मराठी

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 - Mathematics

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प्रश्न

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.

उत्तर

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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पाठ 19: Arithmetic Progression - Exercise 19.7 [पृष्ठ ४९]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 19 Arithmetic Progression
Exercise 19.7 | Q 6 | पृष्ठ ४९

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