question_answer

Divide 184 into two parts such that one third of one part may exceed one seventh of other part by 8.

Answer:

Let one part of 184 be x.   ∴ Other part be (184 ? x)   Now, according to question, \[\Rightarrow \frac{1}{3}x-\frac{1}{7}(184-x)=8\] \[\Rightarrow \frac{x}{3}+\frac{x}{7}-\frac{184}{7}=8\] \[\Rightarrow \,\,\,\,\,\,\,\frac{7x+3}{21}=8+\frac{184}{7}\] \[\Rightarrow \,\,\,\,\,\,\,\,\frac{10x}{21}=\,\,\frac{56+184}{7}\]   \[\frac{10x}{21}=\,\,\frac{240}{7}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,x=\frac{21\times 240}{7\times 10}\]   = 72 Hence, parts are 72 and 184 ? 72 = 112.