Mathematics GCSE Higher Tier 2011
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Question 7 (June 2011 – 1380/3H)
a)
b) this is a vector translation as the shape P is simply picked and moved so many places horizontally
and so many places vertically. It doesn’t turn or change size.
It has moved 6 places to the left (-6) and 1 place down (-1)
Vector translation www.chattertontuition.co.uk 0775 950 1629
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Mathematics GCSE Higher Tier 2011
1380 Taster Pages
Question 18 (June 2011 – 1380/3H)
a) the modal class interval is the one with the highest frequency
modal class interval is 90 m 100
b)
Time (m minutes)
70 m 80
70 m 90
70 m 100
70 m 110
70 m 120
70 m 130
Cumulative Frequency
4
16
50
82
108
120
Check that the final cumulative frequency is the same as the total frequency – which it is.
c)
To find the median draw a horizontal line across from a cumulative frequency of 60 to meet the
curve. Where this line meets the curve drop it down to meet the time axis. This will be the median.
Median = 103 minutes
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Mathematics GCSE Higher Tier 2011
1380 Taster Pages
Question 19 (June 2011 – 1380/3H)
4x + y = 10 equation 1
2x – 3y = 19 equation 2
We can solve by elimination or by substitution
Elimination method
We need to have either the same number of x or the same number of y
Multiply equation 1 by 3 so that we end up with 3y
12x + 3y = 30 equation 3
2x – 3y = 19 equation 2 (unchanged)
Add the two equations together to make the y terms disappear
14x = 49
Divide both sides by 14
x = = = 3.5
Now we have x we can put this back into equation 1 to get y
(4 x 3.5) + y = 10
14 + y = 10
Subtract 14 from both sides
y = -4
x = 3.5, y = -4
Now check the answers by putting both back into equation 2
(2 x 3.5) – (3 x -4) = 7 - -12 = 7 + 12 = 19
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Mathematics GCSE Higher Tier 2011
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Substitution Method
We can easily rearrange equation 1 to get y as the subject
y = 10 – 4x
Now substitute this into the second equation replacing the y with 10 – 4x
2x – (3 x (10 – 4x)) = 19
2x – (30 – 12x) = 19
2x – 30 + 12x = 19
Grouping terms
14x – 30 = 19
Add 30 to both sides
14x = 49
Divide both sides by 14
x = = = 3.5
Now we have x we can put this back into the rearranged equation 1 to get y
Y = 10 - (4 x 3.5)
Y = 10 - 14
y = -4
x = 3.5, y = -4
Now check the answers by putting both back into equation 2
(2 x 3.5) – (3 x -4) = 7 - -12 = 7 + 12 = 19
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Mathematics GCSE Higher Tier 2011
1380 Taster Pages
Question 3 (June 2011 – 1380/4H)
a)
45
15
3
5
3
so 45 = 3 x 3 x 5 = 32 x 5
b) we can do the same to 30
30
10
3
5
23
30 = 2 x 3 x 5
45 = 3 x 3 x 5
Common factors are 3 and 5
3 x 5 = 15
HCF = 15
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Mathematics GCSE Higher Tier 2011
1380 Taster Pages
Question 24 (June 2011 – 1380/4H)
1st pen
2nd pen
4
9
red
3
9
red
5
10
2
9
3
10
green
5
9
blue
green
5
9
green
red
2
9
blue
2
9
2
10
blue
red
3
9
1
9
blue
green
Note that Gary is not replacing the first pen so the probabilities change for the second pen
Prob (red, red) = x = (normally I would cancel down at this stage but we are going to have to
add other fractions so best not to simplify yet)
Prob (blue, blue) = x = Prob (green, green) =
x =
Prob (same colour) = + + = = www.chattertontuition.co.uk 0775 950 1629
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Mathematics GCSE Higher Tier 2011
1380 Taster Pages
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