Question

if the numerator of a fraction is increased by 15% and its denominator is diamond by 8% the value of the fraction is 15 by 16 find the original value​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

If the numerator of a fraction is increased by 15% and it's denominator is decreased by 8%,the value of the fraction is 15/16.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The original number.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the numerator be r

Let the denominator be m

$\therefore\sf{The\:original\:fraction\:is\:=\dfrac{r}{m} }$

$\large{\underline{\tt{1st\:case\::}}}}$

$\mapsto\sf{r+15\%\:of\:r}\\\\\mapsto\sf{r+\dfrac{15}{100} \times r}\\\\\mapsto\sf{\dfrac{100r+15r}{100} }\\\\\mapsto\sf{\red{\dfrac{115r}{100}}}$

$\large{\underline{\tt{2nd\:case\::}}}}$

$\mapsto\sf{m-8\%\:of\:m}\\\\\mapsto\sf{m-\dfrac{8}{100} \times m}\\\\\mapsto\sf{\dfrac{100m-8m}{100} }\\\\\mapsto\sf{\red{\dfrac{92m}{100}}}$

A/q

$\longrightarrow\sf{\dfrac{\frac{115r}{100} }{\frac{92m}{100} } =\dfrac{15}{16} }\\\\\\\longrightarrow\sf{\dfrac{115r}{\cancel{100}} \times \dfrac{\cancel{100}}{92m} =\dfrac{15}{16} }\\\\\\\longrightarrow\sf{\dfrac{115r}{92m} =\dfrac{15}{16} }\\\\\\\longrightarrow\sf{\dfrac{r}{m} =\dfrac{15}{16} \times \dfrac{92}{115} }\\\\\\\longrightarrow\sf{\dfrac{r}{m} =\cancel{\dfrac{1380}{1840} }}\\\\\\\longrightarrow\sf{\green{\dfrac{r}{m} =\dfrac{3}{4} }}$

Thus;

The original number is 3/4 .

Answer 2

QUESTION :

if the numerator of a fraction is increased by 15% and its denominator is diamond by 8% the value of the fraction is 15 by 16 find the original value.

SOLUTION :

Suppose the original fraction is X / Y

New Fraction :

New Numerator : 115% of X

New Denominator : 92 % of Y

New Fraction :

115X / 92 Y = 5X / 4Y

This is equal to 15 / 16

=> X / Y = 3 / 4

Hence the innitial fraction is 3 / 4.