QUESTION 7 An engineering working for the North Dakota DOT is making $59,000/yr.
ID: 2440677 • Letter: Q
Question
QUESTION 7
An engineering working for the North Dakota DOT is making $59,000/yr. Due to state budget cuts no raises will be given at the end of this year, or the following year. Inflation has been 2.5% for this region. Note- the current year salary (59K) is collected at n=1. The salary then grows by 0% at n=2, and 0% again for n=3.
A) What is the Year 0 dollars (purchasing power) equivalent of the year 3 salary?
B) The engineer was also given a sign on bonus in year 0, of $10,000. If the engineer had deposited this into an account earning 10%, what is the account balance after 10 years, in Year 0 Dollars?
C) How many Lottery Tickets can the engineer purchase at the end of year 10 with the savings account, knowing Lottery Tickets cost $10 each currently. Note only whole lottery tickets can be purchased.
Explanation / Answer
A) For finding out the purchasing power of Year 3 salary, we will find out by using the following formula:
Y(n)= Amount *[1/(1+ rate of inflation)]^n
Y(3)= 59,000 * [1/ (1+ 2.5%)] ^ 3
= 59,000 * [ 1/ (1.025)] ^3
= $ 54, 787.365
So, the purchasing power in year 0 =$ 59,000
the purchasing power in year 3 = $54,787.365
B) We will consider two cases under this question:
CASE 1: When the interest rate is taken in nominal terms i.e. nominal interest rate.
CASE 2: When interest rate is taken in real terms i.e. real interest rate. (taking into account the inflation rate for the region)
So, Bonus that he received = $10,000
nominal interest rate= 10%
Assume that the interest rate is given on per annum basis;
therefore interest provided for one year:
0.1*10,000= $1000
so, for 10 years the interest will be:
10*1000= $10,000
So, after 10 years the account will have the following balance:
$10,000+$10,000 (amount deposited + interest for 10 years)
=$20,000
Therefore, ignoring the inflation rate, he will have $20,000 after 10 years.
CASE 2: Since the rate of inflation is also given, we will consider the next case:
According to Fischer's equation:
nominal interest rate= real interest rate + rate of inflation
or, real interest rate= nominal interest rate - rate of inflation
or, real interest rate= 0.1-0.025 ( nominal interest rate= 10% , inflation rate = 2.5%)
= 0.075
= 7.5%
Therefore ,real interest for one year = 0.075 * 10,000
=$ 750
and real interest for 10 years= 750*10
=$ 7500
So,real balance in the account after 10 years is =$10,000 +$7500
=$ 10,750
C) current price of the lottery ticket = $10
Number of lotteries purchased if we consider CASE 1 ( Savings account has 20,000 in balance)
= 20,000/100
= 2000 tickets
Number of lotteries purchased if we consider CASE 2 (Savings account has real balance of $10,750)
=10,750/10
= 1075 tickets (on taking account the rate of inflation)
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