Question

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator, if the numerator and denominator are increased by 3, they are in the ratio. Determine the fraction by choosing the correct option:
A) $\dfrac{1}{7}$
B) $\dfrac{2}{5}$
C) $\dfrac{3}{7}$
D) $\dfrac{5}{9}$

Answer 1

Hint: Denote the unknown quantity by some letters such as numerator of a fraction as $x$ and denominator as $y$, then translate the statements of the problem into mathematical statements and by using the condition(s) given in the problem, form an equation.

Complete step-by-step answer:
According to the question, we let the numerator of a fraction as $x$ and denominator as $y$.
So let the required fraction be $\dfrac{x}{y}$.
As the sum of the numerator and denominator of a fraction is 4 more than twice the numerator.
$\therefore x + y = 4 + 2x$
After transposing $2x$ on the L.H.S, we get
$ \Rightarrow y - x = 4$………………….(i)
The numerator and denominator are increased by 3. So the new numerator= $x + 3$ and new denominator=$y + 3$.
Also given a new numerator: new denominators are in the ratio $2:3$.
$\therefore \dfrac{{x + 3}}{{y + 3}} = \dfrac{2}{3}$
By cross multiplication we get,
$(x + 3) \times 3 = (y + 3) \times 2$
On further solving we get,
$3x + 9 = 2y + 6$
Now transposing the variables on one side and constants on other, we can rewrite the equation as
$3x - 2y = - 9 + 6$
$ \Rightarrow 3x - 2y = - 3$………………………(ii)
Multiplying (i) by 3 and subtracting (ii), we get
$3y - 2y = 12 - 3$
$ \Rightarrow y = 9$
Now substituting $y = 9$ in (i), we get
$9 - x = 4$\[\]
$ \Rightarrow 9 - 4 = x$
\[ \Rightarrow x = 5\]
Hence, the required fraction is $\dfrac{5}{9}$.

$\therefore $ Option D is the correct answer.

Note: A number of the form $\dfrac{p}{q}$ or a number which can be expressed in the form $\dfrac{p}{q}$, where $p$ and $q$ are integers and $q \ne 0$, is called a fraction.
In the fraction $\dfrac{p}{q}$, integer $p$ is known as the numerator and non-zero integer $q$ is called the denominator.