solutionsnpmaavt99+part+i ii

8/3/2019 SolutionsNPMaAVT99+Part+I II

1/17

Skolverket: NpMaA VT99 Part I-II NV-College - Sjdalsgymnasiet

[email protected] to use for free educational purposes. page 1 of 17Not for sale!

(National Agency for Education)

National Test in

MATHEMATICS

Course A

Spring 1999

Stockholm Institute of Education

Skolverket 1999


2/17

Skolverket: Suggested Solutions NpMaA VT99 Part I-II NV-College

The National Agency for Education, referring to 4 kap 3 Sekretesslagen, emphasisesthat this material is to be kept confidential. This material is confidential until theend of November 1999

National Test in

MATHEMATICS COURSE A Part I

National Test in

MATHEMATICS COURSE A Part I

Spring 1999

Instructions

Spring 1999

Instructions

Test time 120 minutes for part I and II without a break maximum of 30 minutesfor part I

Test time 120 minutes for part I and II without a break maximum of 30 minutesfor part I

Resources The formula sheet is attached to the test. No Calculator is allowed forResources The formula sheet is attached to the test. No Calculator is allowed forpart I

Test material The test material should be handed in with your solutions.

The test The answers to the part I must be submitted before having any accessto a calculator.Part I is made up of 24 problems.For all of the problems in part I Only an answer is required

After each question the maximum number of points possible is shown.

For example (0.5/1) indicates that the question can give 0.5 g-points

and 1 vg-points.

Try all of the problems. It can be relatively easy, even at the end of thetest, to earn some points for a partial solution or presentation.

Name: ________________________ School: ________________________________

Adult education/secondary school program: ____________________________________

[email protected] to use for free educational purposes.Not for sale!page 2 of 17


3/17



1. Which number is the smallest? Ring your answer. (0.5/0)1. Which number is the smallest? Ring your answer. (0.5/0)

02.111.1101.1002.101.1 02.111.1101.1002.101.1

Answer: 1 002.

2. How many minutes is h25.0 ? (0.5/0)

Answer: min15min6025.025.0 ==h

3. How many percent of the figure is shaded? Ring your answer. (0.5/0)

18

5

18

4

9

5

9

4

8

4

Answer: 18

5

4. Study the number series and give the missing number. (0.5/0)

3023___12853

Answer: 17

5. A fee of kr60 is %15 . Find the new fee. (0.5/0)

krkrkrkrx 693660 =++= Answer:kr69

6. The insect is photographed in the scale 1:5 . How largeis the insect in the real life? (0.5/0)

mmmm

x 75

35== Answer: 7 mm

7. Give two integer numbers less than ten such that when divided by one another using acalculator the following result is produced. (0.5/0)

Answer: 33333333.14

3 3 ==4


4/17



8. Calculate8. Calculate 163 ? (0.5/0)

1243163 == Answer:

1243163 ==

9. What is one- third of6103.3 ? (0.5/0)

66

6 101.13

103.3103.3

3

1=

= Answer: 66 101.1103.3

3

1=

10. Which one of the following calculations produces largest result? (0.5/0)

94.030094.0/30098.0/30030098.0

Answer: 300 94.0/

11. Kalles way to the school is kma . Fredrik must walk km2 longer than Kalle to go to

his school. Write an expression for Fredriks way to the school. (0.5/0)

Answer: kma 2+

12. Solve the equation . (0.5/0)( ) 3674 =+ x

( ) 279974

3673674 ===+=+=+ xxxxx Answer: 2=x

13. 3=a and 2=b . Find the value ofa. ba +5 (0.5/0)

b.3

ba (0.5/0)

a. 172152355 =+=+=+ ba Answer: 175 =+ b a

b. 248 Answer:323 33 ===ba 243 =ba

14. Macaronis are going to be packed in plastic bags containing kg75.0 each. Which one

of the following calculations would you use to find the number of kg75.0 packages

that would be sufficient to pack kg6 macaronis? Ring your answer. (0/0.5)

75.0675.06675.06/75.075.0/6 +

Answer: 75.0/6

15. Based on CPI with how many percent have prices increased between 1980 and 1989?(0/0.5)

Year 1980 1989


5/17



CPI (consumer price index) 100 259

Answer: 159 %16. The price of the apple is proportional to its weight. Find the values ofa and b . (0/1)

Weight (kg) 3 5 b

Price (kr) 27 a 72

kgkr/93

27=

kra 45953

275 === Answer: kra 45=

kgb 89

72== Answer: kgb 8=

17.

Perfumes are sold in bottles that hold ml5 . To how many of these bottles lit1 perfume is sufficient? (

bottlesn 2005

1000

105

13

==

=

Answer: bottlesn 200=

18. In the figure below AB is an straight line. The angle x is twice as large as the angle y .

How large is the angle y ? (1/0)

=

===+

=

=+60

3

18018031802

2

180yyyyy

yx

yxAnswer: = 60y

19. The diagram below illustrates the distribution of the grades in a class. How many

percent of students got the grade MVG? (1/0)

0

2

4

6

8

10

12

IG G VG MVG

Grades

Numberofstudents

Answer: There are 25 students in the class. Only 4 students got MVG.

Grade Number of students Percent

IG 2 8

G 11 44%

VG 8 32%

MVG 4 16%

Sum 25

%


6/17



Therefore: %1616.025

4== got MVG. Answer: 16%


7/17



0

100

200

300

400

500

600

700

800

9001000

1100

1200

1300

0 2 4 6 8 10 12 14 16 18 20

Weight (kg)

Price(kr)

20. Write a negative number in all of the parentheses so that the equality is valid.

a. (1/0)( ) ( ) 14=+b. (0/1)( ) ( ) 6=

Answers:a. ( ) ( ) 14410 =+b. ( ) ( ) 6104104 =+=

21. How many percent is the base of the triangle, in the figure below, larger than its height?

(0/1)

Answer: %20

22. Ring in the one which is the smallest. (0/1)

kgofppmkgofkgof 100051%21%3.0 o

Answer:

kgkgkgof 03.0100

31%3.0 ==

kgkgkgof 002.01000

21%2 ==o

kgkgkgofppm 005.00000001

000510005 ==

Answer: kgkgkgof 002.01000

21%2 ==o is the smallest.

23. The diagram below illustrates the price of two different products as a function of theweight of the product. How large is the difference in the price per one kilogram of the

products?

Answer:

( ) 190190 11==

yxy ( ) 170170 22 == yxy

kgkryy

/2070901

21 ==

kgkryy /2021 =

h2.1

h


8/17



24. With the help of the figure below find the value of 75sin . (0/1)24. With the help of the figure below find the value of 75sin . (0/1)

97.00.1

97.075sin == Answer 97.0

0.1

97.075sin ==

75

15

dm0.1

dm26.0

dm97.0


9/17


The National Agency for Education, referring to 4 kap 3 Sekretesslagen, emphasisesthat this material is to be kept confidential. This material is confidential until theend of November 1999

National Test in

MATHEMATICS COURSE A

National Test in

MATHEMATICS COURSE A

Spring 1999 Part II

Instructions

Spring 1999 Part II

Instructions

Test time 120 minutes for part I and II without a break.Test time 120 minutes for part I and II without a break.

Resources Calculator and formula sheet. The formula sheet is attached to the test.Resources Calculator and formula sheet. The formula sheet is attached to the test.

Test material The test material should be handed in with your solutions.Test material The test material should be handed in with your solutions.

The test The test is made up of 11 problems.The test The test is made up of 11 problems.There are different versions of problem 9. Your teacher is going toinform you which one of those you are going to solve. You are goingto solve just the one your teacher has chosen and none of the otherversion of problem 9.

There are different versions of problem 9. Your teacher is going toinform you which one of those you are going to solve. You are goingto solve just the one your teacher has chosen and none of the otherversion of problem 9.

Most of the problems are long-answer problems, where a short answeris not sufficient, but it is requiredMost of the problems are long-answer problems, where a short answeris not sufficient, but it is required

that you write down what you do that you write down what you do

that you explain your train of thought that you explain your train of thought that you draw figures when necessary. that you draw figures when necessary.Some of the problems (where it is stated Answer requiredSome of the problems (where it is stated Answer requiredonly) need only an answer.only) need only an answer.


For example (2/3) indicates that the question can give 2 g-points and

3 vg-points.


For example (2/3) indicates that the question can give 2 g-points and

3 vg-points.

Try all of the problems. It can be relatively easy, even at the end of thetest, to earn some points for a partial solution or presentation.Try all of the problems. It can be relatively easy, even at the end of thetest, to earn some points for a partial solution or presentation.

It is possible to earn a maximum of 61 points in the test.It is possible to earn a maximum of 61 points in the test.

Name: ________________________ School: ________________________________ Name: ________________________ School: ________________________________

Adult education/secondary school program: ____________________________________Adult education/secondary school program: ____________________________________



10/17



1. An admission to the Malins Gym costs SKr80 . On the other hand, a season card

costs SKr900 .

1. An admission to the Malins Gym costs SKr80 . On the other hand, a season card

costs SKr900 .

Calculate minimum number of times attending the Gym such that buying the season-card is financially sound. (2/0)Calculate minimum number of times attending the Gym such that

buying the season-card is financially sound. (2/0)

Suggested solutions:Suggested solutions:

25.1180

900= .

If one is planning to attend the Gym more 12 times per season, then it isbeneficial to purchase the season card. Answer:12 (2/0)

2. Rain water is collected in a cylindrical barrel. The height of the barrel

is m1.1 and its diameter is cm60 . All measurements are inner-measure. How many litter water may a completely full barrel hold.

Suggested solutions:

dmcmcmr 3302

60===

dmmh 111.1 == (1/0)

( ) litterdmhrV 31099110.3 322 === (0/1)

Answer: litter311V (1/0)

3. En film varar lngre p bio n p TV. P bio visas film medhastigheten 24 bilder per sekund och p TV med 25 bilder persekund. Filmen Titanic r 175 minuter lng p bio.

How many minutes is Titanic shorter on TV? (2/1)


min168sec080102425

60175

==

=t (1/0)

min7168175 ==t (1/0)

Answer: It is 7 minutes shorter on TV.

(0/1)


11/17



4. The length of a rectangle increases by 10%, simultaneously its width decreases by10%.

4. The length of a rectangle increases by 10%, simultaneously its width decreases by10%.

Only one of the following alternatives true!Only one of the following alternatives true!Which one? Motivate your choice by calculations and figures.Which one? Motivate your choice by calculations and figures. Its area does not change. Its area does not change.

Depending on the original size of the width and length of the rectangle, its area mayincrease or decrease.

Depending on the original size of the width and length of the rectangle, its area mayincrease or decrease.

Its area decreases. Its area decreases. Its area always gets larger. (2/2) Its area always gets larger. (2/2)


Lets denote the length of the original

rectangle by , and its width by . The

dimensions of the new rectangle then

will be 1 and .

Lets denote the length of the original

rectangle by , and its width by . The

dimensions of the new rectangle then

will be 1 and .

l w

l10. w90.0

The area of the original rectangle is:The area of the original rectangle is:

wA = l

The area of the new rectangle is:The area of the new rectangle is:

( ) ( wAnew = 90.010.1 l )

l w

l10. w90.0

wA = l

( ) ( )wAnew = 90.010.1 l

AwAnew

AwAnew == 99.099.0 l

Answer: The area of the new rectangle is always less than the original

one: (2/2)AAnew = 99.0

l

l10.1

w

w90.0


12/17



5. A childs sleeping need may be approximated by the formula5. A childs sleeping need may be approximated by the formula2

15n

S = where S is

the number of sleeping hours in 24-hours and n is the age of the child in years.

a. Anton is four years old. How many hours per day, according

to the formula, does he need to sleep?b. Using the formula plot a graph that may be used to find achilds sleeping need as a function of his/her age.

c. In what age-interval may the formula be valid? Motivate yourchoice.

d. Describe in everyday language the meaning of the formula. (3/4)


a. h13= Answer: AntonSn

S 2152

415

215 ===

needs 13 hours of sleep in 24 hours. (1/0)

b. See the graph below. (1/2)c. h

nS

215= is valid in the interval 014 > n . The lower

boundary is obvious; the upper boundary of 14 yearsis based on the general scientific minimum dailyrecommendation of 8 hours sleep for everybody.

(1/1)

d. According to hn

S2

15= the need for sleep for a child

decreases as the child gets older. A new born baby

needs to sleep 15 hours per day, two years old childneeds 14 hours of sleep, etc. The data is also presented in the tablebelow. (0/1)

8,0

9,0

10,0

11,0

12,0

13,0

14,0

15,0

0 2 4 6 8 10 12

Age (years)

Sleepne

ed(h)

14

Age Sleep need (h)

0 1

1 14,5

2 13 13,5

4 1

5 12,5

6 1

7 11,5

8 1

9 10,5

10 10

11 9,5

12 9

13 8,5

14 8

5

4

3

2

1

n hn

S2

15 =


13/17



6. The weight of a newborn baby, from the moment of being born up to his/her tenthweek, changes according to the figure below.

6. The weight of a newborn baby, from the moment of being born up to his/her tenthweek, changes according to the figure below.

e. What was the babys minimum weight? Only an answer is requirede. What was the babys minimum weight? Only an answer is requiredf. How large was the babys average increase in weight from the point she/he was

two weeks old, to the point she/he is 10 weeks old.

f. How large was the babys average increase in weight from the point she/he was

two weeks old, to the point she/he is 10 weeks old.g. Write a formula that expresses the weight of the baby in the interval 2-10 weeks as

a function of time. (3/3)g. Write a formula that expresses the weight of the baby in the interval 2-10 weeks as

a function of time. (3/3)

3500

3700

3900

4100

4300

4500

47004900

5100

5300

5500

5700

5900

6100

0 1 2 3 4 5 6 7 8 9 10 11 12

Age (weeks)

Weight

(g)


a. Answer: Babys minimum weight was: g7003 . (1/0)

b. wgwgw

g

w

g/230/225

8

8001

210

80036005==

Answer: Babys average increase in weight in the interval 2-10 weekswas: . (2/1)wg /230

c. Answer: 2x (0/2)( ) 1038002230 += gxywhere gram is the weight of the baby in the time interval 2-10

weeks, and

y

x is the age of the baby in weeks.


14/17



7. The average salary of the 15 employees of a company is month/15 , and their

median salary is month/16 . After two new people were hired the average

monthly salary of the companys workers increased to month/16 even

though the salary of one of the new hired worker was less than month/15 .

7. The average salary of the 15 employees of a company is month/15 , and their

median salary is month/16 . After two new people were hired the average

monthly salary of the companys workers increased to month/16 even

though the salary of one of the new hired worker was less than month/15 .

kr800

000

000

800

kr800

kr000

kr000

kr800

kr

kr

kr

h. Suggest reasonable monthly salaries of the newly hired employees.h. Suggest reasonable monthly salaries of the newly hired employees.i. The median monthly salary of the employees of the company did not change after

the two newly hired workers joined the company. Explain why. (2/3)i. The median monthly salary of the employees of the company did not change after

the two newly hired workers joined the company. Explain why. (2/3)


a. Assuming the monthly salary of the newly hired employees asa. Assuming the monthly salary of the newly hired employees as x and

month :kr/00015 ( ) ( )0001617000158001515 =++ x ( ) ( )8001515000150001617 =x monthkrx /00020=

Answer: The individual monthly salaries of the two newly hiredemployees could be and . (1/2)monthkr/00015 monthkr/00020

Second method:The total salary of the two newly employed workers must be:

monthkryx /0003580015150001617 ==+

Therefore the monthly salaries of the newly hired ones could be couldbe and . (1/2)monthkr/00015 monthkr/00020

b. Answer: Due to the fact the monthly salary of one of the newly hiredworkers were less than the median salary of month , and the

other one were more than that the median salary did not change afterhiring the new workers. (1/1)

kr/00016

8. Every day under September 1998 the amount ofrain in a region in North Jmtland is measured. Inthe diagram below the results of themeasurements are presented

a. Under how many days rained more than mm10 ?

Only an answer is requiredb. Somebody mistakenly claims that The diagram

illustrates that under the first few days most ofrain rained. Describe what type of error is madein this statement.

c. How much did rain in September 1998 in the of

the interest in North Jmtland? (2/3)Suggested solutions:

a. Answer: Under 7 days rained more mm10 (1/0)

b. The person is reading the diagram incorrectly. He/She possiblyidentifies the horizontal axis by mistake days of the month. The x-axis is the amount of the rain not days of the month. Moreover,he/she identifies the y-axis as the amount of rain. The vertical axisis the frequency (number) of the days that a certain amount of rainrained. (1/0)

c.

mmV 1955.225.1265.785.215=+++=

(0/3)Answer: rained in September 1998 in the region of the interest.mm195


15/17



9. A The first episode of the new TV-soap-opera Scum was watched by

00080 people. It has become a successful program and the

number of its watchers increases at the rate of %6 per week.

9. A The first episode of the new TV-soap-opera Scum was watched by

00080 people. It has become a successful program and the

number of its watchers increases at the rate of %6 per week.j. How many people watch the program after two weeks?j. How many people watch the program after two weeks?

k.

With how many percent has the number of people who watchthe soap-opera increased after five weeks?k.

With how many percent has the number of people who watchthe soap-opera increased after five weeks?l. After how many weeks has the number of people who watch

the soap-opera doubled? (4/3)l. After how many weeks has the number of people who watch

the soap-opera doubled? (4/3)


a. 0009080 a. 0009080 ( ) 8888906.1000 2 = ( ) 8888906.1000 2 =Answer: After two weeks the number of people who watch the TV-soap-opera increased to about people. (2/0)00090

b. First method:

( ) 34.1338.106.1 5 = Answer: The number of people who watch the soap-opera increasedafter five weeks is increased by . (2/1)%34

Second method:

( ) 05810705810706.100080 5 =

%3434.0338.000080

05827

00080

00080058107===

c. ( ) 206.1 =n

First method:(May be ignored for now)The analytical method of solving this problem involves logarithms

which is part of mathematics course C.( ) 206.1 =n

( ) 2log06.1log =n ( ) 2log06.1log =n

( )12896.11

06.1log

2log=n 12=n

Second method: Graphicalmethod

( ) 206.1 =n may be solved

graphically. To solve it using agraphic calculator (TI-83 or higher) , we may plot ( )nY 06.1:1 , andon the same frame, and using the facilities

2:2Y

nd2 calc intersect find the

solution of ( ) . According to the graph illustrated above206.1 =n 12=n Answer: After 12 weeks number of Scum-watchers are doubled.Third method: Numerical Method

To solve ( ) , we may do the following: write206.1 =n 06.1 ENTER 06.1Ans ENTER ENTER ENTER .tills the calculator shows a number

just above 2, i.e. . The number of012196.2 ENTER s is the solution of

the problem, i.e. 12=n . (0/2)

0,0

0,5

1,0

1,5

2,0

2,5

0 2 4 6 8 10 12 1

n (Weeks)

Numberofthe"Scum"Watchers

4

( ) 206.1 =n


16/17



9:B Jonny bought a second-hand Harley Davidson for . The salesman

claimed that the of value of the motorcycle increases in value at the rate per year.

9:B Jonny bought a second-hand Harley Davidson for . The salesman

claimed that the of value of the motorcycle increases in value at the rate per year.

kr00060

%4

a. How much is it value after two years?a. How much is it value after two years?

b.

With how many percent does its value increaseafter five years?b.

With how many percent does its value increaseafter five years?

c. After how many weeks has its value doubled?c. After how many weeks has its value doubled?(4/3)(4/3)


a. kr00060 a. kr00060 ( ) 658966404.1000 2 =

kr00060

%4

( ) 658966404.1000 2 =Answer: After two years the value of the Harley Davison is increasedto . (2/0)kr00065

b. First method:

( ) 22.1217.104.1 5 = Answer: After five years the value of the Harley Davison is increasedby . (2/1)%22

Second method:

( ) kr000739997204.100060 5 =

%2222.0217.000060

99912

00060

0006099972===

c. ( ) 204.1 =n

First method:(May be ignored for now)

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

( ) 204.1 =n

( ) 2log04.1log =n ( ) 2log04.1log =n

( )18673.17

04.1log

2log=n 18=n


( ) 204.1 =n

may be solvedgraphically. To solve it using a

graphic calculator (TI-83 or higher) , we may plot ( )nY 04.1:1 , andon the same frame, and using the facilities

2:2Y


solution of ( ) . According to the graph illustrated above204.1 =n 18=n Answer: After 18 years the Harley Davison is doubled in value.Third method: Numerical Method


just above 2, i.e. . The number of02582.2 ENTER s is the solution of theproblem, i.e. 18=n . (0/2)

0,0

0,5

1,0

1,5

2,0

2,5

0 2 4 6 8 10 12 14 16 18 20

n (Years)

Valueof"HarleyDavison"(SKr) ( ) 204.1 =n


17/17



9:D Maria borrows 30 to starts her own business. She is not going to pay back

any part of her loan for 20 years when it is expected that the business is fullyestablished and profitable. The annual interest on her loan is .

9:D Maria borrows 30 to starts her own business. She is not going to pay back

any part of her loan for 20 years when it is expected that the business is fullyestablished and profitable. The annual interest on her loan is .

kr000

%6

a. How much is Marias loan after two years?a. How much is Marias loan after two years?

b. With how many percent has her loan increased after fiveyears?

b. With how many percent has her loan increased after fiveyears?

c. How long does it take for her loan to be doubled? (4/3)c. How long does it take for her loan to be doubled? (4/3)


a. kr700 a. kr700 ( ) 337083306.100030 2 =

kr000

%6

( ) 337083306.100030 2 =Answer: After two years Maria owes 33 . (2/0)kr700

b. First method:

( ) 34.1338.106.1 5 = Answer: After five years the amount Maria owes is increased by .%34

Second method:

( ) kr150401474006.100030 5 =

%3434.0338.000030

14710

00030

0003014740===

(2/1)

a. ( ) 206.1 =n

First method:(May be ignored for now)The analytical method of solving this problem involves logarithms

which is part of mathematics course C.( ) 206.1 =n

( ) 2log06.1log =n ( ) 2log06.1log =n

( )12896.11

06.1log

2log=n 12=n


( ) 206.1 =n may be solved

graphically. To solve it using agraphic calculator (TI-83 or higher) , we may plot ( )nY 06.1:1 , andon the same frame, and using the facilities

2:2Y


solution of ( ) . According to the graph illustrated above206.1 =n 12=n Answer: After 12 years Maria owes doubled as much. (0/2)Third method: Numerical Method


just above 2, i.e. . The number of012196.2 ENTER s is the solution of

the problem, i.e. 12=n .

0,0

0,5

1,0

1,5

2,0

2,5

3,0

3,5

0 2 4 6 8 10 1 2 1 4 16 1 8 2 0 2 2

n (Years)

(TheamountMaia'sloan)/(originalloan)

( ) 206.1 =n

solutionsnpmaavt99+part+i ii

Documents