2014 Paper 2 [176 marks]

[1 mark]1a.

As part of his IB Biology field work, Barry was asked to measure the circumference of trees, in centimetres, that were growing at

different distances, in metres, from a river bank. His results are summarized in the following table.

State whether distance from the river bank is a continuous or discrete variable.

Markschemecontinuous (A1)[1 mark]

1b. [4 marks]On graph paper, draw a scatter diagram to show Barry’s results. Use a scale of 1 cm to represent 5 m on the x-axis and 1 cm

to represent 10 cm on the y-axis.

Markscheme

(A1)(A1)(A1)(A1)

Notes: Award (A1) for labelled axes and correct scales; if axes are reversed award (A0) and follow through for their points. Award

(A1) for at least 3 correct points, (A2) for at least 6 correct points, (A3) for all 9 correct points. If scales are too small or graph paper

has not been used, accuracy cannot be determined; award (A0). Do not penalize if extra points are seen.

[4 marks]

1c. [2 marks]Write down

(i) the mean distance, , of the trees from the river bank;

(ii) the mean circumference, , of the trees.

Markscheme(i) 26 (m) (A1)(ii) 65 (cm) (A1)[2 marks]

x

y

[2 marks]1d. Plot and label the point on your graph.

Markschemepoint labelled, in correct position (A1)(A1)(ft)

Notes: Award (A1)(ft) for point plotted in correct position, (A1) for point labelled or . Follow through from their answers to

part (c).

[2 marks]

M( , )x y

M

M ( , )x y

1e. [4 marks]Write down

(i) the Pearson’s product–moment correlation coefficient, , for Barry’s results;

(ii) the equation of the regression line on , for Barry’s results.

Markscheme(i) (G2) Note: Award (G2) for . Award (G1) for .

Award (A1)(A0) if minus sign is omitted.

(ii) (G2) Notes: Award (A1) for , (A1) for . If the answer is not given as an equation, award a maximum of (A1)(A0). [4 marks]

r

y x

−0.988 (−0.988432…)

−0.99 −0.990

y = −0.756x + 84.7 (y = −0.756281… x + 84.6633…)

−0.756x 84.7

[2 marks]1f. Draw the regression line on on your graph.

Markschemeregression line through their (A1)((ft)regression line through their (accept ) (A1)(ft) Notes: Follow through from part (d). Award a maximum of (A1)(A0) if the line is not straight. Do not penalize if either the line does

not meet the y-axis or extends into quadrants other than the first.

If is not plotted or labelled, then follow through from part (c).

Follow through from their y-intercept in part (e)(ii).

[2 marks]

y x

M

(0,85) 85 ± 1

M

[2 marks]1g. Use the equation of the regression line on to estimate the circumference of a tree that is 40 m from the river bank.

Markscheme (M1)

(A1)(ft)(G2)

Notes: Accept ( ) for use of 3 sf. Accept from use of and .

Follow through from their equation in part (e)(ii) irrespective of working shown; the final answer seen must be consistent with

that equation for the final (A1) to be awarded.

Do not accept answers taken from the graph.

[2 marks]

y x

−0.756281(40) + 84.6633

= 54.4 (cm) (54.4120…)

54.5 54.46 54.3 −0.76 84.7

[4 marks]2a.

A group of tourists went on safari to a game reserve. The game warden wanted to know how many of the tourists saw Leopard ( ),

Cheetah ( ) or Rhino ( ). The results are given as follows.

5 of the tourists saw all three

7 saw Leopard and Rhino

1 saw Cheetah and Leopard but not Rhino

4 saw Leopard only 3 saw Cheetah only 9 saw Rhino only

Draw a Venn diagram to show this information.

Markscheme

(A1)(A1)(A1)(A1)

Note: Award (A1) for rectangle and three labelled intersecting circles (the rectangle need not be labelled), (A1) for 5, (A1) for 2 and

1, (A1) for 4, 3 and 9.

[4 marks]

L

C R

Markscheme (M1)

Notes: Award (M1) for their seen even if total is greater than .

Do not award (A1)(ft) if their total is greater than .

(A1)(ft)(G2)

[2 marks]

25 − (5 + 2 + 1 + 4 + 3 + 9)

5 + 2 + 1 + 4 + 3 + 9 25

25

= 1

2c. [6 marks]There were 25 tourists in the group and every tourist saw at least one of the three types of animal.

Calculate the probability that a tourist chosen at random from the group

(i) saw Leopard;

(ii) saw only one of the three types of animal;

(iii) saw only Leopard, given that he saw only one of the three types of animal.

Markscheme(i) (A1)(ft)(A1)(G2)

Notes: Award (A1)(ft) for numerator, (A1) for denominator.

Follow through from Venn diagram.

(ii) (A1)(A1)(G2)

Notes: Award (A1) for numerator, (A1) for denominator.

There is no follow through; all information is given.

(iii) ) (A1)(A1)(ft)(G2)

Notes: Award (A1) for numerator, (A1)(ft) for denominator.

Follow through from part (c)(ii) only.

[6 marks]

(0.48, 48%)1225

(0.64, 64%)1625

(0.25, 25%)416

2d. [2 marks]There were 25 tourists in the group and every tourist saw at least one of the three types of animal.

If a tourist chosen at random from the group saw Leopard, find the probability that he also saw Cheetah.

Markscheme (A1)(A1)(ft)(G2)

Notes: Award (A1) for numerator, (A1)(ft) for denominator.

Follow through from Venn diagram.

[2 marks]

(0.5, 50%)612

[3 marks]3a.

Consider the sequence where

The sequence continues in the same manner.

Find the value of .

Markscheme (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least

6 correct terms seen, award (A1) for at least 20 correct terms seen.

(A1)(G3)

[3 marks]

, , , … , , …u1 u2 u3 un

= 600, = 617, = 634, = 651.u1 u2 u3 u4

u20

600 + (20 − 1) × 17

= 923

[3 marks]3b. Find the sum of the first 10 terms of the sequence.

Markscheme (M1)(A1)

Note: Award (M1) for substituted arithmetic series formula, (A1) for their correct substitutions. Follow through from part (a). For

consistent use of geometric series formula in part (b) with the geometric sequence formula in part (a) award a maximum of

(M1)(A1)(A0) since their final answer cannot be an integer.

OR

(M1) (M1)

Note: Award (M1) for their correctly substituted arithmetic sequence formula, (M1) for their correctly substituted arithmetic series

formula. Follow through from part (a) and within part (b).

Note: If a list is used, award (M1) for at least 10 correct terms seen, award (A1) for these terms being added.

(accept ) (A1)(ft)(G2)

[3 marks]

[2 × 600 + (10 − 1) × 17]102

= 600 + (10 − 1)17 = 753u10

= (600 + their )S10102

u10

= 6765 6770

3c. [3 marks]Now consider the sequence where

This sequence continues in the same manner.

Find the exact value of .

, , , … , , …v1 v2 v3 vn

= 3, = 6, = 12, = 24v1 v2 v3 v4

v10

Markscheme (M1)(A1)

Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least

6 correct terms seen, award (A1) for at least 8 correct terms seen.

(A1)(G3)

Note: Exact answer only. If both exact and rounded answer seen, award the final (A1). [3 marks]

3 × 29

= 1536

3d. [3 marks]Now consider the sequence where

This sequence continues in the same manner.

Find the sum of the first 8 terms of this sequence.

Markscheme (M1)(A1)(ft)

Note: Award (M1) for substituted geometric series formula, (A1) for their correct substitutions. Follow through from part (c). If a list

is used, award (M1) for at least 8 correct terms seen, award (A1) for these 8 correct terms being added. For consistent use of

arithmetic series formula in part (d) with the arithmetic sequence formula in part (c) award a maximum of (M1)(A1)(A1).

(A1)(ft)(G2)[3 marks]

, , , … , , …v1 v2 v3 vn

= 3, = 6, = 12, = 24v1 v2 v3 v4

3×( −1)28

2−1

= 765

3e. [3 marks] is the smallest value of for which is greater than .

Calculate the value of .

k n vn un

k

Markscheme (M1)

Note: Award (M1) for their correct inequality; allow equation.

Follow through from parts (a) and (c). Accept sketches of the two functions as a valid method.

(may be implied) (A1)(ft)

Note: Award (A1) for seen. The GDC gives answers of and to the inequality; award (M1)(A1) if these are

seen with working shown.

OR

(M1) (M1)

Note: Award (M1) for and both seen, (M1) for and both seen.

(A1)(ft)(G2)

Note: Award (G1) for and seen as final answer without working. Accept use of .

[3 marks]

3 × > 600 + (k − 1)(17)2k−1

k > 8.93648…

8.93648… −34.3 8.936

= 384v8 = 719u8

= 768v9 = 736u9

v8 u8 v9 u9

k = 9

8.93648… −34.3 n

[3 marks]4a.

ABC is a triangular field on horizontal ground. The lengths of AB and AC are 70 m and 50 m respectively. The size of angle BCA is

78°.

Find the size of angle .

Markscheme (M1)(A1)

Note: Award (M1) for substituted sine rule, (A1) for correct substitution.

( ) (A1)(G3)

Note: If radians are used the answer is , award at most (M1)(A1)(A0). [3 marks]

ABC

=70sin 78

50

sin A CB

A C =B 44.3∘ 44.3209...

0.375918...

[4 marks]4b. Find the area of the triangular field.

Markscheme (A1)(M1)(A1)(ft)

Notes: Award (A1)(ft) for their seen, (M1) for substituted area formula, (A1)(ft) for correct substitution.

Follow through from part (a).

(A1)(ft)(G3)

Notes: The answer is , units are required. if 3 sf used.

If radians are used the answer is , award (A1)(ft)(M1)(A1)(ft)(A1)(ft)(G3). [4 marks]

area ΔABC = × 70 × 50 × sin(57.6790…)12

57.6790…

= 1480 m2 (1478.86…)

1480 m2 1479.20…

1554.11… m2

4c. [3 marks] is the midpoint of .

Find the length of .

Markscheme (M1)(A1)(ft)

Notes: Award (M1) for substituted cosine rule, (A1)(ft) for correct substitution. Follow through from their angle in part (b).

(A1)(ft)(G2)

Notes: If the 3 sf answer is used the answer is .

If radians are used the answer is , award (M1)(A1)(ft)(A1)(ft)(G2). [3 marks]

M AC

BM

B = + − 2 × 70 × 25 × cos(57.6790…)M2 702 252

BM = 60.4 (m) (60.4457…)

60.5 (m)

62.5757… (m)

4d. [5 marks]A vertical mobile phone mast, , is built next to the field with its base at . The angle of elevation of from is . is

the midpoint of the mast.

Calculate the angle of elevation of from .

TB B T M 63.4∘ N

N M

Markscheme (M1)

Note: Award (M1) for their correctly substituted trig equation.

(A1)(ft)

Notes: Follow through from part (c). If 3 sf answers are used throughout,

If is seen without working, award (A2).

(A1)(ft)(M1)

Notes: Award (A1)(ft) for their divided by seen, (M1) for their correctly substituted trig equation.

Follow through from part (c) and within part (d).

(A1)(ft)(G3) Notes: If 3 sf are used throughout, answer is .

If radians are used the answer is , and if full working is shown, award at most (M1)(A1)(ft)(A1)(ft)(M1)(A0). If no working is shown for radians answer, award (G2). OR

(M1)

(A1)(M1)

Note: Award (A1) for seen.

(M1)

(A1)(G3) Notes: If radians are used the answer is , and if full working is shown, award at most (M1)(A1)(M1)(M1)(A0). If no

working is shown for radians answer, award (G2). [5 marks]

tan =63.4∘ TB60.4457…

TB = 120.707…

TB = 120.815…

TB = 120.707…

tanN B =M( )120.707…

2

60.4457…

TB 2

N B =M 45.0∘ (44.9563 … )

45∘

0.308958…

tanN B =M NBBM

tan =63.4∘ 2×NBBM

2 × NB

tanN B = tanM 12

63.4∘

N B =M 45.0∘ (44.9563…)

0.308958…

[2 marks]5a.

A group of candidates sat a Chemistry examination and a Physics examination. The candidates’ marks in the Chemistry examination

are normally distributed with a mean of and a standard deviation of .

Draw a diagram that shows this information.

60 12

Markscheme

(A1)(A1)

Notes: Award (A1) for rough sketch of normal curve centred at , (A1) for some indication of as the standard deviation eg, as

diagram, or with and shown on the horizontal axis in appropriate places, or for and shown on the horizontal axis in

appropriate places.

[2 marks]

60 12

72 48 96 24

[1 mark]5b. Write down the probability that a randomly chosen candidate who sat the Chemistry examination scored at most 60 marks.

Markscheme (A1)

Note: Accept only the exact answer.

[1 mark]

0.5( , 50% )12

5c. [2 marks]Hee Jin scored 80 marks in the Chemistry examination.

Find the probability that a randomly chosen candidate who sat the Chemistry examination scored more than Hee Jin.

Markscheme (G2)

Note: Award (G1) for , award (M1)(G0) for diagram with correct area shown but incorrect answer.

[2 marks]

0.0478 (0.0477903...)

0.952209…

5d. [2 marks]The candidates’ marks in the Physics examination are normally distributed with a mean of and a standard deviation of .

Hee Jin also scored marks in the Physics examination.

Find the probability that a randomly chosen candidate who sat the Physics examination scored less than Hee Jin.

Markscheme (G2)

Note: Award (G1) for , award (M1)(G0) for diagram with correct area shown but incorrect answer.

[2 marks]

63 10

80

0.955 (0.955434...)

0.044565…

5e. [2 marks]The candidates’ marks in the Physics examination are normally distributed with a mean of and a standard deviation of .

Hee Jin also scored marks in the Physics examination.

Determine whether Hee Jin’s Physics mark, compared to the other candidates, is better than her mark in Chemistry. Give a reason for

your answer.

Markscheme (R1)

Notes: Award (R1) for correct comparison seen. Accept alternative methods, for example, (their answer to part (c)) used in

comparison or a comparison based on scores.

the Physics result is better (A1)(ft) Notes: Do not award (R0)(A1). Follow through from their answers to part (c) and part (d).

[2 marks]

63 10

80

0.0446 < 0.0478

1–

z

5f. [3 marks]To obtain a “grade A” a candidate must be in the top of the candidates who sat the Physics examination.

Find the minimum possible mark to obtain a “grade A”. Give your answer correct to the nearest integer.

Markscheme (G3)

Notes: Award (G1) for , award (G2) for .

Award (M1)(G0) for diagram with correct area shown but incorrect answer.

[3 marks]

10%

76

75.8155… 75

[3 marks]6a.

A lobster trap is made in the shape of half a cylinder. It is constructed from a steel frame with netting pulled tightly around it. The

steel frame consists of a rectangular base, two semicircular ends and two further support rods, as shown in the following diagram.

The semicircular ends each have radius and the support rods each have length .

Let be the total length of steel used in the frame of the lobster trap.

Write down an expression for in terms of , and .

r l

T

T r l π

Markscheme (A1)(A1)(A1)

Notes: Award (A1) for (“ ” must be seen), (A1) for , (A1) for . Accept equivalent forms. Accept .

Award a maximum of (A1)(A1)(A0) if extra terms are seen.

[3 marks]

2πr + 4r + 4l

2πr π 4r 4l T = 2πr + 4r + 4l

6b. [3 marks]The volume of the lobster trap is .

Write down an equation for the volume of the lobster trap in terms of , and .

Markscheme (A1)(A1)(A1)

Notes: Award (A1) for their formula equated to , (A1) for substituted into volume of cylinder formula, (A1) for volume of

cylinder formula divided by .

If “ ” not seen in part (a) accept use of or greater accuracy. Award a maximum of (A1)(A1)(A0) if extra terms are seen.

[3 marks]

0.75 m3

r l π

0.75 = π lr2

2

0.75 l

2

π 3.14

6c. [2 marks]The volume of the lobster trap is .

Show that .

Markscheme (A1)(ft)(A1)

(AG)

Notes: Award (A1)(ft) for correct rearrangement of their volume formula in part (b) seen, award (A1) for the correct substituted

formula for . The final line must be seen, with no incorrect working, for this second (A1) to be awarded.

[2 marks]

0.75 m3

T = (2π + 4)r + 6πr2

T = 2πr + 4r + 4 ( )1.5πr2

= (2π + 4)r + 6πr2

T

6d. [3 marks]The volume of the lobster trap is .

Find .

Markscheme (A1)(A1)(A1)

Note: Award (A1) for , (A1) for , (A1) for .

Accept 10.3 (10.2832…) for , accept for . Award a maximum of (A1)(A1)(A0) if extra terms are

seen.

[3 marks]

0.75 m3

dT

dr

= 2π + 4 −dT

dr

12πr3

2π + 4 −12π

r−3

2π + 4 – 3.82 – 3.81971… −12π

6e. [2 marks]The lobster trap is designed so that the length of steel used in its frame is a minimum.

Show that the value of for which is a minimum is , correct to three significant figures.r T 0.719 m

Markscheme OR (M1)

Note: Award (M1) for setting their derivative equal to zero.

OR OR OR (A1)

(AG) Note: The rounded and unrounded or formulaic answers must be seen for the final (A1) to be awarded. The use of gives an

unrounded answer of .

[2 marks]

2π + 4 − = 012πr3

= 0dT

dr

r = 0.718843… 0.371452…− −−−−−−−−−√3 12π(2π+4)

− −−−−−√3 3.8197110.2832…

− −−−−−−√3

r = 0.719 (m)

3.14

r = 0.719039…

6f. [2 marks]The lobster trap is designed so that the length of steel used in its frame is a minimum.

Calculate the value of for which is a minimum.

Markscheme (M1)

Note: Award (M1) for substituting into their volume formula. Follow through from part (b).

(A1)(ft)(G2)

[2 marks]

l T

0.75 = π× l(0.719)2

2

0.719

l = 0.924 (m) (0.923599…)

6g. [2 marks]The lobster trap is designed so that the length of steel used in its frame is a minimum.

Calculate the minimum value of .

Markscheme (M1)

Notes: Award (M1) for substituting in their expression for . Accept alternative methods, for example substitution of their

and into their part (a) (for which the answer is ). Follow through from their answer to part (a).

(A1)(ft)(G2)

T

T = (2π + 4) × 0.719 + 6

π(0.719)2

0.719 T l

0.719 11.08961024

= 11.1 (m) (11.0880…)

7a. [3 marks]

Tomek is attending a conference in Singapore. He has both trousers and shorts to wear. He also has the choice of wearing a tie or not.

The probability Tomek wears trousers is . If he wears trousers, the probability that he wears a tie is .

If Tomek wears shorts, the probability that he wears a tie is .

The following tree diagram shows the probabilities for Tomek’s clothing options at the conference.

Find the value of

(i) ;

(ii) ;

(iii) .

Markscheme(i) (A1)

(ii) (A1)

(iii) (A1)

[3 marks]

0.3 0.8

0.15

A

B

C

0.7( , , 70% )70100

710

0.2( , , , 20% )20100

210

15

0.85( , , 85% )85100

1720

7b. [8 marks]Calculate the probability that Tomek wears

(i) shorts and no tie;

(ii) no tie;

(iii) shorts given that he is not wearing a tie.

Markscheme(i) (M1) Note: Award (M1) for multiplying their values from parts (a)(i) and (a)(iii).

(A1)(ft)(G1)

Note: Follow through from part (a).

(ii) (M1)(M1) Note: Award (M1) for their two products, (M1) for adding their two products.

(A1)(ft)(G2)

Note: Follow through from part (a). (iii) (A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) for correct numerator, (A1)(ft) for correct denominator. Follow through from parts (b)(i) and (ii).

(A1)(ft)(G2)

[8 marks]

0.7 × 0.85

= 0.595 ( , 59.5% )119200

0.3 × 0.2 + 0.7 × 0.85

= 0.655 ( , 65.5% )131200

0.5950.655

= 0.908 (0.90839… , , 90,8% )119131

7c. [2 marks]The conference lasts for two days.

Calculate the probability that Tomek wears trousers on both days.

Markscheme (M1)

(A1)(G2)

[2 marks]

0.3 × 0.3

= 0.09( ,9%)9100

7d. [3 marks]The conference lasts for two days.

Calculate the probability that Tomek wears trousers on one of the days, and shorts on the other day.

Markscheme (M1)

OR (M1) Note: Award (M1) for their correct product seen, (M1) for multiplying their product by 2 or for adding their products twice.

(A1)(ft)(G2)

Note: Follow through from part (a)(i). [3 marks]

0.3 × 0.7

0.3 × 0.7 × 2 (0.3 × 0.7) + (0.7 × 0.3)

= 0.42( , ,42%)42100

2150

[2 marks]8a.

A cross-country running course consists of a beach section and a forest section. Competitors run from to , then from to and

from back to .

The running course from to is along the beach, while the course from , through and back to , is through the forest.

The course is shown on the following diagram.

Angle is .

It takes Sarah minutes and seconds to run from to at a speed of .

Using ‘distance = speed time’, show that the distance from to is metres correct to 3 significant figures.

Markscheme (A1)

Note: Award (A1) for or equivalent seen.

(A1)

(AG) Note: Both unrounded and rounded answer must be seen for the final (A1) to be awarded.

[2 marks]

A B B C

C A

A B B C A

ABC 110∘

5 20 A B 3.8 ms−1

× A B 1220

3.8 × 320

320

= 1216

= 1220 (m)

8b. [1 mark]The distance from to is metres. Running this part of the course takes Sarah minutes and seconds.

Calculate the speed, in , that Sarah runs from to .

Markscheme (A1)(G1)

[1 mark]

B C 850 5 3

ms−1 B C

(m ) (2.81, 2.80528… )850303

s−1

8c. [3 marks]The distance from to is metres. Running this part of the course takes Sarah minutes and seconds.

Calculate the distance, in metres, from to .

B C 850 5 3

C A

Markscheme (M1)(A1)

Note: Award (M1) for substitution into cosine rule formula, (A1) for correct substitutions.

(A1)(G2)

Notes: Accept .

[3 marks]

A = + − 2(1220)(850)cosC2 12202 8502 110∘

AC = 1710 (m) (1708.87 … )

1705 (1705.33…)

8d. [2 marks]The distance from to is metres. Running this part of the course takes Sarah minutes and seconds.

Calculate the total distance, in metres, of the cross-country running course.

Markscheme (M1) (A1)(ft)(G1)

Notes: Award (M1) for adding the three sides. Follow through from their answer to part (c). Accept .

[2 marks]

B C 850 5 3

1220 + 850 + 1708.87…

= 3780 (m) (3778.87 … )

3771 (3771.33…)

8e. [3 marks]The distance from to is metres. Running this part of the course takes Sarah minutes and seconds.

Find the size of angle .

Markscheme (M1)(A1)(ft)

Notes: Award (M1) for substitution into sine rule formula, (A1)(ft) for correct substitutions. Follow through from their part (c).

(A1)(ft)(G2)

Notes: Accept .

OR

(M1)(A1)(ft)

Notes: Award (M1) for substitution into cosine rule formula, (A1)(ft) for correct substitutions. Follow through from their part (c).

(A1)(ft)(G2)

Notes: Accept .

[3 marks]

B C 850 5 3

BCA

=sin C

1220sin 110∘

1708.87…

C = (42.1339 … )42.1∘

, , ,41.9∘ 42.0∘ 42.2∘ 42.3∘

cosC = 1708.87 + −…2 8502 12202

2×1708.87…×850

C = (42.1339 … )42.1∘

, ,41.2∘ 41.8∘ 42.4∘

8f. [3 marks]The distance from to is metres. Running this part of the course takes Sarah minutes and seconds.Calculate the area of the cross-country course bounded by the lines , and .

B C 850 5 3AB BC CA

Markscheme (M1)(A1)(ft)

OR (M1)(A1)(ft)

OR (M1)(A1)(ft)

Note: Award (M1) for substitution into area formula, (A1)(ft) for correct substitution.

(A1)(ft)(G2)

Notes: The answer is , units are required.

Accept .

If workings are not shown and units omitted, award (G1) for .

Follow through from parts (c) and (e).

[3 marks]

× 1220 × 850 × sin12

110∘

× 1708.87… × 850 × sin 42.133912

…∘

× 1220 × 1708.87… × sin 27.866112

…∘

= 487 000 (487230 … )m2 m2

487 000 m2

486 000 (485633 … )m2 m2

487 000 or 486 000

[1 mark]9a.

A survey was conducted to determine the length of time, , in minutes, people took to drink their coffee in a café. The information is

shown in the following grouped frequency table.

Write down the total number of people who were surveyed.

Markscheme (A1)

[1 mark]

t

60

[1 mark]9b. Write down the mid-interval value for the group.

Markscheme (A1)

[1 mark]

10 < t ⩽ 15

12.5

[2 marks]9c. Find an estimate of the mean time people took to drink their coffee.

Markscheme (M1)

Note: Award (M1) for an attempt to substitute their mid-interval values (consistent with their answer to part (b)) into the formula for

the mean.

Award (M1) where a table is constructed with their (consistent) mid-interval values listed along with the frequencies.

(A1)(ft)(G2)

Note: Follow through from their answer to part (b).

[2 marks]

3×2.5+5×7.5+…+10×27.560

= ( , 17.9, 17.9166…)107560

21512

9d. [2 marks]The information above has been rewritten as a cumulative frequency table.

Write down the value of and the value of .

Markscheme (A1)(A1)

[2 marks]

a b

a = 34, b = 60

9e. [4 marks]This information is shown in the following cumulative frequency graph.

For the people who were surveyed, use the graph to estimate

(i) the time taken for the first people to drink their coffee;

(ii) the number of people who take less than minutes to drink their coffee;

(iii) the number of people who take more than minutes to drink their coffee.

Markscheme(i) (A1) Note: Accept .

Accept any answer between and .

(Accept 21.5, but do not accept 21.)

(ii) (A1) Note: Accept . Do not accept .

Answer must be an integer.

(iii) (M1)

(A1)(G2) Notes: Award (M1) for subtraction from . Accept .

Answer must be an integer.

[4 marks]

40

8

23

⩽ 21.25 minutes

21.25

21 21.5

5

< 6 6

60 − 45

= 15

60 15 ± 1

[2 marks]10a.

Give your answers to parts (a) to (e) to the nearest dollar.On Hugh’s 18th birthday his parents gave him options of how he might receive his monthly allowance for the next two years.

Option A each month for two years

Option B in the first month, in the second month, in the third month, increasing by each month for two years

Option C in the first month and increasing by each month for two years

Option D Investing at a bank at the beginning of the first year, with an interest rate of per annum, compoundedmonthly.

Hugh does not spend any of his allowance during the two year period.

If Hugh chooses Option A, calculate the total value of his allowance at the end of the two year period.

MarkschemeThe first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

(M1) Note: Award (M1) for correct product.

(A1)(G2)

[2 marks]

$60

$10 $15 $20 $5

$15 10%

$1500 6%

60 × 24

= 1440

10b. [5 marks]If Hugh chooses Option B, calculate

(i) the amount of money he will receive in the 17th month;

(ii) the total value of his allowance at the end of the two year period.

MarkschemeThe first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.(i) (M1)(A1) Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitution.

(A1)(G2)

(ii) (M1)

OR (M1)

Note: Award (M1) for correct substitution in arithmetic series formula.

(A1)(ft)(G1)

Note: Follow through from part (b)(i).

[5 marks]

10 + (17 − 1)(5)

= 90

(2(10) + (24 − 1)(5))242

(10 + 125)242

= 1620

10c. [5 marks]If Hugh chooses Option C, calculate(i) the amount of money Hugh would receive in the 13th month;

(ii) the total value of his allowance at the end of the two year period.

MarkschemeThe first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.(i) (M1)(A1) Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions.

(A1)(G2)

Note: Award (M1)(A1)(A0) for .

Award (G1) for if workings are not shown.

(ii) (M1)

Note: Award (M1) for correct substitution in geometric series formula.

(A1)(ft)(G1)

Note: Follow through from part (c)(i).

[5 marks]

15(1.1)12

= 47

47.08

47.08

15( −1)1.124

1.1−1

= 1327

[3 marks]10d. If Hugh chooses Option D, calculate the total value of his allowance at the end of the two year period.

Markscheme

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

(M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.

OR

(A1)(M1) Note: Award (A1) for seen, (M1) for other correct entries.

OR

(A1)(M1)

Note: Award (A1) for seen, (M1) for other correct entries.

(A1)(G2)

[3 marks]

1500(1 + )6100(12)

12(2)

N = 2

I% = 6

PV = 1500

P/Y = 1

C/Y = 12

C/Y = 12

N = 24

I% = 6

PV = 1500

P/Y = 12

C/Y = 12

C/Y = 12

= 1691

10e. [1 mark]State which of the options, A, B, C or D, Hugh should choose to give him the greatest total value of his allowance at the end of

the two year period.

MarkschemeThe first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.Option D (A1)(ft) Note: Follow through from their parts (a), (b), (c) and (d). Award (A1)(ft) only if values for the four options are seen and only if their

answer is consistent with their parts (a), (b), (c) and (d).

[1 mark]

10f. [3 marks]Another bank guarantees Hugh an amount of after two years of investment if he invests $1500 at this bank. The interest

is compounded annually.

Calculate the interest rate per annum offered by the bank.

$1750

Markscheme (M1)(A1)

Note: Award (M1) for substituted compound interest formula equated to , (A1) for correct substitutions into formula.

OR

(A1)(M1)

Note: Award (A1) for seen, (M1) for other correct entries.

(A1)(G2)

[3 marks]

1750 = 1500(1 + )r

1002

1750

N = 2

PV = 1500

FV = −1750

P/Y = 1

C/Y = 1

FV = 1750

= 8.01% (8.01234 … %, 0.0801)

[1 mark]11a.

A parcel is in the shape of a rectangular prism, as shown in the diagram. It has a length cm, width cm and height of cm.

The total volume of the parcel is .

Express the volume of the parcel in terms of and .

Markscheme OR (A1)

[1 mark]

l w 20

3000 cm3

l w

20lw V = 20lw

[2 marks]11b. Show that .

Markscheme (M1)

Note: Award (M1) for equating their answer to part (a) to .

(M1)

Note: Award (M1) for rearranging equation to make subject of the formula. The above equation must be seen to award (M1). OR

(M1) Note: Award (M1) for division by on both sides. The above equation must be seen to award (M1).

(AG)

[2 marks]

l = 150w

3000 = 20lw

3000

l = 300020w

l

150 = lw

20

l = 150w

11c. [2 marks]The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Show that the length of string, cm, required to tie up the parcel can be written as

Markscheme (M1)

Note: Award (M1) for setting up a correct expression for .

(M1)

Notes: Award (M1) for correct substitution into the expression for . The above expression must be seen to award (M1).

(AG)

[2 marks]

S

S = 40 + 4w + , 0 < w ⩽ 20.300w

S = 2l + 4w + 2(20)

S

2 ( ) + 4w + 2(20)150w

S

= 40 + 4w + 300w

11d. [2 marks]The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Draw the graph of for and , clearly showing the local minimum point. Use a scale of cm to represent

units on the horizontal axis (cm), and a scale of cm to represent units on the vertical axis (cm).

S 0 < w ⩽ 20 0 < S ⩽ 500 2 5

w 2 100 S

Markscheme

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct scales, window and labels on axes, (A1) for approximately correct shape, (A1) for minimum point in

approximately correct position, (A1) for asymptotic behaviour at .

Axes must be drawn with a ruler and labeled and .

For a smooth curve (with approximately correct shape) there should be one continuous thin line, no part of which is straight and

no (one-to-many) mappings of .

The -axis must be an asymptote. The curve must not touch the -axis nor must the curve approach the asymptote then deviate

away later.

[4 marks]

w = 0

w S

w

S S

11e. [3 marks]The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find .dS

dw

Markscheme (A1)(A1)(A1)

Notes: Award (A1) for , (A1) for , (A1) for or . If extra terms present, award at most (A1)(A1)(A0).

[3 marks]

dw

4 − 300w2

4 −300 1w2

w−2

11f. [2 marks]The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find the value of for which is a minimum.

Markscheme OR OR (M1)

Note: Award (M1) for equating their derivative to zero.

(A1)(ft)(G2)

Note: Follow through from their answer to part (e).

[2 marks]

w S

4 − = 0300w2

= 4300w2

= 0dS

dw

w = 8.66 ( , 8.66025…)75−−√

11g. [1 mark]The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Write down the value, , of the parcel for which the length of string is a minimum.

Markscheme (A1)(ft)

Note: Follow through from their answer to part (f).

[1 mark]

l

17.3( , 17.3205…)15075√

11h. [2 marks]The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.

Find the minimum length of string required to tie up the parcel.

Markscheme (M1)

Note: Award (M1) for substitution of their answer to part (f) into the expression for .

(A1)(ft)(G2)

Note: Do not accept .

Follow through from their answers to parts (f) and (g).

[2 marks]

40 + 4 +75−−√ 30075√

S

= 110 (cm) (40 + 40 , 109.282…)3√

109

[1 mark]12a.

The front view of the edge of a water tank is drawn on a set of axes shown below.

The edge is modelled by .

Point has coordinates , point has coordinates and point has coordinates .

Write down the value of .

Markscheme (A1)(G1)

[1 mark]

y = a + cx2

P (−3,1.8) O (0,0) Q (3,1.8)

c

0

[2 marks]12b. Find the value of .a

Markscheme (M1)

OR (M1)

Note: Award (M1) for substitution of or and their value of into equation. may be implied.

(A1)(ft)(G1)

Note: Follow through from their answer to part (a).

Award (G1) for a correct answer only.

[2 marks]

1.8 = a(3 + 0)2

1.8 = a(−3 + 0)2

y = 1.8 x = 3 c 0

a = 0.2 ( )15

[1 mark]12c. Hence write down the equation of the quadratic function which models the edge of the water tank.

Markscheme (A1)(ft)

Note: Follow through from their answers to parts (a) and (b).

Answer must be an equation.

[1 mark]

y = 0.2x2

12d. [2 marks]The water tank is shown below. It is partially filled with water.

Calculate the value of y when .

Markscheme (M1)

(A1)(ft)(G1) Notes: Award (M1) for correctly substituted formula, (A1) for correct answer. Follow through from their answer to part (c). Award (G1) for a correct answer only. [2 marks]

x = 2.4 m

0.2 × (2.4)2

= 1.15 (m) (1.152)

12e. [2 marks]The water tank is shown below. It is partially filled with water.

State what the value of and the value of represent for this water tank.

Markscheme is the height (A1)

positive value of is half the width (or equivalent) (A1)[2 marks]

x y

y

x

12f. [2 marks]The water tank is shown below. It is partially filled with water.

Find the value of when the height of water in the tank is m.

Markscheme (M1)

Note: Award (M1) for setting their equation equal to .

(A1)(ft)(G1)

Note: Accept . Award (G1) for a correct answer only.

[2 marks]

x 0.9

0.9 = 0.2x2

0.9

x = ±2.12 (m) (± , ± , ± 2.12132…)32

2√ 4.5−−−√

2.12

Printed for Victoria Shanghai Academy

© International Baccalaureate Organization 2015 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

12g. [2 marks]The water tank is shown below. It is partially filled with water.

When the water tank is filled to a height of m, the front cross-sectional area of the water is .

(i) Calculate the volume of water in the tank.

The total volume of the tank is .

(ii) Calculate the percentage of water in the tank.

Markscheme(i) (M1) Note: Award (M1) for correct substitution in formula.

(A1)(G2)

[2 marks] (ii) (M1)

Note: Award (M1) for correct quotient multiplied by .

(A1)(ft)(G2)

Note: Award (G2) for .

Follow through from their answer to part (g)(i).

[2 marks]

0.9 2.55 m2

36 m3

2.55 × 5

= 12.8 ( )m3 (12.75 ( ))m3

× 10012.7536

100

= 35.4(%) (35.4166…)

35.6(%)(35.5555… (%))