Question

8. The numerator of a fraction is 1 less than denominator. If the denominator is added to the
numerator and the mumerater is subtracted from the denominator, we get 11. Find the fraction​

Answer 1

Given :-  

• The numerator of a fraction is 1 less than denominator.  

Condition :-  

• If the denominator is added to the numerator and the numerator is subtracted from the denominator, we get 11.  

To Find :-  

• Find the fraction

Solution :-  

~Let the denominator of the fraction be ‘ x ’ then the numerator of the fraction will be ‘ x - 1 ’

$\sf \bullet \;\; \dfrac{x-1}{x}$  

ATQ ::  

$\sf \implies \dfrac{ x -1 + ( x ) }{ x - ( x - 1 ) } = 11$

$\sf \implies \dfrac{ x + x -1 }{ x - x + 1 } = 11$

$\sf \implies \dfrac{ 2x -1 }{1} = 11$

$\sf \implies 2x - 1 = 11 \times 1$

$\sf \implies 2x -1 = 11$

$\sf \implies 2x = 11 + 1$

$\sf \implies 2x = 12$

 

$\sf \implies x = \dfrac{12}{2}$

$\sf \implies x = 6$

_____________

Therefore ,

the fraction will be  :

$\sf \bullet \;\; \dfrac{5}{6}$

_____________

Verification :-  

~Let’s put the value of x in the given equation to check the answer.  

$\sf \implies \dfrac{ 6 - 1 + ( 6 ) }{ 6 - ( 6 -1 )} = 11$

 

$\sf \implies \dfrac{ 12 -1 }{ 6 - 6 + 1 } = 11$

$\sf \implies \dfrac{11}{1} = 11$

 

$\sf \implies 11 = 11 \times 1$

$\sf \implies 11 = 11$

LHS = RHS  

Hence, the answer is correct

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Answer 2

Given: The numerator of a fraction is 1 less than denominator. If the denominator is added to the

numerator and the numerator is subtracted from the denominator, we get 11.

Need To Find: The fraction .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's consider the denominator be x

Given that,

⠀⠀⠀⠀⠀The numerator of a fraction is 1 less than denominator.

Then ,

$\therefore$ The Numerator is x - 1 .

Then ,

$\qquad \quad\implies \sf{\bf{Original \:Fraction \:: \dfrac{x-1}{x} }}$

Given that ,

⠀⠀⠀⠀⠀If the denominator is added to the numerator and the numerator is subtracted from the denominator, we get 11.

⠀⠀⠀⠀⠀⠀$\underline {\sf{\bf{\star\:According \: To\: The \: Question \: \::}}}$

$\qquad \quad:\implies \sf{\bf{\Bigg( \dfrac{x-1+(x)}{x -(x+1)} =11\Bigg) \qquad\quad\qquad ..Eq.\:1}}$

⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{ \dfrac{x-1+(x)}{x+(x-1)} =11}$

⠀⠀⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{ \dfrac{x-1+x}{x+x-1} =11}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{\dfrac{x+x-1}{1}=11}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{ \dfrac{2x-1}{1}=11}$

By Cross- Multiplication :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{ 2x - 1=11}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{ 2x =11+1}$

⠀⠀⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{ 2x =12}$

⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{ x =\dfrac{\cancel{12}}{\cancel{2}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  x = 6 \: }}}}$

Therefore,

• The Numerator of the Fraction is (x - 1 ) = 6 - 1 = 5

• The Denominator of the Fraction is x = 6

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore\:{\pink{ \mathrm {  The\:Fraction\:is\:\dfrac{5}{6}}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad\quad\boxed{\bf{\mid{\overline{\underline{\blue{\bigstar\: Verification \: :}}}}}\mid}$

⠀⠀⠀⠀⠀⠀$\underline {\sf{\bf{\star\:Now \: By \: Substituting \: the \: Found \: Values \:in\:Eq.\:1:}}}$

$\qquad \quad:\implies \sf{\bf{\Bigg( \dfrac{x -1+(x)}{x+(x-1)}=11\Bigg) \qquad\quad\qquad ..Eq.\:1}}$

⠀⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{ x =6\:\quad \longrightarrow Putting \:in\:Eq.\:1}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{\dfrac{6-1+(6)}{6 - (6+1)}=11}$

⠀⠀⠀⠀⠀⠀⠀$\qquad \quad:\implies \sf{\dfrac{12 -1}{6-6+1}=11}$

⠀⠀⠀⠀$\qquad \quad:\implies \sf{\dfrac{11}{1}=11}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  11 = 11 \: }}}}$

⠀⠀⠀⠀⠀ $\therefore \sf{\bf{ Hence \;Verified \;}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀