Question

A fraction becomes 9/11, if 2 is added to both the numerator and denominator. If 3 isadded to both the numerator and denominator it becomes5/6, then the fraction



Note:- Dont solve it here Give direct answer

Answer 1

Answer:

• The original fraction is 7/9

Step-by-step explanation:

Given:

• A fraction becomes 9/11, if 2 is added to both the numerator and denominator
• If 3 isadded to both the numerator and denominator it becomes5/6

To Find:

• Find original the fraction

Assumptions:

• Let the numerator be x
• Let the denominator be y
• Let the fraction be x/y

Solution:

✰ According to the question,

$\rightarrow \tt \qquad \dfrac{x+2}{y+2}= \dfrac{9}{11}$

✰ By Cross Multiplying,

$\rightarrow \tt \qquad 11(x+2) = 9(y+ 2)$

$\rightarrow \tt \qquad 11x+22 = 9y+ 18$

$\rightarrow \tt \qquad 11x = 9y+ 18 - 22$

$\rightarrow \tt \qquad 11x = 9y - 4$

$\rightarrow \tt \qquad x = 9y-4/11 ---(1)$

✰ As per 2nd condition,

$\rightarrow \tt \qquad \dfrac{x+3}{y+3}= \dfrac{5}{6}$

✰ By Cross Multiplying we get,

$\rightarrow \tt \qquad 6(x+3) = 5(y+ 3)$

$\rightarrow \tt \qquad 6x + 18 = 5y+ 15$

✰ Substituting the value of x we get,

$\rightarrow \tt \qquad \dfrac{6(9y-4)}{11} + 18 = 5y+ 15$

$\rightarrow \tt \qquad \dfrac{6(9y-4)}{11} = 5y+ 15 - 18$

$\rightarrow \tt \qquad \dfrac{6(9y-4)}{11} = 5y - 3$

$\rightarrow \tt \qquad 6(9y-4) = 55y-33$

$\rightarrow \tt \qquad 54y- 24 = 55y-33$

$\rightarrow \tt \qquad 54y- 55y = -33 + 24$

$\rightarrow \tt \qquad -y = - 9$

$\rightarrow \tt \qquad {\pink{\boxed{\frak{y = 9}}}\purple\bigstar}$

✰ Now let's find out the value of x

$\rightarrow \tt \qquad x = 9y-4/11$

$\rightarrow \tt \qquad x = 9(9)-4/11$

$\rightarrow \tt \qquad x = 81 -4/11$

$\rightarrow \tt \qquad x = 77 /11$

$\rightarrow \tt \qquad {\pink{\boxed{\frak{x= 7}}}\purple\bigstar}$

Therefore:

• ${\pink{\boxed{\tt{Original \; Fraction = \dfrac{7}{9}}}}}$

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Answer 2

Answer:

7/9

Step-by-step explanation:

As per the provided information in the given question, we have :

• A fraction becomes 9/11, if 2 is added to both the numerator and denominator.
• If 3 is added to both the numerator and denominator it becomes 5/6.

Firstly, let's assume the numerator of the fraction be x & denominator of the fraction be y.

According to the question :

$\begin{gathered} \longmapsto \rm { \frac{x _{(Numerator)} + 2}{y_{(Denominator)} + 2} = \frac{9}{11} } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm { \frac{x _{(Numerator)} + 3}{y_{(Denominator)} + 3} = \frac{5}{6} } \\\end{gathered}$

In order to find the fraction, Let's solve these,

$\begin{gathered} \longmapsto \rm { \frac{x + 2}{y+ 2} = \frac{9}{11} } \\\end{gathered}$

By cross multiplication,

$\begin{gathered} \longmapsto \rm { 11(x + 2)} = {9(y+ 2)}\\\end{gathered}$

Multiplying,

$\begin{gathered} \longmapsto \rm { 11x + 22} = {9y+ 18}\\\end{gathered}$

Transposing 18 to LHS, Changing it's sign,

$\begin{gathered} \longmapsto \rm { 11x + 22 - 18} = {9y}\\\end{gathered}$

$\begin{gathered} \longmapsto \rm { 11x + 4} = {9y} _{(EQ - 1..) }\\\end{gathered}$

_

$\begin{gathered} \longmapsto \rm { \frac{x _{(Numerator)} + 3}{y_{(Denominator)} + 3} = \frac{5}{6} } \\\end{gathered}$

By cross multiplication,

$\begin{gathered} \longmapsto \rm {6(x + 3) =5(y + 3) } \\\end{gathered}$

Multiplying,

$\begin{gathered} \longmapsto \rm {6x + 18=5y + 15 } \\\end{gathered}$

Transposing 15 to LHS, Changing it's sign,

$\begin{gathered} \longmapsto \rm {6x + 18 - 15=5y } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {6x + 3=5y } _{(EQ - 2..) } \\\end{gathered}$

From EQ – 1..,

$\begin{gathered} \longmapsto \rm { 11x + 4} = {9y} \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {9y} = { 11x + 4} \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {y} = \frac{{ 11x + 4}}{9} \\\end{gathered}$

Substituting the value of y in EQ - 2...,

$\begin{gathered} \longmapsto \rm {6x + 3=5y } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {6x + 3=5 \bigg(\frac{{ 11x + 4}}{9} \bigg)} \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {6x + 3= \frac{55}{9} x + \frac{20}{9} } \\\end{gathered}$

Subtracting 55/9x from both sides,

$\begin{gathered} \longmapsto \rm {6x + 3 - \frac{55}{9}x = \frac{55}{9} x + \frac{20}{9} - \frac{55}{9}x } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm { \frac{ - 1}{9}x + 3 = \frac{20}{9} } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm { \frac{ - 1}{9}x = \frac{20}{9} - 3 } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm { \frac{ - 1}{9}x = \frac{ - 7}{9} } \\\end{gathered}$

Comparing LHS & RHS,

$\begin{gathered} \longmapsto \rm { - 1x = - 7 } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm { x = \cancel{\frac{ - 7}{ - 1} } } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm { x = 7} \\\end{gathered}$

« Putting x = 7 in Equation 2...,

$\begin{gathered} \longmapsto \rm {6x + 3=5y } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {6(7) + 3=5y } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {42 + 3=5y } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {5y = 45 } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {y = \cancel{\frac{45}{5} } } \\\end{gathered}$

$\begin{gathered} \longmapsto \rm {y = 9 } \\\end{gathered}$

∴ x = 7 & y = 9, So, The fraction is 7/9.