INTERMEDIATE 2 ADDITIONAL QUESTION BANK

1
INTERMEDIATE 2 ADDITIONAL QUESTION BANK
Straight Line
Percentages
UNIT 1
Volumes
Brackets
The Circle
Factors
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2
INTERMEDIATE 2 ADDITIONAL QUESTION BANK
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Straight Line
UNIT 1
Please choose a question to attempt from the
following
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Question
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5
STRAIGHT LINE Question 1
Find the equation of the line shown in this
diagram.
Get hint
Reveal answer only
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6
STRAIGHT LINE Question 1
Find the equation of the line shown in this
diagram.
2. Find gradient using gradient formula.
3. Note where line cuts y-axis
4. Substitute into y mx c
1. Identify two points on the line.
(0, 9)
What would you like to do now?
Reveal answer only
Go to full solution
Go to Comments
(3, 0)
Go to Straight Line Menu
EXIT
7
STRAIGHT LINE Question 1
Find the equation of the line shown in this
diagram.
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Try another like this
Go to full solution
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Go to Straight Line Menu
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8
Question 1
1. Identify two points on the line.
Find the equation of this line
2. Find gradient using gradient formula.
(0, 9)
Using (0, 9) and (3, 0)
- 3
(3, 0)
3. Note where line cuts y-axis.
Cuts y-axis at (0, 9) so c 9
4. Substitute into y mx c .
Continue Solution
y mx c
Try another like this
becomes
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Straight Line Menu
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9
Comments

• Must learn gradient formula

1. Identify two points on the line.
2. Find gradient using gradient formula.
Using (0, 9) and (3, 0)
- 3
mAB
3. Note where line cuts y-axis.
Cuts y-axis at (0, 9) so c 9
4. Substitute into y mx c .
y mx c
Next Comment
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10
Comments

• Must learn equation of the line

1. Identify two points on the line.
2. Find gradient using gradient formula.
Y mx c
Using (0, 9) and (3, 0)
gradient
y - intercept
- 3
3. Note where line cuts y-axis.
Cuts y-axis at (0, 9) so c 9
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
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11
Comments

• Line slopes downward from
• left to right

1. Identify two points on the line.
Negative gradient
2. Find gradient using gradient formula.
m -ve
Using (0, 9) and (3, 0)
- 3

• Line slopes upwards from
• left to right

Positive gradient
3. Note where line cuts y-axis.
m ve
Cuts y-axis at (0, 9) so c 9
4. Substitute into y mx c .
y mx c
Next Comment
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12
STRAIGHT LINE Question 1B
Find the equation of the line shown in this
diagram.
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
13
STRAIGHT LINE Question 1B
Find the equation of the line shown in this
diagram.
2. Find gradient using gradient formula.
3. Note where line cuts y-axis
4. Substitute into y mx c
1. Identify two points on the line.
What would you like to do now?
Reveal answer only
(2, 0)
Go to full solution
Go to Comments
(0,-1)
Go to Straight Line Menu
EXIT
14
STRAIGHT LINE Question 1B
Find the equation of the line shown in this
diagram.
What would you like to do now?
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
15
Question 1B
1. Identify two points on the line.
Find the equation of this line
2. Find gradient using gradient formula.
Using (0, -1) and (2, 0)
(2, 0)
(0,-1)
3. Note where line cuts y-axis.
Cuts y-axis at (0, -1) so c -1
4. Substitute into y mx c .
Begin Solution
y mx c
Continue Solution
becomes
Comments
Straight Line Menu
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16
Comments

• Must learn gradient formula

1. Identify two points on the line.
2. Find gradient using gradient formula.
Using (0, -1) and (2, 0)
mAB
3. Note where line cuts y-axis.
Cuts y-axis at (0, -1) so c -1
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
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17
Comments

• Must learn equation of the line

1. Identify two points on the line.
2. Find gradient using gradient formula.
Y mx c
Using (0, -1) and (2, 0)
gradient
y - intercept
3. Note where line cuts y-axis.
Cuts y-axis at (0, -1) so c -1
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
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18
Comments

• Line slopes downward from
• left to right

1. Identify two points on the line.
Negative gradient
2. Find gradient using gradient formula.
m -ve
Using (0, -1) and (2, 0)

• Line slopes upwards from
• left to right

Positive gradient
3. Note where line cuts y-axis.
Cuts y-axis at (0, -1) so c -1
m ve
4. Substitute into y mx c .
y mx c
Next Comment
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19
STRAIGHT LINE Question 2
Find the equation of the line shown in this
diagram.
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
20
STRAIGHT LINE Question 2
Find the equation of the line shown in this
diagram.
(6, 60)
What would you like to do now?
Reveal answer only
Go to full solution
Go to Comments
2. Find gradient using gradient formula.
1. Identify two points on the line. (NOTE SCALES)
3. Note where line cuts y-axis
4. Substitute into y mx c
(0, 30)
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EXIT
21
STRAIGHT LINE Question 2
Find the equation of the line shown in this
diagram.
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Go to full solution
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Go to Straight Line Menu
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22
Question 2
1. Identify two points on the line.
Find the equation of this line
2. Find gradient using gradient formula.
(6, 60)
Using (0, 30) and (6, 60)
5
(0, 30)
3. Note where line cuts y-axis.
Cuts y-axis at (0, 30) so c 30
4. Substitute into y mx c .
Continue Solution
y mx c
Try another like this
becomes
Comments
Straight Line Menu
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23
Comments

• Must learn gradient formula

1. Identify two points on the line.
2. Find gradient using gradient formula.
Using (0, 30) and (6, 60)
5
mAB
3. Note where line cuts y-axis.
Cuts y-axis at (0, 30) so c 30
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
becomes
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24
Comments

• Must learn equation of the line

1. Identify two points on the line.
2. Find gradient using gradient formula.
Y mx c
Using (0, 30) and (6, 60)
gradient
y - intercept
5
3. Note where line cuts y-axis.
Cuts y-axis at (0, 30) so c 30
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
becomes
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25
Comments

• Line slopes downward from
• left to right

1. Identify two points on the line.
Negative gradient
2. Find gradient using gradient formula.
m -ve
Using (0, 30) and (6, 60)
5

• Line slopes upwards from
• left to right

Positive gradient
3. Note where line cuts y-axis.
Cuts y-axis at (0, 30) so c 30
m ve
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
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26
STRAIGHT LINE Question 2B
Find the equation of the line shown in this
diagram.
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
27
STRAIGHT LINE Question 2B
Find the equation of the line shown in this
diagram.
1. Identify two points on the line. (NOTE SCALES)
3. Note where line cuts y-axis
2. Find gradient using gradient formula.
4. Substitute into y mx c
What would you like to do now?
Reveal answer only
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
28
STRAIGHT LINE Question 2B
Find the equation of the line shown in this
diagram.
What would you like to do now?
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
29
Question 2B
1. Identify two points on the line.
Find the equation of this line
2. Find gradient using gradient formula.
(0, 1)
Using (0, 1) and (40, -4)
(40, -4)
3. Note where line cuts y-axis.
Cuts y-axis at (0, 1) so c 1
4. Substitute into y mx c .
Begin Solution
y mx c
Continue Solution
becomes
Comments
Straight Line Menu
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30
Comments

• Must learn gradient formula

1. Identify two points on the line.
2. Find gradient using gradient formula.
Using (0, 1) and (40, -4)
mAB
3. Note where line cuts y-axis.
Cuts y-axis at (0, 1) so c 1
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
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31
Comments

• Must learn equation of the line

1. Identify two points on the line.
2. Find gradient using gradient formula.
Y mx c
Using (0, 1) and (40, -4)
gradient
y - intercept
3. Note where line cuts y-axis.
Cuts y-axis at (0, 1) so c 1
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
becomes
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32
Comments

• Line slopes downward from
• left to right

1. Identify two points on the line.
Negative gradient
2. Find gradient using gradient formula.
m -ve
Using (0, 1) and (40, -4)

• Line slopes upwards from
• left to right

Positive gradient
3. Note where line cuts y-axis.
m ve
Cuts y-axis at (0, 1) so c 1
4. Substitute into y mx c .
y mx c
Next Comment
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33
STRAIGHT LINE Question 3
Find the equation of the blue line which is
parallel to the red line shown in this diagram.
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
34
STRAIGHT LINE Question 3
Find the equation of the blue line which is
parallel to the red line shown in this diagram.
2. Find gradient using gradient formula.
3. Note where blue line cuts y-axis
4. Substitute into y mx c, remembering
parallel lines have same gradient.
1. Identify two points on the red line.
What would you like to do now?
Reveal answer only
(6, 6)
Go to full solution
(0, 3)
Go to Comments
Go to Straight Line Menu
EXIT
35
STRAIGHT LINE Question 3
Find the equation of the blue line which is
parallel to the red line shown in this diagram.
What would you like to do now?
Try another like this
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
36
Question 3
1. Identify two points on the red line.
Find the equation of blue line
2. Find gradient using gradient formula.
(6, 6)
Using (0, 3) and (6, 6)
(0, 3)
3. Note where blue line cuts y-axis.
Cuts y-axis at (0, -2) so c -2
4. Substitute into y mx c .
Continue Solution
y mx c
Try another like this
becomes
Comments
Straight Line Menu
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37
Comments

• Must learn gradient formula

1. Identify two points on the red line.
2. Find gradient using gradient formula.
Using (0, 3) and (6, 6)
mAB
3. Note where blue line cuts y-axis.
Cuts y-axis at (0, -2) so c -2
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
becomes
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38
Comments

• Must learn equation of the line

1. Identify two points on the red line.
2. Find gradient using gradient formula.
Y mx c
Using (0, 3) and (6, 6)
gradient
y - intercept
3. Note where blue line cuts y-axis.
Cuts y-axis at (0, -2) so c -2
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
becomes
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39
Comments

• Line slopes downward from
• left to right

1. Identify two points on the red line.
Negative gradient
2. Find gradient using gradient formula.
m -ve
Using (0, 3) and (6, 6)

• Line slopes upwards from
• left to right

Positive gradient
3. Note where blue line cuts y-axis.
Cuts y-axis at (0, -2) so c -2
m ve
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
becomes
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40
Comments

• Parallel lines have equal
• gradients

1. Identify two points on the red line.
2. Find gradient using gradient formula.
m1 m2
Using (0, 3) and (6, 6)
m1
m2
3. Note where blue line cuts y-axis.
Cuts y-axis at (0, -2) so c -2
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
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41
STRAIGHT LINE Question 3B
Find the equation of the blue line which is
parallel to the red line shown in this diagram.
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
42
STRAIGHT LINE Question 3B
Find the equation of the blue line which is
parallel to the red line shown in this diagram.
2. Find gradient using gradient formula.
3. Note where blue line cuts y-axis
4. Substitute into y mx c, remembering
parallel lines have same gradient.
1. Identify two points on the red line.
What would you like to do now?
Reveal answer only
Go to full solution
(0, 80)
Go to Comments
Go to Straight Line Menu
(8, 0)
EXIT
43
STRAIGHT LINE Question 3B
Find the equation of the blue line which is
parallel to the red line shown in this diagram.
What would you like to do now?
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
44
Question 3 B
1. Identify two points on the red line.
Find the equation of blue line
2. Find gradient using gradient formula.
Using (0, 80) and (8, 0)
(0, 80)
3. Note where blue line cuts y-axis.
(8, 0)
Cuts y-axis at (0, 120) so c 120
4. Substitute into y mx c .
Begin Solution
y mx c
Continue Solution
becomes
Comments
Straight Line Menu
What would you like to do now?
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45
Comments

• Must learn gradient formula

1. Identify two points on the red line.
2. Find gradient using gradient formula.
Using (0, 80) and (8, 0)
mAB
3. Note where blue line cuts y-axis.
Cuts y-axis at (0, 120) so c 120
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
becomes
Back to Home
46
Comments

• Must learn equation of the line

1. Identify two points on the red line.
2. Find gradient using gradient formula.
Y mx c
Using (0, 80) and (8, 0)
gradient
y - intercept
3. Note where blue line cuts y-axis.
Cuts y-axis at (0, 120) so c 120
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
becomes
Back to Home
47
Comments

• Line slopes downward from
• left to right

1. Identify two points on the red line.
Negative gradient
2. Find gradient using gradient formula.
m -ve
Using (0, 80) and (8, 0)

• Line slopes upwards from
• left to right

Positive gradient
3. Note where blue line cuts y-axis.
Cuts y-axis at (0, 120) so c 120
m ve
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
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48
Comments
1. Identify two points on the red line.

• Parallel lines have equal
• gradients

2. Find gradient using gradient formula.
m1 m2
Using (0, 80) and (8, 0)
m1
m2
3. Note where blue line cuts y-axis.
Cuts y-axis at (0, 120) so c 120
4. Substitute into y mx c .
y mx c
Next Comment
Straight Line Menu
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49
STRAIGHT LINE Question 4
The graph below shows the cost , C , of hiring a
chain-saw for n days. Find the formula for C in
terms of n.
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
50
STRAIGHT LINE Question 4
The graph below shows the cost , C , of hiring a
chain-saw for n days. Find the formula for C in
terms of n.
1. Identify two points on the line. (NOTE SCALES)
3. Note where line cuts vertical axis
2. Find gradient using gradient formula.
4. Substitute into y mx c, using C and n.
What would you like to do now?
Reveal answer only
(10, 70)
Go to full solution
Go to Comments
Go to Straight Line Menu
(0, 30)
EXIT
51
STRAIGHT LINE Question 4
The graph below shows the cost , C , of hiring a
chain-saw for n days. Find the formula for C in
terms of n.
What would you like to do now?
Try another like this
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
52
Question 4
1. Identify two points on the line.
Find the equation of this line
2. Find gradient using gradient formula.
Using (0, 30) and (10, 70)
(10, 70)
(0, 30)
3. Note where line cuts y-axis.
Cuts y-axis at (0, 30) so c 30
4. Substitute into y mx c .
Continue Solution
C mn c
Try another like this
becomes
Comments
Straight Line Menu
What would you like to do now?
53
Comments

• Must learn gradient formula

1. Identify two points on the line.
2. Find gradient using gradient formula.
Using (0, 30) and (10, 70)
mAB
3. Note where line cuts y-axis.
Cuts y-axis at (0, 30) so c 30
4. Substitute into y mx c .
C mn c
Next Comment
Straight Line Menu
becomes
Back to Home
54
Comments

• Must learn equation of the line

1. Identify two points on the line.
2. Find gradient using gradient formula.
Y mx c
Using (0, 30) and (10, 70)
gradient
y - intercept
3. Note where line cuts y-axis.
Cuts y-axis at (0, 30) so c 30
4. Substitute into y mx c .
C mn c
Next Comment
Straight Line Menu
becomes
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55
Comments

• Line slopes downward from
• left to right

1. Identify two points on the line.
Negative gradient
2. Find gradient using gradient formula.
m -ve
Using (0, 30) and (10, 70)

• Line slopes upwards from
• left to right

Positive gradient
3. Note where line cuts y-axis.
Cuts y-axis at (0, 30) so c 30
m ve
4. Substitute into y mx c .
C mn c
Next Comment
Straight Line Menu
becomes
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56
Comments

• Whatever the context of the
• question remember

1. Identify two points on the line.
2. Find gradient using gradient formula.
Change along y-axis
Gradient
Using (0, 30) and (10, 70)
Change along x-axis

• Gradient tells you rate of change so in this
  question

3. Note where line cuts y-axis.
Gradient rate of hire
4
per hour
Cuts y-axis at (0, 30) so c 30
4. Substitute into y mx c .
C mn c
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57
STRAIGHT LINE Question 4B
A fuel tank holds 120 litres but develops a leak.
The volume of fuel, V litres, remaining after t
hours is shown in the graph below. Find the
formula for V in terms of t.
Get hint
Reveal answer only
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
58
STRAIGHT LINE Question 4B
A fuel tank holds 120 litres but develops a leak.
The volume of fuel, V litres, remaining after t
hours is shown in the graph below. Find the
formula for V in terms of t.
What would you like to do now?
1. Identify two points on the line. (NOTE SCALES)
3. Note where line cuts vertical axis
Reveal answer only
2. Find gradient using gradient formula.
4. Substitute into y mx c, using V and t.
(0, 120)
Go to full solution
Go to Comments
Go to Straight Line Menu
(8, 0)
EXIT
59
STRAIGHT LINE Question 4B
A fuel tank holds 120 litres but develops a leak.
The volume of fuel, V litres, remaining after t
hours is shown in the graph below. Find the
formula for V in terms of t.
What would you like to do now?
Go to full solution
Go to Comments
Go to Straight Line Menu
EXIT
60
Question 4B
1. Identify two points on the line.
Find the equation of this line
2. Find gradient using gradient formula.
Using (0, 120) and (8, 0)
(0, 120)
3. Note where line cuts y-axis.
(8, 0)
Cuts y-axis at (0, 120) so c 120
4. Substitute into y mx c .
Begin Solution
V mt c
Continue Solution
becomes
Comments
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61
Comments

• Must learn gradient formula

1. Identify two points on the line.
2. Find gradient using gradient formula.
Using (0, 120) and (8, 0)
mAB
3. Note where line cuts y-axis.
Cuts y-axis at (0, 120) so c 120
4. Substitute into y mx c .
V mt c
Next Comment
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62
Comments

• Must learn equation of the line

1. Identify two points on the line.
2. Find gradient using gradient formula.
Y mx c
Using (0, 120) and (8, 0)
gradient
y - intercept
3. Note where line cuts y-axis.
Cuts y-axis at (0, 120) so c 120
4. Substitute into y mx c .
V mt c
Next Comment
Straight Line Menu
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63
Comments

• Line slopes downward from
• left to right

1. Identify two points on the line.
Negative gradient
2. Find gradient using gradient formula.
m -ve
Using (0, 120) and (8, 0)

• Line slopes upwards from
• left to right

Positive gradient
3. Note where line cuts y-axis.
m ve
Cuts y-axis at (0, 120) so c 120
4. Substitute into y mx c .
V mt c
Next Comment
Straight Line Menu
becomes
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64
Comments

• Whatever the context of the
• question remember

1. Identify two points on the line.
2. Find gradient using gradient formula.
Change along y-axis
Gradient
Using (0, 120) and (8, 0)
Change along x-axis

• Gradient tells you rate of change so in this
  question

3. Note where line cuts y-axis.
Cuts y-axis at (0, 120) so c 120
4. Substitute into y mx c .
V mt c
Next Comment
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65
INTERMEDIATE 2 ADDITIONAL QUESTION BANK
You have chosen to study
Percentages
UNIT 1
Please choose a question to attempt from the
following
1
3
4
5
2
Back to Unit 1 Menu
EXIT
66
PERCENTAGES Question 1

• The speed of a car is 72mph. This falls by 12.5
  as the traffic builds up. What speed is it
  travelling at now?

(b) As the congestion eases the speed increases
again by 6. Find the new speed to the
nearest mph.
Get hint
Reveal answers only
Go to full solution
Go to Comments
Go to Percentages Menu
EXIT
67
PERCENTAGES Question 1

• The speed of a car is 72mph. This falls by 12.5
  as the traffic builds up. What speed is it
  travelling at now?

(b) As the congestion eases the speed increases
again by 6. Find the new speed to the
nearest mph.
What would you like to do now?
Reveal answers only
1. Remember to use multiplication factors
2. For a decrease the factor is less than 1.
Go to full solution
3. For increases factor is more than 1
Go to Comments
Go to Percentages Menu
EXIT
68
PERCENTAGES Question 1

• The speed of a car is 72mph. This falls by 12.5
  as the traffic builds up. What speed is it
  travelling at now?

63 mph
(b) As the congestion eases the speed increases
again by 6. Find the new speed to the
nearest mph.
67 mph
What would you like to do now?
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Go to full solution
Go to Comments
Go to Percentages Menu
EXIT
69
Question 1(a)
1. Find multiplier
The speed of a car is 72mph. This falls by 12.5
as the traffic builds up. What speed is it
travelling at now?
12.5 off leaves 87.5 or 0.875.
2. Use this multiplier to calculate new speed
New speed 0.875 x 72mph
63mph
Begin Solution
Continue Solution
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Percentages Menu
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70
Question 1(b)
1. Find multiplier
As the congestion eases the speed increases
again by 6. Find the new speed to the nearest
mph.
6 more is 106 or 1.06.
2. Use this multiplier to calculate new
speed, using your answer to (a).
New speed 1.06 x 63mph
66.78mph
67mph
Continue Solution
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Percentages Menu
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71
Comments
Question 1(a)
1. Find multiplier
12.5 off leaves 87.5 or 0.875.
Finding a Percentage 12.8 of 12500
2. Use this multiplier to calculate new speed
x
12500
New speed 0.875 x 72mph
or 0.128 x 12500 (Preferred method)
63mph
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72
Comments
Percentage Increase 10 increase Multiplier
1.10 5.6 increase Multiplier 1.056
Question 1(b)
1. Find multiplier
Finding a Percentage 6 of 63
6 more is 106 or 1.06.
2. Use this multiplier to calculate new
speed, using your answer to (a).
x
63
New speed 1.06 x 63mph
or 0.06 x 63 (Preferred method)
66.78mph
67mph
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73
PERCENTAGES Question 1B

• The speed of a car is 60mph. This falls by 15
  as the traffic builds up. What speed is it
  travelling at now?

(b) As the congestion eases the speed increases
again by 16. Find the new speed to the
nearest mph.
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74
PERCENTAGES Question 1B

• The speed of a car is 60mph. This falls by 15
  as the traffic builds up. What speed is it
  travelling at now?

(b) As the congestion eases the speed increases
again by 16. Find the new speed to the
nearest mph.
What would you like to do now?
Reveal answers only
1. Remember to use multiplication factors
2. For a decrease the factor is less than 1.
Go to full solution
3. For increases factor is more than 1
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75
PERCENTAGES Question 1B

• The speed of a car is 60mph. This falls by 15
  as the traffic builds up. What speed is it
  travelling at now?

51 mph
(b) As the congestion eases the speed increases
again by 16. Find the new speed to the
nearest mph.
59 mph
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76
Question 1B(a)
1. Find multiplier
The speed of a car is 60mph. This falls by 15
as the traffic builds up. What speed is it
travelling at now?
15 off leaves 85 or 0.85.
2. Use this multiplier to calculate new speed
New speed 0.85 x 60mph
51mph
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77
Question 1B(b)
1. Find multiplier
As the congestion eases the speed increases
again by 16. Find the new speed to the nearest
mph.
16 more is 116 or 1.16.
2. Use this multiplier to calculate new
speed, using your answer to (a).
New speed 1.16 x 51mph
59.16mph
59mph
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78
Comments
Question 1B(a)
1. Find multiplier
15 off leaves 85 or 0.85.
Finding a Percentage 12.8 of 12500
2. Use this multiplier to calculate new speed
12500
New speed 0.85 x 60mph
or 0.128 x 12500 (Preferred method)
51mph
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79
Comments
Percentage Increase 10 increase Multiplier
1.10 5.6 increase Multiplier 1.056
Question 1B(b)
1. Find multiplier
Finding a Percentage 16 of 51
16 more is 116 or 1.16.
2. Use this multiplier to calculate new speed,
using your answer to (a).
x
51
New speed 1.16 x 51mph
or 0.16 x 63 (Preferred method)
59.16mph
59mph
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80
PERCENTAGES Question 2

• The value of a car falls by 17 per annum. The
  car is worth 14000 when new.
• Find its value after 3 years to 3 significant
  figures.

(b) How many more years will it take until the
value is less than half of its original
value?
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81
PERCENTAGES Question 2

• The value of a car falls by 17 per annum. The
  car is worth 14000 when new.
• Find its value after 3 years to 3 significant
  figures.

(b) How many more years will it take until the
value is less than half of its original
value?
What would you like to do now?
Reveal answers only
1.Make repeated use of multiplication factor
2. For a decrease the factor is less than 1.
Go to full solution
3. Keep multiplying by factor until target is
reached.
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82
PERCENTAGES Question 2

• The value of a car falls by 17 per annum. The
  car is worth 14000 when new.
• Find its value after 3 years to 3 significant
  figures.

8010
(b) How many more years will it take until the
value is less than half of its original
value?
1 more year
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83
Question 2(a)
1. Find multiplier
The value of a car falls by 17 per annum. The
car is worth 14000 when new. Find its value
after 3 years to 3 significant figures.
17 off leaves 83 or 0.83.
2. Use this multiplier repeatedly for number of
years indicated.
Value after 3 years 0.83 x
0.83 x 0.83 x 14000 OR
(0.83)3 x 14000
8005.018
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8010 to 3 sig figs.
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84
Question 2(b)
1. Find target value
How many more years will it take until the
value is less than half of its original value?
Half of original is 7000.
2. Keep multiplying answer to (a) until value is
below 7000.
Value after 4 years 0.83 x 8005.018
6644.16
Below target
Hence time required is 1 more year.
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85
Comments

• Calculating compound
• Appreciation and Depreciation

Question 2(a)
Calculation can be dealt with year by
year YR 0 Value 14000 1 Value
0.83 x 14000 11620 2 Value 0.83 x
11620 9644.6 3 Value 0.83 x 9644.6

8005.018 Value After 3 years 8010
(correct to 3 sig. fig.)
1. Find multiplier
17 off leaves 83 or 0.83.
2. Use this multiplier repeatedly for number of
years indicated.
Value after 3 years 0.83 x
0.83 x 0.83 x 14000 OR
(0.83)3 x 14000
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8005.018
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86
Comments

• Significant Figures
• (Large Numbers)

Question 2(a)
1. Find multiplier
4567100 rounded to 3 sig figs
17 off leaves 83 or 0.83.
3rd sig. fig.
1st sig. fig.
2. Use this multiplier repeatedly for number of
years indicated.

Hence 456 7100
4570000
Value after 3 years 0.83 x
0.83 x 0.83 x 14000 OR
(0.83)3 x 14000
More than 5000
Answer must be of the same order as the original
number
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8005.018
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87
PERCENTAGES Question 2B

• The value of a PC falls by 21 per annum. The PC
  costs 1200 when new.
• Find its value after 4 years to 2significant
  figures.

(b) A company replaces its PCs when their value
is less than one quarter of the original
value. How much longer will they keep
this machine?
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88
PERCENTAGES Question 2B

• The value of a PC falls by 21 per annum. The PC
  costs 1200 when new.
• Find its value after 4 years to 2significant
  figures.

(b) A company replaces its PCs when their value
is less than one quarter of the original
value. How much longer will they keep
this machine?
What would you like to do now?
1.Make repeated use of multiplication factor
2. For a decrease the factor is less than 1.
Reveal answers only
3. Keep multiplying by factor until target is
reached.
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89
PERCENTAGES Question 2B

• The value of a PC falls by 21 per annum. The PC
  costs 1200 when new.
• Find its value after 4 years to 2significant
  figures.

470
(b) A company replaces its PCs when their value
is less than one quarter of the original
value. How much longer will they keep
this machine?
2 more years
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90
Question 2B(a)
1. Find multiplier
The value of a PC falls by 21 per annum. The
PC costs worth 1200 when new. Find its value
after 4 years to 2 significant figures.
21 off leaves 79 or 0.79.
2. Use this multiplier repeatedly for number of
years indicated.
Value after 4 years 0.79 x
0.79 x 0.79 x 1200 OR
(0.79)4 x 1200
467.40.
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470 to 2 sig figs.
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91
Question 2B(b)
1. Find target value
A company replaces its PCs when their value is
less than one quarter of the original value.
How much longer will they keep this machine?
¼ of original value is 300.
2. Keep multiplying answer to (a) until value is
below 300.
Value after 5 years 0.79 x 467.40
369.25
Above target
Value after 6 years 0.79 x 369.25
Below target
291.71
Continue Solution
Hence PC is kept for 2 more years.
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92
Comments

• Calculating compound
• Appreciation and Depreciation

Question 2B(a)
1. Find multiplier
Calculation can be dealt with year by
year YR 0 Value 1200 1 Value
0.79 x 1200 948 2 Value 0.79 x 948
748.92 3 Value 0.79 x 748.92 591.64
4
Value 0.79 x 591.64 467.40. Value After 4
years 470 (correct to
3 sig. fig.)
21 off leaves 79 or 0.79.
2. Use this multiplier repeatedly for number of
years indicated.
Value after 4 years 0.79 x
0.79 x 0.79 x 1200 OR
(0.79)4 x 1200
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467.40.
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470 to 2 sig figs.
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93
Comments

• Significant Figures
• (Large Numbers)

Question 2B(a)
1. Find multiplier
5632800 rounded to 3 sig figs
21 off leaves 79 or 0.79.
3rd sig. fig.
1st sig. fig.
2. Use this multiplier repeatedly for number of
years indicated.

Hence 563 2800
5630000
Value after 4 years 0.79 x
0.79 x 0.79 x 1200 OR
(0.79)4 x 1200
Less than 5000
Answer must be of the same order as the original
number
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94
PERCENTAGES Question 3
Financial experts predict that house prices are
about to rise by 8.5 per annum for the next few
years. A house is currently valued at 93000.
How much is it likely to be worth in 3 years time
if this rate holds? Give the answer to 2
significant figures!
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95
PERCENTAGES Question 3
Financial experts predict that house prices are
about to rise by 8.5 per annum for the next few
years. A house is currently valued at 93000.
How much is it likely to be worth in 3 years time
if this rate holds? Give the answer to 2
significant figures!
What would you like to do now?
Reveal answers only
1. Use multiplication factor repeatedly
2. For increases factor is more than 1
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96
PERCENTAGES Question 3
Financial experts predict that house prices are
about to rise by 8.5 per annum for the next few
years. A house is currently valued at 93000.
How much is it likely to be worth in 3 years time
if this rate holds? Give the answer to 2
significant figures!
120 000
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97
Question 3
1. Find multiplier
Financial experts predict that house prices are
about to rise by 8.5 per annum for the next few
years. A house is currently valued at 93000.
How much is it likely to be worth in 3 years
time if this rate holds? Give the answer to 2
significant figures!
8.5 more gives 108.5 or 1.085.
2. Use this multiplier repeatedly for number of
years indicated.
Value in 3 years 1.085 x 1.085 x
1.085 x 93000 OR (1.085)3 x
93000
118787.88
Continue Solution
120 000 to 2 sig figs.
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98
Comments

• Calculating compound
• Appreciation and Depreciation

1. Find multiplier
Calculation can be dealt with year by
year YR 0 Value 93 000 1 Value 1.085
x 93000 100905 2 Value 1.085 x 100905
109481.92 3
Value 1.085 x 1909481.92..
118787.88
Value After 3
years 120 000
(correct to 2 sig. fig.)
8.5 more gives 108.5 or 1.085.
2. Use this multiplier repeatedly for number of
years indicated.
Value in 3 years 1.085 x 1.085 x 1.085
x 93000 OR
(1.085)3 x 93000
118787.88
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99
Comments

• Significant Figures
• (Large Numbers)

1. Find multiplier
118787.88 rounded to 2 sig figs
8.5 more gives 108.5 or 1.085.
2nd sig. fig.
2. Use this multiplier repeatedly for number of
years indicated.
1st sig. fig.

Hence 11 8787.88
120 000
Value in 3 years 1.085 x 1.085 x 1.085
x 93000 OR
(1.085)3 x 93000
More than 5000
Answer must be of the same order as the original
number
118787.88
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100
PERCENTAGES Question 3B
It has been predicted that the population of a
town will rise by 7.3 per annum for the next
few years. The population is currently 32000.
How big will this become in 4 years time if this
rate stays the same? Give the answer to 3
significant figures!
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101
PERCENTAGES Question 3B
It has been predicted that the population of a
town will rise by 7.3 per annum for the next
few years. The population is currently 32000.
How big will this become in 4 years time if this
rate stays the same? Give the answer to 3
significant figures!
What would you like to do now?
Reveal answers only
1. Use multiplication factor repeatedly
2. For increases factor is more than 1
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102
PERCENTAGES Question 3B
It has been predicted that the population of a
town will rise by 7.3 per annum for the next
few years. The population is currently 32000.
How big will this become in 4 years time if this
rate stays the same? Give the answer to 3
significant figures!
42 400
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103
Question 3B
1. Find multiplier
It has been predicted that the population of a
town will rise by 7.3 per annum for the next few
years. The population is currently 32000. How
big will this become in 4 years time if this rate
stays the same? Give the answer to 3 significant
figures!
7.3 more gives 107.3 or 1.073.
2. Use this multiplier repeatedly for number of
years indicated.
Value in 4 years 1.073 x 1.073 x 1.073 x
1.073 x 32000 OR (1.073)4 x 32000
42 417.8
Continue Solution
42 400 to 3 sig figs.
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104
Comments

• Calculating compound
• Appreciation and Depreciation

1. Find multiplier
Calculation can be dealt with year by year YR 0
Value 32 000 1 Value 1.073 x 32000
34336 2 Value 1.073 x 34336
36842.52.. 3 Value
1.073 x 36842.52..
39532.03.. 4 Value 1.073 x
39532.03..
42417.87 Value after 4 years 42 400
(correct to 3 sig. fig.)
7.3 more gives 107.3 or 1.073.
2. Use this multiplier repeatedly for number of
years indicated.
Value in 4 years 1.073 x 1.073 x 1.073 x
1.073x32000 OR (1.073)4 x 32000
42 417.8
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105
Comments

• Significant Figures
• (Large Numbers)

1. Find multiplier
118787.88 rounded to 2 sig figs
7.3 more gives 107.3 or 1.073.
2nd sig. fig.
2. Use this multiplier repeatedly for number of
years indicated.
1st sig. fig.

Hence 11 8787.88
120 000
Value in 4 years 1.073 x 1.073 x 1.073 x
1.073 x 32000 OR (1.073)4 x 32000
More than 5000
Answer must be of the same order as the original
number
42 417.8
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106
PERCENTAGES Question 4
A police force announced that its annual crime
figures were 11130. This was a 16 fall on the
previous years figure. What would the previous
years figure have been?
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107
PERCENTAGES Question 4
A police force announced that its annual crime
figures were 11130. This was a 16 fall on the
previous years figure. What would the previous
years figure have been?
What would you like to do now?
Reveal answers only
1. For a decrease the factor is less than 1.
2. Express new figure in terms of old using this
factor.
Go to full solution
3. Substitute in known value and solve.
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108
PERCENTAGES Question 4
A police force announced that its annual crime
figures were 11130. This was a 16 fall on the
previous years figure. What would the previous
years figure have been?
13 250
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109
Question 4
1. Find multiplier
A police force announced that its annual crime
figures were 11130. This was a 16 fall on the
previous years figure. What would the previous
years figure have been?
16 fall gives 84 or 0.84.
2. Using this multiplier, express new figure in
terms of old figure.
New figure 0.84 x old figure
3. Substitute known value and solve.
Old figure 11130 ? 0.84
13250
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110
Comments
Alternative approach
1. Find multiplier
After a 16 reduction
16 fall gives 84 or 0.84.
84 represents 11130 Require 100
2. Using this multiplier, express new figure in
terms of old figure.
Using simple proportion
New figure 0.84 x old figure
3. Substitute known value and solve.
Old figure 11130 ? 0.84
13250
Previous figure was 13250
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111
PERCENTAGES Question 4B
After receiving a 6.5 pay rise a nurse now earns
15336 per annum. What was her previous income?
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112
PERCENTAGES Question 4B
After receiving a 6.5 pay rise a nurse now earns
15336 per annum. What was her previous income?
What would you like to do now?
Reveal answers only
1. For an increase the factor is more than 1.
2. Express new figure in terms of old using this
factor.
Go to full solution
3. Substitute in known value and solve.
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113
PERCENTAGES Question 4B
After receiving a 6.5 pay rise a nurse now earns
15336 per annum. What was her previous income?
14 400
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114
Question 4B
1. Find multiplier
After receiving a 6.5 pay rise a nurse now earns
15336 per annum. What was her previous income?
6.5 increase gives 106.5 or 1.065.
2. Using this multiplier, express new figure in
terms of old figure.
New figure 1.065 x old figure
3. Substitute known value and solve.
Old figure 15 336 ? 1.065
14 400
Continue Solution
Previous income was 14 400
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115
Comments
Alternative approach
1. Find multiplier
After a 6.5 increase
6.5 increase gives 106.5 or 1.065.
106.5 represents 15336 Require 100
2. Using this multiplier, express new figure in
terms of old figure.
Using simple proportion
New figure 1.065 x old figure
3. Substitute known value and solve.
Old figure 15 336 ? 1.065
14 400
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116
PERCENTAGES Question 5
Between 2002 2003 the grain production for a
particular country rose from 4.3million tonnes to
5.1million tonnes. What percentage increase is
this?
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117
PERCENTAGES Question 5
Between 2002 2003 the grain production for a
particular country rose from 4.3million tonnes to
5.1million tonnes. What percentage increase is
this?
What would you like to do now?
Reveal answers only
1. Find the actual increase.
2. Make this a fraction of the original and
convert to a decimal.
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118
PERCENTAGES Question 5
Between 2002 2003 the grain production for a
particular country rose from 4.3million tonnes to
5.1million tonnes. What percentage increase is
this?
18.6
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119
Question 5
1. Find actual increase
5.1m 4.3m 0.8m
Between 2002 2003 the grain production for
a particular country rose from 4.3million tonnes
to 5.1million tonnes. What percentage increase
is this?
2. Make this a fraction of the original and
convert to a decimal.
0.1860
0.8/4.3 0.8 ? 4.3
0.186 to 3 sig figs
3. Multiply by 100 to change to .
0.186 x 100 18.6
Continue Solution
So increase is 18.6
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120
Comments
Using formula to express a as a percentage of
b
1. Find actual increase
5.1m 4.3m 0.8m
x 100
2. Make this a frac