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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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.

Similar Questions

If the sum of the series $54 + 51 + 48 + \dots$ is $513$,then the number of terms is:

If the four angles of a quadrilateral are in an arithmetic progression and their common difference is $10^{\circ}$,what are the measures of the angles of the quadrilateral?

If the ratio of the sum of $n$ terms of two $A.P.s$ is $(7n + 1):(4n + 27)$,then the ratio of their $11^{th}$ terms is:

Consider an $A$.$P$.: $a_1, a_2, \dots, a_n$, with $a_1 > 0$. If $a_2 - a_1 = -\frac{3}{4}$, $a_n = \frac{1}{4} a_1$, and $\sum_{i=1}^n a_i = \frac{525}{2}$, then $\sum_{i=1}^{17} a_i$ is equal to:

If the sum of $n$ terms of an $A.P.$ is $(pn + qn^2),$ where $p$ and $q$ are constants,find the common difference.

Vedclass Test Series

Exam Paper Generator

Online Exam Module