Let $S = \{4, 6, 9\}$ and $T = \{9, 10, 11, \ldots, 1000\}$. If $A = \{a_{1} + a_{2} + \ldots + a_{k} : k \in N, a_{1}, a_{2}, \ldots, a_{k} \in S\}$,then the sum of all the elements in the set $T - A$ is equal to:

$205$ students take an examination,of whom $105$ pass in English,$70$ students pass in Mathematics,and $30$ students pass in both. How many students fail in both subjects?

In a city,$25\%$ of families have a telephone and $15\%$ have a car,while $65\%$ of families have neither a telephone nor a car. If $2000$ families have both a car and a telephone,then consider the following statements:
$1.$ $10\%$ of families have both a car and a telephone.
$2.$ $35\%$ of families have either a car or a telephone.
$3.$ $40,000$ families live in the city.
Which of these statements are true?

For three non-impossible events $A$,$B$,and $C$,$P(A \cap B \cap C) = 0$,$P(A \cup B \cup C) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{3}$,and $P(C) = \frac{1}{6}$. The probability that exactly one of $A$ or $B$ occurs but $C$ does not occur is:

Two newspapers $A$ and $B$ are published in a city. It is known that $25\%$ of the city population reads $A$ and $20\%$ reads $B$,while $8\%$ reads both $A$ and $B$. Further,$30\%$ of those who read $A$ but not $B$ look into advertisements,$40\%$ of those who read $B$ but not $A$ look into advertisements,and $50\%$ of those who read both $A$ and $B$ look into advertisements. The percentage of the population who look into advertisements is:

$A$ number is chosen from the first $100$ natural numbers. The probability that the number is even or divisible by $5$ is: