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A Man is Employed to Count Rs 10710. He Counts at the Rate of Rs 180 per Minute for Half an Hour. After this He Counts at the Rate of Rs 3 Less Every Minute than the Preceding Minute. Find the Time

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Question

A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.

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Solution

It is given that the man counts Rs 180 per minute for half an hour.
∴ Sum of money the man counts in 30 minutes = Rs 180

\[\times\] 30 = Rs 5400

Total money counted by the man = Rs 10710
∴ Money left for counting after 30 minutes = Rs (10710 − 5400) = Rs 5310

It is given that after 30 minutes, he counts at the rate of Rs 3 less every minute than the preceding minute.

Therefore, it would be an A.P. where a = 177 and d = −3.
Let the time taken to count Rs 5310 be n minutes.

\[5310 = \frac{n}{2}\left[ 2 \times 177 + \left( n - 1 \right) \times - 3 \right]\]

\[ \Rightarrow 10620 = 354 n - 3 n^2 + 3n\]

\[ \Rightarrow 3 n^2 - 357n + 10620 = 0\]

\[ \Rightarrow n^2 - 119n + 3540 = 0\]

\[ \Rightarrow n^2 - 59n - 60n + 3540 = 0\]

\[ \Rightarrow n\left( n - 59 \right) - 60\left( n - 59 \right) = 0\]

\[ \Rightarrow \left( n - 59 \right)\left( n - 60 \right) = 0\]

\[ \therefore n = 59 \text { or } 60\]

Thus, the time taken to count Rs 5310 would be 59 minutes or 60 minutes.
Hence, the total time taken to count Rs 10710 would be (30 + 59) minutes or (30 + 60) minutes, i.e. 89 minutes or 90 minutes, respectively.

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Chapter 19: Arithmetic Progression - Exercise 19.7 [Page 49]

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R.D. Sharma Mathematics [English] Class 11
Chapter 19 Arithmetic Progression
Exercise 19.7 | Q 6 | Page 49

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