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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 then the preceding minute. Find the entire time taken by him to count the entire amount.
A. 56 minutes
B. 70 minutes
C. 89 minutes
D. 103 minutes

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Last updated date: 20th Jun 2024
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Answer
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Hint: Here we will first calculate the amount counted by him in the first half hour and subtract it from the total amount. Then by using the information, we will form a series of amounts calculated after the first half an hour in AP. We will use the formula of the sum of an AP series and simplify it to get a quadratic equation. We will then use the quadratic formula to get the required time taken by the man.

Complete step-by-step answer:
The total amount to be counted \[ = {\rm{Rs}}.10710\]
He counted at a rate of Rs. 180 per minute for half an hour.
Now we know that an hour has 60 minutes, so half an hour will have 30 minutes.
So, total amount counted by him in half hour \[ = {\rm{Rs}}.180 \times 30 = {\rm{Rs}}.5400\]
Now amount remaining for him to count \[ = {\rm{Rs}}.10710 - {\rm{Rs}}.5400 = {\rm{Rs}}.5310\]
Next, it is given that after half an hour he will count at a rate of 3 less per minute.
$\therefore $ Rate of counting per minute after that \[ = \left( {180 - 3} \right),\left( {180 - 3 - 3} \right),\left( {180 - 3 - 3 - 3} \right)\]…….
Subtracting the terms, we get
\[ \Rightarrow \] Rate of counting per minute after that \[ = \left( {177} \right),\left( {174} \right),\left( {171} \right)\]…….
As we can see that the above value has a relation where the first term is 177 and the difference between each term is 3 and the sum of the above relation will be equal to Rs.5310.
Now, we will use the formula of the sum of an A.P.
Substituting \[a = 177\], \[d = - 3\] and 5310 as sum in the formula Sum of A.P \[ = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], we get
\[5310 = \dfrac{n}{2}\left[ {2 \times 177 + \left( {n - 1} \right)\left( { - 3} \right)} \right]\]
Multiplying the terms, we get
\[ \Rightarrow 5310 = \dfrac{n}{2}\left[ {354 - 3n + 3} \right]\]
Multiplying by 2 on both the sides, we get
\[\begin{array}{l} \Rightarrow 5310 \times 2 = n\left( {357 - 3n} \right)\\ \Rightarrow 10620 = 357n - 3{n^2}\end{array}\]
Rewriting the terms, we get
\[ \Rightarrow 3{n^2} - 357n + 10620 = 0\]
Dividing the above equation by 3 we get,
\[ \Rightarrow {n^2} - 119n + 3540 = 0\]
Now we will solve the above equation using quadratic equation formula which states:
For equation \[a{x^2} + bx + c = 0\] we use formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] to get the value of \[x\].
Substituting \[a = 1,b = - 119\] and \[c = 3540\] in the above formula, we get,
\[x = \dfrac{{ - \left( { - 119} \right) \pm \sqrt {{{\left( { - 119} \right)}^2} - 4 \times 1 \times 3540} }}{{2 \times 1}}\]
Simplifying the equation, we get
\[\begin{array}{l} \Rightarrow x = \dfrac{{119 \pm \sqrt {14161 - 14160} }}{2}\\ \Rightarrow x = \dfrac{{119 \pm \sqrt 1 }}{2}\end{array}\]
Taking the square root, we get,
\[x = \dfrac{{119 \pm 1}}{2}\]
So, we get two values of \[x\].
Now
\[x = \dfrac{{119 + 1}}{2} = \dfrac{{120}}{2} = 60\]
\[x = \dfrac{{119 - 1}}{2} = \dfrac{{118}}{2} = 59\]
As \[59 < 60\]
$\therefore n=59$

The total time taken by the man to count the entire amount \[ = 30 + 59 = 89\] minutes
Hence, option (C) is correct.


Note:
 We have obtained the series in which the man counts the money to be in arithmetic progression. An arithmetic progression is defined as a series or progression in which the consecutive terms differ by a common difference. Here, we can make a mistake by subtracting Rs 3 from the next minute only. It is given that he counts Rs 3 less ‘every’ minute which means we have to subtract 3 from each minute after the first half an hour.