# SGPE Summer School 2016: Quiz 2 Answers

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## 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: 15 x − 3 y = 60 (A) slope= 5, y -intercept= − 20, and x -intercept= 4 (B) slope= 6, y -intercept= − 22, and x -intercept= 5 (C) slope= 1 5 , y -intercept= 20, and x -intercept= − 4 (D) slope= 5 .

5, y -intercept= − 20, and x -intercept= 8 (E) None of the above

### Answer:

for the x (A) Given the form of the equation that we are given, it makes sense to go ahead and solve and y intercepts. Solving for the y -intercept requires setting x = 0 in the above equation and then solving for x − y . Thus 15(0) -intercept by setting y − 3 y = 60 which implies that = 0 and solving for y = − 20. Similarly we can find the x , which yields 15 x − 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: 15 x − 3 y = 60 − 3 y = − 15 x + 60 y = 5 x − 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 = − 12 x + 33 (C) y = 12 x − 33 (D) y = 12 x − 1 33 (E) None of the above

### Answer:

y (C) Simply substituting = 12 x − 33.

m = 12 and b = − 33 into the slope intercept form of a line yields

### Question 3:

Find the equation of a straight line passing through the point (-2, 7) and perpendicular to a line with equation 24 x + 6 y = 30.

(A) y = 4 x + 13 2 1

(B) y = − 1 4 x + 15 2 (C) y = − 4 x + 15 (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 24 x + 6 y = 30. What is the slope of this line? Solving for the slope intercept form yields y = − 4 x + 5. Thus the slope is − 4, which means that the slope of (i.e., the negative reciprocal). Now we need to find the our equation is 1 4 y -intercept. To do this we first write down the point-slope form of the line: ( y − y 1 ) = m ( x − x 1 ). Now substitute for m , (which we just calculated) and the point (which we were given). This yields: x 1 and y 1 using the slope y − 7 = 1 4 ( x − ( − 2)) Now we simply re-arrange this equation to slope-intercept form to get the answer: y − 7 = y − 7 = y = 1 4 ( x 1 4 x − + 1 4 x + 1 ( 2 15 2 − 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:

us y Setting y = 0 gives intercept which is y x = c/b .

intercept. So, ax − 0 = c and x = c/a . Similarly, setting x = 0 gives

### 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) = .

11 x + .

15(100 , 000 − x ) = 15 , 000 − 0 .

04 x Given that our broker wants to achieve an expected return of r as follows: = 14%, we simply need to solve for x .

14(100 , 000) =15 , 000 − 0 .

04 x 14 , 000 =15 , 000 − .

04 x .

04 x = 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:

x = 0 and A line can be represented y y = 1 gives us n = 1 and = mx + n . One point the line passes through is (0 , 1). So putting y = mx + 1. Similarly, the line passes through (2 , 3), therefore 3 = 2 m + 1 and m = 1. Finally, y = x + 1.

## Solving Quadratic Equations

### Question 7:

Solve the following quadratic equation: x 2 + 6 x + 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: 5 x 2 + 47 x + 18 = 0 (A) x = − 2 5 and x = − 9 (B) x = − 2 5 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: − 47 ± p 47 2 − 4(5)(18) x = − 47 ± 2(5) √ 2209 − 360 = − 47 ± √ 10 1849 = − 47 ± 10 43 = = − 4 10 = 10 − 2 5 and − 90 10 = − 9

### Question 9:

Consider the equation the quadratic formula: ax 2 + bx + c = 0 with a = 0. Follow the steps given below to get (i) Multiply both sides of equation 1 /a .

(ii) Add − c a to both sides of the equation.

(iii) Add b 2 a 2 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 x 2 + ( b a ) x + ( c a ) = 0 .

Step ( ii ) yields x 2 + ( b a ) x + ( c a ) − ( c a ) = − ( c a ) which can be simplified to x 2 + ( b a ) x = − ( c a ) .

Step ( iii ) gives us x 2 + ( b a ) x + ( 2 b a ) 2 = − ( c a ) + ( 2 b a ) 2 Now we need to obtain a full square such as ( x + d ) 2 if you look carefully left hand side of the last equality given above you see that it can also be written as ( x + b 2 a ) 2 (If you are not convinced expand this expression to see they are equal.). Thus ( x + 2 b a ) 2 = − c a + ( 2 b a ) 2 = − c a b 2 + 4 a 2 = b 2 − 4 ac 4 a 2 So b x + 2 a = ± r b 2 − 4 a 2 4 ac = ± √ b 2 − 2 a 4 ac or, equivalently, 4

x = − b 2 a ± √ b 2 − 2 a 4 ac = − b ± √ b 2 2 a − 4 ac .

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 V t years is ( t ) = 100000 e − 0 .

10 t ). 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:

in year t , (D) We want to find after how many years t will the scrap value of the car exceed its value V ( t ). To solve this question we need to solve the following inequality: ln 2000 ≥ 100000 e − 0 .

10 t 1 ≥ 50 e − 0 .

10 t 1 50 50 1 ≥ e − 0 .

10 t ≥ ln e − 0 .

10 t ln(1) − ln(50) ≥ − 0 .

10 t 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 + 50 Q and that Q ( t ) = 16 t − 1 4 t 2 . 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 + 300 t − 12 t 2 ; C (1) = 2188; C (10) = 1000 (B) 1900 + 800 t − 25 2 t 2 ; C (1) = 2687 .

5; C (10) = 40650 (C) 1900 + 800 t − 10 t 2 ; C (1) = 2690; C (10) = 31900 (D) 1900 + 300 t − 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 cost function as follows: Q ( t ) in for Q in the C ( t ) = C ( Q ( t )) = 1900 + 50(16 t − 1 4 t 2 ) = 1900 + 800 t − 25 t 2 2 5

Now we need only evaluate this composite function 1900 + 800 − 25 2 = 2700 − 12 .

5 = 2687 .

5; C C ( t ) = C (10) = 1900 + 800(10) ( − Q 25 2 ( t )) at (10) 2 t = 1 and t = 10.

C (1) = = 1900 + 8000 − 1250 = 8650.

## Systems of Equations

### Question 12:

Suppose that supply and demand are described by the following set of equations:

### Supply: Demand:

Q s = 2 P − 2 Q d = − 8 5 P + 16 By equating supply and demand find the market clearing price P e and quantity Q e .

(A) P e = 6; Q e = 9 (B) P e = 4; Q e = 7 (C) P e = 5; Q e = 8 (D) P e = 8; Q e = 5 (E) None of the above

### Answer:

(C) To find the market clearing price, supplied and solve for P . Thus...

P e , simply equate quantity demanded with quantity (Supply) Q s = 2 P e − 2 = − 8 5 P e + 16 = Q d 2 P + 8 5 P e 18 P e 5 P e = 16 + 2 = 18 = 5 (Demand) To find Q e , simply plug P e = 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 Q d = 2(5) − 2 = 8, and Q s = − 8 5 (5) + 16 = 8.

And we have found Q e = Q s ( P e ) = Q d ( P e ) = 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:

where t Suppose the rabbit catches the turtle after is the time to cover d d meter. Then for the turtle we have meters. For the rabbit, however, we have d = 1 .

5( t − d = 0 .

01 3) since it takes 3 t minutes for the rabbit to stop. Then we have 0 .

01 t = 1 .

5( t − 3) = ⇒ 1 .

49 t = 4 .

5 = ⇒ t = 3 .

02

### Question 14:

interest rate, i Solve the following system of equations for the equilibrium levels of income, (round your answers to the nearest hundredth): Y and 0 .

1 Y + 80 i − 75 = 0 0 .

4 Y − 124 i − 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 .

1 Y + 80 i − 75 = 0 Y + 800 i − 750 = 0 Y = − 800 i + 750 Substituting this expression for Y of the interest rate.

into the second equation allows us to solve for the equilibrium level 0 .

4( − 800 i e + 750) − 124 i − 260 = 0 − 320 i + 300 − 124 i − 260 = 0 − 444 i + 40 = 0 40 i = 444 ≈ 0 .

09 or 9% (1) Now using i = 40 444 , we can substitute it into either (or both!) of the above equation to find the equilibrium level of income Y . Thus 0 .

4 Y − 124 40 444 − 260 = 0 and 0 .

1 Y + 80 40 444 − 75 = 0, both equations imply that Y ≈ 677 .

93.

### Question 15:

Solve the following system of equations for the the equilibrium level of income, terms of given levels of government spending, G = G 0 , and investment, I = I 0 .

Y , in Y = C + I + G C = C 0 + bY where 0 < b < 1 (A) Y = 1 b − 1 ( C 0 + I 0 + G 0 ) (B) Y = 1 1 − b ( C 0 + I 0 + G 0 ) (C) Y = 1 − b ( C 0 + I 0 + G 0 ) (D) Y = (1 − b )( C 0 + I 0 + G 0 ) (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 = ( C 0 + bY ) + I 0 + G 0 Y − bY = C 0 + I 0 + G 0 Y (1 − b ) = C 0 + I 0 + G 0 Y = 1 1 − b ( C 0 + I 0 + G 0 )

### Question 16:

Solve the following system of equations: x − 2 y = 14 x + 3 y = 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 -x -2y -3y -5y =14 =-9 =5 Thus, y = − 1 and by using second equation we find x = 12.

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