Question

The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes $\dfrac{1}{2}$. Find the fraction.

Answer 1

Hint:
We will suppose the fraction to be $\dfrac{x}{y}$ where x is the numerator and y is the denominator and then we will form equations from the given data. The first equation will be $x + y = 2y – 3$ and the other will be $\dfrac{{x - 1}}{{y - 1}} = \dfrac{1}{2}$ . Upon solving these two equations, we will get the values of x and y and hence, the value of the fraction will be obtained.

Complete step by step solution:
Let us assume that the fraction is $\dfrac{x}{y}$ where x is the numerator and y is the denominator.
We are given that the sum of numerator and denominator of a fraction is 3 less than twice the denominator.
Writing this data in form of an equation, we get
$ \Rightarrow x + y = 2y – 3 $
Or, this can be written as:
$ \Rightarrow x – y = – 3 \Rightarrow x = y – 3$ – equation (1)
Also, if the numerator and denominator is decreased by 1, the fraction becomes $\dfrac{1}{2}$
This can be written in form of an equation as:
$ \Rightarrow \dfrac{{x - 1}}{{y - 1}} = \dfrac{1}{2}$
Or, this can be written as:
$ \Rightarrow 2\left( {x - 1} \right) = y - 1$ – equation (2)
Substituting the value of x from equation (1) in (2), we get
$ \Rightarrow 2(x – 1) = y – 1$
$ \Rightarrow 2x – 2 = y – 1 $
$ \Rightarrow 2 (y - 3) – 2 = y – 1 $
$ \Rightarrow 2y - 6 = y + 1$
$ \Rightarrow y = 7$
Substituting this value of y in equation (1), we get
$ \Rightarrow x = 7 – 3 = 4$
So, the values of x and y are 4 and 7 respectively.

Hence, the required fraction will be $\dfrac{x}{y}$= $\dfrac{4}{7}$ .

Note:
In this question, you may get confused while forming the equations from the given word problem. You may go wrong while solving for the value of x and y from both the equations as we have used the value of x from equation (1) in equation (2) to get the value of y. By naming the equation, we don’t need to find again which equation we have to use.