A person is to count 4500 currency notes. Let | vedclass

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(A) Notes counted in the first $10$ minutes $= 150 	imes 10 = 1500$.Remaining notes to be counted $= 4500 - 1500 = 3000$.Let $n$ be the number of min

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(A) Notes counted in the first $10$ minutes $= 150 	imes 10 = 1500$.Remaining notes to be counted $= 4500 - 1500 = 3000$.Let $n$ be the number of minutes after the first $10$ minutes.The sequence of notes counted from the $11^{th}$ minute onwards is an $A.P.$ with first term $a = 148$ and common difference $d = -2$.The sum of $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.$3000 = \frac{n}{2}[2(148) + (n-1)(-2)]$.$3000 = \frac{n}{2}[296 - 2n + 2] = \frac{n}{2}[298 - 2n] = 149n - n^2$.$n^2 - 149n + 3000 = 0$.$(n - 24)(n - 125) = 0$.Since $a_{10+n} = 148 + (n-1)(-2) = 150 - 2n$,for $n=125$,$a_{135} = 150 - 250 = -100$,which is impossible as the number of notes cannot be negative.Thus,$n = 24$.Total time $= 10 + 24 = 34$ minutes.

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