A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour

In the given problem, the total amount = Rs 10710.

For the first half and hour (30 minutes) he counts at a rate of Rs 180 per minute. So,

The amount counted in 30 minutes

So, amount left after half an hour 

After 30 minutes he counts at a rate of Rs 3 less every minute. So,

At 31st minute the rate of counting per minute = 177.

At 32nd minute the rate of counting per minute = 174.

So, the rate of counting per minute for each minute will form an A.P. with the first term as 177 and common difference as −3.

So, the total time taken to count the amount left after half an hour can be calculated by using the formula for the sum of n terms of an A.P,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

We get,

$5310=\frac{n}{2}[2(177)+(n-1)(-3)] \ldots(1)$

$5310(2)=n[354-3 n+3]$

$10620=n(357-3 n)$

$10620=357 n-3 n^{2}$

So, we get the following quadratic equation,

$3 n^{2}-357 n+10620=0$

$n^{2}-119 n+3540=0$

Solving the equation by splitting the middle term, we get,

$n^{2}-60 n-59 n+3540=0$

$n(n-60)-59(n-60)=0$

$(n-60)(n-59)=0$

So,

$n-59=0$

$n=59$

Or

$n-60=0$

$n=60$

Now let $\mathrm{n}=60$ then finding the last term, we get

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[\mathrm{a}+\mathrm{l}]$

$5310=\frac{60}{2}[177+1]$

$177=177+1$

$1=0$

It means the work will be finesh in 59 th minute only because 60 th term is 0 . So, we will take $\mathrm{n}=59$

Therefore, the total time required for counting the entire amount $=30+59$ minutes $=89$ minutes

So, the total time required for counting the entire amount is 89 minutes.