Solutions+VG MVG Level+V1MaA2NVCO09+Algebra

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Suggested solutions VG/MVG-Level V1MaA2NVCO09 Algebra NV-College

© [email protected] ☺ Free to use for educational purposes. 1⊗ Not for sale.

Suggested solutions MaA2NVCO09 Algebra

Part II: VG/MVG-Level

WARNING: THERE ARE DIFFERENT VERSIONS OF THE TEST

Instructions

Test period 110 minutes: 8:10-10:00 Friday November 6, 2009

Resources Formula sheet, your personalised formula booklet, ruler, protractor, agraphic calculator, and a lexicon.

The test For all 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.

Score and The maximum score is 43 points, 29 of them VG points. There are 4 ¤-

problemsmark levels

The maximum number of points you can receive for each solution isindicated after each problem. If a problem can give 2 ”Pass”-points and 1

”Pass with distinction”-point this is written [2/1]. Some problems aremarked with ¤ , which means that they more than other problems offer

opportunities to show knowledge that can be related to the criteria for ”Passwith Special Distinction” in Assessment Criteria 2000.

Lower limit for the mark on the test

G: Pass: 14 points

VG: G+ on the G-level test and at least 30 in this test.

MVG: G+ on the G-level test and at least 35 in this test. High quality of calculationsand deep level of understanding must be illustrated through out the test,

especially in problems 8-11 marked ¤. Alternative: Minimum of 25 points inthis test where 17 points are at least VG points an high quality solutions

marked ¤. Note: Those who passed the G-level test in G-/G level, in addition to the conditions above

must show good knowledge and understanding of the subject.

Name: ________________________ Student Number: NVC09 Sjödalsgymnasiet

1 2 3 4 5 6 7 8¤ 9¤ 10¤ 11¤ Sum: 43 Grade

G 0 2 1 2 0 0 1 0 1 4 3 14

VG¤ 2 2 2 2 3 3 2 2¤ 3¤ 4¤ 4¤ 29 ¤¤¤¤

G

VG¤





Useful formulas

Order of Operations:

[], (), exponents (powers), −+÷× ,,,

y x y x aaa+=⋅ Note: Both terms share identical base .a

y x

y

x

aa

a −=

( ) y x y xaa

⋅=

y

y

y

ya

aa

a=⇔=

−

− 11

10 =a

ab

b

a=1

( ) x x xabba = Note: Both a and b share identical power, x .

( ) z y z x z y xbaba

⋅⋅ ⋅=

( )

( )⎪⎩

⎪⎨⎧

−=−

=−

number odd:

number even:

naa

naa

nn

nn

( )( ) abba =−−

( )( ) abba −=−

cbd ad

c

b

a⋅=⋅⇔=

c

c

b

a

b

a

b

a⋅=⋅= 1

bd

bcad

d

c

b

a +=+

bca

bac ⋅=⋅

With addition and subtraction we may add the individual margin of errors!

( ) ( ) 5.15.4325.05.51427 ±=±+±=+ ba

( ) ( ) 5.15.4215.05.51427 ±=±−±=− ba





Glossary-English-Swedish-Lexicon Mathematics Course A:

English Swedish Comments

quotient, ratio kvot

fraction bråk

integer number heltal

simplify förenkla

express uttrycka

surd (exact) form exakt form

quantities kvantitet, mängd, storhet

exchange places byta plats

investigate UTFORSKA

base bas

power, exponent potens, exponent

percent procent per thousand romille

ppm: part per millions ppm

scientific notation grundpotensform

significant figure värdesiffror, gällande siffror

round off avrunda

changing factor förändringsfaktor,

tillväxtfaktor, ändringsfaktor

factorize faktorisera

abbreviating, simplify förkortning

extend, expand förlängning Numerator täljare

Denominator nämnare

quotient, ratio kvot

Factor faktor

Difference differens

real numbers reella talen

rational numbers rationella tal

irrational numbers irrationella tal

prime numbers primtal

annual interest årsräntan





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

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

1. A bacteria culture is doubled in numbers every 24-hours. Calculate, in percent, how many percent more is there after one hour. [0/2]

Suggested Solution:

224= x

029.12 24

1

== x

Answer: It is growing at the rate of per hour.%9.2

2. If an object of mass m is dropped from the height h above the

ground level, its speed just before hitting the ground, ignoringthe air resistance, may be calculated using the conservation of

energy principle according to

2

2

1mvmgh =

where g = is the gravitational acceleration.2/82.9 sm

How fast (in ) is a bullet dropped from top of the Empire

State Building moving just before hitting the ground of the F

Avenue just below it? [0/2]

How f

sm /

ifth

ast is it in [2/0]

mpire e

.

uggested Solution:

hkm / ?

The roof of the E State Building is m0.381 above th

ground

S

21mvmgh =

2

Divide both sides of the equation by :m2

2

1vmghm /=/ ⇔ 2

2

1vgh =

Multiply both sides of the equation by 2 :2

2

122 vgh

/⋅/=⋅ ⇔ ghv ⋅= 22

Take square root of both sides ghv ⋅±= 2

The bullet is moving at speed smghv /2 ⋅= just before hitting the

ground:

smsmsmghv 2 ⋅= /5.86/0.38182.92/ =⋅⋅= Answer: smv /5.86=

Using , andmkm 10001 = sh 60031 = :

sms

m=

s

mhkm /

36

10

36

10

0063

0001/1 =

//

//= ⇔ hkmsm /6.3/1 =

hkmhkmsmv /311/6.35.86/5.86 =⋅== Answer: hkmv /311=





3. The table below shows number of visitors at some swimming facilities in 2002, and

changes from 2001.

Swimming facility Location Number of visitors 2002

Change from previous year

Eriksdalsbadet Stockholm 1 106 000 199 000Fyrishov, bad Uppsala 700 000 51 800

Eyrabadet Örebro 641 000 156 400

Aq-Va-Kul Malmö 627 000 –7 000

Gustavsvik, bad Örebro 554 200 –16 900

Valhallabadet Göteborg 507 319 –24 630

Rosenlundsbadet Jönköping 50 100 –3 219

Högevallsbadet Lund 483 925 17 092

Source: Turistdelegationen

The number of visitors has increased at boththe Eriksdal and the Eyra facilities. Jakob

claims that the increase is greatest at Eyra

while Anna says that the increase at Eriksdal

is greatest. Explain how they might have

reasoned. Present your solution with

explanations and calculations. [1/2] (1/2)

Suggested solution:

Due to the fact that number of visitors to Eriksdal increased by 199 000

as oppose to 156 400, Jakob concludes that the increase at Eriksdal is

greatest.

On the other hand, due to the fact that the percentage increase of visitors

to Eyrabadet (32%) is larger than that of the Eriksdalsbadet (22%) Anna

claims that the increase is greatest at Eyra:

Eriksdalsbadet %222194.00001990001061

000199+=≈

///−///

///=Δ N

Eyrabadet %323227.0004156000641

004156 +≈=//−//

//=Δ [1/2] N




© [email protected] ☺ Free to use for educational purposes. 6

)

⊗ Not for sale.

4. The length of a rectangle increases by 10%, simultaneously its width decreases by 10%.

Only one of the following alternatives true!

Which one? Motivate your choice by calculations and figures.• Its area does not change.

• Depending on the original size of the width and length of the rectangle, its area may

increase or decrease.• Its area decreases.

• Its area always gets larger. (2/2)

Suggested solutions:Lets denote the length of the originalrectangle by , and its width by . The

dimensions of the new rectangle thenwill be and .

l w

l⋅10.1 w90.0

The area of the original rectangle is:w A ⋅= l

The area of the new rectangle is:( ) ( w Anew ⋅⋅= 90.010.1 l

Aw Anew⋅=⋅= 99.099.0 l

Answer: The area of the new rectangleis always less than the original one:

A Anew ⋅= 99.0 (2/2)

5. How many Hydrogen atoms, H are there in a figure who has n Carbon, C atom: [0/3]

1=n 2=n 3=n

H

H

C

H

H −−

|

|

H

H

C

H

H

C

H

H −−−

|

|

|

|

H

H

C

H

H

C

H

H

C

H

H −−−−

|

|

|

|

|

|

Suggested Solution:There are 2 ( 1)+⋅ n Hydrogen atoms in the figure n.

Check:( ) ( ) 4112121 =+⋅=+⋅⇔= nn

( ) ( ) 6122122 =+⋅=+⋅⇔= nn

( ) ( ) 8132123 =+⋅=+⋅⇔= nn

l

l⋅10.1

w

w90.0





6. Solve the equation 03

=−x

analytically. Give the answer with four significant figures.

[0/3]

2 2

3 x

Suggested Solution:

03

22

3 =−x

x ⇔ 3

22

3

x

x =

Cross product: 3223 ⋅=⋅ x x

Use: :mnmn aaa+=⋅ 623 =+

x ⇔ 65 = x

Take both sides to the power5

1. Use ( ) mnmn

aa⋅= : 431.165

1

≈= x

Answer: 431.165

1

≈= x

7. Simplify

3

3

2

⎟⎟ ⎞⋅

b

a

[1/2]

45

6

32

⎠⎜⎜⎝

⎛ −⋅

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ ⋅−

−

b

a

Suggested Solution:

32

212132930151253

9

123

155

30

33

343

535

563

3

45

3

63

3

45

6

3

222

2

2

2

2

2

2

22

a

bbaba

b

a

a

b

b

a

a

b

b

a

a

b

b

a

b

a

⋅=⋅⋅=⋅⋅=

⋅⋅

⋅+=

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⋅−⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛

⋅−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅−⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛

⋅−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅−⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅−

−−−−−

⋅

⋅

⋅

⋅−

⇔

Answer:3

213

3

45

6

3

4

22

a

b

b

a

b

a

⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅−⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅−

−





At the aspect assessment of your work with problems 8-11 your teacher will consider

• what mathematical knowledge you have demonstrated and how well you have carriedthrough the task

• how well you have explained your work and motivated your conclusions• how well you have written your solutions.

8. Solve the equation 00.5= x . [0/2/¤]3 ⋅ x

Suggested Solution:

Rewrite the equation using the fact that 2

1

x x = : 52

1

3 =⋅ x x

Use: :mnmn aaa+=⋅ 52

13

=+

x

Use:2

7

2

16

2

1

2

32

2

13 =+=+⋅=+ : 52

7

= x

Take both sides to the power7

2. Use ( ) mnmn

aa⋅= :

58.1555 7

2

7

27

2

2

7

2

7

≈=⇔=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⇔= x x x [0/2/¤] Answer: 58.157

2

≈= x

9. Write the number ....75 as a fraction45671234561234567123.b

a. [1/3/¤]

Suggested Solution:

We may recognize that 1234567 are repeated unlimitedly, and therefore

the period is a 7 digit number. We may multiply both sides of

....745671234561234567123.5= x by 10 :7

....4567123456712345671234567123.51234567....745671234561234567123.51010 77 =⋅=⋅ x

....745671234561234567123.5....4567123456712345671234567123.5123456700000010 −=−⋅ x x

512345629999999 =⋅ x

Answer:9999999

56223451....745671234561234567123.5 =

Check: Using a calculator: ....745671234561234567123.59999999

56223451= OK!





0/2/¤]

uggested solutions:nual interest rate:

1103

=

⊗ Not for sale.

10. Frida borrows kr 00090 to starts her own business. She is not going to pay back any part

of her loan for 15 years when it is expected that the business is fully established and

profitable. The annual interest on her loan is %7 .

a. How much is Frida’s loan after three years? [2/0]

b. With how many percent has her loan increased after fiveyears? [2/2]

c. How long does it take for her loan to be doubled? [

SData: Loan kr 00090 ; an %7

a. ( kr 00011025307.100090 ≈⋅ )Answer: After three years Frida owes . [2/0]

b.

[2/2]

kr 000110

First method:

( ) 40.1403.107.1

5≈=

Answer: After five unt Fri owes is creased byears the amo da in y %40 .

29 ≈

Second method:

( ) 212607.1000905

=⋅ kr 000126

%4040.0403.000090

22936

00090

00090229126=≈==

− [2/2]

a.

First method:(May be ignored for now)

n

log07.1log =⋅n

( ) 207.1 =n

The analytical method of solving this problem involves logarithmswhich is part of mathematics course C.

( ) 207.1 =n

( ) 2log07.1log =

( ) 2

( )1024.10

07.1log

2log≈≈=n 10≈n

Second method: “Graphical

ay be solved

graphically. To solve it using a er) , we may plot

method”

( ) 207.1 =n

m

graphic calculator (TI-83 or high ( )nY 07.1:1 , and 2:2Y

on the same frame, and using the facilities nd 2 calc intersect find th

solution of ( ) 207.1 =n

. According to the graph illustrated above

e

10≈n

Answer: After 10 years Frida owes doubled as much. [0/2/¤]

ollowing: write

Third method: “Numerical Method”

To solve ( ) 207.1 =n

, we may do the f 06.1 ENTER ×

06.1∗ Ans ENTER ENTER ENTER ….tills the calculator u

should count how many times you hit

the

shows a n mber

You just above 2, i.e. 10485.2 .

ENTER . The number of ENTER s hit is the solution of the problem:

i.e. 10≈n .(Note that in everyday-language we say 11 years instead.)

0,0

0,5

1,0

1,5

2,0

2,5

3,0

a n )

0 1 2 3 4 5 6 7 8 9 1 0 11 12 13 14 15

n (Years)

( T h e a m o u n t F r i d a ' s L o a n ) / ( o r i g i n a l l o ( ) 207.1 =

n





11.

⊗ Not for sale.

In a classical physics experiment in order to measure the ratio of the charge of electron to

its mass,m

, we accelerate electron beam to

a high velocity v , which could be

determined. The electron beam enters a

region of uniform magnetic field

e

B . Themagnetic field B is perpendicular to v . A

result electrons in the beam move at the

constant speed v in a circular path whose

radius of curvature may be measuredexperimentally. According to laws of phys

the force experienced by a moving ele

velocity v may be expressed as evBF

s a

ics

ron atct

= . On

the other hand according to Newton’s law of

motion an object of mass m , moving at

velocity v in a circular path of radius r experience a forcer

vmF

2

= .

a) User

vmF

2

= and evBF = andm

ein terms of Br v and,, . [0/3/¤]

b) Findm

eof an electron that is moving at constant speed smv /100.2 7⋅= in path

curved of radius mr 2101.1 −⋅= in a plane perpendicula T 010.0

c) Calculate

r to a B =

magnetic field. [1/1]

m

eusing the fact that: for an electron: C e

191060.1 −⋅= and

kgm311011.9 −⋅= . Check your results against th ound in part be value f , and

calculate the experimental error in the experiment. [2/0]

Suggested Solution:

r r

mF

vmevBv2 =⇔

⎪⎨

evBF 2⎧ =

⎪⎩

=

Divide both sides of the equation by v :r

vmeB

vr

vm

v

Bve=⇔

/=

/

//2

Divide both sides of the equation by :m r

v

Bm

e

r

v

m

m

Bm

e

=⇔⋅/

/

=

Divide both sides of the equation by B :

Br m Br Bm=⇔=

/ [0/3/¤]

vev Be / Answer:

Br

v

m

e=

kgC /108.1 11

2⋅=

− [1/1] Answer:

Br

v

m

e

101.1010.0

100.2 7

⋅⋅

⋅== kgC

m/

,

e108.1 11⋅=

C e191060.1 −⋅= kgm

311011.9 −⋅= : ⇔ kgC mm

/101011.9

11⋅⋅

[1/0]C e

76.11060.1

31

19

=⋅

=−

−

%3.

76.

20227.01076.1

101108.111

1111

≈=⋅

⋅−⋅

=Δ m

e

[1/0] Answer: %3.2≈ Δ m

e

Solutions+VG MVG Level+V1MaA2NVCO09+Algebra

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