TALLER DE REPASO PARA EXAMEN COORDINADO CON RESPUESTAS 1. · TALLER DE REPASO PARA EXAMEN...

IB Questionbank Maths SL 1

TALLER DE REPASO PARA EXAMEN COORDINADO CON RESPUESTAS

1. El siguiente grafico es un diagrama de caja y bigotes.

El rango intercuartìlico es 20 y el rango es 40.

(a) Escriba el valor de la mediana.

(b) Encuentre el valor de

(i) a;

(ii) b.

(4) (Total 5 marks)

R1. (a) 18

(b) (i) 10

(ii) 44

2. La gràfica de y = p cos qx + r, para –5 ≤ x ≤ 14, se muestra a continuación.

Hay un punto mínimo en (0, –3) y un punto máximo en (4, 7).

(a) Encuentre el valor de

(i) p;

(ii) q;

(iii) r.

(6) (Total 6 marks)


R2. (a) (i) evidence of finding the amplitude

e.g. , amplitude = 5

p = –5

(ii) period = 8

q = 0.785

(iii) r =

r = 2

3. (a) Factorize the expression 3 sin2

x – 11 sin x + 6.

(b) Consider the equation 3 sin2

x – 11 sin x + 6 = 0.

(i) Find the two values of sin x which satisfy this equation,

(ii) Solve the equation, for 0° x 180°.

(Total 6 marks)

R3. (a) (3 sin x – 2)(sin x – 3)

(b) (i) (3 sin x – 2)(sin x – 3) = 0

sin x = sin x = 3

(ii) x = 41.8°, 138°

4. The diagram below shows a quadrilateral ABCD. AB = 4, AD = 8, CD =12, B D = 25,

=.

2

37

4

π

8

π2

2

37

3

2

C

DAB


(a) Use the cosine rule to show that BD =

. (2)

Let = 40.

(b) (i) Find the value of sin .

(ii) Find the two possible values for the size of .

(iii) Given that is an acute angle, find the perimeter of

ABCD. (12)

(c) Find the area of triangle

ABD. (2) (Total 16 marks)

R4. (a) For correct substitution into cosine rule

BD =

For factorizing 16, BD =

=

(b) (i) BD = 5.5653 ...

sin = 0.911 (accept 0.910, subject to AP)

(ii) = 65.7

Or = 180 their acute angle

= 114

(iii) = 89.3

(or cosine rule)

BC = 13.2 (accept 13.17…)

Perimeter = 4 + 8 + 12 + 13.2

= 37.2

(c) Area = 4 8 sin 40 = 10.3

5. In a school of 88 boys, 32 study economics (E), 28 study history (H) and 39 do not study either

subject. This information is represented in the following Venn diagram.

cos454

DBC

DBC

DBC

θcos84284 22

θcos4516

θcos454

5653.5

25sin

12

DBCsin

DBC

DBC

DBC

CDB

65.7sin

12

89.3sin

BCor

25sin

5.5653

89.3sin

BC

2

1


(a) Calculate the values a, b,

c. (4)

(b) A student is selected at random.

(i) Calculate the probability that he studies both economics and history.

(ii) Given that he studies economics, calculate the probability that he does not study

history.

(3)

(c) A group of three students is selected at random from the school.

(i) Calculate the probability that none of these students studies economics.

(ii) Calculate the probability that at least one of these students studies

economics. (5) (Total 12 marks)

R5. (a)

n (E H) = a + b + c = 88 – 39 = 49

n (E H) = 32 + 28 – b = 49

60 – 49 = b = 11

a = 32 – 11 = 21

c = 28 – 11 = 17

(b) (i) P(E H) =

(ii) P(HE) =

a b c

39

H (28)E (32)

U (88)

a b c

E(32) H(28)

39

U(88)

8

1

88

11

88

3288

21

P

'P

E

EH


= (= 0.656)

OR

Required probability =

(c) (i) P(none in economics) =

= 0.253

(ii) P(at least one) = 1 – 0.253

= 0.747

OR

3

= 0.747

6. The graph of a function of the form y = p cos qx is given in the diagram below.

(a) Write down the value of p.

(b) Calculate the value of q. (Total 6 marks)

R6. (a) p = 30

(b) METHOD 1

Period =

=

q = 4

32

21

32

21

868788

545556

86

30

87

31

88

32

86

56

87

31

88

323

86

55

87

56

88

32

40

30

20

10

–10

–20

–30

–40

/2 x

y

q2

2


METHOD 2

Horizontal stretch of scale factor =

scale factor =

q = 4

7. The following frequency distribution of marks has mean 3.6

Mark 1 2 3 4 5 6

Frequency 4 6 12 x 8 4

(a) Find the value of

x. (4)

(b) Write down the standard

deviation. (2) (Total 6 marks)

R7. (a) Σfx = 1(4) + 2(6) + ... + 6(4), Σfx = 116 + 4x (seen anywhere)

evidence of substituting into mean =

correct equation

e.g. 116+4𝑥

34+𝑥 = 3.6, 116 +4x = 3.6(34 + x)

x = 16

(b) σ = 1.39

8. Solve the equation 3 sin2

x = cos2

x, for 0° x 180°.

(Total 4 marks)

R8. tan2 x =

tan x =

x = 30° or x = 150°

9. A box contains 100 cards. Each card has a number between one and six written on it.

The following table shows the frequencies for each number.

Number 1 2 3 4 5 6

q1

4

1

f

fx

3

1

3

1


Frequency 26 10 20 k 29 11

(a) Calculate the value of

k. (2)

(b) Find

(i) the median;

(ii) the interquartile

range. (5) (Total 7 marks)

R9. (a) evidence of using ∑ fi = 100

k = 4

(b) (i) evidence of median position

e.g. 50th

item, 26 + 10 + 20 = 56

median = 3

(ii) Q1 = and Q3 = 5

interquartile range = 4 (accept 1 to 5 or 51, etc.)

10. Given that sin x = , where x is an acute angle, find the exact value of

(a) cos x;

(b) cos 2x.

(Total 6 marks)

R10. (a) x is an acute angle => cos x is positive.

cos2 x + sin

2 x = 1 => cos x =

=> cos x =

= (= )

(b) cos 2x = 1 – 2 sin2

x = 1 – 2

=

11. A test marked out of 100 is written by 800 students. The cumulative frequency graph for the

marks is given below.

3

1

x2sin–12

3

1–1

9

8

3

22

2

3

1

9

7


(a) Write down the number of students who scored 40 marks or less on the

test. (2)

(b) The middle 50 % of test results lie between marks a and b, where a < b.

Find a and

b. (4) (Total 6 marks)

R11.


(a) Lines on graph

100 students score 40 marks or fewer.

(b) Identifying 200 and 600

Lines on graph

a = 55, b = 75

12. The following diagram shows a triangle ABC, where is 90, AB = 3, AC = 2 and

is .

(a) Demuestre que sin = .

(b) Demuestre que sin 2 = .

(c) Find the exact value of cos 2. (Total 6 marks)

R12.(a) Evidence of correct approach

eg sin =

sin =

(b) Evidence of using sin 2 = 2 sin cos

=

=

(c) Evidence of using an appropriate formula for cos 2

eg

BCA CAB

3

5

9

54

523BC,AB

BC 22

3

5

3

2

3

52

9

54

81

801,

9

521,1

9

42,

9

5

9

4


cos 2 =

13. The following is the cumulative frequency curve for the time, t minutes, spent by 150 people in

a store on a particular day.

(a) (i) How many people spent less than 5 minutes in the store?

(ii) Find the number of people who spent between 5 and 7 minutes in the store.

(iii) Find the median time spent in the

store. (6)

(b) Given that 40 of the people spent longer than k minutes, find the value of

k. (3)

(c) (i) On your answer sheet, copy and complete the following frequency table.

9

1

150

140

130

120

110

100

90

80

70

60

50

40

30

20

10

0

cum

ula

tive

freq

uen

cy

121110987654321

time ( )t


t (minutes) 0 t 2 2 t 4 4 t 6 6 t 8 8 t 10 10 t 12

Frequency 10 23 15

(ii) Hence, calculate an estimate for the mean time spent in the

store. (5) (Total 14 marks)

R13. (a) (i) 50 (accept 49, “fewer than 50”)

(ii) Cumulative frequency (7) = 90

90 50

= 40

(iii) 75th or 75.5th person

median = 6.25 (min), 6 min 15 secs

(b) Evidence of finding 40% (60%) of 150

Number spending less than k minutes is (150 60) = 90

k = 7

(c) (i)

t (minutes) 0 t < 2 2 t < 4 4 t < 6 6 t < 8 8 t < 10 10 t < 12

Frequency 10 23 37 38 27 15

(ii) Evidence of using all correct mid-interval values (1, 3, 5, 7, 9, 11)

mean =

= 6.25 (min), 6 min 15 secs

14. The number of hours of sleep of 21 students are shown in the frequency table below.

Hours of sleep Number of students

4 2

5 5

6 4

7 3

8 4

10 2

12 1

Find

(a) the median;

150

1511279387375233101


(b) the lower quartile;

(c) the interquartile range.

(Total 6 marks)

R14.

x f f

4 2 2

5 5 7

6 4 11

7 3 14

8 4 18

10 2 20

12 1 21

(a) m = 6

(b) Q1 = 5

(c) Q3 = 8

IQR = 8 – 5

= 3 (accept 5 – 8 or [5, 8])

15. Two fair 4-sided dice, one red and one green, are thrown. For each die, the faces are labelled 1,

2, 3, 4. The score for each die is the number which lands face down.

(a) List the pairs of scores that give a sum of

6. (3)

The probability distribution for the sum of the scores on the two dice is shown below.

Sum 2 3 4 5 6 7 8

Probability p q r

(b) Find the value of p, of q, and of

r. (3)

Fred plays a game. He throws two fair 4-sided dice four times. He wins a prize if the sum is 5

on three or more throws.

(c) Find the probability that Fred wins a

prize. (6) (Total 12 marks)

R15. (a) three correct pairs

e.g. (2, 4), (3, 3), (4, 2), R2G4, R3G3, R4G2

16

3

16

4

16

3

16

1


(b) p = , q = , r = A1A1A1 N3 3

(c) let X be the number of times the sum of the dice is 5

evidence of valid approach (M1)

e.g. X ~ B(n, p), tree diagram, 5 sets of outcomes produce a win

one correct parameter (A1)

e.g. n = 4, p = 0.25, q = 0.75

Fred wins prize is P(X ≥ 3) (A1)

appropriate approach to find probability M1

e.g. complement, summing probabilities, using a CDF function

correct substitution (A1)

e.g. 1– 0.949…,

probability of winning = 0.0508 A1 N3 6

16. The histogram below represents the ages of 270 people in a village.

(a) Use the histogram to complete the table below.

Age range Frequency Mid-interval

value

0 age 20 40 10

20 ≤ age 40

40 ≤ age 60

16

1

16

2

16

2

256

1

256

12,00390625.0046875.0,

256

2431

256

13


60 ≤ age 80

80 ≤ age ≤100

(2)

(b) Hence, calculate an estimate of the mean

age. (4) (Total 6 marks)

R16. (a)

Age range Frequency Mid - interval value

0 age < 20 40 10

20 age < 40 70 30

40 age < 60 100 50

60 age < 80 50 70

80 age 100 10 90

A1A1 N2

(b) For attempting to find (M1)

Correct substitution (A1)

eg 40 10 + ... + 10 90 = 11900

For dividing by 270 (M1)

eg

Mean = 44.1 A1 N4

17. The diagram below shows the probabilities for events A and B, with P(A′) = p.

xf

270

11900


(a) Write down the value of

p. (1)

(b) Find

P(B).

(3)

(c) Find P(A′ |

B). (3) (Total 7 marks)

R17. (a) p = A1 N1

(b) multiplying along the branches (M1)

e.g.

adding products of probabilities of two mutually exclusive paths (M1)

e.g.

P(B) = A1 N2

(c) appropriate approach which must include A′ (may be seen on diagram) (M1)

e.g.

P(A′│B) = (A1)

P(A′│B) = A1 N2

18. The letters of the word PROBABILITY are written on 11 cards as shown below.

Two cards are drawn at random without replacement.

Let A be the event the first card drawn is the letter A.

Let B be the event the second card drawn is the letter B.

(a) Find

P(A).

(1)

5

4

40

12,

4

1

5

1

40

12

20

1,

8

3

5

4

4

1

5

1

20

7

40

14

)(P

)P(accept not do

)(P

)(P

B

BA

B

BA

20

78

3

5

4

7

6

14

12


(b) Find

P(B│A). (2)

(c) Find P(A ∩

B). (3) (Total 6 marks)

R18. (a) P(A) = A1 N1

(b) P(B│A) = A2 N2

(c) recognising that P(A ∩ B) = P(A) × P(B│A) (M1)

correct values (A1)

e.g. P(A ∩ B) =

P(A ∩ B) =

A1 N3

19. If A is an obtuse angle in a triangle and sin A = , calculate the exact value of sin 2A.

(Total 4 marks)

R19. sin A = cos A = (A1)

But A is obtuse cos A = – (A1)

sin 2A = 2 sin A cos A (M1)

= 2 ×

= – (A1) (C4)

20. Two boxes contain numbered cards as shown below.

Two cards are drawn at random, one from each box.

(a) Copy and complete the table below to show all nine equally likely outcomes.

11

1

10

2

10

2

11

1

110

2

135

13

5

13

12

13

12

13

12

13

5

169

120


3, 9

3, 10

3, 10

(2)

Let S be the sum of the numbers on the two cards.

(b) Write down all the possible values of

S. (2)

(c) Find the probability of each value of

S. (2)

(d) Find the expected value of

S. (3)

(e) Anna plays a game where she wins $50 if S is even and loses $30 if S is odd.

Anna plays the game 36 times. Find the amount she expects to have at the end of the 36

games.

(3) (Total 12 marks)

R20. (a)

3, 9 4, 9 5, 9

3, 10 4, 10 5, 10

3, 10 4, 10 5, 10

A2 N2

(b) 12, 13, 14, 15 (accept 12, 13, 13, 13, 14, 14, 14, 15, 15) A2 N2

(c) P(12) = , P(13) = , P(14) = , P(15) = A2 N2

(d) correct substitution into formula for E(X) A1

e.g. E(S) = 12 ×

E(S) = A2 N2

(e) METHOD 1

correct expression for expected gain E(A) for 1 game (A1)

e.g.

E(A) =

amount at end = expected gain for 1 game × 36 (M1)

= 200 (dollars) A1 N2

9

1

9

3

9

3

9

2

9

215

9

314

9

313

9

1

9

123

309

550

9

4

9

50


METHOD 2

attempt to find expected number of wins and losses (M1)

e.g.

attempt to find expected gain E(G) (M1)

e.g. 16 × 50 – 30 × 20

E(G) = 200 (dollars) A1 N2

21. There are 20 students in a classroom. Each student plays only one sport. The table below gives

their sport and gender.

Football Tennis Hockey

Female 5 3 3

Male 4 2 3

(a) One student is selected at random.

(i) Calculate the probability that the student is a male or is a tennis player.

(ii) Given that the student selected is female, calculate the probability that the student

does not play

football. (4)

(b) Two students are selected at random. Calculate the probability that neither student plays

football.

(3) (Total 7 marks)

R21. (a) (i) correct calculation (A1)

e.g.

P(male or tennis) = A1 N2

(ii) correct calculation (A1)

e.g.

P(not football | female) = A1 N2

(b) P(first not football) = , P(second not football) = A1

P(neither football) = A1

369

5,36

5

4

20

3324,

20

2

20

5

20

9

5

3

20

12

11

33,

20

11

20

6

11

6

20

11

19

10

20

11

19

10


P(neither football) = A1 N1

22. Given that sin θ = , cos θ = – and 0° ≤ θ ≤ 360°,

(a) find the value of θ;

(b) write down the exact value of tan θ. (Total 4 marks)

23. (a)

Acute angle 30° (M1)

Note: Award the (M1) for 30° and/or quadrant diagram/graph

seen.

2nd quadrant since sine positive and cosine negative

= 150° (A1) (C2)

(b) tan 150° = –tan 30° or tan 150° = (M1)

tan 150° = – (A1) (C2)

23. Las calificaciones de diez estudiantes en sus exámenes coordinados de Matemáticas y Biología

aparecen en la siguiente tabla:

Matemàticas (x) 1 4.5 3 2.5 2 1 5 4.5 3 2

Bioklogìa (y) 3.6 3.2 4.5 3.4 5 6 2.4 2.5 3 4.5

(a) Escriba la recta de regresión lineal de x sobre y.

(b) Estime la calificación en Biología de quien haya obtenido 4 en Matemáticas.

(a) 𝑦 = −0.61𝑥 + 5.5 Correcto uso de la calculadora.

(b) 𝑦 = −0.61(4) + 5.5 = 3.11

38

11

380

110

21

23

30º

2

3

2

1

3

1

TALLER DE REPASO PARA EXAMEN COORDINADO CON RESPUESTAS 1. · TALLER DE REPASO PARA EXAMEN...

Documents

Transcript of TALLER DE REPASO PARA EXAMEN COORDINADO CON RESPUESTAS 1. · TALLER DE REPASO PARA EXAMEN...