ALGEBRA PROBLEMS

(WITH ANSWERS)

William Duncombe

Department of Public Administration

Fall 2009

PROBLEM 1

Assume you are a budget analyst for New York

State and collected the following information on

income tax revenues and the consumer price

index.

Year Personal Income Tax Revenue (billions) Consumer Price Index

2000 21.5 88.6

2001 26.9 90.7

2002 27.4 92.1

2003 23.7 94.1

2004 24.1 96.8

2005 28 100

2006 30.8 103.3

2007 34.6 106.2

2008 36.6 108.5

PROBLEM 1

a) What is the percent change in income tax

revenue from 2000 to 2008?

Answer: Using the formula for a percent change:

((36.6-21.5)/21.5)*100 =

70.2% increase

b) What is annual rate of inflation from:

(1) 2007 to 2008?

Answer: Since a price deflator measures the

change in prices for a fixed market basket, we

can use the same formula as percent change:

((108.5-106.2)/106.2)*100 = 2.16% inflation rate

PROBLEM 1

b) What is annual rate of inflation from: 2000 to 2008?

Answer: There are several ways you could get an

annual average, each which tells you something a

little different.

Approach 1: Take (((CPI in 2008/CPI in 2000)-

1)/8)*100 =

(((108.5/88.6)-1)/8)*100 = 2.81%

Approach 2: Calculate the inflation rate for each

year and take the average of these inflation rates.

You should be able to recreate using the table that

this average is 2.57%.

Approach 3: Use compound interest formula (use

approach in problem 6c).

PROBLEM 1

c) How do you remove inflation (deflate) from the

personal income tax revenue?

Answer: Converting financial information from

nominal dollars to inflation-adjusted (real)

dollars involves dividing the nominal value by

the price deflator divided by 100.

Income tax revenue in 2008 in 2002 dollars=

$36.6 billion/(108.5/100) = $33.73 billion.

We know that the deflator is in 2002 dollars

because the index value for 2002 is 100.

PROBLEM 1

d) How do you express income tax revenue in 2008 dollars?

Answer: To readjust a deflator to a new base, divide all deflator values by the deflator for the new base year. For example, to make 2008 the base, divide the CPI index by (108.5/100). For example, the CPI for 2007 would be:

106.2/(108.5/100) = 97.9 Income tax revenue in 2007 in 2008 dollars would be:

34.6/(97.9/100)= $35.3 billion

PROBLEM 1

e) What is the percent change in inflation-

adjusted income tax revenue from 2000

to 2008?

Answer: Now we are in position to look at

percent change in inflation-adjusted

dollars.

2008: $36.6 b. /(108.5/100) = $33.7 b

2000: $21.5 b. /(88.6/100) = $24.3 b

% change = ((33.7-24.3)/24.3) *100 =

38.7% increase in „real‟ tax revenue.

PROBLEM 2

Assume that you are trying to determine the

amount of revenue you are going to raise if you

raise user fees on bus tickets and water

consumption. You have collected the following

information on past fees (in inflation adjusted

dollars) and consumption.

2006 2008 2006 2008

Bus fare 1.5 1.9 5 3.5

Water consumption 0.1 0.13 175 150

Prices (inflation adjusted) Consumption

(millions of tickets)

(millions of gallons)

(per ticket)

(per gallon)

PROBLEM 2

Use this information to answer the following questions:

a) What is the “price elasticity” for water consumption and bus ridership?

Answer: a price elasticity measures the percent change in consumption over the percent change in price. For bus service this is:

((3.5-5)/5) / ((1.9-1.5)/1.5) = -30% / 26.7% = -1.125 A 1% increase in price of bus service is associated with a 1.125% drop in number of tickets sold.

For water:

((150-175)/175) / ((.13-.1)/.1) = -14.3% / 30% = -.476. A 1% increase in price of water is associated with a .476% decrease in water consumption.

PROBLEM 2b) Based on these elasticities, what do you think happened

to total revenue from for water and bus service after the fee increases in 2008? Calculate.

Answer: Revenue for user fees are simply the fee times consumption. For bus in 2008: 3.5 x 1.9 = $6.65

2006: 5 x 1.5 = $7.5 Revenue dropped because the reduction in consumption was faster than the increase in fare. This is consistent with a price elasticity less than -1 (price elasticity<-1).

For water: 2008: 150 x 0.13 = $19.5

2006: 175 x $0.1 = $17.5

Revenue increase because the drop in consumption was less than increase in price, which is consistent with a price elasticity greater than -1 (price elasticity>-1).

PROBLEM 3

a) What is the shape that best describes the

relationships between the following sets of

variables:

i. The amount of water you put on your garden and the

number of tomatoes you get?

Answer: It seems likely that initially when you add water you

will see an increase in yield but that eventually you will add

too much water and yield will go down an inverted U shape.

ii. The number of years since World War II and the

population of the United States?

Answer: Population during normal time generally grows at a

fairly constant growth rate, commonly called exponential

growth. In other words, for each year there is a larger

increase in the population than the previous year.

PROBLEM 3

iii. Your performance on a statistics exam and

the amount of studying you do (in hours)?

Answer: I would like to think that your

performance on the exam will go up with

more hours of studying. It is likely, however,

that the gains from each additional hour of

studying will get smaller and smaller. What

is commonly called “diminishing marginal

returns.”

PROBLEM 3

b) Based on the graph that you came up with, what type of algebraic expression best fits it:

i. The amount of water you put on your garden and the number of tomatoes you get?

ii. The number of years since World War II and the population of the United States?

iii. Your performance on a statistics exam and the amount of studying you do (in hours)?

Answer: See the following sheets that show the algebraic expressions and geometric graphs associated with these three examples. These graphs are linked to spreadsheets (click on them). I encourage you to change the constants and see what happens.

Constant a= b1= b2=

5 2 -0.25

X X2

b1 X b2 X2

Y=a+ b1 X+ b2 X2

1 1 2 -0.25 6.75

2 4 4 -1 8

3 9 6 -2.25 8.75

4 16 8 -4 9

5 25 10 -6.25 8.75

6 36 12 -9 8

7 49 14 -12.25 6.75

Algebraic Expression: Y= a + b1 X + b2 X2

Quadratic Example

0

1

2

3

4

5

6

7

8

9

10

1 2 3 4 5 6 7

YX

Quadratic Example

General form: Y = a e(b t)

Coefficient b1 a

0.01 50

X Y = a e(b1 X)

0 50.00

100 135.91

200 369.45

300 1004.28

400 2729.91

500 7420.66

600 20171.44

1%

1%

1%

Exponential Model Example

Annual Percent

Change in Y

1%

1%

1%

0.00

5000.00

10000.00

15000.00

20000.00

25000.00

0 100 200 300 400 500 600

Po

pu

lati

on

(Y

)

time (t)

Exponential Model Example

General form: eY = aX

b

Coefficient b a

0.05 1

X Y

Percent

Change

in X

Change

in Y

1 0.000

3.00 0.055 100.0% 0.05

9.00 0.110 100.0% 0.05

27.00 0.165 100.0% 0.05

81.00 0.220 100.0% 0.05

243.00 0.275 100.0% 0.05

729.00 0.330 100.0% 0.05

Semi-log Model Example

0.000

0.050

0.100

0.150

0.200

0.250

0.300

0.350

1 201 401 601 801

Y

X

Semi-log Model Example

PROBLEM 3

c) How would you modify this algebraic

expression so that it was a linear

expression?

i. The amount of water you put on your garden

and the number of tomatoes you get?

Answer: We would use a quadratic function:

Y = a + b1 X + b2 X2

Quadratic functions are already expressed in

linear form. If we call Z = X2 then this can be

rewritten as:

Y = a + b1 X + b2 Z

PROBLEM 3

c) How would modify this algebraic expression so that it was a linear expression?

ii. The number of years since World War II and the population of the United States?

Answer: You would use an “exponential function”, which is of the form: Y = a ebt

Take the natural log of both sides to get linear function:

ln(Y) = ln(a) + bt

iii. Your performance on a statistics exam and the amount of studying you do (in hours)?

Answer: You would use a “semi-log” function, which is of the form: eY = a Xb

Take the natural log of both sides to get a linear function:

Y = ln(a) + b ln(X)

PROBLEM 4:

BASIC ALGEBRAIC LAWS

Which of these is true?

1) x + y + c = c + y + x

2) x – y – c = c – y – x

3) x / (c ∙ y) = (c ∙x) / y

4) x ∙ y ∙ c = (c ∙ x) ∙ y

5) x - (c ∙ y) = (c ∙x) - y

6) x (y ∙ c) = (x ∙ y) c

7) x(b+c) = xb + xc

8) (x – b) (x + b) = x2 - b2

9) (x – b) (x + c) = x2 - bc

PROBLEM 4:

BASIC ALGEBRAIC LAWS

Which of these is true?

1) x + y + c = c + y + x T

2) x – y – c = c – y – x F

3) x / (c ∙ y) = (c ∙x) / y F

4) x ∙ y ∙ c = (c ∙ x) ∙ y T

5) x – (c ∙ y) = (c ∙x) – y F

6) x (y ∙ c) = (x ∙ y) c T

7) x(b+c) = xb + xc T

8) (x – b) (x + b) = x2 - b2 T

9) (x – b) (x + c) = x2 – bc F

Commutative law of

addition (not subtraction)

Commutative law of

multiplication (not division)

Associative law of

multiplication

Distributive law

PROBLEM 5: CALCULATING THE EFFECTS

OF A FARE INCREASE

You are a analyst for the Metropolitan Transit

Department. Because of a drop in other revenues,

the City Budget Office has asked you to raise fares

by 20%. Before responding to this request, the

Director wants you to estimate what will happen

to the demand for service and total revenue if this

fare increase is implemented. Before doing the

analysis, you talk one of your former Econ

professors, who tells you that the price elasticity of

demand for public transit is approximately -0.75.

Bus fares last year were $2.0 per trip and there

were 50,000 bus trips taken.

PROBLEM 5: CALCULATING THE EFFECTS

OF A FARE INCREASE

Answer: You are given the following information on price

(P), quantity (Q) and price elasticity (E):

P1 =$2, P2= 1.2 x 2 = $2.4, Q1 = 50,000, E = -0.75, R1 = P1 x

Q1 = $100,000. You are asked to solve for: Q2 and R 2

We can use the formula for a price elasticity:

(Q2 – Q1)/ Q1)/((P2 - P1)/ P1) = E, Solve for Q2 by:

(1) Multiplying both sides by ((P2 - P1)/ P1)

(Q2 – Q1)/ Q1)= E ((P2 - P1)/ P1).

(2) Multiplying by both sides by Q1 (Q2 – Q1)= E ((P2 - P1)/ P1) Q1

(3) Adding to both sides by Q1 Q2 = E ((P2 - P1)/ P1) Q1 + Q1

Now fill in with the numbers:, where ((P2 - P1)/ P1) = 20%

Q2 = -0.75 (.20) 50,000 + 50,0000 = $50,000 - $7,500 =

$42,500 and R 2 = P 2 x Q 2 = $42,500 x $2.4 = $102,000

PROBLEM 6: FUTURE VALUE, PRESENT

VALUE, AND COMPOUND INTEREST,

a) Suppose you invested $1000 in the bank this

year and were guaranteed an interest rate of 5%

per year. What is this investment worth in 2

years and in 20 years?

Answer: This problem is an application of the

concept of compound interest. Let‟s label,

PV=$1,000, r=5%, and FV is the value in the

future. We can express this question as:

2 years: FV = PV (1 + r)2= $1,000 (1.05)2 =

$1,102.5

20 years: FV = PV (1 + r)20= $1,000 (1.05)20 =

$2,653.3

PROBLEM 6: FUTURE VALUE, PRESENT

VALUE, AND COMPOUND INTEREST,

b) Assume that I offered you $10,000 in 10 years if you give me $7,500 now. Should you take this offer?

Answer: This is taking the previous problem and flipping it on its head. Instead of solving for the FV you want to know the PV of $10,000 given you 10 years in the future. If we assume r=5%, then:

FV = PV (1 + r)10= PV(1.05)10 = $10,000

Divide both sides by (1.05)10 :

PV = $10,000/ (1.05)10 = $10,000/1.63 = $6,139. In other words, $10,000 given to you 10 years in the future is only worth $6,139 today if interest rates are 5%. You should not take this offer!

PROBLEM 6: FUTURE VALUE, PRESENT

VALUE, AND COMPOUND INTEREST,

c) Going back to problem 1, what is compound annual growth rate in income tax revenue from 2000 to 2008? (Revenue is $21.5 billion in 2000 and $36.6 billion in 2008):

Answer: We are going to continue using the basic formula. Now we are given PV, and FV, and asked to solve for r. PV = $21.5 and FV=$36.6

FV = PV (1+r)8 .

(1) Divide both sides by PV (FV/PV) = (1+r)8

(2) Take both sides to the power of (1/8) (FV/PV)1/8

= (1+r)

(3) Subtract 1 from both sides (FV/PV)1/8 -1 = r

Plugging in the numbers in this case:

r = (36.6/21.5)1/8 -1 = .0688 = 6.88%

PROBLEM 7: FINDING THE MARKET

EQUILIBRIUM PRICE AND QUANTITY

a) Assume that the supply and demand curve for bread can be represented as:

Demand: X = 4,000 – 10 P

Supply: P = (X + 500)/20

Solve for the equilibrium price (P) and quantity (X). Plot the demand curve and supply curve and find the equilibrium.

Answer: The following are the steps you could take to solve this problem:

(1) Substitute the supply equation into the demand equation for P. X=4,000 – 10((X + 500)/20) and simplify: X=4,000 – (1/2)X -250= 3,750 -.5 X

(2) Add 0.5 X to both sides 1.5X=3750

(3) Divide both sides by 1.5 X = 2,500

(4) Substitute into the supply equation and solve:

P=(2,500 + 500)/20 = 150.

PROBLEM 7A) GRAPH OF SUPPLY AND

DEMAND EQUATIONS

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400

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1000

1200

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2800

3000

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3800

4000

Pric

e

Consumption

Demand

Supply

To calculate the graphs we need to express everything in the

form: P = f(X). You should be able to work out that the supply

and demand equations can be re-expressed as:

Demand: P = 400 - .1 X

Supply: P = 25 + 0.05 X

PROBLEM 7: FINDING THE MARKET

EQUILIBRIUM PRICE AND QUANTITY

b) Assume that the supply and demand curve for water can be represented as:

Demand: P = 5000 X-.75

Supply: X = 10000 P.5

Solve for the equilibrium price (P) and quantity (X).

Answer: You can find P and X with the following steps:

(1)Substitute the supply equation into the demand equation for X P = 5000 (10000 P.5) -.75

(2) Simplify: P = 5000 (10000) -.75(P.5) -.75 = 5 P-.375

(3) Multiply both sides by P-.375 P P.375 = 5 P-.375 P.375

P1.375 = 5 P(-.375 + .375)= 5 since P0 =1

(4) Take both sides to the power of (1/1.375) P = 5(1/1.375) = $3.22

(5) Substitute P into the supply equation

X = 10000 (3.22).5 = 17,944

PROBLEM 7B) GRAPH OF SUPPLY AND

DEMAND EQUATIONS

To calculate the graphs we need to express everything in the

form: P = f(X). You should be able to work out that the supply

and demand equations can be re-expressed as:

Demand: P = 5000 X-.75

Supply: P = (X /10000) 2

0

5

10

15

20

1

1100

2200

3300

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5500

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8800

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12100

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18700

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20900

22000

23100

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25300

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Pric

e

Consumption

Demand

Supply

PROBLEM 8: WORD PROBLEMS

a) The Zerocuse City School District hires

teachers and teaching assistants (TAs) to

provide instruction in their schools. In

2009, they used 100 teachers and 50 TAs,

who got paid $40,000 and $20,000 per year,

respectively. The City has limited the

growth in the district salary budget to 5% in

2010. Assuming that teacher salaries in

2010 are set by contract to be $45,000, what

would the salary of TAs have to be in 2010

to avoid laying off any staff? Is this

realistic?

PROBLEM 8: WORD PROBLEMS

a) Answer:

(1) List all of the information you are given on the number of teachers (T), teaching assistants (TA), their salaries (ST and STA), and the budget last year.

T1 = 100, TA1 = 50, ST1 =$40,000, STA1 = $20,000 and

B1 = T1 ST1 + TA1 STA1 = 100 ($40,000) + 50 ($20,000) = $4,000,000 + $1,000,000 = $5,000,000

(2) List what you are given for this year:

ST1 =$45,000, B2 = B1 (1.05) = $5,250,000, T2 = 100, TA2 = 50. You are asked to solve for STA2

(3) Set up the budget constraint for this year:

B2 = T2 ST2 + TA2 STA2 = 100 ($45,000) + 50 (STA2 ) = $4,500,000 + 50 (STA2 ) = $5,250,000

(4) Solve for STA2 : Subtract $4,500,000 from both sides and divide both sides by 50 ($5,250,000 - $4,500,000)/50 = STA2 = $15,000. This seems unrealistic because this would require a 25% pay cut for TAs.

PROBLEM 8: WORD PROBLEMS

b) You have taken over as the director of ZerocuseZoo. The City is facing a budget crisis and is asking the Zoo to breakeven. Costs for the Zoo can be broken down into fixed costs (a) and variable costs (with a variable cost rate of „b‟ per unit of output, X). Revenue for the Zoo is from zoo admissions (ticket price=p) and donations (D). Donations are negatively related to the ticket price. If the price goes up by $1, the donations drop by a fraction (e). If the ticket price was zero the level of donations would be „c‟. Using this information, find general formulas for the breakeven ticket price and level of donations assuming no reduction in ticket sales (X).

PROBLEM 8: WORD PROBLEMS

b) Answer:

(1) This question is set up deliberately without

any numbers. You should use symbols to

represent variables and constants. For

example, you are given the following symbols:

P = ticket price, D = Donations, a = fixed costs, b=

variable cost rate, X = number of tickets, c = level of

donations with ticket price equal to zero, e = fraction drop

in donations for $1 increase in ticket price.

(2) Using this you can set up the following

equations:

Breakeven (BE) equation: P X + D = a + b X

Donation equation: D = c – e P

PROBLEM 8: WORD PROBLEMS

b) Answer:

(3) Substitute the donation equation into the BE equation for D

PX + (c – eP) = a + bX P(X-e) + c = a + bX

(4) Subtract c from both sides

P(X-e) = a + b X – c

(5) Divide both sides by (X-e)

P = (a + bX – c)/(X-e)

(6) Substitute P back into the donation equation:

D = c – e (a + bX – c)/(X-e)).

While this is messy looking, everything on the right hand side you have available.