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