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If the difference between the reciprocal of a positive proper fraction and the fraction itself be $\dfrac{9}{20}$ then the fraction is
A) $\dfrac{3}{5}$
B) $\dfrac{3}{10}$
C) $\dfrac{4}{5}$
D) $\dfrac{5}{4}$

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Last updated date: 25th Jun 2024
Total views: 405.3k
Views today: 11.05k
Answer
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Hint: First, find the equation from the data provided in the question. Then solve further, a quadratic equation will be formed. Now solve the equation by either factoring or by quadratic formula $\left( x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right)$ to get the fraction. The fraction is the desired result.

Complete step by step answer:
Given: Difference between the reciprocal and the fraction is $\dfrac{9}{20}$.
Let the required fraction be $x$.
Since the difference between the reciprocal and the fraction is $\dfrac{9}{20}$.
$\Rightarrow \dfrac{1}{x}-x=\dfrac{9}{20}$
Take LCM on the left side,
$\Rightarrow \dfrac{1-{{x}^{2}}}{x}=\dfrac{9}{20}$
Cross multiply the term,
$\Rightarrow 20-20{{x}^{2}}=9x$
Move all terms to one side,
$\Rightarrow 20{{x}^{2}}+9x-20=0$
Factorization can be done in 2 ways.

Method 1:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Put $a=20$, $b=9$ and $c=-20$,
$\Rightarrow x=\dfrac{-9\pm \sqrt{{{9}^{2}}-4\times 20\times (-20)}}{2\times 20}$
Multiply and square the terms and add them in the square root,
$\Rightarrow x=\dfrac{-9\pm \sqrt{1681}}{40}$
Then,
$\Rightarrow x=\dfrac{-9\pm 41}{40}$
Then,
$\Rightarrow x=-\dfrac{50}{40},\dfrac{32}{40}$
Since there is no option with a negative value. So,
$\Rightarrow x=\dfrac{32}{40}$
Cancel out the common factors,
$\Rightarrow x=\dfrac{4}{5}$

Hence, the number of oranges is 12.

Additional information:
A quadratic equation is a polynomial equation of degree 2. A quadratic equation has two solutions. Either two distinct real solutions, one double real solution, or two imaginary solutions.
There are several methods you can use to solve a quadratic equation:
- Factoring
- Completing the Square
- Quadratic Formula
- Graphing
All methods start with setting the equation equal to zero.

Note:
Alternative method:
$\Rightarrow 20{{x}^{2}}+9x-20=0$
We can write 9 as \[\left( 25-16 \right)\],
$\Rightarrow 20{{x}^{2}}+\left( 25-16 \right)x-20=0$
Open the brackets and multiply the terms,
$\Rightarrow 20{{x}^{2}}+25x-16x-20=0$
Take common factors from the equation,
$\Rightarrow 5x\left( 4x+5 \right)-4\left( 4x+5 \right)=0$
Take common factors from the equation,
$\Rightarrow \left( 4x+5 \right)\left( 5x-4 \right)=0$
Then,
$\Rightarrow x=-\dfrac{5}{4},\dfrac{4}{5}$
Since there is no option with a negative value. So,
$\Rightarrow x=\dfrac{4}{5}$

Hence, option (C) is the correct answer.