Question

In a fraction twice the numerator is 2 more than the denominator. If 3 is added to the numerator and the denominator, the new fraction is $\dfrac{2}{3}$ .Find the original fraction.

Answer 1

Hint: Word problem with two conditions given in the question. There would be 2 equations and definitely two unknown variables.

Complete step-by-step answer:
Let the denominator of the original fraction be x.
Let the numerator of the original fraction be y.
According to question we have,
Twice the numerator is 2 more than the denominator. Interpreting the statement in the form of equation, we get,
$2y=x+2$
$\Rightarrow 2y-x=2.................(i)$
The second condition given to us in the question is the below statement:
If 3 is added to the numerator and the denominator, the new fraction is $\dfrac{2}{3}$. Interpreting in form of equation, we get,
$\therefore \dfrac{y+3}{x+3}=\dfrac{2}{3}$
On cross-multiplication, we get;
$3\left( y+3 \right)=2\left( x+3 \right)$
$\Rightarrow 3y+9=2x+6$
$\Rightarrow 2x-3y=3...............(ii)$
Now, we will multiply equation (i) by 3 and equation (ii) by 2. So, we get:
$3\times (i)$ : $3(2y-x)=3\times 2$
$2\times (ii)$ : $2(2x-3y)=3\times 2$
Adding both equations we get:
$3(2y-x)+2(2x-3y)=3\times 2+3\times 2$
$\Rightarrow 6y-3x+4x-6y=6+6$
$\therefore x=12$
From equation (i), we have 2y - x = 2
On taking x to the other side of the equation and solving it by substituting the value of x, we get,
$2y=2+x$
$\Rightarrow 2y=2+12$
$\Rightarrow 2y=14$
$\therefore y=7$
Hence, the original fraction is $\dfrac{y}{x}=\dfrac{7}{12}$.

Note: We can also interpret the statement “If 3 is added to the numerator and the denominator, the new fraction is $\dfrac{2}{3}$ ” as:
$y+3=2k$
And
$x+3=3k$
However, we need to take a constant value k with Right hand side of both the equations as $\dfrac{2}{3}$ is the simplest form and some factors might have been cancelled from both numerator and denominator to reach this form.
For instance let a fraction be $\dfrac{9}{12}$ ;
So, its simplest form comes out to be:
\[\dfrac{9}{12}=\dfrac{3}{4}\]
3 is cancelled from both numerator and denominator but the original fraction was $\dfrac{9}{12}$ but in simplest form it is represented as $\dfrac{3}{4}$ .