Question

# The sum of the numerator and denominator of a fraction is $3$ less than twice the denominator. If the numerator and denominator are decreased by $1$ , the numerator becomes half the denominator. Determine the fraction.

Hint: This question can be solved by solving the equation formed from the statements given in the question.

The given problem is based upon a linear equation with two variables.
Let $x$ and $y$ be the numerator and denominator respectively. Thus the required equation is $\dfrac{x}{y}$
Now,
Sum of fraction$= x + y$
And it is given that the sum of numerator and denominator of a fraction is 3 less than twice the denominator,
$\Rightarrow x + y = 2y - 3 \\ {\text{or }}x + y - 2y = - 3 \\ {\text{or }}x - y + 3 = 0 - - - - - - - - \left( i \right) \\$
Also given that, if the numerator and denominator are decreased by 1, the numerator becomes half the denominator
$\Rightarrow x - 1 = \dfrac{1}{2}\left( {y - 1} \right) \\ {\text{or }}2x - 2 = y - 1 \\ {\text{or }}2x - y - 1 = 0 - - - - - - - \left( {ii} \right) \\$
Subtracting $\left( {ii} \right){\text{ and }}\left( i \right)$ gives,
$x - y + 3 - \left( {2x - y - 1} \right) = 0 \\ {\text{or }}x - y + 3 - 2x + y + 1 = 0 \\ {\text{or }} - x + 4 = 0 \\ {\text{or }}x = 4 \\$
Substituting $x$ value in $\left( i \right)$ gives,
$4 - y + 3 = 0 \\ {\text{or }} - {\text{y + 7 = 0}} \\ {\text{or }}y = 7 \\$
Therefore, the required numerator and denominator are
$x = 4{\text{ and }}y = 7$
And the required fraction is: $\dfrac{4}{7}$

Note: These types of questions can be solved by converting statements into equations and then solving the equations. In this question we simply convert the statements into equations and then solve them to get the value of numerator and denominator. Thus, we get our required fraction.