Math, asked by soubeyritheg, 1 year ago

The sum of numerator and denominator of a fraction is 8?if 3 is added to both numerator nd denominator the fraction becomes 3/4..find the fraction???

Answers

Answered by mysticd
8
let the numerator = x
denominator =8-x
the fraction =x/8-x ------(1)
if 3 is added to both the fraction=3/4
(x+3)/(8-x+3)=3/4

(x+3)/(11-x)=3/4
4(x+3)=3(11-x)
4x+12=33-3x
4x+3x=33-12
7x= 21

x=21/7
x=3
plug the x=3 in (1)

required fraction =x/(8-x)

=3/(8-3)

=3/5
Answered by TRISHNADEVI
4

 \huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \:   \: SOLUTION \:  \: } \mid}}}}}

 \underline{ \mathfrak{ \:  \: Given, \:  \: }}\\  \\  \text{The sum of the numerator and denominator  = 8} \\ \\    \text{If 3 is added to both numerator and denominator, } \\  \mathtt{then \:  \:  the \:  \:  fraction  \:  \: becomes \:   \:  \: \frac{3}{4} } \\  \\  \underline{ \mathfrak{ \:  \: To  \:  \: find : \mapsto \: }}  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \text{The \:  fraction = ? }

 \underline{ \mathfrak{ \:  \: Suppose, \:  \: }} \\  \\   \:  \:  \:  \:  \:  \:  \: \text{The numerator = x } \\  \text{And, } \\  \:  \:  \:  \:  \:  \:  \:  \text{The denominator = y}  \:  \\  \\    \mathtt{ \therefore \:  \:  The \:  \:  \:  fraction \:  \:  \:  is = \frac{x}{y} }

 \underline{ \bold{ \:  \: A.T.Q.,  \:  \: }} \\  \\   \:  \:  \:  \:  \:  \: \mathtt{x + y = 8 \:  \:  \:  -  -  -  -  -  -  > (1)} \\  \\  \mathtt{And,} \\  \\ \:  \:  \:  \:  \:  \: \mathtt{ \frac{x + 3}{y + 3}  =  \frac{3}{4} } \\   \\ \mathtt{\Longrightarrow \:4( x + 3)  = 3(y + 3) } \\  \\ \mathtt{\Longrightarrow \:4x + 12 = 3y + 9 } \\  \\ \mathtt{\Longrightarrow \: 4x  - 3y = 9 - 12} \\  \\ \mathtt{\Longrightarrow \:4x - 3y =  - 3 \:  \:  \:  -  -  -  -  -  -  > (2) } \\  \\  \mathfrak{Again,} \\  \\ \mathtt{(1) \times 3\Longrightarrow \:3x + 3y  = 24 \:  \:  \:  \:   -  -  -  -  -  > (3)}

 \mathfrak{Now, } \\  \\  \mathtt{(2) + (3) \Rightarrow 7x = 21 } \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \mathtt{\Longrightarrow x = \frac{21}{7} }   \\  \\  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \mathtt{\therefore x = 3} </p><p>

\underline{ \text{ Putting the value of x in eq. (1) , we get}}  \\  \\  \mathtt{(1) \Rightarrow 3 + y = 8} \\  \\   \:  \:  \:  \:  \mathtt{\Longrightarrow \: y = 8 - 3} \\  \\  \:   \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \mathtt{ \therefore\:  \: y = 5}

\mathtt{ \therefore \:  \:  The \:  \:  \:  fraction \:  \:  \:  is = \frac{x}{y} } \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \red{\mathtt{ = \frac{3}{5}  }}

 \huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \:   \: VERICATION \:  \: } \mid}}}}}

  \underline{\text{ \: Numerator = 3  \:  \: and \:  \: Denominator = 5 \: } } \\  \\   :  \rightsquigarrow \: \text{Sum of the numerator and denominator = 3 + 5 = 8 } \\  \\  \mathfrak{Again,} \\  \\    :  \rightsquigarrow \:  \text{When 3 is added to both the numerator and denominator,} \\  \mathtt{ then \:  \:  the \:  \:  fraction \:  \:  becomes  =  \frac{3 + 3}{5  + 3 } =  \frac{6}{8} =  \frac{3}{4} }

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