Question

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we add 1 to the denominator. What is the fraction?

Answer 1

Answer:

3/5

Step-by-step explanation:

numerator =x

denominator =y

(x+1)/(y-1) =1

x+1=y-1

y-x=2

from second statement x/(y+1)= 1/2

2x=y+1

y-2x=-1

solving two equation we get x=3,y=5

Answer 2

Given:

• If we add 1 to the Numerator and Subtract 1 from the Denominator, the fraction reduces to 1. & it becomes ½ if we add 1 to the Denominator.

Need to find:

• The fraction?

⠀

❍ Let's Consider that the Numerator and Denominator of the fraction be x and y respectively.

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$\underline{\bigstar\:\boldsymbol{According\;to\;the\; Question\; :}}$

⠀⠀

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If we add 1 to the Numerator and Subtract 1 from the Denominator, the fraction reduces to 1.

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$\begin{gathered}:\implies\sf \Bigg(\dfrac{Numerator + 1}{Denominator - 1}\Bigg) = 1\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf \Bigg(\dfrac{x + 1}{y - 1}\Bigg)= 1\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf x + 1 = y -1 \\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf x - y = - 1 - 1\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf x - y = -2\qquad\qquad\quad\sf\Bigg\lgroup eq^{n}\;(1)\Bigg\rgroup\\\\\\\end{gathered}$

Also,

⠀

The fraction becomes ½ if we add 1 to the Denominator.

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$\begin{gathered}:\implies\sf \Bigg(\dfrac{Numerator}{Denominator + 1}\Bigg) = \Bigg(\dfrac{1}{2}\Bigg)\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf \Bigg(\dfrac{x}{y + 1}\Bigg) = \Bigg(\dfrac{1}{2}\Bigg)\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 2x = y + 1\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 2x - y = 1\qquad\qquad\quad\sf\Bigg\lgroup eq^{n}\;(2)\Bigg\rgroup\\\\\end{gathered}$

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⠀

$\begin{gathered}\underline{\bf{\dag} \:\mathfrak{So, Using\;eq^n\;(1)\;\&\;eq^{n}\:(2)\: :}}\\\\\end{gathered}$

$\begin{gathered}:\implies\sf x - y = -2\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 2x - y = 1\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf \cancel{-}\;x = \cancel{-}\;3\\\\\\\end{gathered}$

$\begin{gathered}:\implies\underline{\boxed{\pmb{\frak{\pink{x = 3}}}}}\;\bigstar\\\\\end{gathered}$

⠀

⠀⠀

$\begin{gathered}\underline{\bf{\dag} \:\mathfrak{Using\;eq^n\;(1)\: :}}\\\\\end{gathered}$

⠀⠀⠀⠀

$\begin{gathered}:\implies\sf x - y = - 2\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf3 - y = -2\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf 3 - y = -2\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf -y = -2 - 3\\\\\\\end{gathered}$

$\begin{gathered}:\implies\sf \cancel{-}\;y = \cancel{-}\;5\\\\\\\end{gathered}$

$\begin{gathered}:\implies\underline{\boxed{\pmb{\frak{\pink{y = 5}}}}}\;\bigstar\\\\\end{gathered}$

Therefore,

• Numerator of the fraction, x = 3
• Denominator of the fraction, y = 5

⠀

$\therefore{\underline{\sf{Hence,\;the\; required\; fraction\;is\;{\purple{\pmb{\sf{\dfrac{3}{5}}}}.}}}}$