2014 Paper 2
[176 marks]
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.
[1 mark]
1a. State whether distance from the river bank is a continuous or discrete variable.
Markscheme
continuous
(A1)
[1 mark]
1b. 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 [4 marks]
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]
[2 marks]
1c. Write down
(i)
the mean distance, x̄, of the trees from the river bank;
(ii)
the mean circumference, ȳ , of the trees.
Markscheme
(i)
26 (m)
(ii)
65 (cm)
(A1)
(A1)
[2 marks]
1d. Plot and label the point M(x̄, ȳ ) on your graph.
[2 marks]
Markscheme
point M labelled, in correct position
(A1)(A1)(ft)
Notes: Award (A1)(ft) for point plotted in correct position, (A1) for point labelled M or (x̄, ȳ ) . Follow through from their answers to
part (c).
[2 marks]
[4 marks]
1e. Write down
(i)
the Pearson’s product–moment correlation coefficient, r, for Barry’s results;
(ii)
the equation of the regression line y on x, for Barry’s results.
Markscheme
(i)
−0.988 (−0.988432 …)
(G2)
Note: Award (G2) for −0.99 . Award (G1) for −0.990 .
Award (A1)(A0) if minus sign is omitted.
(ii)
y = −0.756x + 84.7
(y = −0.756281 … x + 84.6633 …)
(G2)
Notes: Award (A1) for −0.756x , (A1) for 84.7 . If the answer is not given as an equation, award a maximum of (A1)(A0).
[4 marks]
[2 marks]
1f. Draw the regression line y on x on your graph.
Markscheme
regression line through their M
(A1)((ft)
regression line through their (0,85) (accept 85 ± 1 )
(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 M is not plotted or labelled, then follow through from part (c).
Follow through from their y-intercept in part (e)(ii).
[2 marks]
1g. Use the equation of the regression line y on x to estimate the circumference of a tree that is 40 m from the river bank.
[2 marks]
Markscheme
−0.756281(40) + 84.6633
= 54.4 (cm) (54.4120 …)
(M1)
(A1)(ft)(G2)
Notes: Accept 54.5 (54.46 ) for use of 3 sf. Accept 54.3 from use of −0.76 and 84.7 .
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]
A group of tourists went on safari to a game reserve. The game warden wanted to know how many of the tourists saw Leopard (L),
Cheetah (C ) or Rhino (R). 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
[4 marks]
2a. 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]
Markscheme
25 − (5 + 2 + 1 + 4 + 3 + 9)
(M1)
Notes: Award (M1) for their 5 + 2 + 1 + 4 + 3 + 9 seen even if total is greater than 25.
Do not award (A1)(ft) if their total is greater than 25.
=1
(A1)(ft)(G2)
[2 marks]
2c. There were 25 tourists in the group and every tourist saw at least one of the three types of animal.
[6 marks]
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
12
25
(i)
(0.48, 48%)
(A1)(ft)(A1)(G2)
Notes: Award (A1)(ft) for numerator, (A1) for denominator.
Follow through from Venn diagram.
16
25
(ii)
(0.64, 64%)
(A1)(A1)(G2)
Notes: Award (A1) for numerator, (A1) for denominator.
There is no follow through; all information is given.
(iii)
4
16
(0.25, 25%) )
(A1)(A1)(ft)(G2)
Notes: Award (A1) for numerator, (A1)(ft) for denominator.
Follow through from part (c)(ii) only.
[6 marks]
2d. 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
6
12
(0.5, 50%)
(A1)(A1)(ft)(G2)
Notes: Award (A1) for numerator, (A1)(ft) for denominator.
Follow through from Venn diagram.
[2 marks]
[2 marks]
Consider the sequence u1 , u2 , u3 , … , un , … where
u1 = 600, u2 = 617, u3 = 634, u4 = 651.
The sequence continues in the same manner.
[3 marks]
3a. Find the value of u20 .
Markscheme
600 + (20 − 1) × 17
(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.
= 923
(A1)(G3)
[3 marks]
[3 marks]
3b. Find the sum of the first 10 terms of the sequence.
Markscheme
10
[2 × 600 + (10 − 1) × 17]
2
(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
u10 = 600 + (10 − 1)17 = 753
S10 =
10
(600 + their
2
u10 )
(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.
= 6765 (accept 6770)
(A1)(ft)(G2)
[3 marks]
3c. Now consider the sequence v1 , v2 , v3 , … , vn , …
where
v1 = 3, v2 = 6, v3 = 12, v4 = 24
This sequence continues in the same manner.
Find the exact value of v10 .
[3 marks]
Markscheme
3 × 29
(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.
= 1536
(A1)(G3)
Note: Exact answer only. If both exact and rounded answer seen, award the final (A1).
[3 marks]
3d. Now consider the sequence v1 , v2 , v3 , … , vn , …
where
[3 marks]
v1 = 3, v2 = 6, v3 = 12, v4 = 24
This sequence continues in the same manner.
Find the sum of the first 8 terms of this sequence.
Markscheme
3×(28 −1)
(M1)(A1)(ft)
2−1
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).
= 765
(A1)(ft)(G2)
[3 marks]
3e. k is the smallest value of n for which vn is greater than un .
Calculate the value of k.
[3 marks]
Markscheme
3 × 2k−1 > 600 + (k − 1)(17)
(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.
k > 8.93648 …
(may be implied)
(A1)(ft)
Note: Award (A1) for 8.93648 … seen. The GDC gives answers of −34.3 and 8.936 to the inequality; award (M1)(A1) if these are
seen with working shown.
OR
v8 = 384
u8 = 719
v9 = 768
u9 = 736
(M1)
(M1)
Note: Award (M1) for v8 and u8 both seen, (M1) for v9 and u9 both seen.
k=9
(A1)(ft)(G2)
Note: Award (G1) for 8.93648 … and −34.3 seen as final answer without working. Accept use of n.
[3 marks]
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°.
[3 marks]
4a. Find the size of angle ABC .
Markscheme
70
sin 78
=
50
^C
sin AB
(M1)(A1)
Note: Award (M1) for substituted sine rule, (A1) for correct substitution.
^ C = 44.3∘ (44.3209...)
AB
(A1)(G3)
Note: If radians are used the answer is 0.375918... , award at most (M1)(A1)(A0).
[3 marks]
4b. Find the area of the triangular field.
[4 marks]
Markscheme
area ΔABC = 12 × 70 × 50 × sin(57.6790 …)
(A1)(M1)(A1)(ft)
Notes: Award (A1)(ft) for their 57.6790 … seen, (M1) for substituted area formula, (A1)(ft) for correct substitution.
Follow through from part (a).
= 1480 m2 (1478.86 …)
(A1)(ft)(G3)
Notes: The answer is 1480 m2 , units are required. 1479.20 … if 3 sf used.
If radians are used the answer is 1554.11 … m2 , award (A1)(ft)(M1)(A1)(ft)(A1)(ft)(G3).
[4 marks]
4c. M is the midpoint of AC .
[3 marks]
Find the length of BM.
Markscheme
BM2 = 702 + 252 − 2 × 70 × 25 × cos(57.6790 …)
(M1)(A1)(ft)
Notes: Award (M1) for substituted cosine rule, (A1)(ft) for correct substitution. Follow through from their angle in part (b).
BM = 60.4 (m) (60.4457 …)
(A1)(ft)(G2)
Notes: If the 3 sf answer is used the answer is 60.5 (m) .
If radians are used the answer is 62.5757 … (m) , award (M1)(A1)(ft)(A1)(ft)(G2).
[3 marks]
∘
[5 marks]
4d. A vertical mobile phone mast, TB, is built next to the field with its base at B. The angle of elevation of T from M is 63.4 . N is
the midpoint of the mast.
Calculate the angle of elevation of N from M.
Markscheme
tan 63.4∘ =
TB
60.4457…
(M1)
Note: Award (M1) for their correctly substituted trig equation.
TB = 120.707 …
(A1)(ft)
Notes: Follow through from part (c). If 3 sf answers are used throughout, TB = 120.815 …
If TB = 120.707 … is seen without working, award (A2).
( 120.707… )
2
^B=
tan NM
60.4457…
(A1)(ft)(M1)
Notes: Award (A1)(ft) for their TB divided by 2 seen, (M1) for their correctly substituted trig equation.
Follow through from part (c) and within part (d).
^ B = 45.0∘
NM
(44.9563 … )
(A1)(ft)(G3)
If 3 sf are used throughout, answer is 45∘ .
Notes:
If radians are used the answer is 0.308958 … , 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
^B=
tan NM
∘
tan 63.4 =
NB
BM
2×NB
BM
(M1)
(A1)(M1)
Note: Award (A1) for 2 × NB seen.
^ B = 1 tan 63.4∘
tan NM
(M1)
2
∘
^
NMB = 45.0 (44.9563 …) (A1)(G3)
Notes: If radians are used the answer is 0.308958 … , 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]
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 60 and a standard deviation of 12.
5a. Draw a diagram that shows this information.
[2 marks]
Markscheme
(A1)(A1)
Notes: Award (A1) for rough sketch of normal curve centred at 60, (A1) for some indication of 12 as the standard deviation eg, as
diagram, or with 72 and 48 shown on the horizontal axis in appropriate places, or for 96 and 24 shown on the horizontal axis in
appropriate places.
[2 marks]
5b. Write down the probability that a randomly chosen candidate who sat the Chemistry examination scored at most 60 marks.
[1 mark]
Markscheme
0.5 ( 12 , 50% )
(A1)
Note: Accept only the exact answer.
[1 mark]
5c. Hee Jin scored 80 marks in the Chemistry examination.
[2 marks]
Find the probability that a randomly chosen candidate who sat the Chemistry examination scored more than Hee Jin.
Markscheme
0.0478 (0.0477903...)
(G2)
Note: Award (G1) for 0.952209 … , award (M1)(G0) for diagram with correct area shown but incorrect answer.
[2 marks]
5d. The candidates’ marks in the Physics examination are normally distributed with a mean of 63 and a standard deviation of 10.
Hee Jin also scored 80 marks in the Physics examination.
Find the probability that a randomly chosen candidate who sat the Physics examination scored less than Hee Jin.
Markscheme
0.955 (0.955434...)
(G2)
Note: Award (G1) for 0.044565 … , award (M1)(G0) for diagram with correct area shown but incorrect answer.
[2 marks]
[2 marks]
5e. The candidates’ marks in the Physics examination are normally distributed with a mean of 63 and a standard deviation of 10.
[2 marks]
Hee Jin also scored 80 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
0.0446 < 0.0478
(R1)
Notes: Award (R1) for correct comparison seen. Accept alternative methods, for example, 1– (their answer to part (c)) used in
comparison or a comparison based on z 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]
5f. To obtain a “grade A” a candidate must be in the top 10% of the candidates who sat the Physics examination.
[3 marks]
Find the minimum possible mark to obtain a “grade A”. Give your answer correct to the nearest integer.
Markscheme
76
(G3)
Notes: Award (G1) for 75.8155 … , award (G2) for 75.
Award (M1)(G0) for diagram with correct area shown but incorrect answer.
[3 marks]
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 r and the support rods each have length l.
Let T be the total length of steel used in the frame of the lobster trap.
6a. Write down an expression for T in terms of r, l and π.
[3 marks]
Markscheme
2πr + 4r + 4l
(A1)(A1)(A1)
Notes: Award (A1) for 2πr (“π” must be seen), (A1) for 4r, (A1) for 4l. Accept equivalent forms. Accept T = 2πr + 4r + 4l .
Award a maximum of (A1)(A1)(A0) if extra terms are seen.
[3 marks]
3
6b. The volume of the lobster trap is 0.75 m .
[3 marks]
Write down an equation for the volume of the lobster trap in terms of r, l and π.
Markscheme
2
0.75 = πr2 l
Notes:
(A1)(A1)(A1)
Award (A1) for their formula equated to 0.75 , (A1) for l substituted into volume of cylinder formula, (A1) for volume of
cylinder formula divided by 2.
If “π” not seen in part (a) accept use of 3.14 or greater accuracy. Award a maximum of (A1)(A1)(A0) if extra terms are seen.
[3 marks]
3
6c. The volume of the lobster trap is 0.75 m .
Show that T = (2π + 4)r +
6
πr2
[2 marks]
.
Markscheme
T = 2πr + 4r + 4 ( 1.52 )
= (2π + 4)r +
6
πr2
πr
(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 T . The final line must be seen, with no incorrect working, for this second (A1) to be awarded.
[2 marks]
6d. The volume of the lobster trap is 0.75 m3 .
Find
dT
dr
[3 marks]
.
Markscheme
dT
dr
= 2π + 4 − πr123
(A1)(A1)(A1)
Note: Award (A1) for 2π + 4 , (A1) for
−12
π
, (A1) for r−3 .
Accept 10.3 (10.2832…) for 2π + 4 , accept – 3.82 – 3.81971 … for
−12
π
. Award a maximum of (A1)(A1)(A0) if extra terms are
seen.
[3 marks]
6e. The lobster trap is designed so that the length of steel used in its frame is a minimum.
Show that the value of r for which T is a minimum is 0.719 m , correct to three significant figures.
[2 marks]
Markscheme
2π + 4 − πr123 = 0
OR
dT
dr
=0
(M1)
Note: Award (M1) for setting their derivative equal to zero.
r = 0.718843 …
r = 0.719 (m)
−−
−−−−−
−−−12−−−
−−−−−−−−−
3−
3.81971
3
3
OR √
0.371452 … OR √
OR √
10.2832…
π(2π+4)
(A1)
(AG)
Note: The rounded and unrounded or formulaic answers must be seen for the final (A1) to be awarded. The use of 3.14 gives an
unrounded answer of r = 0.719039 … .
[2 marks]
6f. The lobster trap is designed so that the length of steel used in its frame is a minimum.
[2 marks]
Calculate the value of l for which T is a minimum.
Markscheme
0.75 =
π× (0.719) 2 l
2
(M1)
Note: Award (M1) for substituting 0.719 into their volume formula. Follow through from part (b).
l = 0.924 (m) (0.923599 …)
(A1)(ft)(G2)
[2 marks]
6g. The lobster trap is designed so that the length of steel used in its frame is a minimum.
[2 marks]
Calculate the minimum value of T .
Markscheme
T = (2π + 4) × 0.719 +
6
π(0.719) 2
(M1)
Notes: Award (M1) for substituting 0.719 in their expression for T . Accept alternative methods, for example substitution of their l
and 0.719 into their part (a) (for which the answer is 11.08961024 ). Follow through from their answer to part (a).
= 11.1 (m) (11.0880 …)
(A1)(ft)(G2)
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 0.3 . If he wears trousers, the probability that he wears a tie is 0.8 .
If Tomek wears shorts, the probability that he wears a tie is 0.15 .
The following tree diagram shows the probabilities for Tomek’s clothing options at the conference.
[3 marks]
7a. Find the value of
(i)
A;
B;
(ii)
C.
(iii)
Markscheme
70
0.7 ( 100
,
(i)
(ii)
(iii)
7
, 70% )
10
20
2
0.2 ( 100 , 10 , 15 , 20% )
85
0.85 ( 100
, 17
, 85% )
20
(A1)
(A1)
(A1)
[3 marks]
7b. Calculate the probability that Tomek wears
(i)
shorts and no tie;
(ii)
no tie;
(iii)
shorts given that he is not wearing a tie.
[8 marks]
Markscheme
(i)
0.7 × 0.85
(M1)
Note: Award (M1) for multiplying their values from parts (a)(i) and (a)(iii).
= 0.595 ( 119
, 59.5% )
200
(A1)(ft)(G1)
Note: Follow through from part (a).
(ii)
0.3 × 0.2 + 0.7 × 0.85
(M1)(M1)
Note: Award (M1) for their two products, (M1) for adding their two products.
= 0.655 ( 131
, 65.5% )
200
(A1)(ft)(G2)
Note: Follow through from part (a).
(iii)
0.595
0.655
(A1)(ft)(A1)(ft)
Notes: Award (A1)(ft) for correct numerator, (A1)(ft) for correct denominator. Follow through from parts (b)(i) and (ii).
= 0.908 (0.90839 … ,
[8 marks]
119
,
131
90,8% )
(A1)(ft)(G2)
[2 marks]
7c. The conference lasts for two days.
Calculate the probability that Tomek wears trousers on both days.
Markscheme
0.3 × 0.3
(M1)
9
= 0.09 ( 100
,9%)
(A1)(G2)
[2 marks]
[3 marks]
7d. 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
0.3 × 0.7
(M1)
0.3 × 0.7 × 2
OR (0.3 × 0.7) + (0.7 × 0.3)
(M1)
Note: Award (M1) for their correct product seen, (M1) for multiplying their product by 2 or for adding their products twice.
42 21
= 0.42 ( 100
, 50 ,42%)
(A1)(ft)(G2)
Note: Follow through from part (a)(i).
[3 marks]
A cross-country running course consists of a beach section and a forest section. Competitors run from A to B, then from B to C and
from C back to A.
The running course from A to B is along the beach, while the course from B, through C and back to A, is through the forest.
The course is shown on the following diagram.
Angle ABC is 110∘ .
It takes Sarah 5 minutes and 20 seconds to run from A to B at a speed of 3.8 ms−1 .
8a. Using ‘distance = speed × time’, show that the distance from A to B is 1220 metres correct to 3 significant figures.
[2 marks]
Markscheme
3.8 × 320
(A1)
Note: Award (A1) for 320 or equivalent seen.
= 1216
(A1)
= 1220 (m)
(AG)
Note: Both unrounded and rounded answer must be seen for the final (A1) to be awarded.
[2 marks]
8b. The distance from B to C is 850 metres. Running this part of the course takes Sarah 5 minutes and 3 seconds.
[1 mark]
Calculate the speed, in ms−1 , that Sarah runs from B to C .
Markscheme
850
303
(m s−1 ) (2.81, 2.80528… )
(A1)(G1)
[1 mark]
8c. The distance from B to C is 850 metres. Running this part of the course takes Sarah 5 minutes and 3 seconds.
Calculate the distance, in metres, from C to A .
[3 marks]
Markscheme
AC2 = 12202 + 8502 − 2(1220)(850) cos 110∘
(M1)(A1)
Note: Award (M1) for substitution into cosine rule formula, (A1) for correct substitutions.
AC = 1710 (m) (1708.87 … )
(A1)(G2)
Notes: Accept 1705 (1705.33 …) .
[3 marks]
8d. The distance from B to C is 850 metres. Running this part of the course takes Sarah 5 minutes and 3 seconds.
[2 marks]
Calculate the total distance, in metres, of the cross-country running course.
Markscheme
1220 + 850 + 1708.87 …
= 3780 (m) (3778.87 … )
(M1)
(A1)(ft)(G1)
Notes: Award (M1) for adding the three sides. Follow through from their answer to part (c). Accept 3771 (3771.33 …) .
[2 marks]
8e. The distance from B to C is 850 metres. Running this part of the course takes Sarah 5 minutes and 3 seconds.
[3 marks]
Find the size of angle BCA.
Markscheme
sin C
1220
=
sin 110 ∘
1708.87…
(M1)(A1)(ft)
Notes: Award (M1) for substitution into sine rule formula, (A1)(ft) for correct substitutions. Follow through from their part (c).
C = 42.1∘ (42.1339 … )
(A1)(ft)(G2)
Notes: Accept 41.9∘ , 42.0∘ , 42.2∘ , 42.3∘ .
OR
cos C =
1708.87…2 + 850 2 − 12202
2×1708.87…×850
(M1)(A1)(ft)
Notes: Award (M1) for substitution into cosine rule formula, (A1)(ft) for correct substitutions. Follow through from their part (c).
C = 42.1∘ (42.1339 … )
(A1)(ft)(G2)
Notes: Accept 41.2∘ , 41.8∘ , 42.4∘ .
[3 marks]
8f. The distance from B to C is 850 metres. Running this part of the course takes Sarah 5 minutes and 3 seconds.
Calculate the area of the cross-country course bounded by the lines AB , BC and CA .
[3 marks]
Markscheme
1
2
× 1220 × 850 × sin 110∘
1
2
× 1708.87 … × 850 × sin 42.1339 …∘
1
2
× 1220 × 1708.87 … × sin 27.8661 …∘
(M1)(A1)(ft)
OR
(M1)(A1)(ft)
OR
(M1)(A1)(ft)
Note: Award (M1) for substitution into area formula, (A1)(ft) for correct substitution.
= 487 000 m2 (487 230 … m2 )
(A1)(ft)(G2)
Notes: The answer is 487 000 m2 , units are required.
Accept 486 000 m2 (485 633 … m2 ) .
If workings are not shown and units omitted, award (G1) for 487 000 or 486 000 .
Follow through from parts (c) and (e).
[3 marks]
A survey was conducted to determine the length of time, t, in minutes, people took to drink their coffee in a café. The information is
shown in the following grouped frequency table.
9a. Write down the total number of people who were surveyed.
[1 mark]
Markscheme
60
(A1)
[1 mark]
9b. Write down the mid-interval value for the 10 < t ⩽ 15 group.
[1 mark]
Markscheme
12.5
(A1)
[1 mark]
9c. Find an estimate of the mean time people took to drink their coffee.
[2 marks]
Markscheme
3×2.5+5×7.5+…+10×27.5
60
(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.
=
1075
60
( 215
, 17.9, 17.9166 …)
12
(A1)(ft)(G2)
Note: Follow through from their answer to part (b).
[2 marks]
9d. The information above has been rewritten as a cumulative frequency table.
Write down the value of a and the value of b.
Markscheme
a = 34, b = 60
[2 marks]
(A1)(A1)
[2 marks]
9e. 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 40 people to drink their coffee;
the number of people who take less than 8 minutes to drink their coffee;
(ii)
the number of people who take more than 23 minutes to drink their coffee.
(iii)
Markscheme
⩽ 21.25 minutes
(i)
(A1)
Note: Accept 21.25 .
Accept any answer between 21 and 21.5 .
(Accept 21.5, but do not accept 21.)
5
(ii)
(A1)
Note: Accept < 6 . Do not accept 6.
Answer must be an integer.
(iii)
= 15
60 − 45
(M1)
(A1)(G2)
Notes: Award (M1) for subtraction from 60. Accept 15 ± 1 .
Answer must be an integer.
[4 marks]
[4 marks]
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.
$60 each month for two years
Option A
$10 in the first month, $15 in the second month, $20 in the third month, increasing by $5 each month for two years
Option B
$15 in the first month and increasing by 10% each month for two years
Option C
Investing $1500 at a bank at the beginning of the first year, with an interest rate of 6% per annum, compounded
Option D
monthly.
Hugh does not spend any of his allowance during the two year period.
10a. If Hugh chooses Option A, calculate the total value of his allowance at the end of the two year period.
[2 marks]
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.
60 × 24
(M1)
Note: Award (M1) for correct product.
= 1440
(A1)(G2)
[2 marks]
10b. If Hugh chooses Option B, calculate
(i)
(ii)
[5 marks]
the amount of money he will receive in the 17th month;
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.
10 + (17 − 1)(5)
(i)
(M1)(A1)
Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitution.
= 90
(ii)
OR
(A1)(G2)
24
(2(10) + (24 − 1)(5))
2
24
(10 + 125)
2
(M1)
(M1)
Note: Award (M1) for correct substitution in arithmetic series formula.
= 1620
(A1)(ft)(G1)
Note: Follow through from part (b)(i).
[5 marks]
10c. If Hugh chooses Option C, calculate
(i)
(ii)
the amount of money Hugh would receive in the 13th month;
the total value of his allowance at the end of the two year period.
[5 marks]
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.
15(1.1)12
(i)
(M1)(A1)
Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions.
= 47
(A1)(G2)
Note: Award (M1)(A1)(A0) for 47.08 .
Award (G1) for 47.08 if workings are not shown.
(ii)
15(1.124 −1)
1.1−1
(M1)
Note: Award (M1) for correct substitution in geometric series formula.
= 1327
(A1)(ft)(G1)
Note: Follow through from part (c)(i).
[5 marks]
10d. If Hugh chooses Option D, calculate the total value of his allowance at the end of the two year period.
[3 marks]
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.
1500(1 +
6
100(12)
)
12(2)
(M1)(A1)
Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.
OR
N =2
I% = 6
PV = 1500
P/Y = 1
C/Y = 12
(A1)(M1)
Note: Award (A1) for C/Y = 12 seen, (M1) for other correct entries.
OR
N = 24
I% = 6
PV = 1500
P/Y = 12
C/Y = 12
(A1)(M1)
Note: Award (A1) for C/Y = 12 seen, (M1) for other correct entries.
= 1691
(A1)(G2)
[3 marks]
10e. 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 [1 mark]
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.
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]
[3 marks]
10f. Another bank guarantees Hugh an amount of $1750 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.
Markscheme
r
1750 = 1500(1 + 100
)
2
(M1)(A1)
Note: Award (M1) for substituted compound interest formula equated to 1750, (A1) for correct substitutions into formula.
OR
N =2
PV = 1500
FV = −1750
P/Y = 1
C/Y = 1
(A1)(M1)
Note: Award (A1) for FV = 1750 seen, (M1) for other correct entries.
= 8.01% (8.01234 … %, 0.0801)
(A1)(G2)
[3 marks]
A parcel is in the shape of a rectangular prism, as shown in the diagram. It has a length l cm, width w cm and height of 20 cm.
The total volume of the parcel is 3000 cm3 .
11a. Express the volume of the parcel in terms of l and w .
[1 mark]
Markscheme
20lw OR V = 20lw
(A1)
[1 mark]
11b. Show that l =
150
w
.
[2 marks]
Markscheme
3000 = 20lw
(M1)
Note: Award (M1) for equating their answer to part (a) to 3000.
l=
3000
20w
(M1)
Note: Award (M1) for rearranging equation to make l subject of the formula. The above equation must be seen to award (M1).
OR
150 = lw
(M1)
Note: Award (M1) for division by 20 on both sides. The above equation must be seen to award (M1).
l=
150
w
[2 marks]
(AG)
11c. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
[2 marks]
Show that the length of string, S cm, required to tie up the parcel can be written as
S = 40 + 4w +
300
, 0 < w ⩽ 20.
w
Markscheme
S = 2l + 4w + 2(20)
(M1)
Note: Award (M1) for setting up a correct expression for S.
2 ( 150
) + 4w + 2(20)
w
(M1)
Notes: Award (M1) for correct substitution into the expression for S. The above expression must be seen to award (M1).
= 40 + 4w + 300
w
(AG)
[2 marks]
11d. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
[2 marks]
Draw the graph of S for 0 < w ⩽ 20 and 0 < S ⩽ 500 , clearly showing the local minimum point. Use a scale of 2 cm to represent 5
units on the horizontal axis w (cm), and a scale of 2 cm to represent 100 units on the vertical axis S (cm).
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 w = 0 .
Axes must be drawn with a ruler and labeled w and S.
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 w .
The S-axis must be an asymptote. The curve must not touch the S-axis nor must the curve approach the asymptote then deviate
away later.
[4 marks]
11e. 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
.
[3 marks]
dw
Markscheme
4 − 3002
w
(A1)(A1)(A1)
Notes: Award (A1) for 4, (A1) for −300 , (A1) for
1
w2
or w−2 . If extra terms present, award at most (A1)(A1)(A0).
[3 marks]
11f. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
[2 marks]
Find the value of w for which S is a minimum.
Markscheme
4 − 3002 = 0
w
OR
300
w2
=4
OR
dS
dw
=0
(M1)
Note: Award (M1) for equating their derivative to zero.
−−
w = 8.66 (√75 , 8.66025 …)
(A1)(ft)(G2)
Note: Follow through from their answer to part (e).
[2 marks]
11g. 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, l, of the parcel for which the length of string is a minimum.
Markscheme
17.3 ( 150 , 17.3205 …)
√75
(A1)(ft)
Note: Follow through from their answer to part (f).
[1 mark]
[1 mark]
11h. The parcel is tied up using a length of string that fits exactly around the parcel, as shown in the following diagram.
[2 marks]
Find the minimum length of string required to tie up the parcel.
Markscheme
−−
40 + 4√75 +
300
√75
(M1)
Note: Award (M1) for substitution of their answer to part (f) into the expression for S.
= 110 (cm) (40 + 40√3, 109.282 …)
(A1)(ft)(G2)
Note: Do not accept 109.
Follow through from their answers to parts (f) and (g).
[2 marks]
The front view of the edge of a water tank is drawn on a set of axes shown below.
The edge is modelled by y = ax2 + c .
Point P has coordinates (−3,1.8) , point O has coordinates (0,0) and point Q has coordinates (3,1.8) .
12a. Write down the value of c.
[1 mark]
Markscheme
0
(A1)(G1)
[1 mark]
12b. Find the value of a .
[2 marks]
Markscheme
1.8 = a(3)2 + 0
(M1)
OR
1.8 = a(−3)2 + 0
(M1)
Note: Award (M1) for substitution of y = 1.8 or x = 3 and their value of c into equation. 0 may be implied.
a = 0.2 ( 15 )
(A1)(ft)(G1)
Note: Follow through from their answer to part (a).
Award (G1) for a correct answer only.
[2 marks]
12c. Hence write down the equation of the quadratic function which models the edge of the water tank.
[1 mark]
Markscheme
y = 0.2x2
(A1)(ft)
Note: Follow through from their answers to parts (a) and (b).
Answer must be an equation.
[1 mark]
12d. The water tank is shown below. It is partially filled with water.
Calculate the value of y when x = 2.4 m .
Markscheme
0.2 × (2.4)2
(M1)
= 1.15 (m) (1.152)
(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]
[2 marks]
12e. The water tank is shown below. It is partially filled with water.
[2 marks]
State what the value of x and the value of y represent for this water tank.
Markscheme
y is the height
(A1)
positive value of x is half the width (or equivalent)
(A1)
[2 marks]
12f. The water tank is shown below. It is partially filled with water.
Find the value of x when the height of water in the tank is 0.9 m.
Markscheme
0.9 = 0.2x2
(M1)
Note: Award (M1) for setting their equation equal to 0.9 .
−−
−
x = ±2.12 (m) (± 32 √2, ± √4.5 , ± 2.12132 …)
(A1)(ft)(G1)
Note: Accept 2.12 . Award (G1) for a correct answer only.
[2 marks]
[2 marks]
[2 marks]
12g. The water tank is shown below. It is partially filled with water.
When the water tank is filled to a height of 0.9 m, the front cross-sectional area of the water is 2.55 m2 .
(i)
Calculate the volume of water in the tank.
The total volume of the tank is 36 m3 .
(ii)
Calculate the percentage of water in the tank.
Markscheme
(i)
2.55 × 5
(M1)
Note: Award (M1) for correct substitution in formula.
= 12.8 (m3 ) (12.75 (m3 ))
(A1)(G2)
[2 marks]
(ii)
12.75
36
× 100
(M1)
Note: Award (M1) for correct quotient multiplied by 100.
= 35.4(%) (35.4166 …)
(A1)(ft)(G2)
Note: Award (G2) for 35.6(%)(35.5555 … (%)) .
Follow through from their answer to part (g)(i).
[2 marks]
© International Baccalaureate Organization 2015
International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®
Printed for Victoria Shanghai Academy