$A$ contractor has two teams of workers,team $A$ and team $B$. Team $A$ can complete a project $P$ in $12$ days and team $B$ can complete $P$ in $36$ days. Team $A$ starts working on $P$ and team $B$ joins team $A$ after four days. Team $A$ is withdrawn after another two days and team $B$ is asked to double its efficiency. The number of additional days required for team $B$ to complete $P$ is
- A$6$
- B$8$
- C$15$
- D$16$
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Similar Questions
Suppose $A_1, A_2, A_3, \dots, A_{30}$ are $30$ sets each having $5$ elements and $B_1, B_2, \dots, B_n$ are $n$ sets each with $3$ 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:
Difficult
View SolutionOut of $500$ car owners investigated,$400$ owned car $A$ and $200$ owned car $B$,and $50$ owned both car $A$ and car $B$. Is this data correct?
Medium
View SolutionLet $A = \{x : |x^{2} - 10| \le 6\}$ and $B = \{x : |x - 2| > 1\}$. Then:
In a survey of $60$ people,it was found that $25$ people read newspaper $H$,$26$ read newspaper $T$,$26$ read newspaper $I$,$9$ read both $H$ and $I$,$11$ read both $H$ and $T$,$8$ read both $T$ and $I$,and $3$ read all three newspapers. Find the number of people who read at least one of the newspapers.
Medium
View SolutionIf $P(A) = \frac{2}{5}$,$P(B) = \frac{1}{4}$ and $P(A \cup B) = \frac{1}{2}$,then $P(A' \cup B') = $
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