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Arithmetic Progressions
CH05 Question Bank
CBSE_CH05_QUESTION_BANK
Grade 10
Question:
A thief runs away from a police station with a uniform speed of 100 m/minute. After one minute, a policeman runs after the thief to catch him. He goes at a speed of 100 m/minute in the first minute, and increases his speed by 10 m/minute every succeeding minute. After how many minutes will the policeman catch the thief?
Step-by-Step Solution
Key Concept: The thief's distance grows linearly (100 m per minute of the policeman's chase, plus the 100 m head start); the policeman's cumulative distance is an AP sum. Set these equal and solve.
Let the policeman catch the thief after $n$ minutes (of the policeman's running). In that time, the thief has run for $(n+1)$ minutes total (including the 1-minute head start), covering a distance of $100(n+1)$ m. [1.0 Mark]
The policeman's distance in $n$ minutes is the sum of an AP with $a=100,d=10$: $S_n=\dfrac n2[2(100)+(n-1)(10)]=\dfrac n2[200+10n-10]=\dfrac n2(190+10n)$. [1.5 Marks]
Setting the distances equal: $\dfrac n2(190+10n)=100(n+1)\Rightarrow n(190+10n)=200(n+1)\Rightarrow190n+10n^2=200n+200$. [1.5 Marks]
$10n^2-10n-200=0\Rightarrow n^2-n-20=0\Rightarrow(n-5)(n+4)=0\Rightarrow n=5$ (rejecting the negative root). So the policeman catches the thief after $5$ minutes. [1.0 Mark]
Correct Answer:
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