Question

The denominator of a fraction is 1 more than its numerator. If 1 is deducted from both the numerator and the denominator, the fraction becomes equivalent to \[0.5\], then the fraction is
(a) \[\dfrac{3}{4}\]
(b) \[\dfrac{4}{5}\]
(c) \[\dfrac{2}{3}\]
(d) \[\dfrac{7}{8}\]

Answer 1

Hint:
Here, we need to find the fraction. Let the numerator and the denominator of the fraction be \[x\] and \[y\] respectively. Using the given information, we can form two linear equations in two variables. We will solve these equations to find the values of \[x\] and \[y\], and use these values to find the fraction.

Complete step by step solution:
Let the numerator and the denominator of the fraction be \[x\] and \[y\] respectively.
Therefore, we get the fraction as \[\dfrac{x}{y}\].
Now, it is given that the denominator is 1 more than its numerator.
Thus, we get
\[ \Rightarrow y = x + 1 \ldots \ldots \ldots \left( 1 \right)\]
The fraction obtained by decreasing the numerator and denominator by 1 is equivalent to \[0.5\].
Thus, we get
\[ \Rightarrow \dfrac{{x - 1}}{{y - 1}} = 0.5\]
Multiplying both sides of the equation by \[y - 1\], we get
$ \Rightarrow \left( {\dfrac{{x - 1}}{{y - 1}}} \right)\left( {y - 1} \right) = 0.5\left( {y - 1} \right) \\
   \Rightarrow x - 1 = 0.5y - 0.5 \\ $
Rewriting the equation, we get
\[ \Rightarrow x - 0.5y = - 0.5 + 1\]
Subtracting the terms, we get
\[ \Rightarrow x - 0.5y = 0.5 \ldots \ldots \ldots \left( 2 \right)\]
We can observe that the equations \[\left( 1 \right)\] and \[\left( 2 \right)\] are a pair of linear equations in two variables.
We will solve the equations to find the values of \[x\] and \[y\].
Substituting \[y = x + 1\] from equation \[\left( 1 \right)\] in equation \[\left( 2 \right)\], we get
\[ \Rightarrow x - 0.5\left( {x + 1} \right) = 0.5\]
Multiplying the terms, we get
\[ \Rightarrow x - 0.5x - 0.5 = 0.5\]
Subtracting the like terms, we get
\[ \Rightarrow 0.5x - 0.5 = 0.5\]
Adding \[0.5\] to both sides, we get
$ \Rightarrow 0.5x - 0.5 + 0.5 = 0.5 + 0.5 \\
   \Rightarrow 0.5x = 1 \\ $
Rewriting \[0.5\] as \[\dfrac{1}{2}\], we get
 $ \Rightarrow \dfrac{1}{2}x = 1 \\
   \Rightarrow \dfrac{x}{2} = 1 \\ $
Multiplying both sides of the equation by 2, we get
  $ \Rightarrow \dfrac{x}{2} \times 2 = 1 \times 2 \\
  \therefore x = 2 \\ $
Substituting \[x = 2\] in the equation \[\left( 1 \right)\], we get
\[ \Rightarrow y = 2 + 1\]
Adding 2 and 1, we get
\[\therefore y = 3\]
Now, we will use the values of \[x\] and \[y\] to find the fraction.
Substituting \[x = 2\] and \[y = 3\] in the expression \[\dfrac{x}{y}\], we get
\[ \Rightarrow \]Fraction \[ = \dfrac{2}{3}\]

Therefore, the fraction is \[\dfrac{2}{3}\]. The correct option is option (c).

Note:
We have formed two linear equations in two variables and simplified them to find the fraction. A linear equation in two variables is an equation of the form \[ax + by + c = 0\], where \[a\] and \[b\]are not equal to 0. For example, \[2x - 7y = 4\] is a linear equation in two variables.
We can verify our answer by using the given information.
The numerator of the fraction \[\dfrac{2}{3}\] is 2, and the denominator of the fraction \[\dfrac{2}{3}\] is 3.
Thus, the denominator is 1 more than the numerator.
If 1 is deducted from both the numerator and the denominator of the fraction \[\dfrac{2}{3}\], the fraction becomes \[\dfrac{{2 - 1}}{{3 - 1}} = \dfrac{1}{2}\], which is equivalent to \[0.5\].
Hence, we have verified our answer.