10th Class Mathematics Solved Paper - Mathematics-2016 Delhi Term-II Set-II

  • 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.


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