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?

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

Another approach is depicted below:

enter image description here

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 $$zero =3\times two.\tag 3$$


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 $zero=15.$


The problem with the original approach is that

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

The right equation is


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.

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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.