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Q#1

The Auditor of ABC and Co. employs his assistant to count cash in hand of $4500. At first he counts quickly at the rate of $150 pr minute for 10 minutes only, but at the end of that time he begins to count the rate of $2 less every minute than he could count in the previous minute. Find the minimum time that the assistant takes to count the sum.(by using A.P and G.P method)

Q#2

The annual salary increment of a monthly salaried person is the arithmetic progression. It is known that he drew monthly salary of $10,000 in the year 1978 and $14,000 in the year 1994. He started his service in the year 1965 and shall attain the age of superannuation in the year 2004. Calculate superannuation in the year 2004. Calculate the monthly salary with which he started the job in the year 1965 and also the monthly salary in the year of his superannuation.

Q.1: The formula for an AP is (where An is the n-th element in the sequence)

An = An-1 + d or An = A1 + (n-1)d.

We also have the formula for the sum of n terms of the sequence

Sn = (n/2)(A1 + An) = (n/2)[2A1 + (n-1)d] (you can and should look up the derivation on the internet; note that we substituted the expression for An from above).

The assistant has to count out a total of $4500. For the first part, he counts at 150 $/hr. Without having to go into AP or GP theory, it is easy to see that if he counts for 10 minutes he will have counted 10･150 = $1500, which leaves $3000 left to count.

The first part of the problem can be considered as an AP if, in the formula for an AP, we take d = 0 which makes An = A1. This gives the same answer as above, i.e., Sn = nA1 = 10･150.

Counting the remaining $3000 is more interesting. Using d = 2 as the difference for the AP, the sum formula becomes

-n^2 + (A1+1)n - Sn = 0 or n^2 + n(A1+1) - Sn.

This is a quadratic in n; using the well-known quadratic formula, and setting A1 = 150 and Sn = 3000, we get n = 23.5

Thus the assistant must count for a minimum of 10 + 24 = 34 minutes.

You should confirm the calculation. It would also be a good exercise for you to put together a simple Excel program to plot the sequence.

Q.2: Given the monthly salary for the years 1978 and 1994 (difference of 16 years), we calculate that his monthly rate (MR) increases by $250/year. Extending this backwards and forwards using the AP formula gives

MR1965 = 10000 - (13)250 = 6750 and MR2004 = 14000 + (10)250 = 16500.

This seems too easy so maybe I'm missing something. I also don't know how to interpret "superannuation" for this problem. Please follow-up if you have more info.

Randy

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