Deklerasyon ng mga prinsipyo at patakaran ng mga estado
Answers
Answer:
ok
Explanation:
Appropriate Question :-
The denominator of a fraction is 3 more than the numerator. If 2 is added to the numerator and 5 is added to the denominator, the fraction becomes \bf \dfrac{1}{2}
2
1
. Find the fraction.
[Note : The wrong part in the question is the fraction is ½. ]
Given :-
The denominator of a fraction is 3 more than the numerator.
2 is added to the numerator and 5 is added to the denominator.
The fraction becomes \sf \dfrac{1}{2}
2
1
.
To Find :-
What is the fraction.
Solution :-
Let,
\mapsto \rm{\bold{Numerator =\: x}}↦Numerator=x
\mapsto \rm{\bold{Denominator =\: (x + 3)}}↦Denominator=(x+3)
Then, the original fraction will be :
\leadsto \sf \dfrac{Numerator}{Denominator}⇝
Denominator
Numerator
\leadsto \bf{\dfrac{x}{x + 3}}⇝
x+3
x
\pink{\bigstar\: \: \: {\sf\bold{\underline{According\: to\: the\: question\: :-}}}}★
Accordingtothequestion:−
\begin{gathered}\implies \sf \dfrac{Numerator + 2}{Denominator + 5} =\: New\: Fraction\\\end{gathered}
⟹
Denominator+5
Numerator+2
=NewFraction
\implies \sf \dfrac{x + 2}{x + 3 + 5} =\: \dfrac{1}{2}⟹
x+3+5
x+2
=
2
1
\implies \sf \dfrac{x + 2}{x + 8} =\: \dfrac{1}{2}⟹
x+8
x+2
=
2
1
\pink{\bigstar} \: \: {\bf{By\: doing\: cross\: multiplication\: we\: get\: :-}}★Bydoingcrossmultiplicationweget:−
\implies \sf 2(x + 2) =\: 1(x + 8)⟹2(x+2)=1(x+8)
\implies \sf 2x + 4 =\: x + 8⟹2x+4=x+8
\implies \sf 2x - x =\: 8 - 4⟹2x−x=8−4
\implies \sf\bold{\purple{ x =\: 4}}⟹x=4
Hence, the required original fraction is :
\longrightarrow \sf Original\: Fraction =\: \dfrac{x}{x + 3}⟶OriginalFraction=
x+3
x
\longrightarrow \sf Original\: Fraction =\: \dfrac{4}{4 + 3}⟶OriginalFraction=
4+3
4
\longrightarrow \sf\bold{\red{Original\: Fraction =\: \dfrac{4}{7}}}⟶OriginalFraction=
7
4
{\small{\bold{\underline{\therefore\: The\: fraction\: is\: \dfrac{4}{7}\: .}}}}
∴Thefractionis
7
4
.
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VERIFICATION :-
\leadsto \sf \dfrac{x + 2}{x + 3 + 5} =\: \dfrac{1}{2}⇝
x+3+5
x+2
=
2
1
\leadsto \sf \dfrac{x + 2}{x + 8} =\: \dfrac{1}{2}⇝
x+8
x+2
=
2
1
By putting x = 4 we get,
\leadsto \sf \dfrac{4 + 2}{4 + 8} =\: \dfrac{1}{2}⇝
4+8
4+2
=
2
1
\leadsto \sf \dfrac{\cancel{6}}{\cancel{12}} =\: \dfrac{1}{2}⇝
12
6
=
2
1
\leadsto \sf \bold{\green{\dfrac{1}{2} =\: \dfrac{1}{2}}}⇝
2
1
=
2
1
Hence, Verified.