A fraction becomes $\dfrac{1}{3}$, if 1 is subtracted from both its numerator and denominator. If 3 is added to both the numerator and the denominator, it becomes $\dfrac{1}{2}$. Find the fraction.
Last updated date: 20th Mar 2023
•
Total views: 306k
•
Views today: 3.85k
Answer
306k+ views
Hint- Here, we will be assuming the original fraction which will consist of two unknowns i.e., numerator and denominator. According to the problem, we will obtain two equations in two unknowns and then will use elimination method.
Complete step-by-step answer:
Let us suppose that the original fraction is $\dfrac{x}{y}$ where x is the numerator of the original fraction and y is the denominator of the original fraction.
Now, when 1 is subtracted from both its numerator and denominator of the original fraction, the fraction reduces to $\dfrac{{x - 1}}{{y - 1}}$.
According to problem statement, $\dfrac{{x - 1}}{{y - 1}} = \dfrac{1}{3}$
By applying cross multiplication in the above equation, we get
$
\Rightarrow 3\left( {x - 1} \right) = 1\left( {y - 1} \right) \Rightarrow 3x - 3 = y - 1 \\
\Rightarrow 3x - y - 2 = 0{\text{ }} \to {\text{(1)}} \\
$
Now, when 3 is added to both the numerator and the denominator of the original fraction, the fraction reduces to $\dfrac{{x + 3}}{{y + 3}}$.
According to problem statement, $\dfrac{{x + 3}}{{y + 3}} = \dfrac{1}{2}$
By applying cross multiplication in the above equation, we get
$
\Rightarrow 2\left( {x + 3} \right) = 1\left( {y + 3} \right) \Rightarrow 2x + 6 = y + 3 \\
\Rightarrow 2x - y + 3 = 0{\text{ }} \to {\text{(2)}} \\
$
Following the approach of the Elimination method.
By subtracting equation (2) from equation (1), we get
\[ \Rightarrow 3x - y - 2 - \left( {2x - y + 3} \right) = 0 - 0 \Rightarrow 3x - y - 2 - 2x + y - 3 = 0 \Rightarrow x - 5 = 0 \Rightarrow x = 5\]
By putting \[x = 5\] in equation (1), we get
$ \Rightarrow \left( {3 \times 5} \right) - y - 2 = 0 \Rightarrow 15 - 2 = y \Rightarrow y = 13$
Therefore, the numerator of the original fraction (x) is 5 and the denominator of the original fraction (y) is 13.
Hence, the original fraction is $\dfrac{5}{{13}}$.
Note- In this particular problem, we can also use substitution method instead of elimination method in order to obtain the values of two variables assumed (x and y). Here, we can also verify that the obtained original fraction is correct or not by simply subtracting 1 from both numerator and denominator which gives $\dfrac{4}{{12}} = \dfrac{1}{3}$ and by adding 3 to both numerator and denominator which gives $\dfrac{8}{{16}} = \dfrac{1}{2}$.
Complete step-by-step answer:
Let us suppose that the original fraction is $\dfrac{x}{y}$ where x is the numerator of the original fraction and y is the denominator of the original fraction.
Now, when 1 is subtracted from both its numerator and denominator of the original fraction, the fraction reduces to $\dfrac{{x - 1}}{{y - 1}}$.
According to problem statement, $\dfrac{{x - 1}}{{y - 1}} = \dfrac{1}{3}$
By applying cross multiplication in the above equation, we get
$
\Rightarrow 3\left( {x - 1} \right) = 1\left( {y - 1} \right) \Rightarrow 3x - 3 = y - 1 \\
\Rightarrow 3x - y - 2 = 0{\text{ }} \to {\text{(1)}} \\
$
Now, when 3 is added to both the numerator and the denominator of the original fraction, the fraction reduces to $\dfrac{{x + 3}}{{y + 3}}$.
According to problem statement, $\dfrac{{x + 3}}{{y + 3}} = \dfrac{1}{2}$
By applying cross multiplication in the above equation, we get
$
\Rightarrow 2\left( {x + 3} \right) = 1\left( {y + 3} \right) \Rightarrow 2x + 6 = y + 3 \\
\Rightarrow 2x - y + 3 = 0{\text{ }} \to {\text{(2)}} \\
$
Following the approach of the Elimination method.
By subtracting equation (2) from equation (1), we get
\[ \Rightarrow 3x - y - 2 - \left( {2x - y + 3} \right) = 0 - 0 \Rightarrow 3x - y - 2 - 2x + y - 3 = 0 \Rightarrow x - 5 = 0 \Rightarrow x = 5\]
By putting \[x = 5\] in equation (1), we get
$ \Rightarrow \left( {3 \times 5} \right) - y - 2 = 0 \Rightarrow 15 - 2 = y \Rightarrow y = 13$
Therefore, the numerator of the original fraction (x) is 5 and the denominator of the original fraction (y) is 13.
Hence, the original fraction is $\dfrac{5}{{13}}$.
Note- In this particular problem, we can also use substitution method instead of elimination method in order to obtain the values of two variables assumed (x and y). Here, we can also verify that the obtained original fraction is correct or not by simply subtracting 1 from both numerator and denominator which gives $\dfrac{4}{{12}} = \dfrac{1}{3}$ and by adding 3 to both numerator and denominator which gives $\dfrac{8}{{16}} = \dfrac{1}{2}$.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India
