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National 5 Mathematics

Practice Paper E

Worked Solutions

Paper One: Non-Calculator

N5 Mathematics Practice Paper E Worked Solutions

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1. 2 13

+ 56

× 1 25

Change 1st & 3rd fraction to improper (top heavy): 73

+ 56

× 75

Cancel diagonally the 5′𝑠𝑠 to give 1′𝑠𝑠: 73

+ 16

× 71

BIDMAS (calculate multiply before add): 7 3

+ 76

Multiply the first fraction by 22 : 2

2× 7

3+ 7

6

Denominators now the same: 14 6

+ 76

Add numerators, denominators stay same: 21 6

Change to a mixed number: 3 1

2

Alternative Method

With reference to line 4 above (sometimes known as the ‘Smile & Kiss’

method):

• Multiply the 3 × 6 and place 18 on the denominator

• Multiply 6 × 7 and place 42 on the numerator (top left)

• Multiply 3 × 7 and place 21 on the numerator (top right) as below:

• 7 3

+ 76

= 42 − 2118

= 2118

= 73

= 3 12



2. (4𝑥𝑥 + 2)(𝑥𝑥 − 5) + 3𝑥𝑥 Multiply out the brackets: 4𝑥𝑥(𝑥𝑥 − 5) + 2(𝑥𝑥 − 5) + 3𝑥𝑥 Multiply out the brackets again: 4𝑥𝑥2 − 20𝑥𝑥 + 2𝑥𝑥 − 10 + 3𝑥𝑥 Simplify: 4𝑥𝑥2 − 15𝑥𝑥 − 10 3. Taking two sets of points on the line: (1, 4) & (2, 2)

(𝑥𝑥1,𝑦𝑦1) & ( 𝑥𝑥2,𝑦𝑦2)

Substitute 𝑥𝑥1 = 1, 𝑦𝑦1 = 4, 𝑥𝑥2 = 2,𝑦𝑦2 = 2 into the gradient formula:

𝑚𝑚 = 𝑦𝑦2 − 𝑦𝑦1𝑥𝑥2 − 𝑥𝑥1

= 2 − 42 − 1

= −21

= −2

Take two points on the line (1, 4), let 𝑎𝑎 = 1, 𝑏𝑏 = 4. Substitute 𝑎𝑎, 𝑏𝑏 & 𝑚𝑚 into: 𝑦𝑦 − 𝑏𝑏 = 𝑚𝑚(𝑥𝑥 − 𝑎𝑎) 𝑦𝑦 − 4 = −2(𝑥𝑥 − 1) Multiply out the brackets: 𝑦𝑦 − 4 = −2𝑥𝑥 + 2

Take − 4 to the RHS: 𝑦𝑦 = −2𝑥𝑥 + 2 + 4 Simplify: 𝑦𝑦 = −2𝑥𝑥 + 6

Alternative Method

Plot the points from the table onto a graph, find the 𝑦𝑦 intercept (𝑐𝑐) and

substitute 𝑚𝑚 & 𝑐𝑐 into 𝑦𝑦 = 𝑚𝑚𝑥𝑥 + 𝑐𝑐 to obtain the same answer.



4. 2𝑥𝑥

+ 9 = 16

Multiply by 𝑥𝑥: 2 + 9𝑥𝑥 = 16𝑥𝑥

Take 16𝑥𝑥 to the LHS & 2 to the RHS: −16𝑥𝑥 + 9𝑥𝑥 = −2

Simplify: −7𝑥𝑥 = −2

Multiply by −1: 7𝑥𝑥 = 2

Divide by 7: 𝑥𝑥 = 27

5. Compare: 2𝑥𝑥2 − 2𝑥𝑥 − 1 = 0

with: 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0

Therefore 𝑎𝑎 = 2, 𝑏𝑏 = −2, 𝑐𝑐 = −1. Substitute 𝑎𝑎, 𝑏𝑏 & 𝑐𝑐 into the quadratic

formula given in the Formulae List: 𝑥𝑥 = −𝑏𝑏 ± √𝑏𝑏2 − 4𝑎𝑎𝑎𝑎2𝑎𝑎

Substituting 𝑎𝑎, 𝑏𝑏 & 𝑐𝑐: 𝑥𝑥 =−(−2 )±�(−2)2− 4 × 2 × (−1)

2×2

Simplify: = 4 ± √4 + 84

Simplify again: = 4 ± √124

Substitute √12 = √4 × √3 = 2√3 into the above: = 4 ± 2√34

Factorise 4 ± 2√3: = 2(1 ± √3 )4

Cancel down the 2 & 4: = 1 ± √32



6.

(a) 𝑦𝑦 = 36 − (𝑥𝑥 − 2)2

Swap positions of 36 & −(𝑥𝑥 − 2)2: = −(𝑥𝑥 − 2)2 + 36

Turning point = (2, 36)

(b) 𝑥𝑥 = 2 is the equation of the axis of symmetry

(c) Since the turning point is (2, 36) and 𝑆𝑆 is (6, 20) then the difference

along the 𝑥𝑥-axis between 𝑆𝑆 and the turning point is 6 − 2 = 4



Due to symmetry, there will also be a difference of 4 between the turning

point and 𝑅𝑅

The 𝑥𝑥 coordinate of 𝑅𝑅 will therefore be 2 − 4 = − 2

The 𝑦𝑦 coordinate of 𝑅𝑅 will be the same as the 𝑦𝑦 coordinate of 𝑆𝑆 since the

line is horizontal, 𝑦𝑦 coordinate of 𝑆𝑆 = 20

Coordinates 𝑅𝑅 (−2, 20)

Notes

• In the equation 𝑦𝑦 = −(𝑥𝑥 − 2)2 + 36. The negative at the front indicates

the parabola is a maximum shape (as opposed to mimimum)

• The first number (−2) in the equation is the opposite sign to the 𝑥𝑥 on the

turning point: 𝑥𝑥 = −(−2) = 2

• The second number in the equation (36) is the same sign to the 𝑦𝑦 on

the turning point: 𝑦𝑦 = 36

• The equation of axis of symmetry is the dotted line above: 𝑥𝑥 = 2



7.

With reference to the triangle at the top right in the above shape:

4 cm

𝑥𝑥 5 cm (radius)

Pythagoras Theorem: 𝑥𝑥2 = 52 − 42

Simplify: 𝑥𝑥2 = 25 − 16

Simply again: 𝑥𝑥2 = 9

Take the square root: 𝑥𝑥 = 9

With reference to the triangle at the bottom left in the above shape:

𝑦𝑦2 = 52 − 42

5 cm 4 cm (from 7 - 3) 𝑦𝑦2 = 25 − 16

𝑦𝑦2 = 9

𝑦𝑦 = 3

Width of base = 2 × 𝑦𝑦 = 2 × 3 = 6 𝑐𝑐𝑚𝑚



8. 𝑦𝑦 = 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑥𝑥0

Notes

• 𝑦𝑦 = 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑥𝑥0 represents a sine wave of amplitude 1 (number of front of

the 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑥𝑥0 )

• 2 (number in front of the 𝑥𝑥) is the number of complete cycles in 360°

9. 𝑓𝑓(𝑥𝑥) = 4√𝑥𝑥 + √2

(a) Substitute 𝑥𝑥 = 72 into the equation: 𝑓𝑓(72) = 4√72 + √2

Swap √72 for √36 × √2 = 4 × √36 × √2 + √2

Simplify √36 = 6: = 4 × 6√2 + √2

Simplify again: = 24√2 + √2

Simplify again: = 25√2



(b) Substitute 𝑥𝑥 for 𝑡𝑡: 𝑓𝑓(𝑡𝑡) = 4√𝑡𝑡 + √2

Substitute 𝑓𝑓(𝑡𝑡) for 3√2: 3√2 = 4√𝑡𝑡 + √2

Swap sides: 4√𝑡𝑡 + √2 = 3√2

Take √2 to the RHS: 4√𝑡𝑡 = 3√2 − √2

Simplify: 4√𝑡𝑡 = 2√2

Divide by 4: √𝑡𝑡 = 2√24

Square both sides: 𝑡𝑡 = 2√24

× 2√24

Multiply numerators & denominators: 𝑡𝑡 = 4 × 216

Simplify: 𝑡𝑡 = 12



10.

(2𝑥𝑥 − 5) 𝑐𝑐𝑚𝑚 Area = 7 𝑐𝑐𝑚𝑚2

2𝑥𝑥 𝑐𝑐𝑚𝑚

Area of a triangle: 𝐴𝐴 = 12

× 𝑏𝑏 × ℎ

Substitute 𝐴𝐴, 𝑏𝑏 & ℎ: 7 = 12

× 2𝑥𝑥 × (2𝑥𝑥 − 5)

Swap sides: 12

× 2𝑥𝑥 × (2𝑥𝑥 − 5) = 7

Cancel out the 2′𝑠𝑠: 𝑥𝑥 × (2𝑥𝑥 − 5) = 7

Multiply out the brackets: 2𝑥𝑥2 − 5𝑥𝑥 = 7

Take all terms to the LHS: 2𝑥𝑥2 − 5𝑥𝑥 − 7 = 0

Factorise: (2𝑥𝑥 − 7)(𝑥𝑥 + 1) = 0

Split the brackets: 2𝑥𝑥 − 7 = 0 or 𝑥𝑥 + 1 = 0

Solve: 2𝑥𝑥 = 7 or 𝑥𝑥 = −1 (discard since −′𝑣𝑣𝑣𝑣)

Divide by 2: 𝑥𝑥 = 72

𝑐𝑐𝑚𝑚



National 5 Mathematics

Practice Paper E

Worked Solutions

Paper Two: Calculator



1. 𝐸𝐸 = 𝑚𝑚𝑐𝑐2

Substitute 𝑚𝑚 & 𝑐𝑐: = 3.6 × 10−2 × (3 × 108)2

Simplify: = 3.6 × 10−2 × 9 × 1016

Simplify again: = 3.24 × 1015

Notes

• Use × 10𝑥𝑥 (or equivalent) button on your calculator.

• When writing the answer in scientific natation, place the decimal point

between the first two numbers.

2.

110°

21cm 19 cm

From the Formulae List: Area of a triangle = 12𝑎𝑎𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠𝑎𝑎

= 12

× 21 × 19 × 𝑠𝑠𝑠𝑠𝑠𝑠110

= 187.5 𝑐𝑐𝑚𝑚2 𝑡𝑡𝑡𝑡 1 𝑑𝑑𝑑𝑑



3. Final Amount = Initial value × �100 ± % 100

�n

,𝑠𝑠 = number of hours

Final Temperature = 28 × �100 − 10.4 100

�3

= 20°C (nearest degree)

Alternative Method

• Hour One: Temperature = 28 − 10.4100

× 28 = 25.088°C

• Hour Two: Temperature = 25.088 − 10.4100

× 22.088 = 22.479 °C

• Hour Three: Temperature = 22.479 − 10.4100

× 22.479 = 20.14°

= 20°C (nearest degree)

Notes

• 𝑠𝑠 = 3 because the difference between 8 pm & 11 pm is 3 hours

• Only use the alternative method if you are not confident with the formula

method as the alternative method can be time consuming.



4.

(a) 𝑎𝑎 = (4, 3, 4), 𝐷𝐷 = (6, 2, 2)

(b) 𝐵𝐵 = (6, 4, 2)

Notes

When determining the coordinates, use A (2, 4, 6) as your reference point

and think how many along 𝑥𝑥, along 𝑦𝑦 and along 𝑧𝑧.

𝑧𝑧

𝑦𝑦

𝑥𝑥



5.

V = 200 ml V = 1600 ml

Volume Scale Factor = 1600200

= 8

Volume Scale Factor = (Linear Scale Factor)³

Substitute 8 into the equation: 8 = (Linear Scale Factor)³

Swap sides: (Linear Scale Factor)3 = 8

Take the cubed root : Linear Scale Factor = √83 = 2

Height of salon bottle = Linear Scale Factor × Height of travel bottle

= 2 × 12 = 24 𝑐𝑐𝑚𝑚

Notes

• Calculating an area use: Area Scale Factor = (Linear Scale Factor)²

• Calculating a volume use: Volume Scale Factor = (Linear Scale Factor)³



6.

In the below, the P coefficients have been scaled to have the same value

although it does not matter which is chosen. Let 𝐵𝐵 = 𝐵𝐵𝑣𝑣𝑎𝑎𝑑𝑑, 𝑃𝑃 = 𝑃𝑃𝑣𝑣𝑎𝑎𝑃𝑃𝑃𝑃:

2𝐵𝐵 + 5𝑃𝑃 = 5.2 --- (1) × 2 to give (3) below:

3𝐵𝐵 + 2𝑃𝑃 = 5.6 --- (2) × 5 to give (4) below:

4𝐵𝐵 + 10𝑃𝑃 = 10.4 --- (3)

15𝐵𝐵 + 10𝑃𝑃 = 28 --- (4)

Equation (4) – (3) 11𝐵𝐵 = 17.6

Divide by 11: 𝐵𝐵 = 1.6

Substitute 𝐵𝐵 = 1.6 into equation (1) above: 2 × 1.6 + 5𝑃𝑃 = 5.2

Simplify: 3.2 + 5𝑃𝑃 = 5.2

Take 3.2 to the RHS: 5𝑃𝑃 = 5.2 − 3.2

Simplify: 5𝑃𝑃 = 2

Divide by 5: 𝑃𝑃 = 0.4

The lengths are 1 bead = 1.6 𝑐𝑐𝑚𝑚 & 1 pearl = 0.4 𝑐𝑐𝑚𝑚



7. (a) Volume of a sphere (Formulae List): V = 43

𝜋𝜋r3

= 43

× 𝜋𝜋 × 0.53

= 0.5235987756

= 0.524 𝑐𝑐𝑚𝑚3 (3 SF)

(b) Volume of a cylinder: V = 𝜋𝜋r2ℎ

0.524 = 𝜋𝜋 × 0.72 × ℎ

Simplify: 0.524 = 1.54 × ℎ

Swap sides: 1.54 × ℎ = 0.524

Divide by 1.54: ℎ = 0.34 𝑐𝑐𝑚𝑚 (2 decimal places)

Notes

• The volume of a cylinder (𝑉𝑉 = 𝜋𝜋𝑃𝑃2ℎ) is not given on the Formulae List

• To help remember the formula, think circle area (𝐴𝐴 = 𝜋𝜋𝑃𝑃2) multiplied by

the height since a cylinder is a prism and 𝑉𝑉𝑑𝑑𝑃𝑃𝑠𝑠𝑠𝑠𝑚𝑚 = 𝐴𝐴 × ℎ



8.

V

50° 66°

5 km

50° 66°

A 24° B

Point V

• 50° is the 𝑧𝑧 − 𝑎𝑎𝑠𝑠𝑎𝑎𝑃𝑃𝑣𝑣 from A

• 66° is the 𝑧𝑧 − 𝑎𝑎𝑠𝑠𝑎𝑎𝑃𝑃𝑣𝑣 from B

Point B

• Bearing of V from B = 294° (from the question)

• 360° − 294° = 66°

• Since B is due east of A: 90°− 66° = 24°

Sine Rule

With two sets of opposites (angles & lengths) and only one unknown length

then the sine rule can be used to find the missing length AB



From the Formulae List: 𝑎𝑎sin𝐴𝐴

= 𝑏𝑏sin𝐵𝐵

= 𝑎𝑎𝑆𝑆𝑆𝑆𝑆𝑆 𝐶𝐶

Substitute values in: 𝐴𝐴𝐵𝐵𝑠𝑠𝑆𝑆𝑆𝑆 116

= 5𝑠𝑠𝑆𝑆𝑆𝑆 24

Multiply by sin 116: 𝐴𝐴𝐵𝐵 = 5 × 𝑠𝑠𝑆𝑆𝑆𝑆116𝑠𝑠𝑆𝑆𝑆𝑆24

𝐴𝐴𝐵𝐵 = 11.05 𝑘𝑘𝑚𝑚 (2 𝑑𝑑𝑑𝑑)

The distance between the two hostels is 11.05 𝑘𝑘𝑚𝑚 correct to 2 decimal

places

Notes The Sine Rule is used to calculate angles & lengths in non-right angled

triangles and requires two sets of opposites (angles & lengths) with only

one unknown. To calculate:

• a length use: 𝑎𝑎

𝑆𝑆𝑆𝑆𝑆𝑆 𝐴𝐴= 𝑏𝑏

𝑆𝑆𝑆𝑆𝑆𝑆 𝐵𝐵= 𝑎𝑎

𝑆𝑆𝑆𝑆𝑆𝑆 𝐶𝐶

• an angle use: 𝑆𝑆𝑆𝑆𝑆𝑆 𝐴𝐴𝑎𝑎

= 𝑆𝑆𝑆𝑆𝑆𝑆 𝐵𝐵𝑏𝑏

= 𝑆𝑆𝑆𝑆𝑆𝑆 𝐶𝐶𝑎𝑎



9.

(a) With reference to the triangle above, use SOH-CAH-TOA to find 𝑥𝑥:

𝑐𝑐𝑡𝑡𝑠𝑠𝑥𝑥 = 11.512.5

Take the inverse cos: 𝑥𝑥 = 𝑐𝑐𝑡𝑡𝑠𝑠−1 �11.512.5

� = 23°

The angle the chain makes with the vertical is approximately 23°

(b) Arc Length = Angle Fraction × 𝜋𝜋𝑑𝑑

= � 46360� × 𝜋𝜋× 25 = 10.0356432

= 10.0 𝑚𝑚 (3 𝑆𝑆𝑆𝑆)

The maximum length of arc through which the end of the chain swings is

10.0 m correct to 3 significant figures

Notes

Double 23° to obtain the 46° in the above as this is the sector angle



10. 𝑘𝑘𝑥𝑥2 − 4𝑥𝑥 + 2 = 0

Real roots means that the discriminant is greater than, or equal, to 0

𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 ≥ 0

Compare 𝑘𝑘𝑥𝑥2 − 4𝑥𝑥 + 2

with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐

therefore: 𝑎𝑎 = 𝑘𝑘, 𝑏𝑏 = −4, 𝑐𝑐 = 2

Substitute 𝑎𝑎, 𝑏𝑏 & 𝑐𝑐 into: 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 ≥ 0 (−4)2 − 4 × 𝑘𝑘 × 2 ≥ 0 Simplify: 16 − 8𝑘𝑘 ≥ 0 Take the 16 to the RHS: −8𝑘𝑘 ≥ −16 Multiply both sides by – 1 (inequality changes direction): 8𝑘𝑘 ≤ 16 Divide by 8: 𝑘𝑘 ≤ 2

Note

With reference to the above, multiplying both sides of the inequality by -1

means that the direction of the inequality sign changes.



11.(a) √3𝑠𝑠𝑠𝑠𝑠𝑠 𝑥𝑥° − 1 = 0 0 ≤ 𝑥𝑥 ≤ 360

Take 1 to the RHS; √3𝑠𝑠𝑠𝑠𝑠𝑠 𝑥𝑥° = 1

Divide by √3: 𝑠𝑠𝑠𝑠𝑠𝑠 𝑥𝑥° = 1√3

Take the inverse sign: 𝑥𝑥 = 𝑠𝑠𝑠𝑠𝑠𝑠−1 � 1√3�

= 35.3°

For second angle: Consider the All-Sine-Tan-Cos diagram shown below Since Sin, then the two angles are in Quadrants 1 & 2 Quadrant 2 (below) shows: 2nd Angle = 180° − 𝑥𝑥 = 180 − 35.3 = 𝟏𝟏𝟏𝟏𝟏𝟏.𝟕𝟕° 𝑥𝑥 = 𝟑𝟑𝟑𝟑.𝟑𝟑° & 𝟏𝟏𝟏𝟏𝟏𝟏.𝟕𝟕° correct to 1 decimal place



(b) 𝑡𝑡𝑎𝑎𝑠𝑠𝑥𝑥°𝑐𝑐𝑡𝑡𝑠𝑠𝑥𝑥°

Substitute 𝑡𝑡𝑎𝑎𝑠𝑠𝑥𝑥° = 𝑠𝑠𝑆𝑆𝑆𝑆𝑥𝑥°𝑎𝑎𝑐𝑐𝑠𝑠𝑥𝑥°

into the above: 𝑠𝑠𝑆𝑆𝑆𝑆𝑥𝑥°𝑎𝑎𝑐𝑐𝑠𝑠𝑥𝑥°

× 𝑐𝑐𝑡𝑡𝑠𝑠𝑥𝑥°

Cancel out the 𝑐𝑐𝑡𝑡𝑠𝑠𝑥𝑥°: 𝑠𝑠𝑠𝑠𝑠𝑠 𝑥𝑥°

Therefore: 𝑡𝑡𝑎𝑎𝑠𝑠𝑥𝑥°𝑐𝑐𝑡𝑡𝑠𝑠𝑥𝑥° = 𝑠𝑠𝑠𝑠𝑠𝑠𝑥𝑥°

Notes

The two trig identities to remember (not given in the Formulae List) are:

• 𝑡𝑡𝑎𝑎𝑠𝑠𝑥𝑥 = 𝑠𝑠𝑆𝑆𝑆𝑆 𝑥𝑥𝑎𝑎𝑐𝑐𝑠𝑠 𝑥𝑥

• 𝑠𝑠𝑠𝑠𝑠𝑠2 + 𝑐𝑐𝑡𝑡𝑠𝑠2𝑥𝑥 = 1

It is also important to note that 𝑠𝑠𝑠𝑠𝑠𝑠2𝑥𝑥 + 𝑐𝑐𝑡𝑡𝑠𝑠2𝑥𝑥 = 1 can be written as:

• 𝑠𝑠𝑠𝑠𝑠𝑠2𝑥𝑥 = 1 − 𝑐𝑐𝑡𝑡𝑠𝑠2𝑥𝑥 • 𝑐𝑐𝑡𝑡𝑠𝑠2𝑥𝑥 = 1 − 𝑠𝑠𝑠𝑠𝑠𝑠2𝑥𝑥


national 5 mathematics practice paper e worked · pdf filen5 mathematics practice paper e...

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