SQA National 5 Mathematics
2015 Paper
Worked Solutions Paper One
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and reproduced with permission from SQA
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2015 SQA N5 Past Paper Worked Solutions
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[2]
2015 SQA N5 Past Paper Worked Solutions
1
6 −2
1.
5
31
Change fractions to top heavy:
3
5
Multiply fractions by � � & � �:
3
5
5
3
� �×
3
Denominators now the same:
−
1
3
7
3
31
− ×� �
93
−
5
15
7
3
5
5
35
15
58
Subtract the numerators:
15
Alternative Method
Sometimes known as the ‘Smile & Kiss’ method:
• Multiply the 5 × 3 and place the 15 on the denominator
• Multiply 3 × 31 and place the 93 on the numerator (top left)
• Multiply 5 × 7 and place the 35 on the numerator (top right)
• Subtract the numerators as shown below:
•
31
5
7
− =
3
93 − 35
15
=
58
15
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[3]
(1 mark)
(1 mark)
2015 SQA N5 Past Paper Worked Solutions
11 − 2(1 + 3𝑥𝑥) < 39
2.
11 − 2 − 6𝑥𝑥 < 39
Multiply out the brackets:
(1 mark)
9 − 6𝑥𝑥 < 39
Simplify:
9 to the RHS (remember to change sign):
−6𝑥𝑥 < 39 − 9
Simplify:
Multiply by −1 (change < to >):
−6𝑥𝑥 < 30
(1 mark)
𝑥𝑥 > −5
(1 mark)
6𝑥𝑥 > −30
Solve for 𝑥𝑥:
Alternative Method
As shown above in line 6, multiplying by −1 changes the direction of the
inequality. A method which avoids this step is to place the −6𝑥𝑥 term on the
RHS of the equation and then solve as shown below:
Simplify:
11 − 2 − 6𝑥𝑥 < 39
Take −6𝑥𝑥 to the RHS:
Take 39 to the LHS:
Simplify:
Swap sides:
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9 − 6𝑥𝑥 < 39
9 < 39 + 6𝑥𝑥
−39 + 9 < 6𝑥𝑥
−5 < 𝑥𝑥
𝑥𝑥 > −5
[4]
2015 SQA N5 Past Paper Worked Solutions
3.
Copyright © Scottish Qualifications Authority
Since angle ABO is a right angle then angle OBD = 13°
(1 mark)
Since triangle BDO is isosceles then angle ODB = 13°
Since angle DFE is 90° then angle EDF = 26°
(1 mark)
Angle BDF = 13° + 26° = 39°
(1 mark)
Notes
• The angle at F is right angled (90°)
• Triangle BDO is isosceles (2 equal lengths and angles)
• Two lines at a tangent means that they are at right angles (90°)
• Angles in a triangle add up to 180°
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[5]
2015 SQA N5 Past Paper Worked Solutions
4.
Multiply out the brackets:
(𝑥𝑥 − 4)(𝑥𝑥 2 + 𝑥𝑥 − 2)
𝑥𝑥(𝑥𝑥 2 + 𝑥𝑥 − 2) − 4(𝑥𝑥 2 + 𝑥𝑥 − 2)
Multiply out the brackets again: 𝑥𝑥 3 + 𝑥𝑥 2 − 2𝑥𝑥 − 4𝑥𝑥 2 − 4𝑥𝑥 + 8
Simplify:
𝑥𝑥 3 − 3𝑥𝑥 2 − 6𝑥𝑥 + 8
(2 marks)
(1 mark)
Notes
• In lines 2 to 3 above a ‘negative × negative = positive’ (−4 × −2 = 8)
• No extra marks are awarded for including line 2 above and therefore this
line can be missed out. However, only miss this line out if you are
confident with going from lines 1 to 3 without making a mistake.
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[6]
2015 SQA N5 Past Paper Worked Solutions
5. Mean (average): 𝑥𝑥̅ =
1+2+2+2+8
5
=
15
5
=3
Draw, and fill out a table, with 3 columns & 6 rows as shown below:
𝑥𝑥
1
(𝑥𝑥 − 𝑥𝑥̅ )
-2
(𝑥𝑥 − 𝑥𝑥̅ )2
2
-1
1
2
-1
1
2
1
1
8
5
25
4
∑(𝑥𝑥 − 𝑥𝑥 2 ) =
From the table:
From the question:
Therefore:
𝑠𝑠𝑠𝑠 = �
∑(𝑥𝑥−𝑥𝑥 2 )
𝑛𝑛−1
=�
32
5−1
= �
32
4
32
(1 mark)
= √8
(1 mark)
√𝑎𝑎 = √8
𝑎𝑎 = 8
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(1 mark)
[7]
2015 SQA N5 Past Paper Worked Solutions
6.
𝑦𝑦 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑎𝑎 = 4
(1 mark)
𝑏𝑏 = 3
(1 mark)
Notes
In the equation 𝑦𝑦 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎:
• 𝑎𝑎 is the height from the centre to the peak of the sine wave (amplitude)
• 𝑏𝑏 is the number of complete cycles in 360°
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[8]
2015 SQA N5 Past Paper Worked Solutions
𝑥𝑥 = 2
7.
𝑦𝑦 = (𝑥𝑥 + 𝑎𝑎)2 + 𝑏𝑏
(a) (i) 𝑎𝑎 = − 2
(1 mark)
𝑏𝑏 = − 4
(ii)
(1 mark)
𝑥𝑥 = 2
(b)
(1 mark)
Notes
• The 𝑥𝑥 on the turning point (2, − 4) corresponds to 𝑎𝑎 in
𝑦𝑦 = (𝑥𝑥 + 𝑎𝑎)2 + 𝑏𝑏 and is always the opposite sign: 𝑎𝑎 = −𝑥𝑥 = −2
• The 𝑦𝑦 on the turning point (2, − 4) corresponds to 𝑏𝑏 in
𝑦𝑦 = (𝑥𝑥 + 𝑎𝑎)2 + 𝑏𝑏 and is always the same sign: 𝑏𝑏 = 𝑦𝑦 = −4
• The axis of symmetry, 𝑥𝑥 = 2 is the dotted line shown in the above
diagram
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[9]
2015 SQA N5 Past Paper Worked Solutions
(−2, 5) & (3, 15)
8.
(𝑥𝑥1, 𝑦𝑦1) & (𝑥𝑥2, 𝑦𝑦2)
Compare with:
Substitute 𝑥𝑥1 = −2, 𝑦𝑦1 = 5, 𝑥𝑥2 = 3, 𝑦𝑦2 = 15 into the gradient formula:
𝑚𝑚 =
𝑦𝑦2 − 𝑦𝑦1
𝑥𝑥2 − 𝑥𝑥1
=
15 − 5
3 − (−2)
=
10
5
=2
(1 mark)
Take (3, 15), let 𝑎𝑎 = 3, 𝑏𝑏 = 15 and substitute 𝑎𝑎, 𝑏𝑏 & 𝑚𝑚 into:
Substitute 𝑎𝑎, 𝑏𝑏 & 𝑚𝑚:
Multiply out the brackets:
Take − 15 to the RHS:
Simplify:
𝑦𝑦 − 𝑏𝑏 = 𝑚𝑚(𝑥𝑥 − 𝑎𝑎)
𝑦𝑦 − 15 = 2(𝑥𝑥 − 3)
(1 mark)
𝑦𝑦 − 15 = 2𝑥𝑥 − 6
𝑦𝑦 = 2𝑥𝑥 − 6 + 15
𝑦𝑦 = 2𝑥𝑥 + 9
(1 mark)
Notes
Substituting the other set of coordinates (−2, 5) into 𝑦𝑦 − 𝑏𝑏 = 𝑚𝑚(𝑥𝑥 − 𝑎𝑎)
would have resulted in the same answer as shown below:
Substitute 𝑎𝑎 = −2 & 𝑏𝑏 = 5:
𝑦𝑦 − 5 = 2(𝑥𝑥 − (−2))
Multiply out the brackets:
𝑦𝑦 − 5 = 2𝑥𝑥 + 4
Simplify:
− 5 to the RHS (change sign):
Simplify:
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𝑦𝑦 − 5 = 2(𝑥𝑥 + 2)
𝑦𝑦 = 2𝑥𝑥 + 4 + 5
𝑦𝑦 = 2𝑥𝑥 + 9
[10]
2015 SQA N5 Past Paper Worked Solutions
9.
Correct order is: cos 100°, cos 90°, cos 300°
(1 mark)
y = cos x
In the above graph of 𝑦𝑦 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐, it can be seen that:
• Cos 100° on the 𝑥𝑥 axis corresponds to a negative value on the 𝑦𝑦 axis
• Cos 90° on the 𝑥𝑥 axis corresponds to zero on the 𝑦𝑦 axis
• Cos 300° on the 𝑥𝑥 axis corresponds to a positive value on the 𝑦𝑦 axis
(1 mark)
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[11]
2015 SQA N5 Past Paper Worked Solutions
10. (a) Writing the list in order from the smallest to the largest number:
12
16
17
18
Finding 𝑸𝑸𝑸𝑸
Median (middle number): 𝑄𝑄2 =
Finding 𝑸𝑸𝑸𝑸 & 𝑸𝑸𝑸𝑸
There are 10 numbers in the list:
18
18 + 21
2
10
4
=
21
22
39
= 19.5
2
27
27
(1 mark)
= 2 remainder 2.
This means that there will be 4 lots of 2 with 2 remaining.
The list is therefore grouped as below:
• (12, 16), 𝟏𝟏𝟏𝟏, (18, 18), (21, 22), 𝟐𝟐𝟐𝟐, (27, 27)
From the above: 𝑄𝑄1 = 17 & 𝑄𝑄3 = 26
SIQR =
𝑄𝑄3 − 𝑄𝑄1
2
=
26 − 17
2
9
= = 4.5
(2 marks)
2
(b) Compare medians: Round one = 19.5, Round two = 26
On average the second rounds scores were higher
Compare SIQRs: Round one = 4.5, Round two = 2.5
The second rounds scores were more consistent
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[12]
(1 mark)
(1 mark)
27
2015 SQA N5 Past Paper Worked Solutions
11.
3𝑥𝑥 + 2𝑦𝑦 = 17
2𝑥𝑥 + 5𝑦𝑦 = 4
15𝑥𝑥 + 10𝑦𝑦 = 85
(3) − (4)
4𝑥𝑥 + 10𝑦𝑦 = 8
11𝑥𝑥
--- (1) × 5 to give (3) below:
--- (2) × 2 to give (4) below:
(1 mark)
--- (3)
--- (4)
= 77
𝑥𝑥 =
77
11
𝑥𝑥 = 7
Substitute 𝑥𝑥 = 7 into equation (1):
3𝑥𝑥 + 2𝑦𝑦 = 17 --- (1)
Take 21 to the RHS:
3 × 7 + 2𝑦𝑦 = 17
Simplify:
Divide by 2:
Simplify:
Solution is 𝑥𝑥 = 7, 𝑦𝑦 = −2
2𝑦𝑦 = 17 − 21
2𝑦𝑦 = −4
𝑦𝑦 =
−4
2
𝑦𝑦 = −2
(2 marks)
Notes
Decide whether the 𝑥𝑥 or 𝑦𝑦 coefficient should be scaled in the 2 equations it does not matter which is decided upon. In the above the 𝑦𝑦 coefficients
have been scaled to have same value.
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[13]
2015 SQA N5 Past Paper Worked Solutions
𝑥𝑥2 − 4𝑥𝑥
𝑥𝑥2 + 𝑥𝑥 − 20
12.
𝑥𝑥(𝑥𝑥 − 4)
Factorise both the numerator & denominator:
(𝑥𝑥 − 4)(𝑥𝑥 + 5)
Cancel the (𝑥𝑥 − 4) from the top & bottom of fraction:
(2 marks)
𝑥𝑥
𝑥𝑥 + 5
(1 mark)
Notes
Factorising 𝒙𝒙𝟐𝟐 − 𝟒𝟒𝟒𝟒:
• 𝑥𝑥 is common to both 𝑥𝑥 2 and − 4𝑥𝑥 therefore take 𝑥𝑥 out to give: 𝑥𝑥(𝑥𝑥 − 4)
Factorising 𝒙𝒙𝟐𝟐 + 𝒙𝒙 − 𝟐𝟐𝟐𝟐:
• Write down the trinomial: 𝑥𝑥 2 + 𝑥𝑥 − 20
• Find two numbers which multiply to give – 20 and add to give 1
• The two numbers are −4 & 5
• Put these numbers into two brackets: (𝑥𝑥 − 4)(𝑥𝑥 + 5)
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[14]
2015 SQA N5 Past Paper Worked Solutions
4
13.
Multiply by
√8
√8
:
√8
4
×
8
√
4√8
Simplify (√8 × √8 = 8):
√8
√8
(1 mark)
8
4×√4×√2
8
Substitute √8 for √4 × √2:
(1 mark)
8√2
Simplify (√4 = 2):
8
√2
Simplify again:
Notes
(1 mark)
• Rationalising the denominator means changing √8 on the denominator
to a whole number
• This is achieved by multiplying the original fraction �
leave 8 on the bottom (√8 × √8 = 8)
• Simplify to leave as √2
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[15]
�8
4
� by � � to
�8
�8
2015 SQA N5 Past Paper Worked Solutions
𝑚𝑚
𝑛𝑛
𝑥𝑥 𝑛𝑛 = � √𝑥𝑥�
14. Using the indice law:
5
3
83 = �√8�
𝑚𝑚
5
(1 mark)
= (2)5
(1 mark)
= 32
Notes
5
3
• �√8� means the cubed root of 8 to the power of 5
• Cubed root of 8 𝑖𝑖𝑖𝑖 2 (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 8 = 2 × 2 × 2 )
• 2 to the power of 5 𝑖𝑖𝑖𝑖 32 (since 2 × 2 × 2 × 2 × 2 = 32)
• Laws of Indices:
𝑥𝑥 𝑚𝑚 × 𝑥𝑥 𝑛𝑛 = 𝑥𝑥 𝑚𝑚+𝑛𝑛 ,
1
𝑚𝑚
𝑥𝑥 𝑚𝑚
𝑥𝑥 𝑛𝑛
𝑛𝑛
= 𝑥𝑥 𝑚𝑚−𝑛𝑛 , 𝑥𝑥 0 = 1, 𝑥𝑥 −𝑚𝑚 =
𝑚𝑚
𝑥𝑥 2 = √𝑥𝑥, 𝑥𝑥 𝑛𝑛 = � √𝑥𝑥 � , (𝑥𝑥 𝑚𝑚 )𝑛𝑛 = 𝑥𝑥 𝑚𝑚𝑚𝑚
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[16]
1
𝑥𝑥 𝑚𝑚
, 𝑥𝑥 𝑚𝑚 =
1
𝑥𝑥 −𝑚𝑚
,
2015 SQA N5 Past Paper Worked Solutions
SQA National 5 Mathematics
2015 Paper
Worked Solutions
Paper Two: Calculator
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[17]
2015 SQA N5 Past Paper Worked Solutions
1. Final Amount = Initial Value × �
Predicted Value = 240 000 × �
100 ± % n
100
� , n = number of years
100 + 2.8 2
�
100
Predicted Value = £253,628.16
(2 marks)
(1 mark)
Alternative Method (Longer)
Year One: Predicted Value = 240000 +
2.8
× 240000 = £246,720
100
Year Two: Predicted Value = 246720 +
2.8
× 246720 = £253,618.16
100
Note
Only use the alternative method if you are not confident with the formula
method as the alternative method can be time consuming.
End of Sample Worked Solutions
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[18]
2015 SQA N5 Past Paper Worked Solutions
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[19]
2015 SQA N5 Past Paper Worked Solutions
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[20]
2015 SQA N5 Past Paper Worked Solutions
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[21]
2015 SQA N5 Past Paper Worked Solutions
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[22]