# Please refer to the image here. ?

Mar 10, 2017

$n \left(P ' \cap Q\right) = 22 - x$
$n \left(P \cap Q\right) \setminus \setminus = 13 + x$
$n \left(P \cap Q '\right) = 15 - x$

#### Explanation:

Using just set theory we have:

We are given $n \left(\epsilon\right) = 50$

And, $n \left(P ' \cap Q '\right) = x$

Part (i); $n \left(P ' \cap Q\right)$

$n \left(P ' \cap Q\right) + n \left(P\right) + x = n \left(\epsilon\right)$
$\therefore n \left(P ' \cap Q\right) + 28 + x = 50$
$\therefore n \left(P ' \cap Q\right) = 22 - x$

Part (ii); $P \cap Q$

$n \left(P \cup Q\right) = n \left(P\right) + n \left(Q\right) - n \left(P \cap Q\right)$
$\therefore 50 - x = 28 + 35 - n \left(P \cap Q\right)$
$\therefore n \left(P \cap Q\right) = 13 + x$

Part (iii); $P \cap Q '$

$n \left(P \cap Q '\right) + n \left(Q\right) + x = n \left(\epsilon\right)$
$\therefore n \left(P \cap Q '\right) + 35 + x = 50$
$\therefore n \left(P \cap Q '\right) = 15 - x$

Part (iv): Range
The min value for each of the above is 0;

$n \left(P ' \cap Q '\right) = x \ge 0 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies x \ge 0$
$n \left(P ' \cap Q\right) \setminus = 22 - x \ge 0 \implies x \le 22$
$n \left(P \cap Q\right) \setminus \setminus \setminus = 13 + x \ge 0 \implies 13 + x \ge - 13$
$n \left(P \cap Q '\right) \setminus = 15 - x \ge 0 \implies x \le 15$

Combining we get $0 \le x \le 15$