Week 2 Quiz: Equations and Graphs, Functions, and Systems
of Equations
SGPE Summer School 2016
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
x
12
+ 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
1
(B) y = − 41 x +
15
2
(C)y = −4x + 15
(D) y =
1
x
12
+
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 and y intercepts (crossing points) of the following line in terms of a,b and c:
ax − by = c
Answer: Setting y = 0 gives x intercept. So, ax − 0 = c and x = c/a. Similarly, setting x = 0 gives
us y intercept which is y = c/b.
Question 5: 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
2
Thus our broker should invest $25,000 in the Portuguese bond paying 11% and $75,000 in the Italian
bond paying 15%.
Question 6: Write down the equation of the line given below:
y axis
(2,3)
(0,1)
x axis
Answer: A line can be represented y = mx + n. One point the line passes through is (0, 1). So putting
x = 0 and y = 1 gives us n = 1 and y = mx + 1. Similarly, the line passes through (2, 3), therefore
3 = 2m + 1 and m = 1. Finally, y = x + 1.
Solving Quadratic Equations
Question 7: 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.
Question 8: Solve the following quadratic equation:
5x2 + 47x + 18 = 0
(A) x = − 52 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
3
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 9: Consider the equation ax2 + bx + c = 0 with a 6= 0. Follow the steps given below to get
the quadratic formula:
(i) Multiply both sides of equation 1/a.
(ii) Add − ac to both sides of the equation.
2
b
(iii) Add 2a
to both sides of the equation.
(iv) Try to obtain a full square such as (x + d)2 and solve for x.
Answer: Following step (i) we get
c
b
x2 + ( )x + ( ) = 0.
a
a
Step (ii) yields
c
c
c
b
x2 + ( )x + ( ) − ( ) = −( )
a
a
a
a
which can be simplified to
b
c
x2 + ( )x = −( ).
a
a
Step (iii) gives us
b
c
b
b
x2 + ( )x + ( )2 = −( ) + ( )2
a
2a
a
2a
Now we need to obtain a full square such as (x + d)2 if you look carefully left hand side of the last
b 2
equality given above you see that it can also be written as (x + 2a
) (If you are not convinced expand
this expression to see they are equal.). Thus
(x +
So
b 2
c
b
c
b2
b2 − 4ac
) = − + ( )2 = − + 2 =
2a
a
2a
a 4a
4a2
r
√
b
b2 − 4ac
b2 − 4ac
x+
=±
=
±
2a
4a2
2a
or, equivalently,
4
√
√
b2 − 4ac
b
−b ± b2 − 4ac
x=− ±
=
.
2a
2a
2a
These steps given in the question is a part of a general method called completing the square.
Non-linear Functions
Question 10: 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, 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 11: 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 − 41 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
t;
2
C(1) = 2687.5; C(10) = 40650
(C) 1900 + 800t − 10t2 ; C(1) = 2690; C(10) = 31900
(D) 1900 + 300t −
25 2
t;
2
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
5
Now we need only evaluate this composite function C(t) = C(Q(t)) at t = 1 and t = 10. C(1) =
1900 + 800 − 25
= 2700 − 12.5 = 2687.5; C(10) = 1900 + 800(10) − 25
(10)2 = 1900 + 8000 − 1250 = 8650.
2
2
Systems of Equations
Question 12: Suppose that supply and demand are described by the following set of equations:
Supply: Qs = 2P − 2
8
Demand: Qd = − P + 16
5
By equating supply and demand find the market clearing price Pe and quantity Qe .
(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 13: In a race a turtle can run 0.01 meters every minute and a rabbit can run 1.5 meters
every minute. But it takes 3 minutes for rabbit to stop. How far can the turtle get before the rabbit
catches him? You can round your answer to two decimal places.
Answer: Suppose the rabbit catches the turtle after d meter. Then for the turtle we have d = 0.01t
where t is the time to cover d meters. For the rabbit, however, we have d = 1.5(t − 3) since it takes 3
minutes for the rabbit to stop. Then we have
0.01t = 1.5(t − 3) =⇒ 1.49t = 4.5 =⇒ t = 3.02
Question 14: 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
(A) Y = 677.93; i = 0.09
6
(B) Y = 505.84; i = 0.22
(C) Y = 604.83; i = 0.11
(D) Y = 605.99; i = 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) − 124i − 260 = 0
−320i + 300 − 124i − 260 = 0
−444i + 40 = 0
40
i=
≈ 0.09or9%
444
(1)
40
Now using i = 444
, we can substitute it into either (or both!) of the above equation
to find the
40
40
equilibrium level of income Y . Thus 0.4Y − 124 444 − 260 = 0 and 0.1Y + 80 444 − 75 = 0, both
equations imply that Y ≈ 677.93.
Question 15: Solve the following system of equations for the the equilibrium level of income, Y , 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) Y =
1
(C0
b−1
+ I0 + G0 )
(B) Y =
1
(C0
1−b
+ I0 + G0 )
(C) Y =
1−b
(C0 +I0 +G0 )
(D) Y = (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
Y = (C0 + bY ) + I0 + G0
Y − bY = C0 + I0 + G0
Y (1 − b) = C0 + I0 + G0
1
Y =
(C0 + I0 + G0 )
1−b
Question 16: Solve the following system of equations:
x − 2y = 14
x + 3y = 9
7
Answer: There are different ways to solve this questions for instance substitution, but I will use
elimination method which can be more handy for econometrics or matrix manipulation in general. In
this method you need to eliminate one of the variables. In this question the easiest variable we can
eliminate is x. So, we multiply second equations with −1 and sum the equations to get rid of x. That
is,
+
x -2y =14
-x -3y =-9
-5y =5
Thus, y = −1 and by using second equation we find x = 12.
8