Question

1 )the denominator of a fraction has 5 less than the numerator. if 2 is added to both its numerator and denominator its becomes 2. find the fraction​

Answer 1

Given :

The denominator of a fraction has 5 less than the numerator. If 2 is added to both its numerator and denominator its becomes ⅘. Find the fraction

Solution :

Let us assume :

$\star \: \frak{The \: numerator \: be \: \boxed{ \blue{\frak{x}}}}$

Given that :

$\frak{The \: denominator \: of \: a \: fraction \: has \: \boxed{ \green{\frak{ 5 \: less \: than \: the \: numerator}}}}$

It means that :

The denominator is 5 more than the numerator

We know that :

$\frak{Original \: Fraction = \frac{ \purple{Numerator}}{ \red{Denominator}}}$

$\frak{Original \: Fraction = \frac{ \purple{x}}{ \red{x + 5}}}$

Now, according to the question :

$\boxed{ \red{ \frak{2 \: is \: added \: to \: both }}}\: \frak{the \: numerator \: as \: well \: as \: the \: denominator}$

$\frak{New \: Fraction = \frac{ \green{Numerator}}{ \blue{Denominator}}}$

$\frak{New \: Fraction = \frac{x + 2}{x + 5 + 2}= \frac{ \green{x + 2}}{ \blue{x + 7}}}$

Now, the fraction becomes ⅘

Hence, the equation is :

$\twoheadrightarrow \frak{\frac{x + 2}{x + 7} = \frac{4}{5} }$

Cross multiplying we get

$\twoheadrightarrow \frak{5(x + 2) = 4(x + 7) }$

$\twoheadrightarrow \frak{5x + 10 = 4x + 28}$

$\twoheadrightarrow \frak{5x - 4x = 28 - 10}$

$\twoheadrightarrow \frak{x = 18}$

Now, finding the original fraction

$\frak{Original \: Fraction = \frac{x}{x + 5}}$

Putting x = 18 instead of x we get

$\frak{Original \: Fraction = \frac{18}{18 + 5}}$

$\frak{Original \: Fraction = \frac{18}{23}}$

• Henceforth, the original fraction is 18/23

Answer 2

Answer:

Given :

The denominator of a fraction has 5 less than the numerator. If 2 is added to both its numerator and denominator its becomes ⅘. Find the fraction

Solution :

Let us assume :

\star \: \frak{The \: numerator \: be \: \boxed{ \blue{\frak{x}}}}⋆Thenumeratorbe

x

Given that :

\frak{The \: denominator \: of \: a \: fraction \: has \: \boxed{ \green{\frak{ 5 \: less \: than \: the \: numerator}}}}Thedenominatorofafractionhas

5lessthanthenumerator

It means that :

The denominator is 5 more than the numerator

We know that :

\frak{Original \: Fraction = \frac{ \purple{Numerator}}{ \red{Denominator}}}OriginalFraction=

Denominator

Numerator

\frak{Original \: Fraction = \frac{ \purple{x}}{ \red{x + 5}}}OriginalFraction=

x+5

x

Now, according to the question :

\boxed{ \red{ \frak{2 \: is \: added \: to \: both }}}\: \frak{the \: numerator \: as \: well \: as \: the \: denominator}

2isaddedtoboth

thenumeratoraswellasthedenominator

\frak{New \: Fraction = \frac{ \green{Numerator}}{ \blue{Denominator}}}NewFraction=

Denominator

Numerator

\frak{New \: Fraction = \frac{x + 2}{x + 5 + 2}= \frac{ \green{x + 2}}{ \blue{x + 7}}}NewFraction=

x+5+2

x+2

=

x+7

x+2

Now, the fraction becomes ⅘

Hence, the equation is :

\twoheadrightarrow \frak{\frac{x + 2}{x + 7} = \frac{4}{5} }↠

x+7

x+2

=

5

4

Cross multiplying we get

\twoheadrightarrow \frak{5(x + 2) = 4(x + 7) }↠5(x+2)=4(x+7)

\twoheadrightarrow \frak{5x + 10 = 4x + 28}↠5x+10=4x+28

\twoheadrightarrow \frak{5x - 4x = 28 - 10}↠5x−4x=28−10

\twoheadrightarrow \frak{x = 18}↠x=18

Now, finding the original fraction

\frak{Original \: Fraction = \frac{x}{x + 5}}OriginalFraction=

x+5

x

Putting x = 18 instead of x we get

\frak{Original \: Fraction = \frac{18}{18 + 5}}OriginalFraction=

18+5

18

\frak{Original \: Fraction = \frac{18}{23}}OriginalFraction=

23

18

Henceforth, the original fraction is 18/23