the denominator of a rational number is greater than its numerator by 4. if 3 subtracted from the numerator and 4 added to the denominator, the new number becomes 2/13. find the original number
2 thoughts on “the denominator of a rational number is greater than its numerator by 4. if 3 subtracted from the numerator and 4 added to the den”
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GiveN :-
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The denominator of a rational number is greater than its numerator by 4. If 3 subtracted from the numerator and 4 added to the denominator, the new number becomes 2/13
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GiveN :-
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To FinD :-
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SolutioN :-
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Let the numerator be x and denominator be 4+x
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[tex] \qquad \quad : \implies \sf\dfrac{(x – 3)}{(4 + x) + 4} = \dfrac{2}{13} [/tex]
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[tex] \qquad \qquad : \implies \sf{ \dfrac{(x – 3)}{(8 + x)} = \dfrac{2}{13} }[/tex]
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[tex] \quad \quad : \implies \sf{13(x – 3) = 2(8 + x)}[/tex]
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[tex] \qquad \quad : \implies \sf{13x – 39 = 16 + 2x}[/tex]
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[tex] \qquad \quad : \implies \sf{13x – 2x = 16 + 39}[/tex]
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[tex] \quad \qquad : \implies \sf{11x = 55}[/tex]
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[tex] \qquad \qquad : \implies \sf{x = \dfrac{ 55}{11} }[/tex]
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[tex] \qquad \qquad : \implies \sf{x = 5}[/tex]
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We also know that :-
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[tex] \qquad \quad \begin{gathered}{ \boxed{ \underline{ \bf{ \red{Numerator = x = 5}}}}} \end{gathered}[/tex]
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[tex] \qquad \qquad \begin {gathered}{ \boxed{ \underline{ \bf{ \red{Denominator = 4 + x = 4 + 5 = 9}}}}} \end{gathered}[/tex]
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[tex] \begin{gathered}{ \boxed{ \underbrace{ \sf{ \blue{Therefore, \: the \: fraction \: is \: \frac{5}{9} }}}}} \end{gathered}[/tex]
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[tex] \\ \underline\red{\rm \large{Answer :- }}[/tex]
[tex] \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \implies \: \dfrac{25}{29} \\ [/tex]
[tex] \\ \underline\red{\rm \large{Solution :- }}[/tex]
Let the numerator be x .
So, the denominator of rational number is greater than its numerator by 4.
Then denominator be (4+x)
Now, According to the question
[tex] \\ \sf \: \: \: \: \: \: \: \: \implies \: \dfrac{(x – 3)}{(4 + x) + 4} = \dfrac{2}{3} \\ [/tex]
[tex] \\ \sf \: \: \: \: \: \: \: \: \implies \: \dfrac{(x – 3)}{(8+ x) } = \dfrac{2}{3} \\ [/tex]
[tex] \\ \sf \: \: \: \: \: \: \: \: \implies \:3 {(x – 3)} = 2{(8+ x)}\\ [/tex]
[tex] \\ \sf \: \: \: \: \: \: \: \: \implies \:3x – 9= 16+ 2x \\ [/tex]
[tex] \\ \sf \: \: \: \: \: \: \: \: \implies \:3x – 2x= 16+ 9 \\ [/tex]
[tex] \\ \sf \: \: \: \: \: \: \: \: \implies \: x= 16+ 9 \\ [/tex]
[tex] \\ \sf \: \: \: \: \: \: \: \: \implies \boxed{ \pink{\frak{\: x= 25}}} \\ [/tex]
Now , We know that
[tex] \\ \sf \: \: \: \: \: Numerator \: = x \: = 25 \\ [/tex]
[tex] \\ \sf \: \: \: \: \: Denominator = 4 + x \\ \: \sf \: \: \: \: \: \:\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4 + 25 \\ \: \sf \: \: \: \: \:\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: = \underline\pink{29} \\ [/tex]
[tex] \\ \: \: \: \: \: \: \: \: \: \: \: \green{ \sf \: \: \: So, \: The \: fraction \: is \: \: \dfrac{25}{29} .}[/tex]