Download 2015-16 Semester I IB Review Questions

Math Studies Year 1 –Semester I Final Exam IB Review Questions
1.
S3
, where S is its surface area.
36 π
The surface area of a sphere is 500 cm2.
(a) Calculate the volume of the sphere. Give your answer correct to two decimal places.
The volume of a sphere is V =
(3)
(b)
Write down your answer to (a) correct to the nearest integer.
(1)
(c)
Write down your answer to (b) in the form a × 10n, where 1 ≤ a < 10 and n 
.
(2)
(Total 6 marks)
2.
Consider the following four numbers.
p = 0.00314 ; q = 0.00314 × 102 ; r =
(a)
π
; s = 3.14 × 10–2
1000
One of these numbers is written in the form a × 10k where 1 ≤ a < 10 and k 
number.
. Write down this
(1)
(b)
Write down the smallest of these numbers.
(2)
(c)
Write down the value of q + s.
(1)
(d)
Give your answer to part (c) in the form a × 10k where 1 ≤ a < 10 and k 
.
(2)
(Total 6 marks)
3.
In a television show there is a transparent box completely filled with identical cubes. Participants have to
estimate the number of cubes in the box. The box is 50 cm wide, 100 cm long and 40 cm tall.
(a) Find the volume of the box.
(2)
Joaquin estimates the volume of one cube to be 500 cm3. He uses this value to estimate the number of cubes
in the box.
(b) Find Joaquin’s estimated number of cubes in the box.
(2)
The actual number of cubes in the box is 350.
(c) Find the percentage error in Joaquin’s estimated number of cubes in the box.
(2)
(Total 6 marks)
4.
(3  2.1) 3
.
7  1 .2
(a)
Calculate exactly
(b)
Write the answer to part (a) correct to 2 significant figures.
(c)
Calculate the percentage error when the answer to part (a) is written correct to 2 significant figures.
(d)
Write your answer to part (c) in the form a  10 where 1  a < 10 and k 
(1)
(1)
(2)
k
.
(2)
(Total 6 marks)
5.
Let x = 7.94.
(a)
(b)
(c)
2 x 1
.
x3
(i)
Give your answer correct to three decimal places.
(ii) Write your answer to (b)(i) as a percentage.
Give your answer to part (b)(i) in the form a × 10k, where 1 a 10, k 
Calculate the value of
.
(Total 6 marks)
6.
(a)
Given x = 2.6 × 104 and y = 5.0 × 10–8, calculate the value of w = x × y. Give your answer in the form
a × 10k where 1 ≤ a < 10 and k  .
(b)
Which two of the following statements about the nature of x, y and w above are incorrect?
(i)
x
(ii) y 
(iii) y 
(iv) w < y
(v) x + y 
(vi)
1
<x
w
(Total 8 marks)
7.
Calculate 3.7 × 16.22 – 500, writing your answer
(a) correct to two decimal places;
(b) (i)
correct to three significant figures;
(ii) in the form a × 10k, where 1 ≤ a < 10, k 
.
(Total 4 marks)
8.
The first term of an arithmetic sequence is 0 and the common difference is 12.
(a) Find the value of the 96th term of the sequence.
(2)
th
th
The first term of a geometric sequence is 6. The 6 term of the geometric sequence is equal to the 17 term
of the arithmetic sequence given above.
(b) Write down an equation using this information.
(2)
(c)
Calculate the common ratio of the geometric sequence.
(2)
(Total 6 marks)
9.
(a)
(b)
The first term of an arithmetic sequence is –16 and the eleventh term is 39.
Calculate the value of the common difference.
16
The third term of a geometric sequence is 12 and the fifth term is
.
3
All the terms in the sequence are positive.
Calculate the value of the common ratio.
(Total 8 marks)
10.
The sixth term of an arithmetic sequence is 24. The common difference is 8.
(a) Calculate the first term of the sequence.
The sum of the first n terms is 600.
(b) Calculate the value of n.
(Total 8 marks)
11.
The fourth term of an arithmetic sequence is 12 and the tenth term is 42.
(a) Given that the first term is u1 and the common difference is d, write down two equations in u1 and d
that satisfy this information.
(b)
Solve the equations to find the values of u1 and d.
(Total 8 marks)
12.
The first four terms of an arithmetic sequence are shown below.
1, 5, 9, 13,......
(a) Write down the nth term of the sequence.
(b) Calculate the 100th term of the sequence.
(c) Find the sum of the first 100 terms of the sequence.
(Total 4 marks)
13.
Mr Jones decides to increase the amount of money he spends on food by d GBP every year. In the first year
he spends a GBP. In the 8th year he spends twice as much as in the 4th year. In the 20th year he spends 4000
GBP.
Find the value of d.
(Total 4 marks)
14.
A woman deposits $100 into her son’s savings account on his first birthday. On his second birthday she
deposits $125, $150 on his third birthday, and so on.
(a) How much money would she deposit into her son’s account on his 17th birthday?
(b) How much in total would she have deposited after her son’s 17th birthday?
(Total 4 marks)
15.
The first three terms of an arithmetic sequence are
2k + 3, 5k − 2 and 10k −15.
(a) Show that k = 4.
(3)
(b)
Find the values of the first three terms of the sequence.
(c)
Write down the value of the common difference.
(1)
(1)
th
(d)
Calculate the 20 term of the sequence.
(e)
Find the sum of the first 15 terms of the sequence.
(2)
(2)
(Total 9 marks)
16.
A basketball is dropped vertically. It reaches a height of 2 m on the first bounce. The height of each
subsequent bounce is 90% of the previous bounce.
(a) What height does it reach on the 8th bounce?
(2)
(b)
What is the total vertical distance travelled by the ball between the first and sixth time the ball hits the
ground?
(4)
(Total 6 marks)
17.
A National Lottery is offering prizes in a new competition. The winner may choose one of the following.
Option one:
$1000 each week for 10 weeks.
Option two:
$250 in the first week, $450 in the second week, $650 in the third week,
increasing by $200 each week for a total of 10 weeks.
Option three: $10 in the first week, $20 in the second week, $40 in the third week
continuing to double for a total of 10 weeks.
(a)
Calculate the amount you receive in the tenth week, if you select
(i)
option two;
(ii) option three.
(6)
(b)
What is the total amount you receive if you select option two?
(c)
Which option has the greatest total value? Justify your answer by showing all appropriate calculations.
(2)
(4)
(Total 12 marks)
18.
Give all answers in this question to the nearest whole currency unit.
In January 2008 Larry had 90 000 USD to invest for his retirement in January 2011.
He invested 40 000 USD in US government bonds which paid 4 % per annum simple interest.
(a) Calculate the value of Larry’s investment in government bonds in January 2011.
(3)
Larry changed this investment into South African rand (ZAR) at an exchange rate of
1 USD = 18.624 ZAR.
(b) Calculate the amount that Larry received in ZAR from the exchange.
(2)
He changed the remaining 50 000 USD to South African rand (ZAR) in January 2008.
The exchange rate between USD and ZAR was 1 USD = 10.608 ZAR. There was 2.5 % commission charged
on the exchange.
(c) Calculate the value, in USD, of the commission Larry paid.
(2)
(d)
Show that the amount that Larry had to invest is 517 000 ZAR, correct to the nearest thousand ZAR.
(3)
In January 2008, Larry deposited this money into a bank account that paid interest at a nominal annual rate of
12 %, compounded monthly.
(e) Find the value of the money in Larry’s bank account in January 2011.
(3)
(Total 13 marks)
19.
Annie is starting her first job. She will earn a salary of $26000 in the first year and her salary will increase by
3 every year.
(a) Calculate how much Annie will earn in her 5th year of work.
(3)
Annie spends $24800 of her earnings in her first year of work. For the next few years, inflation will cause
Annie’s living expenses to rise by 5 per year.
(b) (i)
Calculate the number of years it will be before Annie is spending more than she earns.
(ii) By how much will Annie’s spending be greater than her earnings
in that year?
(6)
(Total 9 marks)
20.
The table below shows the number and weight (w) of fish delivered to a local fish market one morning.
weight (kg)
frequency
cumulative frequency
0.50 ≤ w < 0.70
16
16
0.70 ≤ w < 0.90
37
53
0.90 ≤ w < 1.10
44
c
1.10 ≤ w < 1.30
23
120
1.30 ≤ w < 1.50
10
130
(a)
(i)
Write down the value of c.
(1)
(ii)
On graph paper, draw the cumulative frequency curve for this data. Use a scale of 1 cm to
represent 0.1 kg on the horizontal axis and 1 cm to represent 10 units on the vertical axis. Label
the axes clearly.
(4)
(iii)
Use the graph to show that the median weight of the fish is 0.95 kg.
(1)
(b)
(i)
The zoo buys all fish whose weights are above the 90th percentile.
How many fish does the zoo buy?
(2)
(ii)
A pet food company buys all the fish in the lowest quartile. What is the maximum weight of a
fish bought by the company?
(3)
(c)
A restaurant buys all fish whose weights are within 10% of the median weight.
(i)
Calculate the minimum and maximum weights for the fish bought by the restaurant.
(2)
(ii)
Use your graph to determine how many fish will be bought by the restaurant.
(3)
(Total 16 marks)
21.
The following table shows the times, to the nearest minute, taken by 100 students to complete a mathematics
task.
Time (t) minutes
11–15 16–20 21–25 26–30 31–35 36–40
Number of students
7
13
25
28
20
7
(a) Construct a cumulative frequency table. (Use upper class boundaries 15.5, 20.5 and so on.)
(2)
(b)
On graph paper, draw a cumulative frequency graph, using a scale of 2 cm to represent 5 minutes on
the horizontal axis and 1 cm to represent 10 students on the vertical axis.
(3)
(c)
Use your graph to estimate
(i)
the number of students that completed the task in less than 17.5 minutes;
(ii)
the time it will take for
3
of the students to complete the task.
4
(2)
(Total 7 marks)
22.
A random sample of 167 people who own mobile phones was used to collect data on the amount of time they
spent per day using their phones. The results are displayed in the table below.
Time spent per
0  t 15 15  t  30 30  t  45 45  t  60 60  t  75 75  t  90
day (t minutes)
Number of people
21
32
35
41
27
11
(a) State the modal group.
(1)
(b)
Use your graphic display calculator to calculate approximate values of the mean and standard
deviation of the time spent per day on these mobile phones.
(c)
On graph paper, draw a fully labelled histogram to represent the data.
(3)
(4)
(Total 8 marks)
23.
The cumulative frequency graph has been drawn from a frequency table showing the time it takes a number
of students to complete a computer game.
f
200
180
Number of students
160
140
120
100
80
60
40
20
0
(a)
5
10
15
20
25
30
35
Time in minutes
40
45
50
55
60
From the graph find
(i)
the median time;
(ii) the interquartile range.
(5)
The graph has been drawn from the data given in the table below.
Time in minutes
Number of students
0<x≤5
20
5 < x ≤ 15
20
15 < x ≤ 20
p
20 < x ≤ 25
40
25 < x ≤ 35
60
35 < x ≤ 50
q
50 < x ≤ 60
10
(b) Using the graph, find the values of p and q.
(2)
(c)
Calculate an estimate of the mean time taken to finish the computer game.
(4)
(Total 11 marks)
24.
The stem and leaf diagram below shows the lengths of 22 metal components in cm.
Stem
Leaf
1 2, 2, 3, 7
2 4, 4, 4, 8, 9, 9
3 6, 7, 7
4 1, 1, 1, 1, 3, 5, 6
5 0, 1
Key: 1 | 2 means 1.2 cm
(a)
Write down the modal length of the metal components.
(1)
(b)
Find the median length of the metal components.
(2)
(c)
Calculate the interquartile range of the lengths of the metal components.
(3)
(Total 6 marks)
25.
Eight houses in a street are inhabited by different numbers of people, as shown in the table below.
House
Number of inhabitants
A
5
B
4
C
7
D
6
E
4
F
3
(a)
The following statements refer to the number of inhabitants per house.
Write down true (T) or false (F) for each.
(i)
The mean is 5.
(ii) The range is 4.
(iii) The mode is 6.
(iv) The standard deviation is 1.4 correct to 2 significant figures.
(b)
Calculate the interquartile range for the number of inhabitants per house.
G
6
H
4
(4)
(2)
(Total 6 marks)
26.
A survey was conducted of the number of bedrooms in 208 randomly chosen houses. The results are shown
in the following table.
Number of bedrooms
1
2
3
4
5
6
Number of houses
41
60
52
32
15
8
(a) State whether the data is discrete or continuous.
(1)
(b)
Write down the mean number of bedrooms per house.
(c)
Write down the standard deviation of the number of bedrooms per house.
(d)
Find how many houses have a number of bedrooms greater than one standard deviation above the
mean.
(2)
(1)
(2)
(Total 6 marks)
27.
A cumulative frequency graph is given below which shows the height of students in a school.
(a)
Write down the median height of the students.
(1)
(b)
Write down the 25th percentile.
(1)
(c)
Write down the 75th percentile.
(1)
The height of the tallest student is 195 cm and the height of the shortest student is 136 cm.
(d) Draw a box and whisker plot on the grid below to represent the heights of the students in the school.
(3)
(Total 6 marks)
28.
The birth weights, in kilograms, of 27 babies are given in the diagram below.
(a)
1 7, 8, 9
2 1, 2, 2,
3 0, 1, 3,
4 1, 1, 2,
Calculate the mean birth weight.
key 1|7 = 1.7 kg
3,
4,
3,
5,
5,
7
5,
5,
7,
6,
8,
6,
9
7,
9
(2)
(b)
Write down:
(i)
the median weight;
(1)
(ii)
the upper quartile.
(1)
The lower quartile is 2.3 kg.
(c) On the scale below draw a box and whisker diagram to represent the
birth weights.
(2)
1
2
3
Weight (kg)
4
5
(Total 6 marks)
29.
Peter has marked 80 exam scripts. He has calculated the mean mark for the scripts to be 62.1. Maria has
marked 60 scripts with a mean mark of 56.8.
(a) Peter discovers an error in his marking. He gives two extra marks each to eleven of the scripts.
Calculate the new value of the mean for Peter’s scripts.
(b) After the corrections have been made and the marks changed, Peter and Maria put all their scripts
together. Calculate the value of the mean for all the scripts.
(Total 8 marks)
30.
The numbers of games played in each set of a tennis tournament were
9, 7, 8, 11, 9, 6, 10, 8, 12, 6, 8, 13, 7, 9, 10, 9, 10, 11,
12, 8, 7, 13, 10, 7, 7.
The raw data has been organized in the frequency table below.
games
frequency
6
2
7
5
8
n
9
4
10
4
11
2
12
2
13
2
(a)
(b)
(c)
(d)
Write down the value of n.
Calculate the mean number of games played per set.
What percentage of the sets had more than 10 games?
What is the modal number of games?
(Total 8 marks)
31.
The graph below shows the cumulative frequency for the yearly incomes of 200 people.
200
180
160
140
120
100
Cumulative
frequency 80
60
40
20
0
0
5000
10 000
15 000
20 000
25 000
Annual income in British pounds
30 000
35 000
Use the graph to estimate
(a) the number of people who earn less than 5000 British pounds per year;
(b) the median salary of the group of 200 people;
(c) the lowest income of the richest 20% of this group.
(Total 4 marks)
32.
A group of students has measured the heights of 90 trees. The class calculate the mean height to be x = 12.4
m with standard deviation s = 5.35 m. One student notices that two of the measurements, 44.5 m and 43.2 m,
are much too big and must be wrong.
(a) How many standard deviations away from the mean of 12.4 is the value 44.5?
The incorrect measurements of 44.5 m and 43.2 m must be removed from the data.
(b) Calculate the new value of x after removing the two unwanted values.
(Total 8 marks)
33.
The diagram below shows the cumulative frequency distribution of the heights in metres of 600 trees in a
wood.
(a)
Write down the median height of the trees.
(1)
(b)
Calculate the interquartile range of the heights of the trees.
(2)
(c)
Given that the smallest tree in the wood is 3 m high and the tallest tree is 28 m high, draw the box and
whisker plot on the grid below that shows the distribution of trees in the wood.
(3)
(Total 6 marks)
The histogram below shows the amount of money spent on food each week by 45 families. The amounts
have been rounded to the nearest 10 dollars.
frequency
34.
18
16
14
12
10
8
6
4
2
0
150
(a)
(b)
(c)
160
170
$
180
190
Calculate the mean amount spent on food by the 45 families.
Find the largest possible amount spent on food by a single family in the modal group.
State which of the following amounts could not be the total spent by all families in the modal group:
(i) $2430
(ii) $2495
(iii) $2500 (iv) $2520 (v) $2600
(Total 6 marks)
35.
The length and width of 10 leaves are shown on the scatter diagram below.
Relationship between leaf length and width
70
60
50
Width
(mm) 40
30
20
10
0
20
40
60
(a)
80
100
120
140
Length (mm)
Plot the point M(97, 43) which represents the mean length and the mean width.
160
(b)
(c)
Draw a suitable line of best fit.
Write a sentence describing the relationship between leaf length and leaf width for this sample.
(Total 4 marks)
36.
A survey was conducted of the number of bedrooms in 208 randomly chosen houses. The results are shown
in the following table.
Number of bedrooms
1
2
3
4
5
6
Number of houses
41
60
52
32
15
8
(a) State whether the data is discrete or continuous.
(1)
(b)
Write down the mean number of bedrooms per house.
(c)
Write down the standard deviation of the number of bedrooms per house.
(d)
Find how many houses have a number of bedrooms greater than one standard deviation above the
mean.
(2)
(1)
(2)
(Total 6 marks)
37.
In an environmental study of plant diversity around a lake, a biologist collected data about the number of
different plant species (y) that were growing at different distances (x) in metres from the lake shore.
Distance (x)
Plant species (y)
(a)
2
35
5
34
8
30
10
29
13
24
17
19
23
15
35
13
40
8
Draw a scatter diagram to show the data. Use a scale of 2 cm to represent 10 metres on the x-axis and 2
cm to represent 10 plant species on the y-axis.
(4)
(b)
Using your scatter diagram, describe the correlation between the number of different plant species and
the distance from the lake shore.
(1)
(c)
Use your graphic display calculator to write down
x , the mean of the distances from the lake shore;
(i)
y, the mean number of plant species.
(ii)
(2)
(d)
Plot the point ( x, y) on your scatter diagram. Label this point M.
(2)
(e)
Write down the equation of the regression line y on x for the above data.
(2)
(f)
Draw the regression line y on x on your scatter diagram.
(2)
(g)
Estimate the number of plant species growing 30 metres from the lake shore.
(2)
(Total 15 marks)
38.
In an experiment a vertical spring was fixed at its upper end. It was stretched by hanging different weights on
its lower end. The length of the spring was then measured. The following readings were obtained.
Load (kg) x
0
1
2
3
4
5
6
7
8
Length (cm) y
23.5
25
26.5
27
28.5
31.5
34.5
36
37.5
(a) Plot these pairs of values on a scatter diagram taking 1 cm to represent 1 kg on the horizontal axis and
1 cm to represent 2 cm on the vertical axis.
(4)
(b)
(i)
Write down the mean value of the load ( x ).
(ii)
Write down the standard deviation of the load.
(iii)
Write down the mean value of the length ( y ).
(iv)
Write down the standard deviation of the length.
(1)
(1)
(1)
(1)
(c)
Plot the mean point ( x , y ) on the scatter diagram. Name it L.
(1)
It is given that the covariance Sxy is 12.17.
(d) (i)
Write down the correlation coefficient, r, for these readings.
(1)
(ii)
Comment on this result.
(2)
(e)
Find the equation of the regression line of y on x.
(2)
(f)
Draw the line of regression on the scatter diagram.
(g)
(i)
Using your diagram or otherwise, estimate the length of the spring when a load of 5.4 kg is
applied.
(ii)
Malcolm uses the equation to claim that a weight of 30 kg would result in a length of 62.8 cm.
Comment on his claim.
(2)
(1)
(1)
(Total 18 marks)
39.
Several candy bars were purchased and the following table shows the weight and the cost of each bar.
Yummy Chox
Marz
Twin Chunx
Lite
BigC
Bite
Weight (g)
60
85
80
65
95
50
100
45
Cost (Euros)
1.10
1.50
1.40
1.20
1.80
1.00
1.70
0.90
(a)
Given that sx = 19.2, sy = 0.307 and sxy = 5.81, find the correlation coefficient, r, giving your answer
correct to 3 decimal places.
(2)
(b)
Describe the correlation between the weight of a candy bar and its cost.
(1)
(c)
Calculate the equation of the regression line for y on x.
(3)
(d)
Use your equation to estimate the cost of a candy bar weighing 109 g.
(2)
(Total 8 marks)
40.
The Type Fast secretarial training agency has a new computer software spreadsheet package. The agency
investigates the number of hours it takes people of varying ages to reach a level of proficiency using this
package. Fifteen individuals are tested and the results are summarised in the table below.
Age
32 40 21 45 24 19 17 21 27 54 33 37 23 45 18
(x)
Time
(in hours) 10 12
8
15
7
8
6
9
11 16
t
13
9
17
5
(y)
(a) (i)
Given that Sy = 3.5 and Sxy = 36.7, calculate the product-moment correlation coefficient r for this
data.
(4)
(ii)
What does the value of the correlation coefficient suggest about the relationship between the two
variables?
(1)
(b)
Given that the mean time taken was 10.6 hours, write the equation of the regression line for y on x in
the form y = ax + b.
(c)
Use your equation for the regression line to predict
(i)
the time that it would take a 33 year old person to reach proficiency, giving your answer correct
to the nearest hour;
(3)
(2)
(ii)
the age of a person who would take 8 hours to reach proficiency, giving your answer correct to
the nearest year.
(2)
(Total 12 marks)
Math Studies Year 1 –Semester I Final Exam IB Review Questions Mark Scheme
1.
Unit penalty (UP) applies in part (a)
(a)
V=
500 3
36 π
(M1)
Note: Award (M1) correct substitution into formula.
V = 1051.305...
V = 1051.31 cm3
UP
(A1)
(A1)(ft)
(C3)
Note: Award last (A1)(ft) for correct rounding to 2 decimal places of their
answer. Unrounded answer must be seen so that the follow through can be
awarded.
(b)
1051
(c)
1.051 × 103
(A1)(ft)
(C1)
(A1)(ft)(
(C2)
A1)(ft)
Note: Award (A1) for 1.051 (accept 1.05) (A1) for × 10.3
[6]
2.
(a)
3.14 × 102 or s
(b)
0.00314 or 3.14 × 103 or
p
(C2)
Note: Award (M1) for indication of comparing numbers where at least one
of them is converted. The converted number does not have to be correct. A
single converted number is sufficient for (M1) to be awarded.
(c)
0.3454 (0.345)
(d)
3.454 × 10–1 (3.45 × 10–1)
(A1)
(C1)
(M1)(A1)
(A1)
(C1)
(A1)(A1)
(C2)
(ft)
Notes: Follow through from their (c).
Award (A1) for 3.454 (3.45) (A1) for 10–1.
[6]
3.
UP
Unit penalty applies in part (a)
(a) 50 × 100 × 40 = 200000
cm3
(C2)
Note: Award (M1) for correct substitution in the volume formula.
(b)
(M1)(A1)
200000
= 400
500
(M1)(A1)
(C2)
(ft)
Note: Award (M1) for dividing their answer to part (a) by 500.
(c)
400  350
× 100 = 143%
350
(M1)(A1)
(C2)
(ft)
Notes: Award (M1) for correct substitution in the percentage error formula.
Award (A1) for answer, follow through from part (b).
Accept –14.3%
% sign not necessary.
[6]
4.
(a)
29.7675
(A1)
(C1)
(A1)(ft)
(C1)
Note: Accept exact answer only
(b)
(c)
(d)
30
30 – 29.7675
100 % For correct formula with correct substitution.
29.7675
= 0.781
Note: accept 0.78 only if formula seen with 29.7675
as denominator
7.81×10–1 (7.81×10−3 with no percentage sign)
(M1)
(A1)(ft)
(C2)
(A1)(ft)(A1)(ft)
(C2)
[6]
5.
(a)
0.0337
(b)
(i)
(ii)
(c)
3.4×10–2
(M1)(A1)
Note: Award (M1) for substituting into the formula, (A1) for correct answer.
(C2)
(A1)(ft)
(A1)(ft)
(C1)
(C1)
(A1)(ft)(A1)(ft)
Note: Award (A1) for 3.4 and (A1) for –2. (ft) is from candidate’s answer to
part (b)(i) only.
(C2)
0.034
3.4%
[6]
6.
(a)
w = (2.6 × 104) × (5.0 × 10–8)
= 13 × 10–4 or 0.0013 (if written as working)
= 1.3 ×10–
3
(A2)
(A1)(A1)
(C4)
Note: For incorrect answers with no working, award marks as follows:
13 × 10–4: (A1)(A1), 0.0013: (G2), 1.3E-3: (G2), 1.3E3: (G1), 1.3
×103:(A2)(A0).
(b)
Statements (ii) and (iv) are
incorrect.
(C4)
Note: Both correct statements given with at most one extra, allow (A2).
(A2)(A2)
[8]
7.
(a)
471.03
(b)
(i)
471
(A1)
(C1)
(A1)
(C1)
(ii)
4.71×102 or 4.71028×102 or
4.7103×102
(C2)
(A1)(A1)
[4]
8.
(a)
(b)
(c)
u96 = u1 + 95d
= 0 + 95 ×12
= 1140
(M1)
6r5 = 16d
6r5= 16 × 12 (192)
Note: (A1) only, if both terms seen without an equation.
(A1)
(A1)
r5 = 32
(A1)
(C2)
(C2)
(A1)(ft)
Note: (ft) from their (b)
r=2
(A1)(ft)
(C2)
[6]
9.
(a)
(b)
u1 = –16, u1 + 10d = 39
–16 + 10d = 39
Note: Award (M1) for correct formula, (A1) for correct numbers.
10d = 39 + 16 = 55
d = 5.5
(M1)
(A1)
u1r2 = 12
16
u1r4 =
3
(M1)
(A1)
(A1)
(C4)
(A1)
Note: Award (M1) for correct formula, (A1) for correct numbers.
 16 
 
16 4
3
r2     
12
36 9
2
r
3
(M1)
(A1)
(C4)
[8]
10.
(a)
(b)
u6 = u1 + 5d = 24
u1 + 5 × 8 = 24
u1 = 24 – 40
= –16
n
(2 × –16 + (n – 1)8)
2
n
600 = (–32 + 8n – 8)
2
1200 = –40n + 8n2
150 = –5n + n2
(n – 15)(n + 10) = 0
n = 15 or –10
Sn =
(M1)(A1)
(A1)
(C3)
(M1)(A1)
(A1)
(A1)
(A1)
(C5)
Note: Not all the steps of working out need to be shown.
[8]
11.
(a)
u1 + 3d = 12
u1 + 9d =
42
(C4)
(A1)(A1)
(A1)(A1)
Note: Award (A1) for left hand side correct, (A1) for right hand side correct.
(b)
6d = 30
d=5
u1 = –
3
(C4)
(A1)
(A1)
(M1)(A1)
Note: Follow through (ft) from candidate's equations.
[8]
12.
(a)
(b)
4n – 3
397
(c)
S100 =
100
[(2 × 1) + (99 × 4)] or 50(1 + 397)
2
= 19 900
(A1)
(A1)
(M1)
(A1)
[4]
13.
4th term = a + 3d
8th term = a + 7d
20th term = a + 19d
Note: Award (M1) for each correct answer up to a maximum of [2 marks].
a + 7d = 2(a + 3d)
a + 19d = 4000
Note: Award (M1) for any one correct equation.
d = 200
(M2)
(M1)
(A1)
[4]
14.
(a)
(b)
a = 100
d = 25
T17 = 100 + (17 – 1)×25
= $500
(C2)
n
(a + l)
2
17
S17 =
(100 + 500)
2
= $5100
Note: Allow follow through from candidate’s answer for T17, which is l
OR
n
Sn = {2a + (n – 1)d}
2
17
S17 =
{2×100 + (17 – 1)×25}
2
= $5100
(M1)
(A1)
Sn =
(M1)
(A1)
(M1)
(A1)
(C2)
OR
Table
[4]
15.
(a)
(5k – 2) – (2k + 3) = (10k – 15) – (5k – 2)
Note: Award (M1) for equating the common difference
5k – 2 – 2k – 3 = 10k – 15 – 5k + 2
3k – 5 = 5k – 13
–2k = –8 or 2k = 8
k=4
(M1)
(A1)
(A1)
(AG)
OR
(2k + 3 + 10k – 15) ÷ 2 = 5k – 2
(M1)(A1)
Note: Award (M1)(A1) for finding the arithmetic mean with correct
substitutions
2k + 3 + 10k – 15 = 10k – 4
(A1)
k=4
(AG)
(b)
11, 18, 25
(A1)
(c)
7
(A1)
(d)
U20 = 11 + 19  7
(M1)
Note: Award (M1) for using the correct formula with candidate’s values
= 144
(A1)(ft)(G2)
(e)
S15 
15
(M1)
(2  11  14  7)
2
= 900
(A1)(ft)(G2)
Note: Award (M1) for using the correct formula with candidate’s values
[9]
16.
(a)
(b)
un = 2(0.9)7 = 0.957
m
2
Note: Award (M1) for substitution into formula, list or suitable diagram.
Sn =
(M1)(A1)
2(1  (0.9) 5 )
= 8.19 m
1  ( 0 .9 )
(M1)(M1
)(A1)
Note: Award (M1) for substitution into formula, list or suitable diagram.
Total distance travelled = 2 × 8.19 = 16.4 m.
(A1)
4
[6]
17.
(a)
(i)
a = $250
(A1)
d = $200
(A1)
T10 = 250 + 9 × 200
= 2050
(A1)
Note: Award the marks for the values of ‘a’ and ‘d’ if they are correctly
substituted into the formula without being explicitly stated.
(ii)
a = $10
(A1)
r=2
(A1)
T0 = 10 × 29
= 5120
(A1)
Note: Award the marks for the values of ‘a’ and ‘r’ if they are correctly
substituted into the formula without being explicitly stated.
6
(b)
(c)
10
(250 + 2050)
2
= 11500
OR
10
S10 =
{2 × 250 + (10 – 1) × 200}
2
= 11500
S10 =
Option One: $10000
Option Two: $11500
10(210  1)
Option Three: S10 =
2 1
= 10 230
Therefore, Option Two would be best.
(M1)
(A1)
(M1)
(A1)
2
(A1)
(M1)
(A1)
(R1)
4
[12]
18.
Financial Penalty applies in parts (b) and (e).
40 000(4)(3)
(a) I =
100
Note: Award (M1) for substituted simple interest formula.
I = 4800
OR
40 000(4)(3)
40 000 +
100
(M1)
(A1)
(M1)(M1
)
Note: Award (M1) for substituted simple interest formula, (M1) for addition
of 40 000.
Amount = 44 800USD
Note: Award final (A1) for 44 800 only.
(b)
FP
44 800 × 18.624
= 834355 ZAR
(A1)(G2)
(M1)
(A1)(ft)(
G2)
Note: Follow through from (a).
(c)
(d)
2.5
100
= 1250 USD
50 000 ×
(M1)
(A1)(G2)
(50 000 – 1250) × 10.608
(M1)(M1
)
Note: Award (M1) for their 48750 seen or implied, (M1)
for × 10.608
(e)
= 517140
= 517 000 ZAR
Note: Follow through from (c), both unrounded and rounded answers must
be seen for final (A1) to be awarded.
(A1)
(AG)
517 000 × (1.01)36
OR
(M1)(A1)
123

12 

= 517000 1 
 100 (12) 
Note: Award (M1) for substituted compounded interest formula, (A1) for
correct substitutions.
739 707 ZAR
Notes: Accept 739 908 if 517 140 used.
FP
(M1)(A1)
(A1)(G2)
[13]
19.
(a)
tn = arn-1 t5 = 26000(1.034)
(M1)
For formula or list.
(A1)
Correct values substituted.
(A1)(G2)
= $29263.23 (29263 or 29300)
Note: For n – 1 = 5, and working seen, award at most (M1)(A0)(A1)(ft) for
answer of 30141.13.
(b)
(i)
26000(1.03)n−1 = 24800(1.05)n−1
(M1)
For equating appropriate expressions or for comparing list or for sketch
from GDC showing intersection
Correct values appearing in candidate’s method.
(A1)(ft)(A1)(ft)
So a total of 3.46 years after she starts work.
(A1)(ft)(G3)
Note: Allow 3 or 4 years.
(ii)
24800(1.05)3 – 26000(1.03)3
Attempt to take a sensible difference.
(M1)
= $298.20
(A1)(ft)(G2)
Note: If (b)(i) is wrong follow through must be consistent with their decision
to round up or down in (b)(i).
[9]
(a)
(i)
(ii)
c = 97
(A1)
1
(not drawn to scale)
Weights of fish
140
cumulative frequency
20.
120
100
80
60
40
20
0
0.5
(iii)
(b)
(i)
0.7
0.9
1.1
1.3
1.5
weight (kg)
Note: Award (A1) for correct scales and axes labelled, (A2) for 5 correct
points, (A1) for 3 to 4 correct, (A1) for the curve.
median is 0.95 kg
line drawn correctly on diagram
(M1)(AG)
90th percentile ...
130 × 0.9 = 117
The zoo buys 13 fish (±2).
(G2)
2
(A4)
1
(M1)
(A1) or
4
(ii)
(c)
(i)
(ii)
First quartile = 32/33 fish.
Maximum weight = 0.79 kg
(0.03).
3
(M1)
maximum : 0.95 × 1.10 = 1.045 kg (1.05 to 3 s.f.)
minimum : 0.95 × 0.90 = 0.855 kg
number of fish bought by restaurant = 88 – 46 = 42(4)
(A1)
(A1)
(M1)(A1)
2
(M1)(M1
3
)(A1)
[16]
21.
(a)
Time less than (mins) Cumulative frequency
10.5
0
15.5
7
20.5
20
25.5
45
30.5
73
35.5
93
40.5
100
Note: Award (A1) for each correct column
(A2)
2
(b)
100
90
number of students
80
70
60
50
40
c(ii)
30
20
c(i)
10
0
(c)
(i)
(ii)
5
10
15
20
25
30
35
40
time
Note: Award (A1) for the correct scale and labelling.
Award (A2) for plotting 6 or 7 points correctly, (A1) for plotting 4 or 5
points correctly.
12  1 students (allow ft)
31  0.5 minutes (allow ft)
(A3)
3
(A1)
(A1)
2
[7]
22.
Unit penalty (UP) is applicable in question part (b) only.
(a) 45  t < 60
(b)
UP
(c)
42.4 minutes
21.6 minutes
(A1)
1
(G2)
(G1)
3
(A4)
4
[8]
23.
(a)
(b)
(i)
25 minutes (2 minutes)
(A2)
(ii)
Lower quartile = 18 (±1 minute)
Upper quartile = 32 (1 minute)
Interquartile range = 32 – 18 = 14 minutes (±2 minutes)
OR
Accept [18 to 32] as interval for the interquartile range.
p = 20
q = 30
(A1)
(A1)
(A1)
(A3)
5
(A1)
(A1)
2
(c)
Midpoint
2.5
10
17.5
22.5
30
42.5
55
(A1)
Mean =
Frequency
20
20
20
40
60
30
10
Total = 200
M×f
50
200
350
900
1800
1275
550
Total = 5125
(A1)
5125
= 25.625 (exact) or 25.6 (3 s.f.)
200
Note: Not every step needs to be seen to get the marks.
OR
Mean = 25.625 or 25.6 (using GDC)
(M1)(A1)
(G4)
4
[11]
24.
UP
Unit penalty (UP) applies in parts (a) and (b)
(a) 4.1 cm
(b)
UP
Median is midpoint of 11th and 12th entries
3.65 cm
Note: Award (M1) for 3.6 and 3.7 seen.
(A1)
(C1)
(M1)
(A1)
(C2)
(c)
4.1 – 2.4
= 1.7
(A1)(A1)
(A1)(ft)
(C3)
Note: Award (A1) for each quartile. Award (A1)(ft) for correct answer to
subtraction of their quartiles.
[6]
25.
(a)
(b)
(i)
(ii)
(iii)
(iv)
F
T
F
F
(A1)
(A1)
(A1)
(A1)
(C4)
IQR = 6 – 4
(M1)
Note: Award (M1) for identifying correct quartiles.
IQR = 2
(A1)(ft)
(C2)
[6]
26.
(a)
Discrete
(A1)
(C1)
(b)
For attempting to find  fx /  f
2.73
(M1)
(A1)
(C2)
1.34
(A1)
(C1)
(M1)
(A1)(ft)
(C2)
(c)
Notes: for (b) and (c), if both mean and standard
deviation given to 2 significant figures
Award (C1)(C0)(AP) for 2.7. Award (A1)(ft)
for 1.3 ((AP) already deducted)
(d)
Attempt to find their mean + their standard deviation (can be implied)
23, (ft) their mean and standard deviation.
[6]
27.
(a)
170
(A1)
(C1)
(b)
163
(A1)
(C1)
(c)
172
(A1)
(C1)
(d)
(A1)(ft)(
A1)(ft)(A1)
(C3)
Notes: Award (A1)(ft) for correct median, (A1)(ft) for correct quartiles and
box, (A1) for correct end points of whiskers.
Award at most (A1)(A1)(A0) if lines go right through the box.
[6]
28.
UP
Unit penalty (UP) is applicable in question parts (a) and (b) only.
83.6
(a) =
( 3.096296 ...)
27
mean weight = 3.10 kg
(b)
UP
(i)
(ii)
(M1)
(accept 3.1 kg)
(C2)
(A1)
(A1)
(A1)
(A2)(ft)
Note: (A1)(ft) for median and quartiles in correct place. (A1) for whiskers in
correct place and joined to box with straight lines.
(C2)
median = 3.1 kg
upper quartile = 3.7kg
(C2)
(c)
[6]
29.
(a)
(b)
80 × 62.1 + 2 ×11 = 4990
Note: Award (M0)(A0) if 2 × 11 is subtracted and ft the remainder of the
question to answers of 61.825 (or 61.8) and 59.7 respectively.
4990
= 62.375 (or 62.4 to 3
80
s.f.)
(C4)
(M1)(A1)
4990 + 56.8 × 60 = 8398
8398
= 60.0 (3
140
s.f.)
(C4)
Note: An answer of 60 (2 s.f.) with no working receives (G2) or with working
using 4990 receives (M1)(A1)(M1)(A0) AP, however, if 80 × 62.4 is used
then 60 is an exact answer and can receive all the marks.
(M1)(A1)
(M1)(A1)
(M1)(A1)
[8]
30.
(a)
n=4
(b)
Mean number of games is 9.08 (accept
9).
(A2)
(C2)
(M1)(A1)
(C2)
Note: Award (M1) for indicating a sum of games times frequency (possibly
curtailed by dots) or for 227 seen.
(c)
6 100
=

25 1
24%
(C2)
(M1)(A1)
Note: Award (M1)(A0) if 6 is replaced by 10. No other alternative.
(d)
Modal number of games is 7.
(A2)
(C2)
[8]
31.
(a)
(b)
(c)
19 or 20 people
Median salary = 15000 GBP
80% of 200
= 160
23000 ± 500
(A1)
(A1)
(M1)
(A1)
[4]
32.
(a)
(b)
44.5  12.4
=
5.35
6
(C2)
90 × 12.4 = 1116
1116 – 44.5 – 43.2 = 1028.3
1028 .3
=
88
11.7
(C6)
Note: Award (M0)(A0) then ft for 88 × 12.4.
1028.3
Award (M0)(A0) for
.
90
(M1)(A1)
(M1)(A1)
(M1)(A1)
(M1)(A1)
[8]
33.
Unit penalty (UP) applies in part (a) in this question
(a) Median = 11m
(b)
(c)
Interquartile range = 14 – 10
=4
(C2)
Note: (M1) for taking a sensible difference or for both correct quartile
values seen.
(A1)
(C1)
(A1)
(A1)(ft)
correct median
correct quartiles and box
endpoints at 3 and 28, joined to box by straight lines
(A1)(ft)
(A1)(ft)
(A1)
(C3)
Note: Award (A0) if the lines go right through the box.
Award final (A1) if the whisker goes to 20 with an outlier at 28
[6]
34.
(8  150  16  160  11  170  7  180  3  190)
(M1)(M1)
45
Note: Award (M1) for five correct products shown or implied in the numerator,
(M1) for denominator 45.
= $165.78 per week (allow $166)
(A1)
7
Notes: For 165.7 or 165 award (C3) for exact answer.
9
For 165.77 award (C2) and no (AP).
For 165.77 with no working award (C2)(A0)(AP).
(C3)
(b)
$164.99 ($165)
(C1)
(c)
16  $155 = $2480 and 16  164.99 = $2639.84 ($2640)
(M1)
Note: The (M1) is for a sensible attempt to calculate both bounds or for showing
division by 16 of any of the values (i) to (v).
$2430 is not possible
(A1)(ft)
Note: Follow through if wrong modal group is used in (b).
(a)
mean =
(A1)
(C2)
[6]
35.
(a)
(see diagram)
(A1)
(C1)
(b)
Relationship between leaf length and width
70
60
50
Width
(mm) 40
30
M
20
10
0
20
Notes:
40
60
80
100
Length (mm)
120
140
160
(A2)
(C2)
(a) Award (A1) for the point M.
(b) Award (A1) for a line of best fit going through M and (A1) for a
reasonable line of best fit
(c)
leaf length and leaf width are positively correlated
(R1)
(C1)
[4]
36.
(a)
Discrete
(A1)
(C1)
(b)
For attempting to find  fx /  f
2.73
(M1)
(A1)
(C2)
1.34
(A1)
(C1)
(M1)
(A1)(ft)
(C2)
(c)
Notes: for (b) and (c), if both mean and standard
deviation given to 2 significant figures
Award (C1)(C0)(AP) for 2.7. Award (A1)(ft)
for 1.3 ((AP) already deducted)
(d)
Attempt to find their mean + their standard deviation (can be implied)
23, (ft) their mean and standard deviation.
[6]
37.
(a)
(A1)(A3)
Notes: Award (A1) for scales and labels (accept x/y).
Award (A3) for all points correct.
Award (A2) for 7 or 8 points correct.
Award (A1) for 5 or 6 points correct.
Award at most (A1)(A2) if points are joined up.
If axes are reversed award at most (A0)(A3)(ft).
(b)
Negative
(A1)
(c)
(d)
(i)
17
(G1)
(ii)
23
(G1)
Point correctly placed and labelled M
(A1)(ft)(
A1)
Note: Accept an error of ±0.5.
(e)
y = –0.708x + 35.0
Note: Award at most (G1)(G0) if y = not seen. Accept 35.
(f)
Regression line drawn that passes through M and (0, 35)
(G1)(G1)
(A1)(ft)(
A1)(ft)
Note: Award (A1) for straight line that passes through M, (A1) for line
(extrapolated if necessary) that passes through (0, 35) (accept error of ±1).
If ruler not used, award a maximum of (A1)(A0).
(g)
y = –0.708(30) + 35.0
= 14 (Accept 13)
(M1)
(A1)(ft)(
G2)
OR
Using graph: (M1) for some indication on graph of point, (A1)(ft)
for answers. Final answer must be consistent with their graph.
(M1)
(A1)(ft)(
G2)
Note: The final answer must be an integer.
[15]
38.
(a)
(A4)
Note: Award (A1) for correct scales and labels, (A3) for correct points, (A2)
for 7 or 8 correct, (A1) for 5 or 6 correct.
(b)
(i)
4
(G1)
(ii)
(iii)
(iv)
2.58
30
4.78
(G1)
(G1)
(G1)
Note: If wrong version of s.d. used in (ii), can (ft) in (iv) (5.07).
(c)
L correctly plotted on graph and named
(d)
(i)
(ii)
(e)
y = 1.83x + 22.7 (y = 1.825x + 22.7)
Award (G1) for y= 1.83x (1.825x), (G1) for 22.7
(f)
Line drawn on graph.
(A1)(A1)(ft)
Note: Award (A1) for passing through the mean point, (A1) for y intercept
between 22 and 23.
(g)
(i)
r = 0.986 (0.987)
(very) strong positive correlation
(A1)(ft)
(G1)
(R1)(ft)(R1)(ft)
(G1)(G1)
32.6 cm
(A1)(ft)
Note: Allow margin of error of 0.2 from value on candidate’s diagram.
Not possible to find an answer as the value lies too far outside the
given set of data.
(R1)
(ii)
[18]
39.
(a)
S xy
r=
(S x S y )
= 0.986

5.81
(19.2  0.307 )
(M1)
(A1)
2
Strong, positive correlation
y = 0.182 + 0.0158x
OR
5.81
y – 1.325 =
(x – 72.5)
19.2 2
y = 0.0158x + 0.182
(A1)
(G3)
1
(A1)
3
y = 0.0158 × 109 + 0.182
= 1.90 euros.
(M1)
(A1)
2
Note: Award (G2) for 0.985 from GDC.
(b)
(c)
(d)
(M1)(A1)
[8]
40.
(a)
(i)
Sx = 11.2
36.7
r=
11.2  3.5
= 0.936 (3 s.f.)
OR
Sx = 11.6
36.7
r=
11.6  3.5
= 0.904 (3 s.f.)
(ii)
The correlation coefficient suggests a strong positive correlation
between the two variables.
_
(b)
y– y
_
Sxy
(
x

x
)
( Sx) 2
(A1)
(M2)
(A1)
(A1)
(M2)
(A1)
(R1)
5
36.7
(x – 30.4)
11.2 2
y = 0.293x + 1.69 (or y = 0.293x + 1.71) Allow ft from (a) (i))
y – 10.6 =
(c)
(i)
(ii)
(M1)
(A2)
y = 0.293 × 33 + 1.69
= 11.359
= 11 hours
(M1)
8 = 0.293x + 1.69
x = 21.54
= 22 years
(M1)
3
(A1)
(A1)
4
[12]