question_answer

A thief, after committing a theft, runs at a uniform speed of 50 m/ minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?

Answer:

Let total time be n minutes

Since policeman runs after two minutes he will catch the thief in \[(n-2)\] minutes.

Total distance covered by thief \[=50\text{ }m/min\times n\text{ }mm=(50\,n)\text{ }m\]

Now, total distance covered by the policeman \[=(60)+(60+5)+(60+5+5)+.....+(n-2)\] terms

i.e., \[60+65+70+.....+(n-2)\]terms

\[\therefore \,{{S}_{n-2}}=\frac{n-2}{2}[2\times 60+(n-3)5]\]

\[\Rightarrow \frac{n-2}{2}[120+(n-3)5]=50n\]

\[\Rightarrow n-2(120+5n-15)=100n\]

\[\Rightarrow \,120n-240+5{{n}^{2}}-10n-15n+30=100n\]

\[\Rightarrow \,5{{n}^{2}}-5n-210=0\]

\[\Rightarrow \,{{n}^{2}}-n-42=0\]

\[\Rightarrow \,{{n}^{2}}(7-6)n-42=0\]

\[\Rightarrow \,{{n}^{2}}-7n+6n-42=0\]

\[\Rightarrow \,n(n-7)+6(n-7)=0\]

\[\Rightarrow \,(n+6)(n-7)=0\]

\[n=7\] or \[n=-6\] (neglect)

Hence, policeman will catch the thief in \[(7-2)\] i.e., 5 minutes.