Question

If an improper fraction in its lowest form and its reciprocal is subtracted, the difference is $\dfrac{{77}}{{18}}$. Find the improper fraction? What will you get if you divide the improper fraction by its equivalent having $4$ as the denominator?

Answer 1

Hint:
Improper fraction is a fraction whose numerator is greater than the denominator. Assume any variable for the fraction and form a quadratic equation by subtracting its reciprocal from it. Find roots using the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. For equivalent fraction, remember that the equivalent fraction of a fraction $\dfrac{a}{b}$ can be written as $\dfrac{{a \times n}}{{b \times n}}$ . Compare the denominator with four and evaluate the required division.

Complete step by step solution:
Here in this problem, first, we have to find an improper fraction whose difference with its reciprocal is $\dfrac{{77}}{{18}}$ . And then we will divide this improper fraction with its equivalent having $4$ as the denominator.
Before starting with the solution, let’s first try to understand the concepts used here.
An improper fraction is a fraction whose numerator is equal to, larger than, or of equal or higher degree than the denominator. It can be represented as a fraction $\dfrac{a}{b}$ where $'a'$ is a non-zero number greater than a non-zero number $'b'$. It is said to be in its simplest form if $a{\text{ and }}b$ are co-primes.
According to the question, the difference between the improper fraction $\dfrac{a}{b}$ and its reciprocal $\dfrac{1}{{\left( {\dfrac{a}{b}} \right)}} = \dfrac{b}{a}$ is $\dfrac{{77}}{{18}}$.
$ \Rightarrow \dfrac{a}{b} - \dfrac{1}{{\left( {\dfrac{a}{b}} \right)}} = \dfrac{{77}}{{18}}$
Now, we can further simplify this equation to form a quadratic equation as follows:
$ \Rightarrow {\left( {\dfrac{a}{b}} \right)^2} - 1 = \dfrac{{77}}{{18}}\left( {\dfrac{a}{b}} \right) \Rightarrow 18{\left( {\dfrac{a}{b}} \right)^2} - 77\left( {\dfrac{a}{b}} \right) - 18 = 0$
This is a quadratic equation in terms of variable $\dfrac{a}{b}$ , which can be solved using the Quadratic formula.
Given a general quadratic equation of the form $a{x^2} + bx + c = 0$ with $x$ representing an unknown, $a,b{\text{ and }}c$ representing constants with $a \ne 0$ , the quadratic formula is:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here in our case, the constants $a,b{\text{ and }}c$ are given as $18, - 77{\text{ and }} - 18$ and the unknown $x$ is represented by $\dfrac{a}{b}$ respectively.
Therefore, on using the Quadratic formula, we get:
$ \Rightarrow \dfrac{a}{b} = \dfrac{{77 \pm \sqrt {{{77}^2} - 4 \times 18 \times \left( { - 18} \right)} }}{{2 \times 18}} = \dfrac{{77 \pm \sqrt {5929 + 1296} }}{{36}} = \dfrac{{77 \pm \sqrt {7225} }}{{36}}$
So now we can use $7225 = {85^2}$ in the above expression and further simplify it as: $ \Rightarrow \dfrac{a}{b} = \dfrac{{77 \pm 85}}{{36}} = \dfrac{{162}}{{36}}$ or $\dfrac{{ - 8}}{{36}}$
Since $\dfrac{a}{b}$ is an improper fraction in simplest form we can ignore the second negative fraction, therefore:
$ \Rightarrow \dfrac{a}{b} = \dfrac{{162}}{{36}} = \dfrac{9}{2}$
Thus, the required improper fraction is $\dfrac{9}{2}$ .
Now, equivalent to this fraction will be any fraction for which $\dfrac{9}{2}$ is its simplest form, i.e. equivalent fractions of $\dfrac{9}{2}$ will be of the form $\dfrac{{9 \times n}}{{2 \times n}}$ , where $'n'$ can be any non-zero number.
But we already know that the denominator of the equivalent fraction is $4$ , which will only be possible when $n = 2$ .
So, for $n = 2$ , the equivalent fraction $ = \dfrac{{9 \times 2}}{{2 \times 2}} = \dfrac{{18}}{4}$
According to the second part of the question, we need to find the value that we will get n dividing the improper fraction $\dfrac{9}{2}$ by its equivalent $\dfrac{{18}}{4}$
$ \Rightarrow \dfrac{9}{2} \div \dfrac{{18}}{4} = \dfrac{{\dfrac{9}{2}}}{{\dfrac{{18}}{4}}} = \dfrac{{9 \times 4}}{{2 \times 18}} = \dfrac{{1 \times 2}}{{1 \times 2}} = 1$

Thus, we get $1$ on dividing the improper fraction with its equivalent having $4$ as the denominator.

Note:
Notice that the definition of improper fraction played a crucial part in the solution. An alternative approach for this problem can be to assume the required improper fraction as some variable $'x'$ and then finding its value using the Quadratic formula. Also, notice the use of the property $\dfrac{m}{n} \div \dfrac{p}{q} = \dfrac{m}{n} \times \dfrac{q}{p} = \dfrac{{mq}}{{np}}$ in the last step.