A man is employed to count to RS.10710.He counts @180/- per min for 1/2 hrAfter this he counts @3/-less every min than preceding min.time taken?

Let n be the time taken in minutes, after half an
hour.

Then in half an hour he counts @180 /min an amount =
 30*180 and left out amount = 10710 - 30*180 = 5310.

Now he
counts 180-3  i1st min = 177.

In the 2nd  min he counts
180-2*3 = 174 .

 In  the nth minute he cnts = 180 -3n
.

Therefore the sum of the money in n minutes = 177+176+173
+.... 180-3n  which should be equal to 5310 

LHS = 3
(59+58+57+....60-n) , n terms  = 5310.

LHS is an AP  with 
common difference -1. So sum = (1st term +last term)/2

3
(59+60-n)n/2 = 5310

(110-n)n = (5310*2/3 =
3540

110n-n^2 = 3540

n^2 -110n
+3540 = 0

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

n =
60 Or n = 59.

So it take 59 minutes  after half an hour
when the sum  becomes 5310 . And in the 60th minutes after 1/2 hour  he does mot have
any thing to count and his speed of counting  reduces to 0 per
minute.

So the total time require = (1/2) hour+59 min = 89
min.