Algebra> A person is to count 4500 currency notes....

A person is to count 4500 currency notes. Let a n denote the number of notes he counts in the n th minute. If a1 = a 2 = .... = a 10 = 150 and a 10 , a 11 ,... are in an AP with common difference – 2, then the time taken byhim to count all notes is

saket kumar , 10 Years ago

Grade 12

3 Answers

SHAIK AASIF AHAMED

Last Activity: 10 Years ago

suppose he takes n minutes to count 4500 notes.

If a1 = a 2 = .... = a 10 = 150

then a1+a2+........+a10=1500

hence a11+a12+.....+an=3000

$\frac{n-10}{2}(a11+an)=3000$

$\frac{n-10}{2}(148+148+(n-11)(-2))=3000$

n²-169n+4590=0

n=135,34

so all the notes get counted in 34 minutes

Thanks and Regards

Shaik Aasif

askIITians Faculty

Fkr

Last Activity: 6 Years ago

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Rishi Sharma

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

suppose he takes n minutes to count 4500 notes.
If a1 = a 2 = .... = a 10 = 150
then a1+a2+........+a10=1500
hence a11+a12+.....+an=3000
0.5*(n-10)*(a11+an) = 3000
0.5*(n-10)*(148+148+(n-11)(-2)) = 3000
n^2 – 169n + 4590 = 0
n = 135,34
so all the notes get counted in 34 minutes

Thanks and Regards