Question

A numerator of the fraction is 6 less then the denominator if 1 is added to both numerator and denominator the fraction become 1/2 find the fraction

Answer 1

let the required fraction be ' y / x '

let the denominator of fraction = x

Given that the numerator of the fraction is 6 less than the denominator of that fraction.

so then ,the numerator y = x - 6
_______________________________
given that if 1 is added to both numerator and denominator then the fraction become 1 /2

new numerator y = x - 6 + 1 = ( x - 5 )

new denominator = ( x + 1 )

according to this statement :

( x - 5 ) / ( x + 1 ) = 1 / 2

2 ( x - 5 ) = 1 ( x + 1 )

2 x - 10 = x + 1

2 x - x = 1 + 10

x = 11

denominator x = 11

numerator y = x - 6 = 11 - 6 = 5

therefore , required fraction

= y / x = 5 / 11
_______________________________
Your Answer : fraction = 5/ 11
_______________________________

Answer 2

Given:

• The numerator of a fraction is 6 less than the denominator .
• If 1 is added to both, the numerator and denominator the fraction becomes 1/2 .

To Find:

• Original Fraction .

Solution:

$:\implies\sf{Let\:Denominator\:be=x}$

• As Given that The numerator of a fraction is 6 less than the denominator .

 So,

$:\implies\sf{Numerator=x-6}$

Now,

• If 1 is added to both, the numerator and denominator the fraction becomes 1/2

$:\implies \sf{Numerator=x-6+1=x-5}$

$:\implies \sf{Denominator=x+1}$

$\boxed{\boxed{\pmb{\sf{\purple{According\: to\: the\: question}}}}}$

$:\implies \sf{\dfrac{x-5}{x+1}=\dfrac{1}{2}}$

$:\implies \sf{2(x-5)=1(x+1)}$

$:\implies \sf{2x-10=x+1}$

$:\implies \sf{2x-x=1+10}$

$:\implies \underline{\boxed{\frak{x=11}}} \: \blue{ \bigstar}$

• Hence, the value of x is 11 .

Therefore :

$:\implies \sf{Numerator=11-6}$

$:\implies \sf{5}$

$:\implies \sf{Denominator=x}$

$:\implies \sf{11}$

$:\implies \underline{\boxed{\frak{ \frac{5}{11} }}} \: \pink{ \bigstar}$

• The Fraction is ${\bf{ \frac{5}{11}} }$ .