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Divide 184 into two parts such that one-third of one part may exceed one-seventh of the other part by 8 . Hint. Let the two parts be $x$ and $184-x$ . Then, $\frac{1}{3} x-\frac{1}{7}(184-x)=8 $
Given: Given number is 184
To do: Divide 184 in two parts such that one-tird exceeds one-seventh of other by 8
Solution:
Lets divide 184 into two parts , such as $x$ and $184 - x (x + 184 - x = 184 )$
One third of one part = $\frac{1}{3} x$
One seventh of other part = $\frac{1}{7} (184 - x)$
One third of one part exceeds One seventh of other part
So, $\frac{1}{3}x \frac{1}{7}(184-x) = 8$
multiply $\frac{1}{7}$ inside the brackets,
$\frac{1}{3}x- \frac{184}{7}+\frac{1}{7}x = 8$
$\frac{1}{3}x+\frac{1}{7}x = 8+ \frac{184}{7}$
Take LCM on to add,
LCM of 3 , 7 is 21
$\frac{7}{21}x+\frac{3}{21}x = \frac{8 \times7}{7}+ \frac{184}{7}$
$\frac{10}{21}x = \frac{56}{7}+ \frac{184}{7}$
$\frac{10}{21}x = \frac{240}{7}$
$x = \frac{240}{7} \times \frac{21}{10}$
[$(\frac{240}{10} = 24 ; \frac{21}{7} = 3)$]
$x$ = 24 $\times$ 3
$x$ = 72
184 - $x$ = 184 - 72 = 112
So, the parts are 72 and 112
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