The denominator of a fraction is 3 more than numerator. If 2is added to the numerator and 5 is added to thr drnominator then the fraction becomes 1/2. Finf the fraction
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The denominator of a fraction is 3 more than numerator. If 2is added to the numerator and 5 is added to thr drnominator then the fraction becomes 1/2. Finf the fraction
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Given :
To Find :
Solution :
Let the required fraction be [tex]\sf{\dfrac{x}{y}}[/tex] . So that numerator is "x" and denominator is "y"
We are given that denominator of fraction is 3 more than numerator this mean , y = x + 3 ...........(1)
Now numerator and denominator when 2 and 5 are added are , x + 2 and y + 5.
We are given that the fraction when numerator and denominator are added with 2 and 5 as 1/2. So ,
[tex] \\ : \implies \sf \: \dfrac{x + 2}{y + 5} = \dfrac{1}{2} \\ \\ [/tex]
From eq(1) ,
[tex] \\ : \implies \sf \: \dfrac{x + 2}{x + 3 + 5} = \dfrac{1}{2} \\ \\ [/tex]
[tex] \\ : \implies \sf \: \dfrac{x + 2}{x + 8} = \dfrac{1}{2} \\ \\ [/tex]
[tex] \\ : \implies \sf \: 2(x + 2) = 1(x + 8) \\ \\ [/tex]
[tex] \\ : \implies \sf \: 2x + 4 = x + 8 \\ \\ [/tex]
[tex] \\ : \implies \sf \: 2x - x = 8 - 4 \\ \\ [/tex]
[tex] \\ : \implies {\underline{\boxed{\pink{ \mathfrak{x = 4}}}}} \: \bigstar \\ \\ [/tex]
Now , From eq(1) ;
[tex] \\ : \implies \sf \: y = 4 + 3 \\ \\ [/tex]
[tex] \\ : \implies{\underline{\boxed{\pink{\mathfrak{y = 7}}}}} \: \bigstar \\ \\ [/tex]
Now the fraction is [tex]\sf{\dfrac{x}{y}}[/tex] = [tex]\sf{\dfrac{4}{7}}[/tex]
Hence ,
Verified answer
Answer:
Given :-
To Find :-
Solution :-
Let, the numerator be x
Then, the fraction will be [tex]\sf\dfrac{x}{x + 3}[/tex]
According to the question,
⇒ [tex]\sf\dfrac{x + 2}{x + 8}[/tex] = [tex]\sf\dfrac{1}{2}[/tex]
By doing cross multiplication we get,
⇒ 2(x + 2) = x + 8
⇒ 2x + 4 = x + 8
⇒ 2x - x = 8 - 4
➠ x = 4
Hence, the required fraction will be,
↦ [tex]\sf\dfrac{x}{x + 3}[/tex]
↦ [tex]\sf\dfrac{4}{4 + 3}[/tex]
➦ [tex]\sf\dfrac{4}{7}[/tex]
[tex]\therefore[/tex] The fraction is [tex]\sf\bold{\boxed{\large{\dfrac{4}{7}}}}[/tex] .
[tex]\\[/tex]
Let's Verify :-
↬ [tex]\sf\dfrac{x + 2}{x + 8}[/tex] = [tex]\sf\dfrac{1}{2}[/tex]
Put x = 4 we get,
↬ [tex]\sf\dfrac{4 + 2}{4 + 8}[/tex] = [tex]\sf\dfrac{1}{2}[/tex]
↬ [tex]\sf\dfrac{6}{12}[/tex] = [tex]\sf\dfrac{1}{2}[/tex]
↬ [tex]\sf\dfrac{\cancel{6}}{\cancel{12}}[/tex] = [tex]\sf\dfrac{1}{2}[/tex]
↬ [tex]\sf\dfrac{1}{2}[/tex] = [tex]\sf\dfrac{1}{2}[/tex]
➠ LHS = RHS
Hence, Verified ✔