Question

The sum of the numerator and denominator of a fraction is 17. On adding 3 to the numerator
and subtracting 3 from the denominator we get the reciprocal of the fraction. Find the fraction.

​

Answer 1

Given :

• The sum of the numerator and denominator of a fraction is 17.

• On adding 3 to the numerator and subtracting 3 from the denominator we get the reciprocal of the fraction.

To Find :

Required Fraction = ?

Answer :

• Let Numerator be x
• Denomintaor be y
• Required fraction = Numerator/Denomintaor = x/y

$\bigstar \: \:\displaystyle \underline{\textsf{According to Question now :}}$

It is given that, The sum of the numerator and denominator of a fraction is 17 :]

$:\implies \sf x + y = 17 \qquad\Bigg\lgroup\textsf{\textbf{Equation (I)}}\Bigg\rgroup$

It is also given that, On adding 3 to the numerator

and subtracting 3 from the denominator we get the reciprocal of the fraction :]

$:\implies \sf \dfrac{x + 3}{y - 3} = \dfrac{y}{x} \qquad\Bigg\lgroup\textsf{\textbf{Equation II}}\Bigg\rgroup$

Now, by cross multiplying both sides we get :

$:\implies \sf x(x + 3) = y(y - 3)$

$:\implies \sf {x}^{2} + 3x = {y}^{2} - 3y$

$:\implies \sf {x}^{2} + 3x - {y}^{2} + 3y = 0$

$:\implies \sf {x}^{2} - {y}^{2} + 3x + 3y = 0$

$:\implies \sf {x}^{2} - {y}^{2} + 3(x + y )= 0$

$:\implies \sf {x}^{2} - {y}^{2} + 3(17)= 0 \qquad\Bigg\lgroup\textsf{\textbf{Taking 17 from Equation (I)}}\Bigg\rgroup$

$:\implies \sf {x}^{2} - {y}^{2} + 51= 0$

$:\implies \sf {x}^{2} - {y}^{2} = - 51$

By using identity a² - b² = (a + b) (a - b) we get :

$:\implies \sf ( x + y) (x - y ) = - 51$

$:\implies \sf 17 \times x - y = - 51$

$:\implies \sf x - y = \dfrac{ - 51}{17}$

$:\implies \sf x - y = - 3\qquad\Bigg\lgroup\textsf{\textbf{Equation II}}\Bigg\rgroup$

By solving equation (I) and equation (II) we get :

$:\implies \sf 2x = 14$

$:\implies \sf x = \dfrac{14}{2}$

$:\implies \underline{ \boxed{\sf x = 7}}$

Now, Substituting the value of x = 7 in equation (I) we get :

$\dashrightarrow\:\:\sf 7 + y = 17$

$\dashrightarrow\:\:\sf y = 17 - 7$

$\dashrightarrow\:\: \underline{ \boxed{\sf y = 10 }}$

$\bigstar\:\:\underline{\textsf{Required Fraction :}}$

$:\implies \sf Required \: Fraction= \dfrac{Numerator}{Denomintaor}$

$:\implies \sf Required \: Fraction= \dfrac{x}{y}$

$:\implies \underline{ \boxed{ \sf Required \: Fraction= \dfrac{7}{10}}}$

$\therefore\:\underline{\textsf{Required Fraction is \textbf{ \dfrac{\text {7}}{\text {10}} }}}.$

Answer 2

$\large{\underline{\bf{Given:-}}}$

✞Sum if Numerator and Denominator = 17

✞After adding 3 to the numerator and subtracting 3 from the denominator we get the reciprocal of the fraction.

$\large{\underline{\bf{Find:-}}}$

✟What will be the Fraction

$\large{\underline{\bf{Solution:-}}}$

Let,

➙Numerator = x

➙Denominator = y

Now, It is said that Sum of numerator and denominator is 17

➮x + y = 17.................➊

Now, It is written that after adding 3 on numerator and subtracting 3 from denomiator fraction becomes reciprocal of original fraction.

➮$\sf \dfrac{x+3}{y-3}=\dfrac{y}{x}$..................➋

Taking Eq. ➊

⇨x + y = 17

⇨x = 17-y

Using this value in eq➋

➤$\sf \dfrac{x+3}{y-3}=\dfrac{y}{x}$

➤$\sf \dfrac{17-y+3}{y-3}=\dfrac{y}{17-y}$

➤$\sf \dfrac{20-y}{y-3}=\dfrac{y}{17-y}$

$\sf \bigstar Corss-multiplication\bigstar$

➤$\sf 20-y(17-y)=y(y-3)$

➤$\sf 340-17y-20y+y^2=y^2 - 3y$

➤$\sf 340-37y+\not{y^2}=\not{y^2} - 3y$

➤$\sf 340-37y= - 3y$

➤$\sf 340= - 3y+37y$

➤$\sf 340= 34y$

➤$\sf \dfrac{340}{34}= y$

➤$\sf 10 =y$

$\small{\sf\therefore y = 10\red\bigstar}$

Substituting Value of y in eq➊

➪x + y = 17

➪x + 10 = 17

➪x = 17-10

➪x = 7

$\small{\sf\therefore x = 7 \pink\bigstar}$

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Thus,

▶Original Fraction = $\bf{\dfrac{x}{y} = \dfrac{7}{10}}$