• # 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?

 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.