Question

The sum of the numerator and the denominator of a fraction is 10 . If 1 is added to both the numerator and the denominator , the fraction becomes 1/2 . Frame equations for the same .

Answer 1

Solution :

Let the numerator place be r & denominator place be m;

$\boxed{\bf{The\:required\:fraction=\frac{r}{m} }}}$

$\underbrace{\tt{According\:\:to\:\:the\:\:question\::}}}$

$\longrightarrow\sf{r+m=10......................(1)}$

&

$\longrightarrow\sf{\dfrac{r+1}{m+1} =\dfrac{1}{2} }\\\\\longrightarrow\sf{2(r+1) = 1(m+1)}\\\\\longrightarrow\sf{2r + 2 =m+1}\\\\\longrightarrow\sf{2r-m=1-2}\\\\\longrightarrow\sf{2r-m=-1}\\\\\longrightarrow\sf{2r=-1+m}\\\\\longrightarrow\bf{r=-1+m/2..........................(2)}$

∴ Putting the value of r in equation (1),we get;

$\longrightarrow\sf{\dfrac{-1+m}{2} +m=10}\\\\\longrightarrow\sf{-1+m+2m=20}\\\\\longrightarrow\sf{-1+3m=20}\\\\\longrightarrow\sf{3m=20+1}\\\\\longrightarrow\sf{3m=21}\\\\\longrightarrow\sf{m=\cancel{21/3}}\\\\\longrightarrow\bf{m=7}$

∴ Putting the value of m in equation (2),we get;

$\longrightarrow\sf{r=\dfrac{-1+7}{2} }\\\\\longrightarrow\sf{r=\cancel{6/2}}\\\\\longrightarrow\bf{r=3}$

Thus;

$\boxed{\bf{The\:fraction=\frac{r}{m}=\frac{3}{7} }}}$

Answer 2

GivEn:-

• The sum of the numerator and the denominator of a fraction is 10.

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

To find:-

• Fraction of numerator and denominator

SoluTion:-

✮ Let's numerator be x.✮ And let's denominator be y.

$\dag\;{\underline{\underline{\bf{\purple{\;\;Now,\;as\;per\;givEn\; Question:-\;\;}}}}}$

✇ The sum of the numerator and the denominator of a fraction is 10.

$:\implies\sf\star\;\;\underline{x + y = 10}$$:\implies\sf x = 10 - y\;\;.....(1)$

✇ If 1 is added to both the numerator and the denominator , the fraction becomes $\sf \dfrac{1}{2}$.

$:\implies\sf \dfrac{x + 1}{y + 1} = \dfrac{1}{2}$

$\small\sf\;\;\dag\;{\underline{\blue{\;\;Cross\; Multiplying\;the\;fraction\;:-\;\;}}}$

$:\implies\sf 2(x + 1) = y + 1$$:\implies\sf 2x + 2 = y + 1$

$:\implies\sf 2x = y + 1 - 2$$:\implies\sf 2x = y - 1$

$:\implies\sf x = \dfrac{y - 1}{2}\;\;.....(2)$

$\small\sf\;\;\dag\; {\underline{\red{\;\; Substituting\;the\;value\;of\;x\;from\;eq(1)\;and\;2\;:-\;\;}}}$

$:\implies\sf 10 - y = \dfrac{y - 1}{2}$

$\small\sf\;\;\dag\; {\underline{\blue{\;\;Cross\; Multiplying\;the\;fraction\;:-\;\;}}}$

$:\implies\sf 2(10 - y) = y - 1$

$:\implies\sf 20 - 2y = y - 1$

$:\implies\sf - 2y - y = - 1 - 20$

$:\implies\sf - 3y = - 21$

$:\implies\sf y = \cancel{ \dfrac{21}{3}}$

$:\implies{\underline{\boxed{\bf{\pink{y = 7}}}}}$

★$\small\sf\;\;\dag\; {\underline{\red{\;\; Substituting\;the\;value\;of\;y\;in\;eq(1)\;:-\;\;}}}$

$:\implies\sf x = 10 - 7$

$:\implies{\underline{\boxed{\bf{\pink{x = 3}}}}}$

★Therefore,

• Numerator of fraction (x) = 3

• Denominator of fraction (y) = 7

$\therefore\sf\; The\;fraction\;becomes = {\bf{\purple{ \dfrac{3}{7}}}}.$

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