if 2 is added to a fraction ,it will become 9÷11 . if 3 is added to the same fraction ,it will be 5÷6 .find the fraction
2 thoughts on “if 2 is added to a fraction ,it will become 9÷11 . if 3 is added to the same fraction ,it will be 5÷6 .find the fraction”
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2 thoughts on “if 2 is added to a fraction ,it will become 9÷11 . if 3 is added to the same fraction ,it will be 5÷6 .find the fraction”
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Answer:
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. Hence, the fraction is 7/9.
Step-by-step explanation:
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[tex]\bf \underline{ \underline{\maltese\:Given} }[/tex]
[tex] \sf A \: fraction \: becomes \: \dfrac{9}{11} \: if \: 2 \: is \:added \: in \: both \: the \: numerator \: and \: denominator. [/tex]
[tex] \sf If \: 3 \: is \: added \: to \: both \: the \: numerator \: and \: denominator \: it \: becomes \: \dfrac{5}{6} [/tex]
[tex]\bf \underline{ \underline{\maltese \: To \: find } }[/tex]
[tex] \sf \implies Original \: fraction = \: ? [/tex]
[tex]\bf \underline{ \underline{\maltese \: Solution } }[/tex]
[tex] \sf Let \: the \: numerator \: be \: x, [/tex]
[tex] \sf Let \: the \: denominator \: be \: y, \: \: fraction = \dfrac{x}{y} [/tex]
[tex] \rm \underline{First \: case \: \: [When \: 2 \: is \: added \: in \: both]}[/tex]
[tex] \sf \implies \dfrac{x + 2}{y + 2} = \dfrac{9}{11} [/tex]
[tex] \rm Solve \: for \: y[/tex]
[tex] \sf \implies 9(y + 2) = 11(x + 2)[/tex]
[tex] \sf \implies 9y + 18= 11x + 22[/tex]
[tex] \sf \implies 9y = 11x + 22 – 18[/tex]
[tex] \sf \implies 9y = 11x +4[/tex]
[tex] \sf \implies y = \dfrac{11x}{9} + \dfrac{4}{9} [/tex]
[tex] \rm \underline{Second \: case \: \: [When \: 3 \: is \: added \: in \: both]}[/tex]
[tex] \sf \implies \dfrac{x + 3}{y + 3} = \dfrac{5}{6} [/tex]
[tex] \rm Solve \: for \: x[/tex]
[tex] \sf \implies 6(x + 3) = 5(y + 3)[/tex]
[tex] \sf \implies 6x +18 = 5y + 15[/tex]
[tex] \sf \implies 5y = 6x + 3[/tex]
[tex] \sf \implies y = \dfrac{6x + 3}{5} [/tex]
[tex] \sf \implies y = \dfrac{6x}{5} + \dfrac{3}{5} [/tex]
[tex] \sf \implies \dfrac{11x}{9} + \dfrac{4}{9} = \dfrac{6x}{5} + \dfrac{3}{5} [/tex]
[tex] \sf \implies \dfrac{11x + 4}{9} = \dfrac{6x + 3}{5}[/tex]
[tex] \sf \implies 5(11x + 4) = 9(6x + 3)[/tex]
[tex] \sf \implies 55x + 20 = 54x + 27[/tex]
[tex] \sf \implies 55x + 20 – 54x = 27[/tex]
[tex] \sf \implies x + 20 = 27[/tex]
[tex] \sf \implies x = 27 – 20[/tex]
[tex] \sf \implies x = 7 \: \: (N umerator)[/tex]
[tex] \sf Put \: the \: value \: of \: x \: in \: \bigg( y = \dfrac{11x}{9} + \dfrac{4}{9} \bigg) \: to \: get \: denominator. [/tex]
[tex]\sf \implies y = \dfrac{11x}{9} + \dfrac{4}{9} [/tex]
[tex]\sf \implies y = \dfrac{11 \times 7}{9} + \dfrac{4}{9} [/tex]
[tex]\sf \implies y = \dfrac{77 + 4}{9} [/tex]
[tex]\sf \implies y = \cancel \dfrac{81}{9} [/tex]
[tex]\sf \implies y = 9 \: \: (Denominator)[/tex]
[tex] \sf \underline{Therefore, \: value \: of \: x = 7 \: and \: y = 9 }[/tex]
[tex] \underline{ \boxed{ \bf \red{Hence, \: original \: fraction = \dfrac{7}{9} }}}[/tex]
Reply
[tex]\bf \underline{ \underline{\maltese\:Given} }[/tex]
[tex] \sf A \: fraction \: becomes \: \dfrac{9}{11} \: if \: 2 \: is \:added \: in \: both \: the \: numerator \: and \: denominator. [/tex]
[tex] \sf If \: 3 \: is \: added \: to \: both \: the \: numerator \: and \: denominator \: it \: becomes \: \dfrac{5}{6} [/tex]
[tex]\bf \underline{ \underline{\maltese \: To \: find } }[/tex]
[tex] \sf \implies Original \: fraction = \: ? [/tex]
[tex]\bf \underline{ \underline{\maltese \: Solution } }[/tex]
[tex] \sf Let \: the \: numerator \: be \: x, [/tex]
[tex] \sf Let \: the \: denominator \: be \: y, \: \: fraction = \dfrac{x}{y} [/tex]
[tex] \rm \underline{First \: case \: \: [When \: 2 \: is \: added \: in \: both]}[/tex]
[tex] \sf \implies \dfrac{x + 2}{y + 2} = \dfrac{9}{11} [/tex]
[tex] \rm Solve \: for \: y[/tex]
[tex] \sf \implies 9(y + 2) = 11(x + 2)[/tex]
[tex] \sf \implies 9y + 18= 11x + 22[/tex]
[tex] \sf \implies 9y = 11x + 22 – 18[/tex]
[tex] \sf \implies 9y = 11x +4[/tex]
[tex] \sf \implies y = \dfrac{11x}{9} + \dfrac{4}{9} [/tex]
[tex] \rm \underline{Second \: case \: \: [When \: 3 \: is \: added \: in \: both]}[/tex]
[tex] \sf \implies \dfrac{x + 3}{y + 3} = \dfrac{5}{6} [/tex]
[tex] \rm Solve \: for \: x[/tex]
[tex] \sf \implies 6(x + 3) = 5(y + 3)[/tex]
[tex] \sf \implies 6x +18 = 5y + 15[/tex]
[tex] \sf \implies 5y = 6x + 3[/tex]
[tex] \sf \implies y = \dfrac{6x + 3}{5} [/tex]
[tex] \sf \implies y = \dfrac{6x}{5} + \dfrac{3}{5} [/tex]
[tex] \sf \implies \dfrac{11x}{9} + \dfrac{4}{9} = \dfrac{6x}{5} + \dfrac{3}{5} [/tex]
[tex] \sf \implies \dfrac{11x + 4}{9} = \dfrac{6x + 3}{5}[/tex]
[tex] \sf \implies 5(11x + 4) = 9(6x + 3)[/tex]
[tex] \sf \implies 55x + 20 = 54x + 27[/tex]
[tex] \sf \implies 55x + 20 – 54x = 27[/tex]
[tex] \sf \implies x + 20 = 27[/tex]
[tex] \sf \implies x = 27 – 20[/tex]
[tex] \sf \implies x = 7 \: \: (N umerator)[/tex]
[tex] \sf Put \: the \: value \: of \: x \: in \: \bigg( y = \dfrac{11x}{9} + \dfrac{4}{9} \bigg) \: to \: get \: denominator. [/tex]
[tex]\sf \implies y = \dfrac{11x}{9} + \dfrac{4}{9} [/tex]
[tex]\sf \implies y = \dfrac{11 \times 7}{9} + \dfrac{4}{9} [/tex]
[tex]\sf \implies y = \dfrac{77 + 4}{9} [/tex]
[tex]\sf \implies y = \cancel \dfrac{81}{9} [/tex]
[tex]\sf \implies y = 9 \: \: (Denominator)[/tex]
[tex] \sf \underline{Therefore, \: value \: of \: x = 7 \: and \: y = 9 }[/tex]
[tex] \underline{ \boxed{ \bf \red{Hence, \: original \: fraction = \dfrac{7}{9} }}}[/tex]
Answer:
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. Hence, the fraction is 7/9.
Step-by-step explanation: