solutions+to+final test v2 mabnvco08

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Solutions to Final Test_V2_MaBNVCO08 NV-College Name: _________________________ Student Number: __________

© [email protected] ☺Free to use for educational purposes. Not for sale! page 1 of 11

Final Test Mathematics Course B

Fall 2008-Spring 2009 : MaBNVC08

Warning: There are more than one versions of the test.Instructions

Test period 11:30-14:00 for Part I and Part II as a whole. We recommend that you use at

the most 30 minutes to work with Part I. You are not allowed to usecalculator in the part I. Only after you hand in your solutions for part I youmay use a calculator.

Resources Formula sheet, your personalised formula booklet, ruler and protractor.

The test For most items a single answer is not enough. It is also expected• that you write down what you do• that you explain/motivate your reasoning

• that you draw any necessary illustrations.

Try all of the problems. It can be relatively easy, even towards the end of the test, to receive some points for partial solutions. A positive evaluation

can be given even for unfinished solutions.

Solving Problems number 8, 9 and 15 are of the greatest importance for

the highest grade ¤.

Score and The maximum score is 54 points, 28 of it VG points.mark levelsThe maximum number of points you can receive for each solution is indicated after each problem. If a problem can give 2 ”Pass”-points and 1 ”Pass with distinction”-

point this is written [2/1]. Some problems are marked with ¤ , which means that theymore than other problems offer opportunities to show knowledge that can be related

to the criteria for ”Pass with Special Distinction” in Assessment Criteria 2000.Lower limit for the mark on the testG: Pass: 18 pointsVG: Pass with distinction: 36 points, at least 9 VG pointsMVG: Pass with special distinction: 40 points, at least 18 VG points You should also

show the highest quality work in the ¤ -problems.

P 1 2 3 4 5 6 7 8 9 Sum

G 2 3 1 1 1 2 10

VG 1 3 3 3 10

MVG ¤ ¤ ¤¤

GVG

MVG

P 10 11 12 13 14 15 Sum Total Name

G 4 4 4 4 16 26 Student Number:

VG 1 3 2 2 5 5 18 28

MVG ¤ ¤ ¤¤¤

G

VG

MVG

Grade

☺ Enjoy it! Have Fun! Behzad





In each case, show how you arrived at your answer by clearly indicating all of the necessary

steps, formula substitutions, diagrams, graphs, charts, etc.

In each case, show how you arrived at your answer by clearly indicating all of the necessary

steps, formula substitutions, diagrams, graphs, charts, etc.

Part IPart IYou are not allowed to use calculator in this part. Write your answer to the firstYou are not allowed to use calculator in this part. Write your answer to the first part on this paper. Only after the submission of your solutions to the part oneyou may be allowed to use your calculator. You may start working with part II before submission of your solutions to part I.

1. Expand and simplify as far as possible: ( ) ( )2255 −−+ x x . [2/0]

Suggested Solution: Answer: ( ) ( ) x x x 205522

=−−+

( ) ( ) ( ) x x x x x x x x x x x 20251025102510251055 222222=−+/−++/=+−−++=−−+

2. In the Cartesian Coordinates below, three straight lines are plotted. Pair three of the fiveequations below with the corresponding lines A , B , andC . Note that two of equations inthe table are not represented in the figure! In such case answer “none”. [3/0]

-15

-10

-5

0

5

10

-6 -4 -2 0 2 4

x

y A

B

C

Only answer is required:

5−= x y 53 += x y 62 +−= x y 53 −−= x y 53 +−= x y

Suggested Solution:

5−= x y 53 += x y 62 +−= x y 53 −−= x y 53 +−= x y

C A None None B





3. Solve the equation for 3. Solve the equation for x

02452 =−+ x x . [1/0]

Suggested Solution: Answer:⎩⎨⎧

−=⇔=+

=⇔=−

808

303

2

1

x x

x x;

02452=−+ x x

( )( ) 083 =+− x x

⎩⎨⎧

−=⇔=+

=⇔=−

808

303

2

1

x x

x x

4. Let . Find ( )1− f [1/0]( ) ( )253 −= x x x f

Suggested Solution: Answer: ( ) 641 −=− f

( ) ( )253 −= x x x f

( ) ( ) ( )( ) ( ) 6485351311 222−=−=−−−=−−−=− f

5. Solve the inequality below for x :

9375 +≥− x x [1/0]

Suggested Solution: Answer: 8≥ x 9375 +≥− x x

7935 +≥− x x

162 ≥ x

2

16≥ x

8≥ x

6. Solve the simultaneous equations below for x and y :

⎩⎨⎧

=+

=+

754

1235

x y

x y [2/1]

Suggested Solution: Answer: 1−= x ; 3= y

We may solve the problem by the “elimination” procedure. To eliminate x

we may multiply both sides of the first equation by 5 , and multiply sides

of the second equation by , and add the resulting equations to eachother:

3−

⎩⎨⎧

=+

=+

754

1235

x y

x y

313

39391321601225

211512

601525=⇔=⇔=⇔−=−⇔

⎩⎨⎧

−=−−

=+ y y y y y

x y

x yAnswer:

3= y

We may then find x by substituting 3= y in any one of the equations, for

example, the second one 754 =+ x y :

754 =+ x y ⇔ ( ) 7534 =+ x ⇔ 7512 =+ x ⇔ 1275 −= x ⇔ 55 −= x ⇔55−= x 1−= x





7. Find the equation of the quadratic function plotted in the figure below: [0/3]7. Find the equation of the quadratic function plotted in the figure below: [0/3]

-40

-32

-24

-16

-8

0

8

-6 -5 -4 -3 -2 -1 0 1 2 3 4

x

y

Suggested Solution: Answer: ( )( )352 −+= x x y

The equation of the quadratic equation may be written as( )( )( 35 −−−= x x A y )

( )( 35 −+= x x A y )

The symmetry line of the equation is 12

2

2

35max −=

−=

+−= x .

The local maximum has coordinates ( )32,1 −− . The constant A in

( )( 35 −+= x x A y ) may be determined by requiring that the coordinates of

the local maximum must satisfy the equation:( )( 35 −+= x x A y )

( )( ) A A 16315132 −=−−+−=−

216

32=

−

−= A ( )( 352 −+= x x y )

8. Solve the equation for x :

53

53

2

−=⎟ ⎠

⎞⎜⎝

⎛ +⋅ x x [0/3/¤]

Suggested Solution: Answer: 01 = x ; 312 = x

Square both sides of the equation:

( )22 59

253 −=⎟

⎠ ⎞⎜

⎝ ⎛ +⋅ x x

25109

2599 2 +−=

/⋅/+⋅ x x x

2510259 2 +−=+⋅ x x x

x x x 109 2 −=⋅

09102 =⋅−− x x x

( ) 019 =− x x

01 = x

019 =− x ⇔ 192 = x





9. If all four vertices of a quadrilateral ABCD lie on the circumference of a circle, the

quadrilateral ABCD is called a “cyclic quadrilateral.”

Prove that the opposite angles of any cyclic quadrilateral are supplementary. [0/3/¤]

9. If all four vertices of a quadrilateral ABCD lie on the circumference of a circle, the

quadrilateral ABCD is called a “cyclic quadrilateral.”

Prove that the opposite angles of any cyclic quadrilateral are supplementary. [0/3/¤]

Suggested Solution: Suggested Solution: As illustrated in the figure below:As illustrated in the figure below:

¤ Draw the radius¤ Draw the radius CO and AO , where O is the

center of the circle.¤ Rename the angle subtended by the arc CBA

at the center of the circle x .

¤ Similarly, rename the angle subtended by themajor arc ADC at the center of the circle y .

Using the fact that the angle subtended by an arcat the center of a circle is twice the angle

subtended at the circumference by the same arc,

we may conclude that2

x ABC =∠ .

Similarly:2

360

2

x y ADC

−°==∠ .

Therefore: °=−°

+=∠+∠ 1802

360

2

x x ADC ABC

Using the fact that the total sum of the internalangles of any quadrilateral is we may

conclude that:

°360

°=°−°=∠+∠ 180180360 ADC ABC .

A

B

C

D

xO

y

2

x

2

y A

B

C

D





Part IIPart II

You may use your calculator in this part. Note that you should submit yoursolutions to the part I before having access to your calculator.

You may use your calculator in this part. Note that you should submit yoursolutions to the part I before having access to your calculator.

10. The table below illustrates the distribution of scores that 100 students received on astandardized test: [4/1]

10. The table below illustrates the distribution of scores that 100 students received on astandardized test: [4/1]a. Construct an cumulative frequency table.a. Construct an cumulative frequency table.

b. Draw a cumulative frequency diagram. b. Draw a cumulative frequency diagram.c. Which interval contains the upper quartile?c. Which interval contains the upper quartile?

d. What percent of students scored less than 51?d. What percent of students scored less than 51?e. What is median score?e. What is median score?f. What is the mode score?f. What is the mode score?

Suggested Solution:Suggested Solution:

Score Cumulative

Frequency

21-30 2

21-40 5

21-50 10

21-60 18

21-70 38

21-80 66

21-90 9221-100 100

Score Cumulative

Frequency

21-30 2

21-40 5

21-50 10

21-60 18

21-70 38

21-80 66

21-90 9221-100 100

c. The intervals 81-90 contain the upper quartile.

d. 10 students, i.e. 10% of students scored less than 51 . e. The median score lies in the interval 8071−

f. The mode score is 8071−

Score Frequency

91-100 8

81-90 26

71-80 28

61-70 20

51-60 8

41-50 5

31-40 2

21-30 3Sum 100

0

10

20

30

40

50

60

70

80

90

100

21-30 21-40 21 -50 21-60 21-70 21-80 21-90 21-100

Scores

C u

m u l a t i v e F r e q u e n c y





11. Two DVD rental stores offer two different membership plans: 11. Two DVD rental stores offer two different membership plans: Store A charges $12 for membership and $2.5 for each DVD that is rented.Store A charges $12 for membership and $2.5 for each DVD that is rented.Store B does not have a membership fee but charges $4 for each DVD that is rented.Store B does not have a membership fee but charges $4 for each DVD that is rented. Write an equation that shows the total cost , C , of renting n DVD from store Write an equation that shows the total cost , C , of renting n DVD from store A .

Write an equation that shows the total cost , C , of renting n DVD from store B .

Construct a simultaneous equation system of the equations obtained above and solveit graphically. Interpret the results. [2/2]

Solve the equation system analytically. Interpret the results. [2/1]

Suggested solution:

Answer: nC ⋅+= 5.212$ , nC ⋅= 4$ ,⎩⎨⎧

⋅=

⋅+=

nC

nC

4$

5.212$, 8=n , 32$$ =C

• The total cost, C , of renting n DVD from store A is nC ⋅+= 5.212$

• The total cost, C , of renting n DVD from store B is nC = 4$ ⋅

• ⎩

⎨⎧

⋅=

=

nC

C

4$

12$ ⋅+ n5.2The intersection of the two lines is the solution of the

equation system: The cost of renting 8 DVDs is the $32 no matterwhich store they are rented from. As illustrated below, the store A ischeaper than store B if more than 8 DVDs are rented. It is, on theother hand more expensive if less than 8 DVDs are rented from. Theresults are consistent with the graphical solution of part C.

04

8

12

16

20

24

28

32

36

40

0 4 8 12 16

Number of rented DVD

C o s t $

B

nC ⋅= 5.3$ A

nC ⋅+= 212$

• ⎩⎨⎧

⋅=

⋅+=

nC

nC

4$

5.212$⇔ nn ⋅+=⋅ 5.2124 ⇔ 125.24 =⋅−⋅ nn ⇔ 125.1 =⋅ n ⇔ 8

5.1

12==n

⎩⎨⎧

=

⋅+=

8

5.212$

n

nC ⇔ 32$201285.212$ =+=⋅+=C The cost of renting 8 DVDs is

the $32 no matter which store they are rented from. As in part (c).





12. 32 −+ 12. 32 −+ ( ) 2= x x x f ( ) 2= x x x f

( )a. Find ( )1− f . [1/0] a. Find ( )1− f . [1/0]

b. Find( )

b. Find( )( )

h

h f 1 f 1−−+−. [2/0]

c. Find( )( )

h

hb f + b f −. [1/1]

d. Find( )( )

h

hb f b f −+for extremely small values of h , i.e. as h approaches zero

in the limits. [0/1]

Suggested Solution:

Answer: ,( ) 41 −=− f ( ) ( )

111

−=−−+−

hh

f h f ,

( ) ( )12 ++=

−+ah

h

a f ha f ;

( ) ( ) 12lim0

+=−+→

ah

a f ha f h

a. ( ) 322 −+= x x x f ⇔ ( ) ( ) ( ) 43 −=− [1/0] 21312112

−=−−+−=− f

b. ( ) ( ) ( ) ( ) ( )h

h

h==

h

hhh

h

hh

h

f h f /+/−+/−+−/=

−−−+−++−=

−−+− 2224322214312111

( ) ( )h

h

f h f =

−−+− 11 [2/0]

c. ( ) ( ) ( ) ( ) ( )h

b 32 −+ [0/1]

bhbhb

h

b f hb f 32 22−−+++

=−+

( ) ( ) ( )h

bhh

h

hbhh

h

bbhbhbhb

h

b f hb f

/

++/=

++=

+−−−++++=

−+ 2222323222 2222

( ) ( ) ( )22

22++=

/

++/=

−+bh

h

bhh

h

b f hb f

d. ( ) ( )( ) 22 += b [0/1] 22limlim

00++=

−+→→

bhh

b f hb f

hh

Check: ( ) ( ) 22 ++=−+ bhh

b f hb f ⇔ ( ) ( ) ( ) hhh

f h f =+−+=−−+− 21211 OK!





13. Malin bought a package of marbles and stored them by colour as illustrated below. Givethe answers in surd form (exact):

13. Malin bought a package of marbles and stored them by colour as illustrated below. Givethe answers in surd form (exact):

a. If one marble is selected at random,what is the probability it will be blue,

yellow, or green? [2/0]

a. If one marble is selected at random,what is the probability it will be blue,

yellow, or green? [2/0]

b. If three marbles are selected atrandom, without replacement, what isthe probability that at least one redmarble is selected? [0/2]

b. If three marbles are selected atrandom, without replacement, what isthe probability that at least one redmarble is selected? [0/2]

c. If three marbles are selected at random, without replacement, what is the probabilitythat neither marble is selected is blue or yellow? [2/0]

c. If three marbles are selected at random, without replacement, what is the probabilitythat neither marble is selected is blue or yellow? [2/0]

Suggested Solution:Suggested Solution:

a. Answer: If one marble is selected atrandom, the probability that it will bered, yellow, or green is 56.0=P .

( ) ( ) ( )greenyellow blue PPPP ++=

56.0100

56

25

14

25

356

25

3

25

5

25

6===

++=++=P

b. Answer: If three marbles are selected at random, withoutreplacement, the probability that at least one red marble is selected

is: ( ) 65.0≈ .575

371 selectedismarbleredoneleastat =P

575

371

2300

1484

23

7

4

17

25

3

23

23

4

7

25

3

23

23

4

4

25

7

23

7

24

17

25

18

24

7

25

18

25

7==⋅⋅+⋅⋅+⋅⋅=⋅⋅+⋅+=P

Second (better) Method:

( )575

204

23

4

1

17

25

3

23

16

24

17

25

18=⋅⋅=⋅⋅=red not P

( ) ( )230

371

575

204575

575

20411selectedismarbleredoneleastat =

−=−=−= red not PP

c. Answer: If three marbles are selected at random, withoutreplacement, the probability that neither marble selected is blue or

red:575

91=P .

%161583.0575

91

2325

713

23

12

24

13

25

14≈==

⋅

⋅=⋅⋅=P

Colour Number of marbles

Red 7

White 4

Blue 6

Yellow 5

Green 3

Sum 25

0

1

2

3

4

5

6

7

8

Red White Blue Yellow Green

Colour

N u m b e r o f M a r b l e s





14. Three sides of a right triangle are14. Three sides of a right triangle are x , 1− x and 2− x incm.

a. Calculate the area of the triangle in 2cm . [0/3]

b. Calculate the measure of the angle BCA∠ . [0/2]

Suggested solutions:

Answer:

⎪⎪⎩

⎪⎪⎨

⎧

=+=+≡

=+=+≡

=≡

5232

4131

3

x BC

x AC

x AB

; ;26 cm A = °≈∠ 37 BCA

( ) ( ) 22221 x x x =−+−

222 4412 x x x x x /=+−++−/

0562 =+− x x

Find two numbers so that when multiplied produces 5 and whenever are

added to each other give 6− . The numbers are 5− and 1− 0562 =+− x x

( )( ) 051 =−− x x ⇔⎩⎨⎧

=⇔=−

=⇔=−

505

Ignored! physical.-non101

x x

x x

Therefore the sides of the triangle are⎪⎩

⎪⎨

⎧

=−=−

=−=−

=

3252

4151

5

x

x

x

“Golden Triangle!”

Area of the triangle is:( )( ) 2

62

43

2

21cm

x x A =

⋅

=

−−

= Answer:2

6 cm A =

3

4

25

15

2

1tan =

−

−=

−

−=

x

xC ⇔ °≈⎟

⎠

⎞⎜⎝

⎛ =∠ − 53

3

4tan 1

BCA ⇔ Answer: °≈∠ 53 BCA

x

A B

C

1− x

2− x





When assessing your work, your teacher will take into consideration:

¤ How well you carry out your calculations

¤ How well you present and comment on your work

¤ How well you justify your conclusions

¤ What mathematical knowledge you show¤ How well you use the mathematical language

¤ How general your solution is

15. A straw is placed into a rectangular box that is cma by cmb by

cmc , as illustrated in the figure below. If the straw fits exactly

into the box diagonally from the bottom left front corner to the topright back corner, [0/5/¤]a. how long is the straw?

b. What is the angle the straw makes with the bottom of the boxif cm , cm and cma 12= b 6= c 20= .

Note: You may, instead choose to solve the problem numerically:Find the length of the straw, to the nearest tenth of a centimetre, if

cm , cm and cmc 20a 10= b 5.7= = . [Instead Only: 1/2]

Suggested solutions: Answer: cmcba222 ++=l ; °≈ 56α ; cm1.24=l

Lets denote the length of the straw by . Due to the fact that the box

is right angled, the length of the diagonal of the bottom (as well as thetop) of the box, , may be calculated using Pythagoras theorem:

cml

cmd 222

bad +=

On the other hand, the diagonal of the bottom of the box, , is

perpendicular to the vertical side of the box, . Therefore the triangle that

is made of the straw, , the side and the diagonal d is a right triangle,

where the straw is its hypotenuse. Using Pythagoras theorem:

d

c

l c

222cd +=l

Therefore: 2222cba ++=l

Answer: The length of the straw is: cmcba222 ++=l´

If we name the angle that the straw makes with the bottom of the box α :

d

c=α tan ⇔

22tan

ba

c

+=α ⇔ Answer: ⎟

⎟ ⎠

⎞⎜⎜⎝

⎛

+= −

22

1tanba

cα

( ) °≈=⎟ ⎠

⎞⎜⎝

⎛ =⎟

⎟ ⎠

⎞⎜⎜⎝

⎛

+= −−− 5649.1tan

4.13

20tan

612

20tan 11

22

1α Answer: °≈ 56α

cmcmcmcba 1.2420612 222222 ≈++=++=l Answer: cm1.24=l

cma 12= , andcmb 6= cmc 20=

cma

cmb

cmc

solutions+to+final test v2 mabnvco08

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