Question

the denominator of a rational number is greater than its numerator by 4 if numerator is increased by 11 and the denominator is decreased by 1 the new number becomes 7 by 3 find the original number​

Answer 1

Answer:

$let \: the \: original \: faction \: be \: \frac{x}{y} \\ \\ as \: per \: the \: question \\ \\ = > x + 4 = y \\ = > x - y = - 4................(1) \\\\and\\ \\ = > \frac{x + 11}{y - 1} = \frac{7}{3} \\ \\ = > 3x + 33 = 7y - 7 \\ \\ = > 3x - 7y = - 40............(2) \\ \\ solving \: (1 ) \: and \: (2) \\ \\ 3x - 3y = - 12 \\ 3x - 7y = - 40 \\ ......................... \\ = > 4y = 28 \\ = > y = 7 \\ \\ from \: eq \:(1) \\ = > x = - 4 + y \\ = > x = - 4 + 7 \\ = > x = 3 \\ \\ therefore \: he \: required \: fraction \: is \: \frac{3}{7}$

Answer 2

Given:

• The denominator of a rational number is greater than it's numerator by 4. If the numerator is increased by 11 and the Denominator is decreased by 1, the new number becomes ⁷⁄₃.

To find:

•  The Original number

Solution:⠀⠀

Let's say, that the numerator of the fraction be x. Then, denominator of the fraction be (x + 4) respectively.

⠀

$\begin{gathered}\sf \orange\bigstar \: \boxed{\sf{According \;to\;the\; Question\; :}}\end{gathered}$

If the numerator is increased by 11 and the Denominator is decreased by 1, the new number becomes ⁷⁄₃.

⠀⠀

$\begin{gathered} : \implies\sf\Bigg\{\dfrac{x + 11}{x + 4 - 1}\Bigg\} = \Bigg\{\dfrac{7}{3}\Bigg\}\\\\ : \implies\sf\Bigg\{\dfrac{x + 11}{x + 3}\Bigg\} = \Bigg\{\dfrac{7}{3}\Bigg\}\\\\ : \implies\sf 3\Big\{x + 11\Big\}=7\Big\{x + 3\Big\}\\\\ : \implies\sf 3x + 33 = 7x + 21 \\\\ : \implies\sf 3x - 7x = 21 - 33\\\\ : \implies\sf -4x = -12\\\\\twoheadrightarrow\sf x = \cancel\dfrac{-12}{-4}\\\\ : \implies\underline{\boxed{{\frak{x = 3}}}}\; \pink\bigstar\end{gathered}$

Therefore,

• Numerator of the fraction, x = 3
• Denominator of the fraction, (x + 4) = (3 + 4) = 7 

$\boxed{\sf{Hence,\: the \;original\; number\; is\;\bf{\dfrac{3}{7}}}}.$

Thank you!!!

@itzshivani