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 \[2:3\]. Determine the fraction.

Answer 1

Hint: At first we’ll assume the values of the numerator and the denominator to be x and y respectively. Then using the given data we’ll create the equations in assumed variables. Solving them we’ll get two linear equations in two variables, then we’ll solve them using the substitution method will get the value of the numerator and the denominator.

Complete step by step Answer:

Given: 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 \[2:3\].
Let, numerator =x and
denominator=y

∴Fraction=​\[\dfrac{{\text{x}}}{{\text{y}}}\]
Given that the sum of numerator and denominator of a fraction is 4 more than twice the numerator,
Using this we get,

\[ \Rightarrow {\text{x + y = 2x + 4}}\]
\[ \Rightarrow {\text{y = x + 4}}...........{\text{(1)}}\]

It is also given that if the numerator and denominator are increased by 3 then they are in ratio \[2:3\],
Using this we get,

\[ \Rightarrow \dfrac{{{\text{x + 3}}}}{{{\text{y + 3}}}}{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}\]
On cross multiplication we get,​

\[ \Rightarrow {\text{3x + 9 = 2y + 6}}\]
\[ \Rightarrow {\text{3x - 2y = - 3}}........(2)\]

Now we substitute the value of y from eq (1) to eq (2), we get,

\[ \Rightarrow {\text{3x - 2(x + 4) = - 3}}\]
\[ \Rightarrow {\text{3x - 2x - 8 = - 3}}\]
\[ \Rightarrow {\text{x = 5}}\]

Now, substituting\[{\text{x = 5}}\]in eq ( 1), we get,

\[ \Rightarrow {\text{y}} = 5 + 4\]
\[ \Rightarrow {\text{y}} = 9\]
Hence, the required fraction is, \[\dfrac{5}{9}\].

Note: We can also solve a pair of linear equation i.e. equation(i) and (ii) by elimination method which is,

2(equation(i)) \[ \Rightarrow 2{\text{y = 2x + 8}}..........{\text{(iii)}}\]
 equation(ii) \[ \Rightarrow {\text{3x - 2y = - 3}}...........{\text{(iv)}}\]

adding equation(iii) and (iv)

\[ \Rightarrow {\text{3x + 2y - 2y = 2x + 8 - 3}}\]
\[ \Rightarrow {\text{3x = 2x + 5}}\]
\[ \Rightarrow {\text{x = 5}}\]

Substituting the value x=5
\[ \Rightarrow {\text{y = x + 4}}\]
\[ \Rightarrow {\text{y = 9}}\]

Therefore, the required fraction is, \[\dfrac{5}{9}\].