Week 2 Quiz: Equations and Graphs, Functions, and Systems of Equations
SGPE Summer School 2014
June 24, 2014
Lines: Slopes and Intercepts
Question 1: Find the slope, y-intercept, and x-intercept of the following line:
15x − 3y = 60
(A) slope= 5, y-intercept= −20, and x-intercept= 4
(B) slope= 6, y-intercept= −22, and x-intercept= 5
(C) slope= 51 , y-intercept= 20, and x-intercept= −4
(D) slope= 5.5, y-intercept= −20, and x-intercept= 8
(E) None of the above
Answer: (A) Given the form of the equation that we are given, it makes sense to go ahead and solve for the x
and y intercepts. Solving for the y-intercept requires setting x = 0 in the above equation and then solving for y.
Thus 15(0) − 3y = 60 which implies that y = −20. Similarly we can find the x−-intercept by setting y = 0 and
solving for x, which yields 15x − 3(0) = 60, or x = 4. The easiest way to find the sole is just to convert the above
equation to slope-intecept form (i.e, of the form y = mx + b). Follow the algebra:
15x − 3y = 60
−3y = −15x + 60
y = 5x − 20
Now we can just read off the slope from this expression. The slope is 5.
Question 2: Find the equation of a straight line with slope of 12, and y-intercept of -33.
(A) y =
1
12 x
+ 33
(B) y = −12x + 33
(C)y = 12x − 33
(D) y = 12x −
1
33
(E) None of the above
Answer: (C) Simply substituting m = 12 and b = −33 into the slope intercept form of a line yields y = 12x − 33.
Question 3: Find the equation of a straight line passing through the point (-2, 7) and perpendicular to a line
with equation 24x + 6y = 30.
(A) y = 4x +
13
2
(B) y = − 14 x +
15
2
(C)y = −4x + 15
1
(D) y =
1
12 x
+
15
2
(E) None of the above
Answer: (E) First we need to figure out the slope of our equation. We are told that it is perpendicular to
a line whose equation is 24x + 6y = 30. What is the slope of this line? Solving for the slope intercept form
yields y = −4x + 5. Thus the slope is −4, which means that the slope of our equation is 14 (i.e., the negative
reciprocal). Now we need to find the y-intercept. To do this we first write down the point-slope form of the line:
(y − y1 ) = m(x − x1 ). Now substitute for m, x1 and y1 using the slope (which we just calculated) and the point
(which we were given). This yields:
1
y − 7 = (x − (−2))
4
Now we simply re-arrange this equation to slope-intercept form to get the answer:
1
y − 7 = (x − (−2))
4
1
1
y−7= x+
4
2
1
15
y = x+
4
2
Question 4: Find the x intercept of the following equation:
−13x + 19y = −39
Answer: Substitute 0 for y:
−13x + 19 ∗ 0 = −39 ⇔ x = 3
Therefore the x intercept is (3, 0).
Question 5: Find the y intercept of the following equation:
y = 79.5x − 250
Answer: Substitute 0 for x:
y = 79.5 ∗ 0 − 250 ⇔ y = −250
Therefore the y intercept is (0, −250).
Question 6: Find the x intercept of the following equation:
1
−5x − y = 55
2
Answer: Substitute 0 for y:
−5x −
1
∗ 0 = 55 ⇔ x = −11
2
Therefore the x intercept is (−11, 0).
Question 7: Derive the equation for a straight line with: a) Slope: − 32 and b) y intercept: (0, 49).
Answer: y = − 23 x + 49.
2
Linear Equations in Business and Economics
Question 8: Find the total cost of producing (a) 25 units, and (b) 40 units of output for a firm that has fixed
costs of $3200 and a marginal cost of $500 per unit.
(A) 15,700 and 23,200
(B) 16,000, and 23, 000
(C) 24,000 and 54,000
(D) 32,000 and 43,500
(E) None of the above
Answer: (A) First need to find the total cost function. Total cost, T C, is the sum of total fixed costs, T F C, and
total variable costs, T V C. Total variable costs, T V C, depend on the level of output, Q, that the firm chooses
to produce. Mathematically, we say that T V C is a function of Q, and write T V C(Q). Since T C is a function of
T V C(Q), this means that T C is also a function of Q. This is denoted by T C(Q). Now we can write a general
total cost functions as follows:
T C(Q) = T F C + T V C(Q)
We are told that T F C = $3200 and that the only type of variable costs are marginal costs, and that these marginal
costs amount to $500 per unit output, which implies that T V C(Q) = $500Q. Substituting into the above equation
yields a total cost function of:
T C(Q) = 3200 + 500(Q)
To answer the question we simply need to evaluate the total cost function we have derived at Q = 25, and Q = 40.
This yields answers of T C = 3200 + 25(500) = 15, 700 and T C = 3200 + 40(500) = 23, 500.
Question 9: Find the profit level of a firm in perfect competition that has a fixed cost of $900, a variable cost of
$80, and a selling price for their output of $90 per unit if they sell (a) 55 units, and (b) 100 units.
(A) 350 and 100
(B) -350 and -100
(C) -350 and 100
(D) 4950 and 9000
(E) None of the above
Answer: (C) First need to find the profit function. Profit (π) is always Revenue (R) less Total Costs (TC).
π = R − TC
From solution to the previous question, we know that T C(Q) = T F C + T V C(Q). Thus T C(Q) = 900 + 80Q.
What about revenue? First note that a firm’s revenue is a function of the quantity of output produced (i.e., R(Q)).
In this case we are told that the firm brings in $90 per unit output, thus their revenue function is R(Q) = 90Q.
Now we have enough information to write the firm’s profits as a function of quantity produced:
π = R(Q) − T C(Q) = 90Q − (900 + 80Q) = 10Q − 900
Now we need only to evaluate the profit function at Q = (a) 55 units, and (b) 100 units. Which yields profits of
π = 10(55) − 900 = −350 (Negative profits indicates firm is operating at a loss!); and π = 10(100) − 900 = 100.
Question 10: Assume that there are no salvage (or liquidation) costs. Find (a), the value of an asset after 6
years and (b), the salvage (sometimes also called the liquidation) value after 100 years of an asset whose current
value A after t years is:
A = 1, 500, 000 − 1, 500t
3
(A) 1,492,000 and 1,360,000
(B) 1,491,000 and 1,350,000
(C) 1,591,000 and 1,356,000
(D) 1,191,000 and 1,400,000
(E) None of the above
Answer: (B) Note that given there are no liquidation costs, the salvage (or liquidation) value of an asset after
t years is simply the current value of the asset after t years. Thus we need only evaluate the expression for
current value after 6 and 100 years respectively. This yields A = 1, 500, 000 − 1, 500(6) = 1, 491, 000 and A =
1, 500, 000 − 1, 500(100) = 1, 350, 000.
Question 11: With $100,000 to invest, how much should a broker invest in a Portuguese bond that pays 11% and
how much in an Italian bond that pays 15% in order to earn an expected return of 14% on the total investment?
(A) 23,000 in Portuguese bond and 77,000 in Italian bond
(B) 50,000 in Portuguese bond and 50,000 in Italian bond
(C) 75,000 in Portuguese bond and 25,000 in Italian bond
(D) 35,000 in Portuguese bond and 65,000 in Italian bond
(E) None of the above
Answer: (E) If x denotes the amount of wealth invested in the Portuguese bond, then (100, 000 − x) denotes the
amount of wealth invested in the Italian bond. We can express the the total annual interest on the portfolio, r as:
r(100, 000) = .11x + .15(100, 000 − x) = 15, 000 − 0.04x
Given that our broker wants to achieve an expected return of r = 14%, we simply need to solve for x as follows:
.14(100, 000) =15, 000 − 0.04x
14, 000 =15, 000 − .04x
.04x = 1, 000
x = 25, 000
Thus our broker should invest $25,000 in the Portuguese bond paying 11% and $75,000 in the Italian bond paying
15%.
Solving Quadratic Equations
Question 12: Solve the following quadratic equation:
x2 + 6x + 9 = 0
(A) x = 3 and x = −3
(B) x = 3
(C) x = −3
(D) x = 2 and x = 3
(E) None of the above
Answer: (C) Easiest way to solve this equation is NOT to use the quadratic formula, but to simply factor the
polynomial on the left-hand-side of the equation. This yields (x + 3)2 = 0, which has a double root of x = −3.
4
Question 13: Solve the following quadratic equation:
5x2 + 47x + 18 = 0
(A) x = − 25 and x = −9
(B) x = − 52 and x = 9
(C) x =
2
5
and x = 9
(D) x =
2
5
and x = −9
(E) None of the above
Answer: (A) Solve by applying the quadratic equation. Follow the algebra:
p
−47 ± 472 − 4(5)(18)
x=
2(5)
√
−47 ± 2209 − 360
=
√10
−47 ± 1849
=
10
−47 ± 43
=
10
4
2
−90
= − = − and
= −9
10
5
10
Question 14: Solve the following quadratic equation:
1
3
x2 + x + = 0
4
8
Answer: Solve by applying the quadratic equation formula:
q 3 2
− 34 ±
− 4(1)
4
x=
2(1)
q
9
− 34 ± 16
− 12
=
2
q
3
1
− 4 ± 16
=
2
− 34 ± 14
=
2
1
1
1
=−2 =− ∨=−
2
4
2
Question 15: Solve the following quadratic equation:
4x2 − 31x − 45 = 0
5
1
8
Answer: Solve by applying the quadratic equation formula:
p
31 ± (−31)2 − 4(4)(−45)
x=
2(4)
√
31 ± 961 + 720
=
√8
31 ± 1681
=
8
31 ± 41
=
2
72
10
=
= 9∨ = − = 1.25
8
8
Question 16: Solve the following quadratic equation:
x2 − 67.5x + 126 = 0
Answer: Solve by applying the quadratic equation formula. Note that the square root is not an exact square in
this case, and has been rounded to 2 decimal places:
p
67.5 ± (67.5)2 − 4(1)(126)
x=
2(1)
√
67.5 ± 4556.25 − 504
=
√ 2
67.5 ± 4052.25
=
2
67.5 ± 63.66
=
2
131.16
3.84
=
= 65.58∨ =
= 1.92
2
2
Non-linear Functions
Question 17: Items such as automobiles are subject to accelerated depreciation whereby they lose value faster
than they do under linear depreciation. Suppose that a car with initial value of $100, 000 value depreciates at 10%
per year (continuously compounded) (i.e., the cars value after t years is V (t) = 100000e−0.10t ). Further, suppose
that there always exists the option to sell the car for scrap to a “chop-shop” for $2000. After how many years will
in be optimal to sell the car for scrap?
(A) 39
(B) 41
(C) Never!
(D) 40
(E) None of the above
Answer: (D) We want to find after how many years t will the scrap value of the car exceed its value in year t,
6
V (t). To solve this question we need to solve the following inequality:
2000 ≥ 100000e−0.10t
1 ≥ 50e−0.10t
1
≥ e−0.10t
50
1
ln
≥ ln e−0.10t
50
ln(1) − ln(50) ≥ −0.10t
t ≥ 10ln(50)
t ≥ 39.12 ≈ 40 years!
So it would be optimal to sell the car for scrap after about 40 years.
Question 18: A factory’s cost function C(Q) is a function of the number (quantity) of units produced; suppose
that the quantity of units produced is itself a function of time, Q(t). Specifically, assume that C(Q) = 1900 + 50Q
and that Q(t) = 16t − 14 t2 . Find the function that expresses the factory’s cost as a function of time, and then find
out the factory’s costs are after t = 1 and t = 10 periods.
(A) 1900 + 300t − 12t2 ; C(1) = 2188; C(10) = 1000
(B) 1900 + 800t −
25 2
2 t ;
C(1) = 2687.5; C(10) = 40650
(C) 1900 + 800t − 10t2 ; C(1) = 2690; C(10) = 31900
(D) 1900 + 300t −
25 2
2 t ;
C(1) = 2187.5; C(10) = 3650
(E) None of the above
Answer: (E) To find the factory’s costs as a function of time, which is a composition of the factory’s cost function
and the quantity function. Simply substitute the quantity function Q(t) in for Q in the cost function as follows:
1
25
C(t) = C(Q(t)) = 1900 + 50(16t − t2 ) = 1900 + 800t − t2
4
2
Now we need only evaluate this composite function C(t) = C(Q(t)) at t = 1 and t = 10. C(1) = 1900 + 800 − 25
2 =
25
2
2700 − 12.5 = 2687.5; C(10) = 1900 + 800(10) − 2 (10) = 1900 + 8000 − 1250 = 8650.
Question 19: The average level of carbon monoxide in the air is given by the function L(n) = (1 + 0.6n) parts
per million (ppm), where n is the number of people (in thousands). Assuming the population in thousands at
time t is N (t) = 400 + 30t + 0.15t2 , a) express the level of carbon monoxide in their air as a function of time, and
b) estimate the level of carbon monoxide at time t = 5.
Answer: First substitute the function N (t) for n in the function L(n), to get the composite function L(N (t)):
L(N (t)) = 1 + 0.6(400 + 30t + 0.15t2 ) = 241 + 18t + 0.09t2 .
Then to solve part b) find L(N (5)) by substituting 5 got t:
L(N (5)) = 241 + 18 ∗ 5 + 0.09 ∗ (5)2 = 333.25ppm.
Systems of Equations
Question 20: Suppose that supply and demand are described by the following set of equations:
Supply: Qs = 2P − 2
8
Demand: Qd = − P + 16
5
7
Find the market clearing price and quantity.
(A) Pe = 6; Qe = 9
(B) Pe = 4; Qe = 7
(C) Pe = 5; Qe = 8
(D) Pe = 8; Qe = 5
(E) None of the above
Answer: (C) To find the market clearing price, Pe , simply equate quantity demanded with quantity supplied and
solve for P . Thus...
8
(Supply) Qs = 2Pe − 2 = − Pe + 16 = Qd (Demand)
5
8
2P + Pe = 16 + 2
5
18
Pe = 18
5
Pe = 5
To find Qe , simply plug Pe = 5 into both the equation for supply and demand. Why both equation? Shouldn’t you
get the same answer by plugging the market clearing price into either equation? YES! But plugging the market
clearing price into both equations is an easy way to check and make sure that you haven’t made a silly maths mistake
somewhere. Thus Qd = 2(5) − 2 = 8, and Qs = − 58 (5) + 16 = 8. And we have found Qe = Qs (Pe ) = Qd (Pe ) = 8.
Question 21: Assume a revenue function of R(Q) = 48Q − 3Q2 and a cost function of C(Q) = 6Q + 120. Find
the break even level (or perhaps levels) of Q.
(A) 3 and 11
(B) 4 and 10
(C) 2 and 9
(D) 6 and 13
(E) None of the above
Answer: (B) There are multiple ways to solve this problem. I prefer to use the revenue and cost functions to
derive the profit function, and then fund those values of Q for which profits are zero. If profits are zero, then
revenue exactly covers costs and the firm is breaking even. Profits, π(Q) = R(Q)−C(Q) = 48Q−3Q2 −6Q−120 =
42Q − 3Q2 − 120. Setting profits equal to zero, and solving the quadratic equation will gives us the break-even
levels of Q. Thus set π(Q) = −3Q2 + 42Q − 120 = 0 and solve.
p
−42 ± 422 − 4(−3)(−120)
Q=
2(−3)
√
−42 ± 324
Q=
−6
−42 ± 18
Q=
−6
Q = 4 and Q = 10
Question 22: Solve the following system of equations for the equilibrium levels of income, Y and interest rate, i
(round your answers to the nearest hundredth):
0.1Y + 80i − 75 = 0
0.4Y − 124i − 260 = 0
8
(A) Ye = 677.93; ie = 0.09
(B) Ye = 505.84; ie = 0.22
(C) Ye = 604.83; ie = 0.11
(D) Ye = 605.99; ie = 0.24
(E) None of the above
Answer: (A) Easiest way to solve the above equation is to use the substitution method. Solving the first equation
for Y yields:
0.1Y + 80i − 75 = 0
Y + 800i − 750 = 0
Y = −800i + 750
Substituting this expression for Y into the second equation allows us to solve for the equilibrium level of the
interest rate.
0.4(−800ie + 750) − 124ie − 260 = 0
−320ie + 300 − 124ie − 260 = 0
−444ie + 40 = 0
40
ie =
≈ 0.09or9%
444
(1)
40
Now using ie = 444
, we can substitute
it into either (or both!) ofthe above equation to find the equilibrium level of
40
40
income Y . Thus 0.4Ye − 124 444
− 260 = 0 and 0.1Ye + 80 444
− 75 = 0, both equations imply that Ye ≈ 677.93.
Question 23: Solve the following system of equations for the the equilibrium level of income, Ye , in terms of
given levels of government spending, G = G0 , and investment, I = I0 .
Y =C +I +G
C = C0 + bY where 0 < b < 1
(A) Ye =
1
b−1 (C0
+ I0 + G0 )
(B) Ye =
1
1−b (C0
+ I0 + G0 )
(C) Ye =
1−b
(C0 +I0 +G0 )
(D) Ye = (1 − b)(C0 + I0 + G0 )
(E) None of the above
Answer: (B) Substitute for C, G, and I in the first equation above, and then solve for Y.
Y =C +I +G
Ye = (C0 + bYe ) + I0 + G0
Ye − bYe = C0 + I0 + G0
Ye (1 − b) = C0 + I0 + G0
1
(C0 + I0 + G0 )
Ye =
1−b
Question 24: Solve the following system:
(
24x − 7y = 37
−6x + 9y = 27
9
Answer: Use substitution:
(
(
24x − 7y = 37
24x − 7y = 37
⇔
−6x + 9y = 27
x = 23 y − 92
(
24 23 y − 92 − 7y = 37
⇔
x = 32 y − 92
(
29y − 145 = 0
⇔
x = 23 y − 92
(
y=5
x = 23 ∗ 5 −
⇔
(
y=5
⇔
x=3
9
2
Question 25: Solve the following system:
(
y = 0.75x − 4
y = 2.5x − 11
Answer: In this case, as both equations as given with y already isolated, simply set the two left hand sides equal
to each other to find x. Then substitute x in either equation to find y.
(
(
y = 0.75x − 4
x=4
⇔ 0.75x − 4 = 2.5x − 11 ⇔ 1.75x = 7 ⇔ x = 4;y = 2.5 ∗ 4 − 11 = −1 ⇔
y = 2.5x − 11
y = −1
Question 26: Solve the following system:
(
18x − 2y = 32
12x + 5y = −23
Answer: Use substitution:
(
(
18x − 2y = 32
y = 9x − 16
⇔
12x + 5y = −23
12x + 5y = −23
(
y = 9x − 16
⇔
12x + 5(9x − 16) = −23
(
y = 9x − 16
⇔
⇔
57x − 57 = 0
(
(
y = 9 ∗ 1 − 16
y = −7
⇔
x=1
x=1
Question 27: Solve the following system:
(
11x + 3y = 53
4x − 18y = 172
Answer: Use substitution:
(
(
11x + 3y = 53
11x + 3y = 53
⇔
4x − 18y = 172
x = 43 + 4.5y
(
11(43 + 4.5y) + 3y = 53
⇔
x = 43 + 4.5y
(
52.5y + 420 = 0
⇔
x = 43 + 4.5y
(
y = −8
x = 43 + 4.5 ∗ −8
10
⇔
(
y = −8
⇔
x=7