Question

Assume that a small and poor DVC has an a population of 17,000 and a total national income of $3.7 million.

a) If the population grows by 4% during the year, while the total national income grows 3%, what will be the new per capita?

b) If the population didn't grow, but income still grew by 3%, what would be the per capita?

Question 1 options:

	a) $218 b) $224
	a) $215, b) $220
	a) $218, b) 220
	a) $215, b) $224

Compare a hypothetical DVC with a hypothetical IAC. In the DVC, average per capita income is $5000 per year. In the IAC, average per capita income is $42,000 per year. If both countries have a savings rate of 8 percent per year, what is the savings amount per person per year?

Question 2 options:

	DVC = $400, IAC = $3360
	DVC = $500, IAC = $4200
	DVC = $5000, IAC = $42,000
	DVC = -$400, IAC = -$3360

Assume a DVC and an IAC currently have real per capita outputs of $3000 and $30,000, respectively. Further assume that both nations have a 2% increase in their real per capita outputs.

a) Find the per capita for each country.

b) Calculate the output gaps before and after the growth.

Question 3 options:

	a)DVC = $3600, IAC = $36,000. b)Before Growth = $27,000, After Growth = $32,400
	a)DVC = $600, IAC = $6000. b)Before Growth = $27,000, After Growth = $5400
	a)DVC = $3060, IAC = $30,600. b)Before Growth = $27,000, After Growth = $660
	a)DVC = $3060, IAC = $30,600. b)Before Growth = $27,000, After Growth = $27,540

Suppose it takes a minimum of 5 units of food to keep a person alive for a year, the population can double itself every 10 years, and the food supply can increase every 10 years by an amount to what it was in the beginning (year 0). In Year 0, the food supply is 300 units, and the population is 15. Create a table to answer the following questions

a) In the 40th year, does the food supply meet the needs of the population?

b) In the 60th year, does the food supply meet the needs of the population?

c) In year 80, what does the size of the population needs to be in order to not run out of food?

Question 4 options:

	a) YES, b) NO, c) 3840 people
	a) YES, b) YES, c) 540 people
	a) YES, b) YES, c) 3840 people
	a) YES, b) NO, c) 540 people

Answer 1

Solution:

1). Income per capita = total national income/population

So, income per capita = 3,700,000/17,000 = $217.65

So, growth rate in income per capita = growth rate in total national income - growth rate in population

Growth rate in income per capita = 3% - 4% = -1% or -0.01

So, new per capita income = 217.65*(1 - 0.01) = 215.47

With no growth in population, and still a 3% increase in total income, growth in per capita = 3% - 0% = 3% = 0.03

So, new per capita income = 217.65*(1 + 0.03) = 224.18

Thus, the correct option is (D) a) 215, b) 224

2). Savings amount per person = savings rate*average per capita income

So, for DVC, savings amount per person = 0.08*5000 = 400

For IAC, savings amount per person = 0.08*42000 = 3360

Thus, the correct option is (A) DVC = $400, IAC = $3360.

3). New per capita = initial per capita*(1 + growth rate in per capita)

Then, with growth rate of 2%:

For DVC, per capita = 3000*(1 + 0.02) = 3060

For IAC, per capita = 30000*(1 + 0.02) = 30600

Output gap before growth = 30000 - 3000 = 27000

Output gap after growth = 30600 - 3060 = 27540

Thus, the correct option is (D) a) DVC = $3060, IAC = $30,600. b) Before Growth = $27,000, After Growth = $27,540.

4). Note that the food supply increases by a particular value every 10 years and the beginning year is 0, so this generates an arithmetic progression. So, value of food supply at nth year (where n is a multiple of 10) = a + (n/10)*d ; where a is the initial value, 300 here, and d is the common difference, which again is 300 here (increment is the food supply at year 0 which is 300 units).

Similarly, population increases by a particular rate or ratio every 10 years and again, the beginning year is 0, so this generates a geometric progression. So, value of population at nth year (where n is a multiple of 10) = a*r^n/10 ; where a is the initial value, 15 here, and r is the common ratio, which is 2 here (the value 'doubles' every 10 years).

Finally, as it takes at least 5 units of food an year to keep a person alive, in order to meet the needs of population, we must have population*5 <= food supply

a) In the 40th year, n = 40

So, food supply = 300 + (40/10)*300 = 1500 units

And, population = 15*2^40/10 = 240

240*5 = 1200 < 1500, so yes the food needs of the population is met.

b) In the 60th year, n = 60

So, food supply = 300 + (60/10)*300 = 2100 units

And, population = 15*2^60/10 = 960

960*5 = 4800 > 2100, so no the food needs of the population is met.

c) In year 80, food supply = 300 + (80/10)*300 = 2700 units

In order to meet the needs of the population, population size <= 2700/5 = 540. So, population size must be at most 540

Thus, the correct option is (D) a) YES, b) NO, c) 540 people