Question

Let * be a binary operation, on the set of all non – zero real numbers, given by \[a\text{ }*\text{ }b\text{ }=\dfrac{ab}{5}\] , for all $a,b\in R-\{0\}$ . Find the value of x, given that $2*(x*5)=10$ .

Answer 1

Hint: To solve this question, what we will do is we will apply the mathematical definition of binary operation given in the question on $2*(x*5)=10$. First we will use binary operation on x*5 and then finally with 2 and hence we will solve for the value of x.

Complete step by step answer:
Before, we solve the question, let see what is binary operation and what does $a,b\in R-\{0\}$means as for solving the given function it is important to know the meaning of binary operation.
Binary operation is a calculation that combines two elements to produce another element and binary operator can be any operation between two numbers.
Let's understand with an example.
We know that 4 + 5 = 9. Now, we can say that two numbers used in binary operation are 4 and 5 and addition ‘ + ‘ is a binary operator and the produced number is 9. This is what binary operation means.
$a,b\in R-\{0\}$, means a , b are any number which belongs to a set of real numbers and which cannot be equal to 0.
Now, in question it is given that * be a binary operation, on the set of all non – zero real numbers, given by \[a\text{ }*\text{ }b\text{ }=\dfrac{ab}{5}\] , for all $a,b\in R-\{0\}$ and $2*(x*5)=10$, so we have to solve for the value of x.
Now, $2*(x*5)=10$
Applying binary operation \[a\text{ }*\text{ }b\text{ }=\dfrac{ab}{5}\] in ( x * 5 ), we get
$2*\left( \dfrac{x\cdot 5}{5} \right)=10$
On solving, we get
$2*x=10$
Now, Applying binary operation \[a\text{ }*\text{ }b\text{ }=\dfrac{ab}{5}\] in ( 2 * x ), we get
\[\dfrac{2x}{5}=10\]
Solving using we get
\[x=25\]
Hence, the value of x is 25.

Note:
For solving such problems, one must know what binary operation actually means. Also, use binary operation on two numbers at a time unless and until other information is given in question as it will reduce the risk of calculation error.