986 Sep 29 '13 at 19:56
  • 1
    $\begingroup$ Now that I have gone through this equation my study guide directed me to read the section on the Inclusion/Exclusion Rule and now I understand these types of equations. The study guide had me attempt these questions without telling me to read this section. Should have been the other way around. $\endgroup$ – user92986 Sep 29 '13 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy