# 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?

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The man counts 180 per minute for 1/2 hr. or 30 minutes. Therefore he counts till 5400. The amount left is 5310.

From the 31st minutes he counts 3 less every minute than he did in the previous. This is a arithmetic progression a+nd where a=177 and d=-3. Let assume he needs n minutes to count 5310. The amount counted in the nth minute is 177-3(n-1). So the sum of all the terms from 177 to 177-3(n-1) is 5310.

The sum of n terms of an AP is (T1+Tn)*(n/2)

Here it is [177+177-3(n-1)]*(n/2).

As this is equal to 5310 we get [177+177-3(n-1)]*(n/2)=5310. Eliminate 3 from both the sides, 2*1770=n*[59+59-n+1]

2*1770=n*[119-n]=119n-n^2=3540

This gives the quadratic equation n^2-119n+3540=0

=>n^2-59n-60n+3540=0

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

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

Or n=either 59 or 60

Substituting 59 in the expression 119n-n^2 we get 3540. Therefore 59 more minutes are required after the initial 30 minutes. This makes the total time equal to 59+30=89 minutes.

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.