 QUESTION

# The sum of a number and its reciprocal is $2\dfrac{1}{30}$. Find the number.

Hint: To solve such types of questions we first of all assume the number as some variable x then the reciprocal of the number would become $\dfrac{1}{x}$. Now use the given conditions in the question to formula a polynomial in terms of the variable x and then calculate the value of the number.

Let the one number be x
So, its reciprocal will be$\dfrac{1}{x}$.
Given that the sum of a number and its reciprocal is $2\dfrac{1}{30}$.
Converting $2\dfrac{1}{30}$ into fraction form we have,
$2\dfrac{1}{30}=\dfrac{61}{30}$
Now, according to the given conditions in the question we have,
$x+\dfrac{1}{x}=\dfrac{61}{30}$
Taking the LCM of the left-hand side of the expression we get,
\begin{align} & x+\dfrac{1}{x}=\dfrac{61}{30} \\ & \Rightarrow \dfrac{\left( x{}^\text{2}\text{ }+\text{ }1 \right)}{x}=\dfrac{61}{30} \\ \end{align}
After cross multiplying the above obtained expression, we get
$\Rightarrow 30x{}^\text{2}\text{ }+\text{ }30=61x$
Making necessary arrangements we get,
$\Rightarrow 30x{}^\text{2}\text{ }-\text{ }61x\text{ }+\text{ }30\text{ }=\text{ }0$
Now we have obtained a quadratic expression. Solving it by splitting middle terms we get,
$\begin{array}{*{35}{l}} {} \\ \Rightarrow 30x{}^\text{2}\text{ }-\text{ }61x\text{ }+\text{ }30\text{ }=\text{ }0 \\ \Rightarrow 30x{}^\text{2}\text{ }-\text{ }36x\text{ }-\text{ }25x\text{ }+\text{ }30=0 \\ \Rightarrow 6x\left( 5x\text{ }-\text{ }6 \right)-5\left( 5x\text{ }-\text{ }6 \right)=0 \\ \Rightarrow \left( 6x\text{ }-\text{ }5 \right)\text{ }\left( 5x\text{ }-\text{ }6 \right)\text{ }=\text{ }0 \\ \end{array}$
$\Rightarrow 6x = 5$ and $5x = 6$
$\Rightarrow x={\dfrac{5}{6}}, {\dfrac{6}{5}}$

Therefore the number is either $\dfrac{5}{6}$ or $\dfrac{6}{5}$

Note: Students can assume both the numbers as different variables which would become complex and while making an equation of the given condition it will be a polynomial or equations in two variables which would be difficult to solve. Therefore, always assume a single variable for such types of equations and try to generalise the second variable in terms of the first one.