Tutorial exercises Preparation for TMA02 Unit 3 exercises – Preparation for TMA02 Unit 3 1. Use a...
Transcript of Tutorial exercises Preparation for TMA02 Unit 3 exercises – Preparation for TMA02 Unit 3 1. Use a...
Tutorial exercises – Preparation for TMA02
Unit 3
1. Use a factor tree to write the following numbers as a product of prime factors.
(a) 96 (b) 255 (c) 770
2. Use prime factorisations to find the lowest common multiple and highest
common factor of each of the following sets of numbers.
(a) 20 and 56 (b) 45 and 150
3. A queen bee may lay about 800 eggs each day. If a queen bee lives for 3 years,
how many eggs has she laid in total? Write your answer in scientific notation
correct to 2 significant figures.
4. Helen is making miniature books for a dolls house. Each book measures 10.4 mm
by 4.5 mm. Convert these measurements to metres, writing your answers in
scientific notation.
5. Carry out the following calculations, without using your calculator.
(a) 3 7
11 11 (b)
2 1
9 12 (c)
7 3
8 5 (d)
2 1
3 5
(e) 2 3
7 4 (f)
1 22 1
3 7 (g)
3 1
8 4 (h)
105
11
6. Find the values of the following, without using your calculator.
(a) 50 (b) 11
9
2 3
2
(c)
12
9
(d) 22
7. Find the following roots of numbers, without using your calculator.
(a) Two square roots of 100.
(b) 5 1
(c) 49
100
(d) 4 81
8. Simplify the following surds, where possible, without using your calculator.
(a) 32
(b) 70
(c) 98
9. Simplify the following surds, where possible, without using your calculator.
(a) 2 5 2
(b) 2 2 18
(c) 4 12 3 2
(d) 27
3
(e) 80
80
(f) 17 71
(g) 7 12 3
10. Find the values of the following powers, without using your calculator.
(a) 1
249 (b) 2
38 (c) 1
327
11. Express the following ratios in their simplest forms.
(a) 5 : 25
(b) 0.2 : 0.8 : 1.4
12. Ebenezer wishes to divide his Christmas gift of £3000 amongst his grandchildren,
Pip, Tim and Betsy, in the ratio 7 : 4 : 1. How much will each get?
Unit 3 review
13. True or false?
(a) All rational numbers are integers.
(b) All integers are rational numbers.
(c) 1.5 is a rational number.
(d) 2 is a rational number.
(e) 27 is a prime number.
(f) 9 3 272 2 2
(g) 21 4 25
Unit 4
1. Classify the investigations below into summarising, comparing or seeking a
relationship.
(a) How many working days are lost because of illness in a year?
(b) Is the number of days lost because of illness related to the type of work
undertaken?
(c) Are more days lost because of illness now than 10 years ago?
(d) Are the heights of children linked to the amount of physical activity they
have?
(e) Are children healthier in the north of Britain or the south of Britain?
(f) What proportion of time do children spend watching television?
2. Which of the following are examples of continuous variables and which are
discrete?
(a) The world ranking of a tennis player.
(b) The distance you walk in a day.
(c) The attendance at a book launch.
(d) The number of times that you phone a helpline before you get a reply.
(e) The time taken to get an answer when you phone a helpline.
(f) The length of an icicle over several days.
(g) Students’ verdicts on the course MU123 on a scale from 1 to 5 (where 1
means highly dissatisfied and 5 means highly satisfied).
3. True or false?
(a) The mean of a dataset is always greater than the median.
(b) The number of cornflakes in a packet is an example of a continuous variable.
(c) The standard deviation of a dataset is a measure of how spread out the data
values are.
(d) About half the values in a dataset lie below the median and about half lie
above the median.
(e) ‘Are men more intelligent then women?’ is an example of a summarising
type of investigation.
(f) Precision describes how close measurements are to each other.
(g) The height and weight, measured for each of a group of children, are paired
data.
4. You are considering going to Puddletown for a holiday next July. So that you are
prepared you would like to know how much rain to expect. You find that rainfall
in July over the last 7 years (in mm) has been
63 55 71 75 59 141 68
Find the mean rainfall in July and the median rainfall. Which of the mean and
median do you think is more typical of rainfall in Puddletown in July? Why?
5. Terry is an OU tutor. He is interested in the way that marks vary between TMAs.
He collects the following data for two TMAs for 11 students in his student group.
Student A B C D E F G H I J K
TMA01 49 77 52 89 57 88 85 80 86 80 95
TMA02 55 76 41 90 37 75 76 89 86 78 80
(a) From Terry’s point of view are these primary or secondary data?
(b) Are the data discrete or continuous?
(c) Are these paired data or are they two-sample data?
(d) Use the following table to compare the location and the spread of marks for
the two TMAs.
TMA01 TMA02
Mean 76.2 71.2
Median 80 76
Range 46 53
Interquartile range 31 31
Standard deviation 15.2 17.6
6. You can measure reaction times for your right and left hands by asking a friend
to drop a ruler (held vertically) without warning, catching it with the relevant
hand as quickly as possible and noting the position on the ruler where it is
caught. If you would like to know more about this or try it yourself, you can find
further information at http://www.bbc.co.uk/dna/h2g2/A33659418
Layla decides to use the experiment to investigate whether her reaction times are
faster with her right hand or with her left hand. She repeats the experiment 15
times for each hand and gets the following reaction times in milliseconds.
Left hand 100 130 200 210 120 100 220 130 170 250 200 130 150 120 150
Right hand 180 120 160 170 190 150 180 230 200 170 200 180 230 180 210
(a) Thinking about the PCAI statistical investigation cycle, what question is
Layla posing?
(b) From Layla’s point of view are these primary or secondary data?
(c) Are these paired data or are they two-sample data?
(d) Scan the data. What do you notice about the data? Are there any values that
you might question?
(e) When the data are entered into Dataplotter, they produce the following
information.
Mark the range and the interquartile range on the dotplots. Use these to
describe the spread of the two datasets. Do the values for standard deviation
fit in with your conclusions about spread?
(f) Compare the mean values for the two datasets and the median values. What
do they tell you about how reaction times compare for Layla’s left and right
hands?
(g) Explain how the data were analysed in this investigation.
(h) Write a brief summary of your conclusions.
Unit 5
Terms
1. Identify the terms in the expression 34 3 27k h k .
2. True or false?
In the expression 24 3 2a a b
(a) The coefficient of b is 2.
(b) a2 has no coefficient.
(c) The coefficient of a2 is 1.
(d) The coefficient of a2 is −1.
(e) 3 is a constant term.
(f) 3 − 2b is a term.
(g) a2 is a term.
(h) 4a and − a2 are like terms.
Simplification
3. Simplify the following expressions by collecting like terms.
(a) k + 3k
(b) 2m + m2 − m
(c) 6xy − 3x − 2x + y
(d) 2ab − a2b − 5ba
(e) 4a + 2A + 3 − 1
(f) 4pq − 3q + pq +3q
4. Multiply out the brackets in the following expressions.
(a) 2a(3a − 6)
(b) 2(x + x2)y
(c) −2(4 − 2a)
(d) a(b + c) + b(a − c)
(e) 2a − (b − a)
5. Expand the following algebraic fractions and then simplify where possible.
(a)
(b)
Linear equations
6. Show (without solving the equation) that x = −2 is a solution to the equation 3 −
5x = 15 + x. Now solve the equation.
2 6
2
A
25 3p p q pq
p
7. Solve the following equations.
(a) 2a = 22 (b) a + 3 = 8
(c) 13
x (d) 2k = 7 + k
(e) y = 12 − 2y (f) B + 13 = 7 − 5B
(g) x − 1 = 2(x + 1) (h) 3
16
x
(i) 3( 7) 142
yy (j)
51
3 6
a a
Puzzles
8. A table has been reduced in price by 25%. It now costs £270. Express this
information as an equation and then solve the equation to find how much the
table cost before the reduction.
How would your equation need to change if the price reduction was 20%?
9. Sara is five times the age she was 28 years ago. How old is Sara?
10. In four years time, Tom will be twice the age he was 11 years ago. How old is
Tom?
11. A number is divided by 7 and then 6 is added to the result giving 15. What
number did you start with?
12. The sum of three consecutive integers is 351. Find the numbers.
13. Nicholas buys two books for his nieces, Amy and Becky, for Christmas. Becky’s
book costs £5 more than Amy’s and together the bill came to £21. How much
were the books?
14. There are four people travelling in a car. Bill and Ben are the same age, Molly is
twice as old as Bill and Pete is twenty years older than Molly. The sum of their
ages is 86. How old is Molly?
Unit 5 review
15. Reviewing Unit 5 – true or false?
(a) In the expression (4 − a) the coefficient of a is −1.
(b) 2 3x xy x x xy
(c) 3(2x − 3) = 6x − 3
(d) (2a + b) − (a + b) = a
(e) x = 2 is a solution of the equation 2x + 1 = 5.
(f) x = 1 is a solution of the equation 2x + 1 = 7 − x.
Unit 3 - solutions
1. (a)
So 96 = 25 × 3
(b)
So 255 = 3 × 5 × 17
(c)
So 770 = 2 × 5 × 7 × 11
2. (a) 20 = 22 × 5
2 3
2 6
2 12
2 24
2 48
96
5 17
3 85
255
7 11
5 77
2 385
770
56 = 23 × 7
The LCM of 20 and 56 is 23 × 5 × 7 = 280.
The HCF of 20 and 56 is 22 = 4.
(b) 45 = 32 × 5
150 = 2 × 3 × 52
The LCM of 45 and 150 is 2 × 32 × 52 = 450.
The HCF of 20 and 56 is 3 × 5 = 15.
3. The queen bee will have laid about 800 × 365 × 3 = 876 000 eggs. In scientific
notation, this is 8.8 × 105 (correct to 2 significant figures).
4. 10.4 mm = 10.4 ÷ 1000 m = 0.0104 m = 1.04 × 10-2 m
4.5 mm = 4.5 ÷ 1000 m = 0.0045 m = 4.5 × 10-3 m
5. (a) 3 7 10
11 11 11
(b) 2 1 8 3 11
9 12 36 36 36
(c) 7 3 35 24 11
8 5 40 40 40
(d) 2 1 2
3 5 15
(e)
(f)
(g)
(h)
6. (a) 50 = 1
(b) 11
11 9 2
9
2 32 3 2 3 12
2
(c)
12 9 1
49 2 2
(d) 2
2
1 12
2 4
7. (a) The two square roots of 100 are 10 and −10.
(b) 5 1 = 1
1
2
2 3 3
7 4 14
1 1
1 3 1 2 7 9
2 1 33 7 3 7
2
1 3 1 3 4 3 1
18 4 8 1 2 2
2
1 10 5 11 11 15 5
11 1 10 2 2
(c) 49 49 7
100 10100
(d) 4 81 3
8. (a) 32 16 2 16 2 4 2
(b) 70 This cannot be simplified. (70 = 2 × 5 × 7)
(c) 98 2 49 2 49 7 2
9. (a) 2 5 2 5 2 2 5 2 10
(b) 2 2 18 2 2 18 2 36 2 6 12
(c) 4 12 3 2 4 3 12 2 12 6 4 12 6 4 12 2 6 24 6
(d) 27 27
9 333
(e) 80 80 80
80 16 5 4 580 80
(f) 17 71 This cannot be simplified.
(g) 7 12 3 7 4 3 3 7 2 3 3 14 3 3 13 3
10. (a) 1
249 49 7
(b) 2
22338 8 2 4
(c) 1
31 3
3
1 1 127
32727
11. (a) 5 : 25 = 1 : 5
(b) 0.2 : 0.8 : 1.4 = 2 : 8 : 14 = 1 : 4 : 7
12. The total number of parts is 7 + 4 + 1 = 12.
Pip gets 7 parts or 7
£3000 £175012
.
Tim gets 4 parts or 4
£3000 £100012
.
Betsy gets 1 part or 1
£3000 £25012
.
How could you check your answers?
13. Review
(a) All rational numbers are integers. False
(b) All integers are rational numbers. True
(c) 1.5 is a rational number. True
(d) 2 is a rational number. False
(e) 27 is a prime number. False
(f) 9 3 272 2 2 False
(g) 21 4 25 False
Unit 4 - solutions
1. Summarising (a) and (f) Comparing (c) and (e) Seeking a relationship
(b) and (d)
2. Continuous (b), (e), (f) Discrete (a), (c), (d) and (g)
3. (a) False
(b) False
(c) True
(d) True
(e) False
(f) True
(g) True
4. The mean rainfall is (63 + 55 + 71 + 75 + 59 + 141 + 68) / 7 = 76 mm
To find the median, put the values in numerical order
55 59 63 68 71 75 141
The median is the middle value, 68 mm.
The median may be a more typical value than the mean here. The very high value
of 141 mm pulls the mean value up, making it higher than all but one of the
values. So if you want a ‘typical’ value for rainfall in Puddletown, you might be
better to use the median.
5. (a) Terry collected the data himself so they are primary data.
(b) The values can only be whole numbers so they are discrete.
(c) The data are paired. We know two TMA marks for each of the eleven
students.
(d) Both the mean and the median are higher for TMA01 than for TMA02, so on
average marks were higher for TMA01. (Note that the median is higher than
the mean for each set of marks. The mean is ‘pulled down’ by the three low
marks.)
The range and the standard deviation are higher for TMA02, while the
interquartile range is the same for both TMAs. The marks for TMA02 are a
little more widely spread than the marks for TMA01.
6. (a) Layla is asking whether her reaction times are faster with her right hand or
with her left hand.
(b) Layla collected the data herself so they are primary data.
(c) These are two-sample data.
(d) All the values appear to have been rounded to the nearest 10 milliseconds
(which is probably sensible for this type of experiment). There are no values
that seem out of line.
(e)
Both the range and the interquartile range are greater for the left hand so data
values are more spread out for the left hand. This is confirmed by the
standard deviation which is 45.3 milliseconds for the left hand compared to
28 milliseconds for the right hand.
(f) The mean value for the left hand is 159 milliseconds compared to 183
milliseconds for the right hand. The median value for the left hand is 150
milliseconds compared to 180 milliseconds for the right hand. For both
measures, the value is lower for the left hand, so, from this experiment, it
appears that Layla’s reaction times are faster with her left hand.
(g) The data were analysed by drawing dotplots and calculating measures of
location and spread.
(h) From this investigation, Layla’s reaction times were faster with her left hand.
However, reaction times were also more spread out for the left hand
compared with the right.
Range
IQR
Unit 5 - solutions
1. The terms are +4, 3k , 3h and 27 k .
2. (a) False
(b) False
(c) False
(d) True
(e) True
(f) False
(g) False
(h) False
3. (a) k + 3k = 4k
(b) 2m + m2 − m = m + m2
(c) 6xy − 3x − 2x + y = 6xy − 5x + y
(d) 2ab − a2b − 5ba = −3ab − a2b
(e) 4a + 2A + 3 − 1 = 4a + 2A + 2
(f) 4pq − 3q + pq + 3q = 5pq
4. (a) 2a(3a − 6) = 6a2 − 12a
(b) 2(x + x2)y = 2xy + 2x2y
(c) −2(4 − 2a) = −8 + 4a (or 4a − 8)
(d) a(b + c) + b(a − c) = ab + ac + ab − bc = 2ab + ac − bc
(e) 2a − (b − a) = 2a − b + a = 3a − b
5. (a)
(b)
6. When x = −2, LHS = 3 − 5 × (−2) = 13 and RHS = 15 + (−2) = 13. LHS = RHS
so −2 is a solution to the equation.
The equation is: 3 − 5x = 15 + x
Add 5x: 3 = 15 + 6x
Swap sides: 15 + 6x = 3
Subtract 15: 6x = −12
Divide by 2: x = −2
The solution is x = −2.
7. (a) The equation is: 2a = 22
Divide by 2: a = 11
The solution is a = 11.
(Check: if a = 11, LHS = 2 × 11 = 22 = RHS; the solution is correct.)
(b) The equation is: a + 3 = 8
Subtract 3: a = 5
The solution is a = 5.
(Check: if x = 5, LHS = 5 + 3 = 8 = RHS; the solution is correct.)
2 6 2 63
2 2 2
A AA
2 25 3 5 35 3
p p q pq p p q pqpq q
p p p p
(c) The equation is: 13
x
Multiply by 3: x = 3
The solution is x = 3.
(Check: if x = 3, LHS = 3/3 = 1 = RHS; the solution is correct.)
(d) The equation is: 2k = 7 + k
Subtract k: k = 7
The solution is k = 7.
(Check: if k = 7, LHS = 2 × 7 = 14 and RHS = 7 + 7 = 14, so LHS = RHS
and the solution is correct.)
(e) The equation is: y = 12 − 2y
Add 2y: 3y = 12
Divide by 3: y = 4
The solution is y = 4.
(Check: if y = 4, LHS = 4 and RHS = 12 − 2 × 4 = 4, so LHS = RHS and the
solution is correct.)
(f) The equation is: B + 13 = 7 − 5B
Add 5B: 6B + 13 = 7
Subtract 13: 6B = −6
Divide by 6: B = −1
The solution is B = −1.
(Check: if B = −1, LHS = −1 + 13 = 12 and RHS = 7 − 5 × (−1) = 12, so
LHS = RHS and the solution is correct.)
(g) The equation is: x − 1 = 2(x + 1)
Multiply out the bracket: x − 1 = 2x + 2
Subtract x: −1 = x + 2
Swap sides: x + 2 = −1
Subtract 2: x = −3
The solution is x = −3.
(Check: if x = −3, LHS = −3 − 1 = −4 and RHS = 2 × (−3 + 1) = −4, so LHS
= RHS and the solution is correct.)
(h) The equation is: 3
16
x
Multiply by 6: x + 3 = 6
Subtract 3: x = 3
The solution is x = 3.
(Check: if x = 3, LHS = 3 3
6
= 1 = RHS; the solution is correct.)
(i) The equation is: 3( 7) 142
yy
Multiply by 2: 6(y − 7) = 28 − y
Multiply out the bracket: 6y − 42 = 28 − y
Add y: 7y − 42 = 28
Add 42: 7y = 70
Divide by 10: y = 10
The solution is y = 10.
(Check: if y = 10, LHS = 3 × (10 − 7) = 9 and RHS = 14 − 10
2 = 9, so LHS =
RHS and the solution is correct.)
(j) The equation is: 5
13 6
a a
Multiply by 6: 2a = a − 5 + 6
Tidy up: 2a = a + 1
Subtract a: a = 1
The solution is a = 1.
(Check: if x = 1, LHS = 1
3 and RHS =
1 5 4 11 1
6 6 3
, so LHS = RHS
and the solution is correct.)
8. Let the original price of the table be £P.
The price has been reduced by 25%, so the new price is 75% of £P, or £0.75P.
The table now costs £270. So 0.75P = 270.
Divide by 0.75: 270
3600.75
P
The cost of the table before the reduction was £360.
(Check: 75% of £360 = 0.75 × £360 = £270.)
If the reduction was 20%, the equation would become 0.8P = 270.
9. Let Sara’s age now be A.
We’re told that Sara’s age 28 years ago × 5 = her age now.
28 years ago, Sara’s age was A − 28.
So the equation is 5 × (A − 28) = A
Multiply out the bracket: 5A − 140 = A
Subtract A: 4A − 140 = 0
Add 140: 4A = 140
Divide by 4: A = 35
So Sara is 35 years old.
(Check: 28 years ago, Sara was 7 years old and 5 times 7 gives 35, her present
age.)
10. Let Tom’s age now be T.
We’re told that in four years time, Tom will be twice the age he was 11 years
ago.
Putting this in a table:
Time Tom’s age
Now T
In 4 years time T + 4
11 years ago T − 11
Create an equation: T + 4 = 2 × (T − 11)
Multiply out the bracket: T + 4 = 2T − 22
Subtract T: 4 = T − 22
Swap sides: T − 22 = 4
Add 22: T = 26
So Tom is 26 years old now.
(Check: in 4 years time Tom will be 30. 11 years ago, he was 15, and twice 15
gives 30.)
11. Let the number you start with be n.
Create an equation: 6 157
n
Multiply by 7: n + 42 = 105
Subtract 42: n = 63
You start with the number 63.
(Check: if the number is 63, divide it by 7 to get 9, then add 6 to get 15.)
12. Let the middle one of the three integers be n. Then the other two will be n − 1
and n + 1.
Adding these gives the equation:
(n − 1) + n + (n + 1) = 351
Tidy up the equation: 3n = 351
Divide by 3: n = 117
The three integers are 116, 117 and 118.
(Check: 116 + 117 + 118 = 351, as expected.)
13. Let the price of Amy’s book be £P. Then Becky’s book cost five pounds more or
£(P + 5).
The total cost was £21, giving the equation:
P + (P + 5) = 21
Tidy up: 2P + 5 = 21
Subtract 5: 2P = 16
Divide by 2: P = 8
So the books were £8 and £13.
(Check: if the books are £8 and £13, then the total cost is £21, as expected.)
14. Although you are asked to find Molly’s age, you may find it easier to start by
letting Bill’s age be b years. Then you will avoid fractions!
If Bill’s age is b years, then Ben’s age is also b years. Molly is twice as old as
Bill so she is 2b years old. Pete is 20 years older than Molly so he is (2b + 20)
years old.
The sum of their ages is 86, so we can create the equation:
b + b + 2b + (2b + 20) = 86
Tidy up: 6b +20 = 86
Subtract 20: 6b = 66
Divide by 6: b = 11
So Bill is 11 years old and from this we can work out that Molly is 22 and Pete is
42.
(Check: Adding the ages together, 11 + 11 + 22 + 42 = 86, as expected.)
15. (a) True
(b) False
(c) False
(d) True
(e) True
(f) False