Solutions+Make+Up+Test+V2+MaA+Mathematical+Reasoning+Ch1 3

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Suggested Solutions Make up Chapter Test V2 MaA: Mathematical Reasoning Ch1-3 NV-College Name: ________________________ Student number:______

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

CHAPTER TEST: Mathematical Reasoning

MATHEMATICS COURSE A

FALL 2008: MaANVC08

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

Instructions

Test period 90 minutes for Part I and Part II as a whole. We recommend that you use atthe most 45 minutes to work with Part I. You must not use the calculator

before you have handed in Part I. Use a separate sheet for part I. Tools Formula sheet and ruler.

Part I 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.

After every item is given the maximum mark your solution can receive.(2/3) means that the item can give 2 g-points (Pass level) and 3 vg-points(Pass with distinction level).

Items marked with (Problems 1f, 4, 5, 7, 15 and 16) give you a possibilityto show MVG-quality (Pass with special distinction quality) This means that

you use generalised methods, models and reasoning, that you analyse your

results and account for a clear line of thought in a correct mathematical

language.

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 evaluationcan be given even for unfinished solutions.

Mark limitsMaximum score: TOTAL 63 out of which 34 vg-points, and 7 MVG pointsG: 21 pointsVG: 42 points/ at least 11 VG points:MVG: 46 points/ at least 23 VG points; MVG-quality work ¤

1 2 3a 3b 3c 3d 3e 3f 4a 4b 4c 4d 4e 4f 4g 4g¤ 5a 5b 5c 5c¤ 6a 6b 6c 6d 7a 7b 7b¤

1 2 1 1 1 1 1 1 1 1 1 1 2 2 2 1 1 1 2 1 1 2 3 2 2 2 1

G vg G G G G G G G G vg G vg vg vg mvg G G vg mvg G G vg vg G vg mvg

8 9 10 11a 11b 12 12 13 13 14 14 14¤ 15 15 15¤ 16 16 16¤ Sum ¤

varav1 1 1 1 1 1 2 1 2 2 2 1 2 4 1 2 2 1 7 34

G G G G G G vg G vg G vg mvg G vg mvg G vg mvg mvg g&vg vg

Minimum G 21

Minimum VG 42 11

Minimum MVG 46 23

Summa

63

Only the marked problems in the box below will be graded.

1 2 3a 3b 3c 3d 3e 3f 4a 4b 4c 4d 4e 4f 4g 5a 5b 5c

6a 6b 6c 6d 7a 7b 8 9 10 11a 11b 12 13 14 15 16 6c





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 I

You are not allowed to use calculator in this part. Write your answer to the first part on a separate paper. Only after the submission of your solutions to the partone you may use your calculator. You may start working with part II beforesubmission of your solutions to part I.

In the multi-choice problems below just circle the correct answer (s) in the test paper andwrite the answer in the space provided as “Answer: _____________”

1. A member of the solution set of 32 <≤− x is (1/0)

a. 5−

b.

4−

c. 1 d. 3

Answer: Alternative: _____________

Suggested solutions:

32 <≤− x means x is larger than or equal to 2− but less than 3 . Therefore only 1 is a

member of − .32 <≤ x

Answer: Alternative (c)

2. Which statement is true when 01 <<−

< x

3

x <

x ? (1/0)

a. 32 x< x x b. x x << 23

c. x x << 2 x

d. 3 x x < 2

Answer: Alternative: _____________

Why? Explain in sufficient detail. (0/2)Suggested solution:

23

3

2

001

1001 x x x

x x

x x <<<⇔

⎪

⎩

⎪⎨⎧

<<<−

<<⇔<<−

1001 2 <<⇔<<− x x . This is due to the fact that the square of any real number is positive!On the other hand if the magnitude of the number is smaller than one, the square of thenumber will be a number smaller than one.On the other hand, if multiply both sides of an equality with a negative number, only the

direction of the inequality reverses: x x x x x x x >>⇔⋅>⋅>⇔<<⇔<<− 322 0101001

Answer: Alternative (d) 23 x x x <<





3. Calculate and answer in surd (exact) form:

a. 3

64 ⋅ (1/0)21

2653 −+×+×−

Suggested Solution:

41421212412153

21

642653 −=−=⋅−++−=⋅−+×+×− ⇔

Answer: 413

21642653 −=⋅−+×+×−

b. 1525

16525 −(1/0)

−

Suggested Solution: 14−10

140

1525

16525=−=

−

−⇔ Answer: 14

1525

16525

−

−−=

c.

( ) ( )− (1/0)

322

235 −−−Suggested Solution:

( ) ( ) 248925235322 =+−=−−−− ⇔ Answer: 5 ( ) ( ) 2423

322 =−−−−

d. 25

45 (1/0)

6

Suggested Solution: 5.7=2

15

4

52

5

6

254

5

6

2

53

=

/

//⋅

/

/= ⇔ Answer: 5.7

4

5 =

25

6

e. 25

81(1/0)

Suggested Solution: 8.1=10

18

5

9

25

81

25

81=== ⇔ Answer: 8.1

2

8

5

1=

f. 3 27 (1/0)

Suggested Solution: 333=327 3

= ⇔ Answer: 33= 27

4. Solve for x :

a. 113 =− (1/0)7 x

Suggested Solution:

1137 =− x ⇔ 3117 += x ⇔ 147 = x ⇔ 27

14== x ⇔ Answer: 2= x





b. 15− x

(1/0)5

3=

Suggested Solution:

155

3=−

x⇔ 51

5

3+=

x⇔ 6

5

3=

x⇔ 30563 =×= x ⇔ 10

3

30== x Answer: 10= x

c. 76 = (0/1)5

4+

x

Suggested Solution:

765

4=+

x⇔ 67

5

4−=

x⇔ 1

5

4=

x⇔ 5 4= x ⇔ 8.0

5

4== x

113 −

Answer: 8.0= x

d. x9 −≥ (1/0) x 7

Suggested Solution:

x x 79113 −≥− ⇔ 11973 +≥+ x x ⇔ 2010 ≥ x ⇔ 10

20

≥ x ⇔ Answer:2≥ x 2≥ x

e. 125= x (0/1)64 3

Suggested Solution:

12564 3 = x ⇔64

1253 = x ⇔3

1

64

125⎟ ⎠

⎞⎜⎝

⎛ = x ⇔

( )

( )3

1

3

1

64

125= x ⇔

4

5= x ⇔ Answer: 25.1= x

3 − x

f. 6453= (0/2)( )

Suggested Solution:

( ) 64533=− x ⇔ ( ) 3

1

6453 =− x ⇔ 453 =− x ⇔ 9543 =+= x ⇔ 3

9= x Answer: 3= x

g. 0≠∀ x . (0/2/¤)0256

4 3

2

=+ x

x

Suggested Solution:

00256

4 3

2

≠∀=+ x x

x⇔

3

2 256

4 x

x−= ⇔ 256432 ×−=⋅ x x ⇔ 45 44×−= x

55 4−= x ⇔ ( )55 4−= x ⇔ 4−= x Answer: 4−= x

5. Simplify the following expressions to their simplest possible form:

a. 757

21a(1/0)

27

ba

b−

Suggested Solution:

5

2527257

75

27 333

7

21

b

ababa

ba

ba−=⋅−=⋅−=

− −−− Answer:5

252

75

27 33

7

21

b

aba

ba

ba−=⋅−=

− −







Answer: 16 300300300 232 −− =×

10

d. 99910− (0/2)1001

Suggested Solution:

( ) ( ) 100099999929999991001 109.91099110010110101010 ×=×=−=−=−

Answer: 10 10009991001 109.910 ×=−

=ax

7. Find the relationship between two real positive numbers a and b such that the equation

( )7− xb 23

a) has the solution 4= x . (2/0) b) does not have any solution. (0/2/¤)Suggested Solution:

a) If 5= x is the solution of ( )723 −= xb , we may replaceax x by 4= x in the equation

which implies:

( ) ( )243 = ba ( 74 − )( ) bba 63212 −=−=

ba −=2

ba2

1−= Answer: ba

2

1−=

=ax

b) If the equation ( )7− xb does not have any solution, that would mean we may

not be able to find a value for

23

x such that the left hand side of the equation is the same

as the right hand side of it. Such situation may arouse if ba 23 = . To prove the claim

we may replace a3 by b2 in the equation ( )73 2 −= xb which will produce:ax( ) ( ) 7722723 −=⇔−=⇔−= x x xbbx xbax

7+= x x does not have any solution! Answer: baba3

22 =⇔=3





Part II

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

8. Simplify ( )3a (2/0)2

927

Suggested Solutions:

( ) ( ) ( ) 66263

233

293

2

3

29 9332727 aaaaa ==== Answer: ( ) 63

29 927 aa =

9. Calculate the side of a cubic box whose volume is 30.100 cm .

(1/0)


cmcmcmacma 64.4642.4100100 3

1

33≈==⇔= Answer: cma 64.4≈

10. Solve the equation 215 = x . Round off the answer to 3 decimal places. (1/0)


047.122 15

1

15 ==⇔= x x Answer: 047.1= x

11. (a) Find the number that is midway between 00000010 and 000000100 (1/0)

(b) Find a number that is greater than but less than . (1/0) 5105.2 −× 4105.2 −×


(a) 7105.5000000552

000000110

2

00000010000000010 ×===

+

(b) Answer:2 :5106. −×

455 105.2106.2105.2 −−− ×<×<×

12. Kevin sold two second hand cars for 00010$ each. On the first one he made %30 , but on

the second car he lost %30 . Did he loose or made money in the transaction? By how

much and how many percent? Explain. (1/2)


A : Original price of the first car: 6927$30.1

00010$30.100010$ ==⇔⋅= A A

B : Original price of the second car: 28614$70.0

00010$70.000010$ ==⇔⋅= B B

Kevin had paid 97821$28614$6927$ =+=+ B A for the cars, but he sold them only for

. Therefore, he lost00020$ 9781$00020$97821$ =− in the deal. This is:

%909.097821$

97821$00020$−≈−=

−

Answer: Kevin lost in the transaction. He lost 9% of his investment.9781$





13. In a survey of 200 seniors at a high school it was found that 75 students play soccer but

not softball, 45 students play softball but not soccer, and 28 students do not participate ineither sport. Of the seniors surveyed, what percent play both soccer and softball? (1/2)

Suggested solutions:We may draw a Venn graph usingCircle A : 200 seniorsCircle B : Students playing soccerCircle C : Students playing softball

From the Venn diagram illustrated we maydeduce that:The number of students who play both soccerand football is:

( ) 52148200284575200 =−=++−=C B

N I

%2626.020052 ===

tot

C B

N

N I

Answer: 26% of students play both soccer and football.

At the aspect assessment of your work with problems 9-11 the 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.

This means that you use generalised methods, models and reasoning , that you analyse your

results and account for a clear line of thought in a correct mathematical language .

14. Choose any arbitrary three-digit-number. Multiply the number by 7. Multiply the result by11 and then by 13. What is the result? Repeat the procedure for any other arbitrary three-digit number. What is the result? Is there any pattern? Why? Generalize and Explain.

(2/2/¤)


Answer: I get a six-digit-number which is my number repeated twice. Ex:12312313117123 =×××

36436413117364 =×××

57957913117579 =×××

99999913117999 =×××

Reason:Assume the number I choose is . Due to the fact that ,

when I multiply by

abc 110013117 =××

abc 110013117 =×× , I get abcabcabc +×=× 10001001 . My

three-digit number multiplied by 1000 looks like and when I add to

it my number: . This is true for any 3-digit number.

000abc

abcabcabcabc =+000

A

B

C 75

45

28





15. Quadratic small tables, that has only three places for three people at each side, is arrangedas a long dining table, as illustrated below:

¤ How many persons, at the most, may be placed at along dining table consisted of five small squaretable?

¤ How many small square tables are needed in order to be able to place 60 persons, at a long dining tablearranged by according to the rules above?

¤ Find a formula that describes the relationship between the number of persons, p , that will be

placed at a dining table arranged as above of n

number of small tables.

¤ Describe your formula above in words. (2/4/¤)


¤ 36 persons, at the most, may be placedat a long dining table consisted of fivesmall square table. This is due to the factthat 9 people will be able to take a seatat each of the two corner tables, theremaining 3 tables have 6 place each:

366392 =×+×=P

¤ 9 small tables are needed to place 60 people around it.This is due to the fact that as mentioned above 9 people have placearound each of the two corner tables. The middle tables have each a

maximum of 6 places available. Therefore 7 middle tables are neededto place 42 people at them.¤ Using the logic above, we may develop a formula for P people and n ,

number of small tables:( ) ( )1666126182692 +=+=−+=−×+×= nnnnP

⇔ 6166

66666 >∀−=

−=⇔−=⇔+= P

PPnPnnP

Answer ( ) 21666 ≥+=+= nnnP and 2;7166

6≥≥∀−=

−= nP

PPn

8

40

6

2

8

4

0

n

2 1

3 24 3

5 3

6 4

7 4

8 5

9 6

10 66

11 72

12 78

13 84

14 90

15 96

16 102

17 108

18 114

19 120

20 126

( ) 21666 ≥+=+= nnnP





16. Arrange the following numbers in order of magnitude without the aid of a calculator:

, , , . (2/2/¤)16525 5564 11064 66243


In order to compare numbers in exponent form either we should find amaximum common exponent, MCE, to all of them and then compare thebases of the numbers, or find a maximum common base and compare theexponents of the numbers with each other. In this case we may choose amaximum common exponent. In the first glance we may assume that theMCE of the numbers 55, 66, 110, and 165 is 11, which is fine, but we mayrewrite the numbers and try to get the smallest possible integer base foreach number using the prime factors of the exponents, and lowest base,as follows:

( ) 115321652165 5525 ×××== using and2525= 1153165 ××=

( ) 11532115655 2264 ××××== using and6264 = 11555 ×=

( ) 1153211532211526110 42264 ×××××××××=== using ,6264 = 422 = and 1152110 ××=

( ) 115321132566 33243 ×××××== using , and53243 = 113266 ××=

It is clear that 3307532 =××× is the maximum common exponent, MCE of

the numbers above. Therefore, we may rearrange the numbers above,changing the base of each exponent, and keeping the common power of

each as 330:( ) 33011532115655 22264 === ××××

( ) 330115321132566 333243 === ×××××

( ) 3301153211532211526110 442264 ==== ×××××××××

( ) 330115321652165 55525 === ××× Answer: 1651106655 256424364 <<<

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