A) \[12.5\,\,\min \]
B) \[135\,\,\min \]
C) \[34\,\,\min \]
D) \[24\,\,\min \]
Correct Answer: C
Solution :
The number of notes counted in first 10 minutes\[=150\times 10=1500\] Suppose, the person counts the remaining \[3000\] currency notes in n minutes. Then, \[3000=\] Sum of n terms of an \[AP\] with first term \[148\]and common difference\[-2\] \[\Rightarrow \] \[3000=\frac{n}{2}\{2\times 148+(n-1)\times (-2)\}\] \[\Rightarrow \] \[3000=n(149-n)\] \[\Rightarrow \] \[{{n}^{2}}-149n+3000=0\] \[\Rightarrow \] \[(n-125)(n-24)=0\] \[\Rightarrow \] \[n=125,\,\,24\] Clearly, \[n=125\] is not possible. Total, time taken\[=(10+24)=34\min \].You need to login to perform this action.
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