Question

The sum of the numerator and denominator of a fraction is 17. On adding 3 to the numerator and subtracting 3 from the denominator we get the reciprocal of the fraction.Find the fraction!​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{\orange{Question}}}}}}$

The sum of the numerator and denominator of a fraction is 17. On adding 3 to the numerator and subtracting 3 from the denominator we get the reciprocal of the fraction . Find the fraction ?

$\Large{\underline{\underline{\mathfrak{\bf{\orange{Solution}}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

• The sum of the numerator and denominator of a fraction is 17.
• adding 3 to the numerator and subtracting 3 from the denominator we get the reciprocal of the fraction .

$\Large{\underline{\mathfrak{\bf{\pink{Find}}}}}$

• Fraction=?

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

Let,

• Numerator = x
• Denominator = y

A/c to question,

( The sum of the numerator and denominator of a fraction is 17 )

$:\mapsto\sf{\:(x + y )\:=\:17......(1)}$

Again,

(adding 3 to the numerator and subtracting 3 from the denominator we get the reciprocal of the fraction)

$:\mapsto\sf{\:\dfrac{(x+3)}{(y-3)}\:=\:\dfrac{y}{x}} \\ \\ :\mapsto\sf{\:x(x+3)\:=\:y(y-3)} \\ \\ :\mapsto\sf{\:x^2+3x-y^2+3y\:=\:0} \\ \\ :\mapsto\sf{\:x^2-y^2+3(x+y)\:=\:0} \\ \\ \small\sf{\green{\:\:\:\:\:keep\:value\:by\:equ(1)}} \\ \\ :\mapsto\sf{\:x^2-y^2+3(17)\:=\:0} \\ \\ :\mapsto\sf{\:(x^2-y^2)\:=\:-51} \\ \\ \small\sf{\green{\:\:\:\:\:\:(a^2-b^2)\:=\:(a+b)(a-b)}} \\ \\ :\mapsto\sf{\:(x+y)(x-y)\:=\:-51} \\ \\ \small\sf{\green{\:\:\:again,\:keep\:value\:by\:equ(1)}} \\ \\ :\mapsto\sf{\:(x-y)\:=\:\cancel{\dfrac{-51}{17}}} \\ \\ :\mapsto\sf{\:(x-y)\:=\:-3.......(2)}$

Add equation (1) and equ(2),

$:\mapsto\sf{\:2x\:=\:14} \\ \\ :\mapsto\sf{\:x\:=\:\cancel{\dfrac{14}{2}}} \\ \\ :\mapsto\sf{\red{\:x\:=\:7}}$

keep value of x in equ(1),

$:\mapsto\sf{\:(7+y)\:=\:17} \\ \\ :\mapsto\sf{\:y\:=\:17-7} \\ \\ :\mapsto\sf{\red{\:y\:=\:10}}$

$\Large{\underline{\mathfrak{\bf{\pink{Thus}}}}}$

$:\mapsto\textsf{\orange{\:value of x = 7}} \\ \\ :\mapsto\textsf{\orange{\:value of y\:=\:10}}$

$\Large{\underline{\mathfrak{\bf{\pink{Hence}}}}}$

$:\mapsto\sf{\blue{\:fraction\:will\:be \:=\:\dfrac{7}{10}}}$

Answer 2

$\huge\bold\green{\underline{Solution:}}$

$\bold\red{\underline{Let:}}$

The numerator be N

The denominator be D

The fraction be N/D

$\large{\underline{\red{\rm{A.T.Q:}}}}$

$\sf{The\:sum\:of\: numerator\:and\: denominator\:is\:17}$

$\sf{\dashrightarrow N+D=17------(i)}$

$\sf{on\: adding\:3\:to\:the\: numerator\:and\: subtracting}$

$\sf{from\: denominator\:we\:get\: reciprocal\:of\:the\: fraction}$

$\sf{\dashrightarrow \frac{N+3}{D-3}=\frac{D}{N}}$

$\sf{\dashrightarrow N^2+3N=D^2-3D}$

$\sf{\dashrightarrow N^2-D^2+3N+3D=0}$

$\sf{\dashrightarrow N^2-D^2+3(N+D)=0}$

$\sf{\dashrightarrow N^2-D^2+3*17=0}$

$\sf{\dashrightarrow N^2-D^2+51=0}$

$\sf{\dashrightarrow N^2-D^2=-51}$

$\sf{Here\: using\: identity}$

$\sf{a^2-b^2=(a+b)(a-b)}$

$\sf{\dashrightarrow (N+D)(N-D)=-51}$

$\sf{\dashrightarrow 17*(N-D)=-51}$

$\sf{\dashrightarrow (N-D)=\frac{-51}{17}}$

$\sf{\dashrightarrow N-D=-3----(ii)}$

$\sf{by\:solving\:equation\:I\:and\:II}$

$\sf{\dashrightarrow N+D=17}$

$\sf{\dashrightarrow N-D=-3}$

$\sf{\dashrightarrow \cancel{2}N=\cancel{14}}$

$\sf{\dashrightarrow N=7}$

$\sf{putting\:the\: value\:of\:N=7\:in\:eq\:1}$

$\sf{\dashrightarrow N+D=17}$

$\sf{\dashrightarrow 7+D=17}$

$\sf{\dashrightarrow D=17-7}$

$\sf{\dashrightarrow D=10}$

Hence,

The numerator=7

The denominator=10

The fraction=7/10

$\large{\underline{\red{\rm{AnswEr:\frac{7}{10}}}}}$