Question

$A=\left\{ \left( a,b \right)/b=2a-5 \right\}$ if $\left( m,5 \right)$ and $\left( 6,n \right)$ are the members of the set $A$ , then $m$ and $n$ are respectively
1) $5$ , $7$
2) $7$ , $5$
3) $2$ , $3$
4) $5$ , $3$

Answer 1

Hint: In this problem we need to calculate the values of $m$ and $n$ where $\left( m,5 \right)$ and $\left( 6,n \right)$ are the members of the set $A$ and the set $A$ is given by $A=\left\{ \left( a,b \right)/b=2a-5 \right\}$. Here we need to find what are the elements of the set and the relation between the elements of the set from the given definition. After that we will compare the given elements with the nominal elements from the definition and apply the relation to the given elements. By simplifying the relation we can have the values of $m$ and $n$.

Complete step by step solution:
Given definition of the set $A$ is $A=\left\{ \left( a,b \right)/b=2a-5 \right\}$.
From the above definition the elements of the $A$ are in the form of $\left( a,b \right)$ and the relation between the values is given by $b=2a-5$ .
If $\left( m,5 \right)$ is one of the element of the set $A$, then there must be a relation between $m$ and $5$ such that
$5=2m-5$
Adding $5$ on both sides of the above equation, then we will have
$\begin{align}
  & 5+5=2m-5+5 \\
 & \Rightarrow 10=2m \\
\end{align}$
Dividing the above equation with $m$ on both sides of the above equation, then we will get
$\begin{align}
  & \dfrac{10}{2}=\dfrac{2m}{2} \\
 & \Rightarrow m=5 \\
\end{align}$
Here the value of $m$ must be equal to $5$.
If $\left( 6,n \right)$ is also a element of the set $A$, then there must be a relation between $6$ and $n$ such that
$n=2\left( 6 \right)-5$
Simplify the above equation by using basic mathematical operations, then we will have
$\begin{align}
  & n=12-5 \\
 & \Rightarrow n=7 \\
\end{align}$
Here the value of $n$ must be equal to $7$.
Hence option 1 is the correct answer.

Note: For this kind of problems the relation that defined between the values of set is important. Based on the relation the value sets may change. While coming to this problem, the substitution of values in the given relation is also important. We have the second value as $\left( 6,n \right)$ where $n$ is in the position of $b$ , so the relation must be like $n=2\left( 6 \right)-5$ but don’t write it as $6=2n-5$ .