Question

ki?
9
(v) A fraction becomes if 2 is added to both the numerator and the denominator.
11
If, 3 is added to both the numerator and the denominator it becomes Find the
fraction
6​

Answer 1

Answer:

-9/1

Step-by-step explanation:

Let the required fraction be say a/b

According to the questions we get 2 equations

a - 2b = 11

a - 6b = 15

on solving those equations we get ,

a = 9

b = -1

For the full solution refer the pic

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Answer 2

$\clubs \: \: { \underline{ \large{ \textsf{ \textbf{ \sf{ \color{purple}{Question : }}}}}}}$

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

$\clubs \: \: { \underline{ \large{ \textsf{ \textbf{ \sf{ \color{purple}{Answer : }}}}}}}$

• The fraction is $\large \pink{ \frac{7}{9} }$

$\clubs \: \: { \underline{ \large{ \textsf{ \textbf{ \sf{ \color{purple}{ Solution: }}}}}}}$

• Let the Numerator be (x)
• & Denominator be (y) .

$\qquad \sf \red{so \: the \: fraction \: is \: (\frac{x}{y}) }$

$\clubs \: { \underline{ \sf{ \color{red}{Given : }}}}$

• If 2 is added to both the numerator and denominator , fraction becomes 9/11

$\qquad \bullet \: \: \sf \green{so \: \: \frac{numerator + 2}{denominator + 2} = \frac{9}{11} }$

$\qquad \bullet \: \: \sf \green{ \: \: \frac{x + 2}{y + 2} = \frac{9}{11} }$

• By cross Multiplication

$\qquad \mapsto\sf{11(x + 2) = 9(y + 2)}$

$\qquad \mapsto\sf{11x + 22= 9y + 18}$

$\qquad \mapsto\sf{11x - 9y = 18 - 22}$

$\qquad \mapsto\sf{11x - 9y = ( - 4)} \: \: \: \: \: ... \tiny{equation \: (1)}$

Also,

$\clubs \: { \underline{ \sf{ \color{red}{Given \: that \: : }}}}$

• If 3 is added to both the numerator and denominator, fraction becomes 5/6

$\qquad \bullet \: \: \sf \green{so \: \: \frac{numerator + 3}{denominator + 3} = \frac{5}{6} }$

$\qquad \bullet \: \: \sf \green{ \: \: \frac{x + 3}{y + 3} = \frac{5}{6} }$

• By cross Multiplication

$\qquad \mapsto\sf{6(x + 3) = 5(y + 3)}$

$\qquad \mapsto\sf{6x + 18 = 5y + 15}$

$\qquad \mapsto\sf{6x - 5y =15 - 18}$

$\qquad \mapsto\sf{6x - 5y =( - 3)} \: \: \: ... \tiny{equation \: (2)}$

• Hence, our 2 equations are :–

$\qquad \leadsto\sf{11x - 9y = ( - 4)} \: \: \: \: \: ... \tiny{equation \: (1)}$

$\qquad \leadsto\sf{6x - 5y =( - 3)} \: \: \: ... \tiny{equation \: (2)}$

• From equation (1)

$\qquad \leadsto\sf{11x = - 9y = ( - 4)}$

$\qquad \leadsto\sf{11x - 9y = ( - 4)}$

$\qquad \leadsto\sf{x = \left(\frac{9y - 4}{11} \right )}$

• putting the values of x in equation (2)

$\qquad \leadsto\sf{6x - 5y =( - 3)}$

• we get,

$\qquad \sf {6 \: \left( \frac{9y - 4}{11} \right) - 5y + 3 = 0}$

• Multiplying both sides by 11

$\qquad \sf {11 \times 6 \: \left( \frac{9y - 4}{11} \right)11 \times - 5y + 3 \times 11= 0 \times 11}$

$\qquad \sf {6(9y - 4) - 55y + 33 = 0}$

$\qquad \sf {6(9y) - 6(4) - 55y + 33= 0}$

$\qquad \sf {54y - 24 - 55y + 33= 0}$

$\qquad \sf {( - y) + (9)= 0}$

$\qquad \sf \pink{( y) = 9}$

• putting y = 9 in equation (1)

$\qquad \leadsto \sf{11x - 9y = ( - 4)}$

$\qquad \leadsto \sf{11x - 9(9) = ( - 4)}$

$\qquad \leadsto \sf{11x - 81 = ( - 4)}$

$\qquad \leadsto \sf{11x= ( - 4) + 81}$

$\qquad \leadsto \sf{11x= 77}$

$\qquad \leadsto \sf{x = \frac{77}{11} }$

$\qquad \leadsto { \pink{\sf{x = (7)}}}$

• Therefore, x = 7 & y = 9
• so, Numerator (x) = 7 and (y) = 9

Hence,

${ \sf{ \color{red}{orignal \: fraction \: = \frac{numerator}{denominator} }}}$

${ \sf{ \color{red}{orignal \: fraction \: = \frac{x}{y} }}}$

${ \sf{ \color{red}{= \frac{7}{9} }}}$