# How to best approach a problem of this kind (problem solving and simple linear equations)

This is a common type of problem that appear on algebra tests to test problem solving abilities (this is not from a real test or homework). I am curious on how to best approach these kind of problems since they can often overwhelm you (like this one has for me).

The problem:
The final test was written by $U = 190$ people.
Of the $190$, $P = 75$ people passed the final test.
Of the $T = 161$ with all three previously completed exams, $74$ passed the final test.
Of the $B = 107$ with full bonus points, $63$ passed the final test.

There was $C = 99$ people who had made all previous exams and had full bonus.
Of those who lacked bonus and had also failed a previous exam there was 21 who did not pass the final test.

How many with full bonus and all three completed exams $C$ passed the final test?

My attempt:
We know that there was $C = 99$ people with full bonus and three completed exams.

Our problem is then to find the relation of $T$ and $B$ in $C$. Since then we can use that factor to multiply with the number of people who has passed the test in both $T$ and $B$.

That relation can be written as: $Tx + By = C \leftrightarrow 161x + 107y =99$.
Now we need another equation for $C$ expressed in $x$ and $y$ to have an equation system. And here I am stuck. What method would you use to find another equation?

## 1 Answer

I would probably use a table and/or Venn diagram to track the various different sub-populations cleanly.

You have three attributes for your universe of 190 people $U$:

• passed the final test $|P|=75, |\bar P|=115$
• passed all three earlier tests $|T|=161, |\bar T|=29$
• got full bonus points $|B|=107, |\bar B|=83$

You want $|P \cap T \cap B|$

You also know $|T \cap P|=74, |B \cap P|=63, |T \cap B|=99,$ and $|\bar P \cap \bar T \cap \bar B|=21$

You therefore know $|\bar T \cap P|=75-74=1$ and $|\bar T \cap \bar P|=29-1=28$

This gives you $|B \cap \bar T \cap \bar P| = |\bar T \cap \bar P| - |\bar B \cap \bar T \cap \bar P| = 28-21=7$

Then $|B \cap \bar P|= 107-63=44$ gives you $|B \cap T \cap \bar P| = 44-7=37$

Finally then you have $|P \cap T \cap B| = |T \cap B|- |B \cap T \cap \bar P| = 99-37 = 62$

• Thank you Joffan! This is indeed correct and it is maybe the best way to do it! – Lukas Arvidsson Jan 15 '15 at 17:20