In a town of 10,000 families,it was found tha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given total families $N = 10,000$.
$n(A) = 40\% 	ext{ of } 10,000 = 4,000$
$n(B) = 20\% 	ext{ of } 10,000 = 2,000$
$n(C) = 10\% 	ext{ of } 1

Vedclass · Accepted Answer

$3300$

Vedclass · Answer

(B) Given total families $N = 10,000$.
$n(A) = 40\% 	ext{ of } 10,000 = 4,000$
$n(B) = 20\% 	ext{ of } 10,000 = 2,000$
$n(C) = 10\% 	ext{ of } 10,000 = 1,000$
$n(A \cap B) = 5\% 	ext{ of } 10,000 = 500$
$n(B \cap C) = 3\% 	ext{ of } 10,000 = 300$
$n(A \cap C) = 4\% 	ext{ of } 10,000 = 400$
$n(A \cap B \cap C) = 2\% 	ext{ of } 10,000 = 200$
To find the number of families that buy newspaper $A$ only,we use the formula:
$n(A 	ext{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$
$n(A 	ext{ only}) = 4,000 - 500 - 400 + 200$
$n(A 	ext{ only}) = 4,000 - 700 = 3,300$

Similar Questions

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

If $X$ and $Y$ are two sets such that $X$ has $40$ elements,$X \cup Y$ has $60$ elements and $X \cap Y$ has $10$ elements,how many elements does $Y$ have?

$A$ number is chosen at random from the set $\{1, 2, 3, \ldots, 2000\}$. Let $p$ be the probability that the chosen number is a multiple of $3$ or a multiple of $7$. Then the value of $500p$ is . . . . . .

$A$ and $B$ are two subsets of set $S = \{1, 2, 3, 4\}$ such that $A \cup B = S$. Then,the number of ordered pairs $(A, B)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module