
The product of a positive number, its square, and its reciprocal is $\dfrac{{100}}{{81}}$. What is the number?
Answer
476.1k+ views
Hint: Here in this question, we need to find the unknown number of given mathematical solutions. For this, first we need to consider the unknown term as $x$ and write down the given question in mathematical format then further simplify by using basic arithmetic operations to get the required solutions.
Complete step by step answer:
Consider the given question: we need to find the final answer of the product of a positive number, its square, and its reciprocal is equal to $\dfrac{{100}}{{81}}$. Let us consider the unknown term i.e., positive number as ‘$x$’.Square of the positive number $x$ is ${x^2}$ and its reciprocal is $\dfrac{1}{x}$. Given, the product of all three is equal to $\dfrac{{100}}{{81}}$.
$ \Rightarrow \,\,\,x \times {x^2} \times \dfrac{1}{x} = \dfrac{{100}}{{81}}$
Now, we have to find the $x$ value.
On multiplication, we have
$ \Rightarrow \,\,\,x \times {x^2} \times \dfrac{1}{x} = \dfrac{{100}}{{81}}$
$ \Rightarrow \,\,\,{x^3} \times \dfrac{1}{x} = \dfrac{{100}}{{81}}$
$ \Rightarrow \,\,\,\dfrac{{{x^3}}}{x} = \dfrac{{100}}{{81}}$
On cancelling the like terms on RHS we have
$ \Rightarrow \,\,\,{x^2} = \dfrac{{100}}{{81}}$
Take the square root of both sides gives
$ \Rightarrow \,\,\,x = \sqrt {\dfrac{{100}}{{81}}} $
By the property $\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}$, then we have
$ \Rightarrow \,\,\,x = \dfrac{{\sqrt {100} }}{{\sqrt {81} }}$
As we know the square root value of $\sqrt {100} = 10$ and $\sqrt {81} = 9$.
$\therefore \,\,\,x = \dfrac{{10}}{9}$
Hence, the required mathematical expression is:
$\left( {\dfrac{{10}}{9}} \right) \times {\left( {\dfrac{{10}}{9}} \right)^2} \times \left( {\dfrac{9}{{10}}} \right) = \dfrac{{100}}{{81}}$
On simplification, we get
$\therefore \dfrac{{100}}{{81}} = \dfrac{{100}}{{81}}$
Therefore, the required positive number is $\dfrac{{10}}{9}$.
Note: For this type of question, we need to consider whether the unknown number has any variable to make the solution easy. Reciprocal means an expression which when multiplied by another expression, gives unity (1) as a result. The reciprocal of any quantity is, one divided by that quantity.
Complete step by step answer:
Consider the given question: we need to find the final answer of the product of a positive number, its square, and its reciprocal is equal to $\dfrac{{100}}{{81}}$. Let us consider the unknown term i.e., positive number as ‘$x$’.Square of the positive number $x$ is ${x^2}$ and its reciprocal is $\dfrac{1}{x}$. Given, the product of all three is equal to $\dfrac{{100}}{{81}}$.
$ \Rightarrow \,\,\,x \times {x^2} \times \dfrac{1}{x} = \dfrac{{100}}{{81}}$
Now, we have to find the $x$ value.
On multiplication, we have
$ \Rightarrow \,\,\,x \times {x^2} \times \dfrac{1}{x} = \dfrac{{100}}{{81}}$
$ \Rightarrow \,\,\,{x^3} \times \dfrac{1}{x} = \dfrac{{100}}{{81}}$
$ \Rightarrow \,\,\,\dfrac{{{x^3}}}{x} = \dfrac{{100}}{{81}}$
On cancelling the like terms on RHS we have
$ \Rightarrow \,\,\,{x^2} = \dfrac{{100}}{{81}}$
Take the square root of both sides gives
$ \Rightarrow \,\,\,x = \sqrt {\dfrac{{100}}{{81}}} $
By the property $\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}$, then we have
$ \Rightarrow \,\,\,x = \dfrac{{\sqrt {100} }}{{\sqrt {81} }}$
As we know the square root value of $\sqrt {100} = 10$ and $\sqrt {81} = 9$.
$\therefore \,\,\,x = \dfrac{{10}}{9}$
Hence, the required mathematical expression is:
$\left( {\dfrac{{10}}{9}} \right) \times {\left( {\dfrac{{10}}{9}} \right)^2} \times \left( {\dfrac{9}{{10}}} \right) = \dfrac{{100}}{{81}}$
On simplification, we get
$\therefore \dfrac{{100}}{{81}} = \dfrac{{100}}{{81}}$
Therefore, the required positive number is $\dfrac{{10}}{9}$.
Note: For this type of question, we need to consider whether the unknown number has any variable to make the solution easy. Reciprocal means an expression which when multiplied by another expression, gives unity (1) as a result. The reciprocal of any quantity is, one divided by that quantity.
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