find the area the following figure p.l
Answers
Answer:
Question :-
The denominator of a rational number is greater than its numerator by 3, if numerator is increased by 14 and denominator is decreased by 3 the new number becomes 11 by 4 what is the rational number ?
\large \dag† Answer :-
\begin{gathered}\red\dashrightarrow\underline{\underline{\sf \green{The \: Original \: Number \: is \: \frac{8}{11} }} }\\\end{gathered}
⇢
The Original Number is
11
8
\large \dag† Step by step Explanation :-
Let Numerator and Denominator of Original Rational Number be :
Numerator = x
As Per the question denominator of the rational number is greater than its numerator by 3 so,
Denominator should be = x + 3
Hence,
\begin{gathered}\text{Original Number = } \frac{\text x}{\text x + 3} \\ \end{gathered}
Original Number =
x+3
x
❒ When numerator is increased by 14 and denominator is decreased by 3 :
\begin{gathered} \rm \text{Number Becomes } : \frac{x + 14}{x} \\ \end{gathered}
Number Becomes :
x
x+14
⏩ According To Question :
\begin{gathered} \large \blue \bigstar \: \: \red{ \bf \frac{ x + 14 }{x} = \frac{11}{4} } \\ \end{gathered}
★
x
x+14
=
4
11
\maltese \:✠ On Cross Multiplying ;
\begin{gathered}:\longmapsto \rm11x = 4(x + 14) \\ \end{gathered}
:⟼11x=4(x+14)
\begin{gathered}:\longmapsto \rm 11x = 4x + 56 \\ \end{gathered}
:⟼11x=4x+56
\begin{gathered}:\longmapsto \rm 11x - 4x = 56 \\ \end{gathered}
:⟼11x−4x=56
\begin{gathered}:\longmapsto \rm 7x = 56 \\ \end{gathered}
:⟼7x=56
\begin{gathered}:\longmapsto \rm x = \cancel \frac{56}{7} \\ \end{gathered}
:⟼x=
7
56
\begin{gathered}\purple{ \large :\longmapsto \underline {\boxed{{\bf x = 8} }}} \\ \end{gathered}
:⟼
x=8
\begin{gathered}\blue\dashrightarrow\underline{\underline{\sf \orange{Numerator \: of \: Original \: Number = 8 }} }\\ \end{gathered}
⇢
Numerator of Original Number=8
☆ As The denominator of the rational number is greater than its numerator by 3
\begin{gathered}\rm\therefore \: Denominator = 8 + 3\\ \end{gathered}
∴ Denominator =8 +3
\begin{gathered}\blue\dashrightarrow\underline{\underline{\sf \orange{Denominator \: of \: Original \: Fraction = 11 }} }\\ \end{gathered}
⇢
Denominator of Original Fraction=11
Therefore,
\large\underline{\pink{\underline{\frak{\pmb{ Original \: Fraction = \dfrac{8}{11} }}}}}
Original Fraction =
11
8
Original Fraction =
11
8
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Step-by-step explanation:
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