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 (in minutes) by him to count the entire amount.

Answer 1

Hint:It is said that in 1 minute Rs. 180 is counted. Now find the money counted in half an hour. Then find the remaining amount to count and thus form the A.P. and find the time taken to count the entire amount.

Complete step-by-step answer:
The total amount the man has to count = Rs. 10710.
In 1 minute, he counts Rs. 180.
It is said that he counts at the rate of Rs. 180 per minute for half an hour, which means that he counts at this rate for 30 minutes.
\[\therefore \]In 30 minutes he counts \[=180\times 30=\] Rs. 5400.
Thus the money he counted in half an hour is Rs. 5400.
Now the money left for counting = 10710 – 5400 = Rs. 5310.
It is said that after half an hour he counted 3 less per minute.
Thus the rate of counting per minute,
\[\begin{align}
  & =\left( 180-3 \right),\left( 180-3-3 \right),\left( 80-3-3-3 \right),..... \\
 & =177,174,171,..... \\
\end{align}\]
Now this forms an arithmetic progression, with first term 177 = a,
common difference, d = 174 – 177 = (-3)
The sum of A.P. = 5310 = \[{{S}_{n}}\].
Thus we have \[{{S}_{n}}=5310,a=177,d=-3\].
Let us substitute these values in the formula to find the sum of n terms and get us the value of n.
\[\begin{align}
  & {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\
 & 5310=\dfrac{n}{2}\left[ 2\times 177+\left( n-1 \right)(-3) \right] \\
 & 10620=n\left[ 354-3n+3 \right] \\
 & 10620=n\left[ 357-3n \right] \\
 & 10620=357n-3{{n}^{2}} \\
 & \Rightarrow 3{{n}^{2}}-357n+10620=0 \\
\end{align}\]
Divide the entire equation by 3.
\[{{n}^{2}}-119n+3540=0\]
Now this is in the form of a quadratic equation \[a{{x}^{2}}+bx+c=0\]. Now let us compare both the equations, we get,
a = 1, b = -119, c = 3540.
Substitute these values in \[n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
\[\begin{align}
  & n=\dfrac{-(-119)\pm \sqrt{{{(-119)}^{2}}-4\times 1\times 3540}}{2\times 1} \\
 & n=\dfrac{119\pm \sqrt{14161-14160}}{2}=\dfrac{119\pm \sqrt{1}}{2}=\dfrac{119\pm 1}{2} \\
\end{align}\]
\[\therefore n=\dfrac{119+1}{2}=\dfrac{120}{2}=60\] and \[n=\dfrac{119-1}{2}=\dfrac{118}{2}=59\].
\[\therefore n=60\] and \[n=59\].
But 59 < 60.
So we take n = 59.
Thus, time taken to count Rs. 5310 = 59 minutes.
Thus the total time taken to count the entire amount = 30 + 59 = 89 minutes
Thus the entire time taken by him to count the entire amount is 89 minutes.

Note:We can also solve the quadratic equation by splitting the middle term,
\[{{n}^{2}}-119n+3450=0\]
The sum of zeroes =119
The product of zeroes \[=3540=\left( -59 \right)\times \left( -60 \right)\]
\[\begin{align}
  & \therefore {{n}^{2}}-59n-60n+3450=0 \\
 & n(n-59)-60(n-59)=0 \\
 & \therefore (n-59)(n-60)=0 \\
 & \therefore n=59,n=60 \\
\end{align}\]