In a college of $300$ students,every student reads $5$ newspapers and every newspaper is read by $60$ students. The number of newspapers is

Let $X$ be a set of $5$ elements. The number $d$ of ordered pairs $(A, B)$ of subsets of $X$ such that $A \neq \phi, B \neq \phi, A \cap B \neq \phi$ satisfies:

If ${A_1}, {A_2}, {A_3}, \dots, {A_{30}}$ are $30$ sets each having $5$ elements and ${B_1}, {B_2}, \dots, {B_n}$ are $n$ sets each having $3$ elements. Let $\bigcup_{i=1}^{30} {A_i} = \bigcup_{j=1}^n {B_j} = S$ and each element of $S$ belongs to exactly $10$ of the $A_i$'s and exactly $9$ of the $B_j$'s,then $n$ is equal to:

In a certain town,$65\%$ of families own cell phones,$15000$ families own scooters,and $15\%$ of families own both. Given that every family owns at least one of the two,the total number of families in the town is:

If $n(X)=700, n(A)=200, n(B)=300,$ and $n(A \cap B)=100$,where $X$ is the universal set and $A$ and $B$ are subsets of $X$,then $n(A' \cap B')=$

Let $S = \{1, 2, 3, \ldots, 40\}$ and let $A$ be a subset of $S$ such that no two elements in $A$ have their sum divisible by $5$. What is the maximum number of elements possible in $A$?