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

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

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(A) The number of notes counted in the first $10$ minutes is $150 	imes 10 = 1500$.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 follows an $A.P.$ starting from $a_{11} = 150 - 2 = 148$ with common difference $d = -2$.The sum of $n$ terms of this $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Substituting the values: $3000 = \frac{n}{2}[2(148) + (n-1)(-2)]$.$3000 = \frac{n}{2}[296 - 2n + 2] = \frac{n}{2}[298 - 2n] = n(149 - n)$.$3000 = 149n - n^2$,which gives the quadratic equation $n^2 - 149n + 3000 = 0$.Factoring the equation: $(n - 24)(n - 125) = 0$.This gives $n = 24$ or $n = 125$.Since the number of notes counted cannot be negative,we check the term $a_{10+n} = 148 + (n-1)(-2)$. For $n=125$,the term becomes negative,which is not possible. Thus,$n = 24$.Total time = $10 + 24 = 34$ minutes.

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