question_answer

A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 W 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 1 minutes so he will catch the thief in \[(n-1)\] minutes,

Total distance covered by thief \[=100\text{ }m/minute\times n\] minute

\[=(100\text{ }n)\text{ }m\]

Now, total distance covered by the policeman \[=(100)m+(100+10)m+(100+10+10)m+.....+(n-1)\] terms

i.e., \[100+110+120+....+(n-1)\] terms

\[\therefore \,\,{{S}_{n-1}}=\frac{n-1}{2}[2\times 100+(n-2)10]\]

\[\Rightarrow \,\frac{n-1}{2}[200+(n-2)10]=100\,n\]

\[\Rightarrow \,\,(n-1)(200+10n-20)=200n\]

\[\Rightarrow \,\,200n-200+10{{n}^{2}}-10n+20-20n=200n\]

\[\Rightarrow \,\,10{{n}^{2}}-30n-180n=0\]

\[\Rightarrow \,\,{{n}^{2}}-3n-18=0\]

\[\Rightarrow \,\,{{n}^{2}}-(6-3)n-18=0\]

\[\Rightarrow \,\,{{n}^{2}}-6n+3n-18=0\]

\[\Rightarrow \,\,n(n-6)+3(n-6)=0\]

\[\Rightarrow \,\,(n+3)(n-6)=0\]

\[\therefore n=6\] or \[n=-3\] (Neglect)

Hence, policeman will catch the thief in \[(6-1)\] i.e., 5 minutes.\