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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 entire amount of time taken by him to count the entire amount.A. 56 minutesB. 70 minutesC. 89 minutesD. 103 minutes

Last updated date: 15th Jul 2024
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Hint: In this question, we will use sum of n terms of an AP i.e. ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$ to calculate the time required by a man to count the time required for the amount left after half an hour.

We know that from the question man is employed to count Rs.10710 and he counts at the rate of Rs.180 per minute for half an hour. So, now we need to calculate money counted by the man in the first half an hour i.e. 30 minutes.
In first half an hour, the man counted at the rate of Rs.180 per minute, then the total money counted by man in 30 minutes is
$\Rightarrow 30\times 180$
$\Rightarrow 5400$
Rs.5400 was counted by man in the first half an hour.
Then the remaining amount to be counted=Total money employed to count -Total money counted in half an hour
$\Rightarrow 10710-5400$
$\Rightarrow 5310$
We know that after half an hour he counts at the rate of Rs.3 less every minute than the preceding minute.
In the 31st min man counts 3 less than preceding minute i.e. 180-3=177
So, the man counted 177 rupees in the 31st minute.
So, in the 32nd min man counts 3 less than preceding minute i.e. 177-3=174
So, the man counts 174 rupees in 32nd minute and so on. This sequence will continue until the money is over. The sequence formed is 177, 174, ……… , which is an Arithmetic progression with first term a=177 and common difference d=174 - 177= -3
We found that amount of money that has to be counted after half an hour is Rs.5310 which is equal to ${{S}_{n}}$ i.e.${{S}_{n}}=5310$.
We know that sum of n terms in an Arithmetic Progression is ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)\cdot \cdot \cdot \cdot \cdot \left( 1 \right)$ ,
where a is the first term of an AP and d is the common difference of an AP and n is number of terms, but in our case it is time taken to count the money.
Substituting a, d, ${{S}_{n}}$ in equation (1), we will get n.
So, $5310=\dfrac{n}{2}\left( 2\times 177+(n-1)(-3) \right)$
Now, simplifying the above we will get
$5310=\dfrac{n}{2}\left( 354+3-3n \right)$
We will simplify further and get
$5310\times 2=n\left( 357-3n \right)$
$\Rightarrow 10620=357n-3{{n}^{2}}$
$\Rightarrow 3{{n}^{2}}-357n+10620=0$
We will find n using formula $n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where a=3, b=-357, c=10620
$n=\dfrac{357\pm \sqrt{{{(-357)}^{2}}-4\times 3\times 10620}}{2\times 3}$
Now simplifying the above one we will get
$n=\dfrac{357\pm 3}{2\times 3}$
$n=\dfrac{357+3}{2\times 3}$ or $n=\dfrac{357-3}{2\times 3}$
$n=\dfrac{360}{6}$ or $n=\dfrac{354}{6}$
$n=60$ or $n=59$
At 60th minute count=0. At n=59 i.e. at 59th minute counting will be ended as money will be completed at this minute. So, there is no chance of counting in 60th minute.
So, n=60 is not possible.
So, the time taken to count remaining money is 59 minutes.
Therefore, the total time taken to count the total amount = 59 + 30 = 89 min.
Hence, the time taken by him to count the total amount is 89 minutes.
Hence, the correct answer is (C).

Note: If we consider 60th minute, 60+30=90 will be obtained. This is wrong as 60 is not true because count will be zero as there will be no money to count at that minute. We can solve the quadratic equation, to obtain the value of n by completing the square method and factorisation method also.