A and B can do a work in 8 days; B and C can do the same work in 12 days; A, B and C together can finish it in 6 days. In how many days A and C together will do the same work?
Answer
Suppose $A, B$ and $C$ alone can do the work in $n_{A}, n_{B}$ and $n_{C}$ days respectively. A and B together can do the work in 8 days. $\Rightarrow \frac{1}{n_A}+\frac{1}{n_B}=\frac{1}{8} \ldots$ (i) B and C together can do the work in 12 days. $\Rightarrow \frac{1}{n_B}+\frac{1}{n_C}=\frac{1}{12} \ldots$ (ii) A, B and C together can do the work in 6 days. $\Rightarrow \frac{1}{n_A}+\frac{1}{n_B}+\frac{1}{n_C}=\frac{1}{6} \ldots$ (iii) Subtracting (i) and (ii) successively from (iii), we get $\frac{1}{n_{C}}=\frac{1}{6}-\frac{1}{8}$ and $\frac{1}{n_{A}}=\frac{1}{6}-\frac{1}{12}$ $\Rightarrow \frac{1}{n_{C}}=\frac{1}{24}$ and $\frac{1}{n_{A}}=\frac{1}{12} \Rightarrow \frac{1}{n_{A}}+\frac{1}{n_{C}}=\frac{1}{12}+\frac{1}{24}=\frac{1}{8}$ Hence, A and C together can do the work in 8 days
Write down all possible subsets of $A=(1,\{2,3\})$.
Answer
Here, we have, A contains two elements, namely 1 and $\{2,3\}$ $\{2,3\}=B$, then $A=\{1, B\}$ $\therefore P(A)=\{\phi,\{1\},\{B\},\{1, B\}\}$ $\Rightarrow P(A)=\{\phi,,\{1\},\{\{2,3\}\},\{1,\{2,3\}\}\}$.
Find the union set of A = {x :x is a natural number and 1 <x ≤ 5} and B = {x : x is a natural number and 5 < x ≤ 10}.
Answer
According to the Question, A = {x: x is a natural number and 1 < x ≤ 5} $\Rightarrow A=\{2,3,4,5\}$ B = {x: x is a natural number and 5 < x ≤ 10} On taking the union of A and B, $A \cup B=\{2,3,4,5\} \cup\{6,7,8,9,10\}=\{2,3,4,5,6,7,8,9,10\}$ Also in set builder form, A ∪ B = {x: x is natural number and 1 < x ≤ 10}
At what time between 2 and 3 O'Clock will the hands of a clock coincide?
Answer
At $2 \mathrm{O}^{\prime}$ clock, the hour hand is at 2 and the minute hand is at 12 , i.e. they are 10 minute spaces apart. To be coincident, the minute hand must gain 10 minute spaces. We know that 55 minute spaces are gained by minute hand over the hour hand in 60 minutes $\therefore 10$ minute spaces will be gained in $\left(\frac{60}{55} \times 10\right)$ minutes $=10 \frac{10}{11}$ minutes Hence, the two hands will coincide in $10 \frac{10}{11}$ minutes
