question_answer

Along a road lie an odd number of stones placed at intervals of \[10\,\,m\]. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man started the job with one of the end stones by carrying them in succession. In carrying all the stones, the man covered a total distance of\[3\,\,km\]. Then the total number of stones is:

A) \[20\]                                   

B) \[25\]

C) \[12\]                                   

D) \[24\]

Correct Answer: C

Solution :

\[\therefore \]  \[[10n+2(n-1)10+2(n-2)10+...\]  \[+1\times 2\times 10]+[1\times 2\times 10+2\times 2\times 10+...\]                \[+2(n-1)\times 10+2n\times 10]\]                 \[=3000\]                                             (given) \[\Rightarrow \]\[10[4(1+2+3+...+n)-n]=3000\] \[\Rightarrow \]               \[4\times \frac{n(n+1)}{2}-n=300\] \[\Rightarrow \]               \[2{{n}^{2}}+2n-n=300\] \[\Rightarrow \]               \[n=\frac{-1\pm \sqrt{1\pm 2400}}{2\times 2}=\frac{-1\pm 49}{4}\] \[\Rightarrow \]               \[n=\frac{48}{4}=12\]