Question

The numerator of a fraction is 4 less than the denominator .If 1
is added to the both the numerator and the denominator it
becomes Find the fraction​

Answer 1

Answer:

Given:

• The numerator of a fraction is 4 less than the denominator.
• If 1  is added to the both the numerator and the denominator it  becomes 1/2.

To Find:

• Fraction.

Let denominator be x.

So, it is given that numerator is 4 less than denominator, Numerator = x - 4.

Now, according to question, if 1  is added to the both the numerator and the denominator it  becomes 1/2. So,

$\rm{\implies \dfrac{x-4+1}{x+1}=\dfrac{1}{2}}$

$\rm{\implies \dfrac{x-3}{x+1}=\dfrac{1}{2}}$

Now, by cross multiply we will find the value of x,

$\rm{\implies 2(x-3)=1(x+1)}$

$\rm{\implies 2x-6=x+1}$

$\rm{\implies 2x-x=6+1}$

$\rm{\implies x=7}$

Therefore, Denominator = 7

Numerator = 7 - 4 = 3.

${\boxed{\boxed{\rm{Hence,\;fraction\;be\;\;\dfrac{3}{7}}}}}$

Answer 2

$\bold{\underline{\underline{\huge{\mathfrak{AnsWer:}}}}}$

$\bold{\tt{\blue{Fraction\:}}}$ = $\bold{\tt{\red{\dfrac{3}{7}}}}$

$\bold{\underline{\underline{\large{\mathfrak{StEp\:by\:stEp\:explanation:}}}}}$

$\bold{\underline{\underline{\red{\tt{GiVeN:}}}}}$

• The numerator of a fraction is 4 less than the denominator.
• If 1 is added to the both the numerator and the denominator it becomes $\sf{\dfrac{1}{2}}$

$\bold{\underline{\underline{\blue{\tt{To\:FiNd:}}}}}$

• The Fraction

$\bold{\underline{\underline{\pink{\tt{SoLuTioN :}}}}}$

Let the numerator of the fraction be x.

Let the denominator of the fraction be y.

Fraction :

• $\sf{\dfrac{x}{y}}$

$\bold{\underline{\underline{\green{\tt{As\:per\:the\:first\:condition:}}}}}$

• The numerator of a fraction is 4 less than the denominator.

$\longrightarrow$ $\tt{x\:=\:y\:-\:4}$

$\longrightarrow$ $\tt{x-y=-4\:--->(1)}$

$\bold{\underline{\underline{\green{\tt{As\:per\:the\:second\:condition:}}}}}$

• If 1 is added to the both the numerator and the denominator it becomes $\sf{\dfrac{1}{2}}$

$\sf{\therefore{Numerator\:=\:x+1}}$

$\sf{\therefore{Denominator\:=\:y+1}}$

$\sf{\therefore{Fraction\:={\dfrac{1}{2}}}}$

$\longrightarrow$ $\tt{\dfrac{x+1}{y+1}}$ = $\tt{\dfrac{1}{2}}$

Cross multiplying,

$\longrightarrow$ $\tt{2(x+1)=1(y+1)}$

$\longrightarrow$$\tt{2x+2\:=\:y+1}$

$\longrightarrow$ $\tt{2x-y=1-2}$

$\longrightarrow$ $\tt{2x-y=-1\:---->(2)}$

Solve equation 1 and equation 2 simultaneously by elimination method.

Subtract equation (2) from equation (1)

$\longrightarrow$ $\tt{2x-y-(x-y)\:=\:-1-(-4)}$

$\longrightarrow$ $\tt{2x-y-x+y=-1+4}$

$\longrightarrow$ $\tt{2x-x=3}$

$\longrightarrow$ $\tt{x=3}$

Substitute x = 3 in equation (1),

$\longrightarrow$ $\tt{x-y=-4\:--->(1)}$

$\longrightarrow$ $\tt{3\:-y\:=\:-4}$

$\longrightarrow$ $\tt{-y=-4-3}$

$\longrightarrow$ $\tt{-y=-7}$

$\longrightarrow$ $\tt{y=7}$

$\bold{\underline{\boxed{\Large{\purple{\rm{Numerator\:=\:x\:=3}}}}}}$

$\bold{\underline{\boxed{\Large{\purple{\sf{Denominator\:=\:y\:=7}}}}}}$

$\bold{\underline{\boxed{\Large{\purple{\tt{Fraction\:=\:{\dfrac{x}{y}\:=\:{\dfrac{3}{7}}}}}}}}}$