Goutham wrote: In a school, more than $90\% $ of the students know both English and German, and more than $90\%$ percent of the students know both English and French. Prove that more than $90\%$ percent of the students who know both German and French also know English. Considering the three languages $E,F,G$ in this order, we can define $8$ categories of students depending on the languages known by the student : If binary representation of $k\in\{0,1,2,3,4,5,6,7\}$ is $\overline{abc}_2$, let $x_k$ be the number of students who speaks $E$ if $a=1$ and $F$ if $b=1$ and $G$ if $c=1$. The conditions are : $x_5+x_7\ge 0.9(x_0+x_1+x_2+x_3+x_4+x_5+x_6+x_7)$ $x_6+x_7\ge 0.9(x_0+x_1+x_2+x_3+x_4+x_5+x_6+x_7)$ Which may be written $x_7\ge 9(x_0+x_1+x_2+x_3+x_4)+9x_6-x_5$ $x_7\ge 9(x_0+x_1+x_2+x_3+x_4)+9x_5-x_6$ And so $x_7\ge 9(x_0+x_1+x_2+x_3+x_4)\ge 9x_3$ and so $10x_7\ge 9(x_3+x_7)$ and so $\frac{x_7}{x_3+x_7}\ge 0.9$ Which means that the ratio between students speaking the three languages and the students speaking at least French and German is not less than $90\%$ Q.E.D.

Replace $90\%$ with $\lambda$. Denote by $x$ the number of students speaking both $E,G$, but no $F$; by $y$ the number of students speaking both $E,F$, but no $G$; by $z$ the number of students speaking both $F,G$, but no $E$; finally, by $t$ the number of students speaking all three $E,F,G$. Let $n$ be the total number of students. The givens are $x+t \geq \lambda n \geq \lambda (x+y+z+t)$, $y+t \geq \lambda n \geq \lambda (x+y+z+t)$. Therefore $x+y+2t \geq 2\lambda (x+y+z+t)$, or $t\geq \dfrac {2\lambda-1} {2}(x+y) + z+t \geq z+t$ if $2\lambda - 1\geq 0$, i.e. if $\lambda \geq 50\%$.