# Solving The Equation (Venn Diagram)

In a community, there are $20$ people who know at least a language(German or italian) and $30$ people who at most a language(German or italian). The people who know no language is $3$ times of the people who know the both language. How many people who know no language are there?

I drew up the venn diagram and given numbers for regions.

$$r(1) = \text {who knows german}$$ $$r(2) = \text {who knows italian}$$ $$r(3) = \text {who knows both}$$ $$r(4) = \text {who knows no language}$$

The equation part

$$r(1) + r (2) + r (3) = 20$$

$$r(1) + r (2) + r (4) = 30$$

$$r (4) = 3\times r (3)$$

Might I get some help?

• I edited it a sec ago. – Morata Nov 4 '17 at 13:08
• What am i missing? Also I am so confused right now. – Morata Nov 4 '17 at 13:18

Another approach is depicted below:

Here $two$, $one$, $zero$, $\color{blue}{at \ most \ one}$, $at\ least\ one$ mean that in the given region people speak that many languages.

We have that $$two+one=20,\tag 1$$ $$zero+one=30,\tag 2$$ and $$zero =3\times two.\tag 3$$

EDIT

Substitute $(3)$ in $(2)$. We have the following two equations now:

$$two+one=20,\tag 4$$ $$3\times two+one=30\tag 5$$

Subtract $(4)$ from $(5)$: $$2\times two=10.$$ So, $two=5$. Then, from $(1)$ $one=15.$

Finally, the solution of the equations is $two=5$, $one=15$, and $zero=15.$

EDIT 2

The problem with the original approach is that

$$r(1)+r(2)+r(3)\not = 20.$$

The right equation is

$$r(1)+r(2)-r(3)=20$$

because $r(1)+r(2)$ contains $r(3)$ twice. Is this enough?

OK, let's step further. The equation $r(1)+r(2)+r(4)=30$ is not correct either. The correct equation is $$r(1)+r(2)-2r(3)+r(4)=30$$ because, as I mentioned above the sum $r(1)+r(2)$ contains $r(3)$ twice. So, $r(1)+r(2)-2r(3)$ is the number of people who speak only one language.

Now, we have three equations:

$$r(1)+r(2)-r(3)=20,$$ $$r(1)+r(2)-2r(3)+r(4)=30,$$$$3r(3)=r(4).$$

Substituting the last equation, we get

$$r(1)+r(2)-r(3)=20,$$ $$r(1)+r(2)+r(3)=30.$$

From here $$r(1)+r(2)=25.$$

And finally, from the first equation, we get that $r(3)=5$ and then $$r(4)=3r(3)=15.$$ But we cannot calculate $r(1)$ and $r(2)$ separately.

• You need to swap one and two. The approach seems true. – kimi Tanaka Nov 4 '17 at 13:52
• More lucid than mine. – kimi Tanaka Nov 4 '17 at 13:53
• @kimiTanaka: Thank you and thank you again. – zoli Nov 4 '17 at 13:54
• Can you be more clear at solution of equation part? – Morata Nov 4 '17 at 14:00
• @Morata: Yes, I'll edit. – zoli Nov 4 '17 at 14:04