0580_s23_qp

Question1

1(a): C1.1
1(b)(i): C1.1
1(b)(ii): C1.1
1(c): C1.9
1(d)(i): C1.3
1(d)(ii): C1.7
1(d)(iii): C1.14
1(e)(i): C1.15
1(e)(ii): C1.13

(a) Write the number three hundred thousand and three in figures.

(b) Write 15896 correct to

(i) the nearest thousand

(ii) the nearest ten.

(c) By writing each number in the calculation correct to 1 significant figure, work out an estimate for the value of $\frac{28.9\times5.49}{0.472+0.97}$

You must show all your working.

(d) Find the value of

(i) $\sqrt{1849}$

(ii) $5^{0}-5^{-1}$

(iii) $\frac{5~sin~30-8}{11}.$

(e) A cyclist travels at a constant speed of 8.5 metres per second.

(i) Work out how long the cyclist takes to travel a distance of 5.27 kilometres.

Give your answer in minutes and seconds.

(ii) The cyclist increases speed from 8.5 m/s to 10.2 m/s.

Work out the percentage increase in speed.

▶️Answer/Explanation

<p(a) 300003
(b)(i) 16000
(b)(ii) 15900
(c) 30 × 5 / (0.5 + 1)
100
(d)(i) 43
(d)(ii) 0.8
(d)(iii) -0.5
(e)(i) 10 (min) 20 (s)
(e)(ii) 20

Detailed Solution

(a)
Three hundred thousand $\rightarrow$ 300,000
And three $\rightarrow$ 3
Three hundred thousand and three means we have 300,000 and 3,
$
300003
$

(b) (i)
The number is 15896.
The hundreds digit is 8, which is ≥ 5,
$
15896 \approx 16000
$

(ii)
The number is 15896.
The last digit is 6, which is ≥ 5,
$
15896 \approx 15890
$

Round each number to 1 significant figure
$ 28.9 \approx 30 $
$ 5.49 \approx 5 $
$ 0.472 \approx 0.5 $
$ 0.97 \approx 1 $
$
\frac{30 \times 5}{0.5 + 1} = \frac{150}{1.5}
$
$
150 \div 1.5 = 100
$

(d) (i) $ \sqrt{1849} $
$
\sqrt{1849} = 43
$

(ii)
$ 5^{0} – 5^{-1} $
$ 5^0 = 1 $ (any number to the power of 0 is 1)
$ 5^{-1} = \frac{1}{5} $
$
1 – \frac{1}{5} = \frac{5}{5} – \frac{1}{5} = \frac{4}{5}
$

(iii)
$ \frac{5 \sin 30 – 8}{11} $
$ \sin 30 = 0.5 $, so
$
5 \times 0.5 = 2.5
$
$
\frac{2.5 – 8}{11} = \frac{-5.5}{11} = -0.5
$

(e)

(i)
km to meters
$
5.27 \text{ km} = 5270 \text{ m}
$
$
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{5270}{8.5}
$
$
= 620 \text{ seconds}
$

Convert to minutes and seconds:
$
620 \text{ sec} = 10 \text{ min } 20 \text{ sec}
$

(ii)
$
\text{Percentage Increase} = \frac{\text{New Speed} – \text{Old Speed}}{\text{Old Speed}} \times 100
$
$
= \frac{10.2 – 8.5}{8.5} \times 100
$
$
= \frac{1.7}{8.5} \times 100
$
$
= 20\%
$

Question2

2(a)(i): C9.4
2(a)(ii): C9.3
2(b)(i): C9.3
2(b)(ii): C8.1
2(c)(i): C9.3
2(c)(ii): C9.3

(a) Mika counts the number of letters in each of the 61 words in a paragraph.

Some of his results are shown in the table and bar chart.

(i) Complete the table and the bar chart.

(ii) Write down the mode.

(b) Grace also counts the number of letters in each word of another paragraph.

Her results are shown in the table.

(i) Work out the mean.

(ii) She picks one of these words at random.

Find the probability that it has more than three letters.

These are her results.

(i) Find the median.

(ii) Find the range.

▶️Answer/Explanation

<p(a)(i) 14 8
Bars with heights 12 and 8
(a)(ii) 4
(b)(i) 2.68
(b)(ii) 13/50 oe
(c)(i) 3.5 cao
(c)(ii) 8

Detailed Solution

(a)(i)

(a)(ii)
1 letter → 7
2 letters → 12
3 letters → 14
4 letters → 15
5 letters → 8
6 letters → 5
The highest frequency is 15, corresponds to 4 letters.

(b)(i)
$
\text{Mean} = \frac{\sum (x \times f)}{\sum f}
$
$
\sum (x \times f) = (10 × 1 = 10) + (18 × 2 = 36) + (9 × 3 = 27
) + (6 × 4 = 24
) + (6 × 5 = 25
) + (2 × 6 = 12
) = 134
$
$
\sum f = 10 + 18 + 9 + 6 + 5 + 2 = 50
$
$
\text{Mean} = \frac{134}{50} = 2.68
$

(b)(ii)
word has more than 3 letters.
selecting words with 4, 5, or 6 letters.
$
6 + 5 + 2 = 13
$
$
P(\text{more than 3 letters}) = \frac{13}{50}
$

Question3

3(a)(i): C2.9
3(a)(ii): C1.11
3(b)(i): C1.15
3(b)(ii): C1.11
3(b)(iii): C1.13

(a)

(i) On the grid, draw a conversion graph between Australian dollars and South African rands.

(ii) A watch costs 1350 South African rands.

Find the cost of this watch in Australian dollars.

(b) (i) A plane leaves Sydney at 2148 local time to fly to Johannesburg.

The flight takes 14 hours 15 minutes.

The local time in Sydney is 8 hours ahead of the local time in Johannesburg.

Find the local time in Johannesburg when the plane arrives.

(ii) On the plane there are 315 people.

The ratio of children : adults = 7 : 8

Work out the number of adults on the plane.

(iii) Another plane has 420 seats.

90% of the seats are occupied.

Work out the number of seats that are occupied.

▶️Answer/Explanation

<p(a)(i) Ruled line joining (0, 0) to (50, 540)
(a)(ii) 125 or 123 to 127
(b)(i) 04 03
(b)(ii) 168
(b)(iii) 378

Detailed Solution

(a)(i)

(a)(ii)
$
50 \text{ AUD} = 540 \text{ ZAR}
$
$
1 \text{ AUD} = \frac{540}{50} = 10.8 \text{ ZAR}
$
1350 ZAR to AUD:
$
\text{AUD} = \frac{1350}{10.8} = 125
$

(b) (i)
Departure time in Sydney: 21:48 (local time)
Flight duration: 14 hours 15 minutes
Sydney is 8 hours ahead of Johannesburg.

$
21:48 + 14\text{h}15\text{m} = 12:03 \text{ (next day, Sydney time)}
$
$
12:03 – 8\text{h} = 04:03
$

(b)(ii)
Total passengers = 315
Ratio of children to adults= 7:8
Total parts in the ratio
$
7 + 8 = 15
$
$
\frac{315}{15} = 21
$
Number of adults
$
8 \times 21 = 168
$

(iii)
Total seats = 420
90% are occupied:
$
90\% \text{ of } 420 = \frac{90}{100} \times 420
$
$
= 0.9 \times 420 = 378
$

Question4

4(a): C9.5
4(b): C9.5
4(c): C9.5
4(d)(i): C9.5
4(d)(ii): C9.5
4(e): C1.13

Fidel gives different amounts of water to some plants.

The scatter diagram shows the height (cm) and the amount of water (ml) for each of 15 plants.

(a) Plot these two results on the scatter diagram.

(b) What type of correlation is shown in the scatter diagram?

On the scatter diagram, put a ring around the point for this plant.

(d) (i) On the scatter diagram, draw a line of best fit.

(ii) Another plant is given 65 ml of water.

Use your line of best fit to estimate the height of this plant.

(e) Find the percentage of these 17 plants that have a height of more than 24 cm.

Give your answer correct to 1 decimal place.

▶️Answer/Explanation

<p(a) Two points accurately plotted
(b) Positive
(c) (60, 7) indicated
(d)(i) Accurate straight line of best fit
(d)(ii) 24 to 32
(e) 47.1

Detailed Solution

(a)

(b)

(c)

;

(d)(i)

(e)

Number of plants with height greater than 24 cm = 8

Total number of plants = 17
$
\text{Percentage} = \left( \frac{\text{Number of plants with height } > 24 \text{ cm}}{\text{Total number of plants}} \right) \times 100
$
$
\text{Percentage} = \left( \frac{8}{17} \right) \times 100
$
$
\frac{8}{17} \approx 0.4705
$
$
0.4705 \times 100 = 47.05\%
$

Question5

5(a): C5.2
5(b): C5.2
5(c): C5.3
5(d): C5.4

(a)

This rectangle has an area of 12 cm² and a perimeter of 16 cm.

This shape is made from six of these rectangles.

Find the area and perimeter of this shape.

(b) Find the area of this triangle.

(d) A cube has a volume of 125m$^3$
Work out the surface area of the cube.

▶️Answer/Explanation

<p(a) [a =] 72
[p =] 52
(b) 67.2
(c) 4.46
(d) 150

Detailed Solution

(a)

Area equation
$
L \times W = 12
$

Perimeter equation
$
2L + 2W = 16
$
$
L + W = 8
$
$
L \times (8 – L) = 12
$
$
L^2 – 8L + 12 = 0
$
$
(L – 6)(L – 2) = 0
$
$
L = 6 \quad \text{or} \quad L = 2
$

Area:
The shape is made of 6 rectangles
total area
$
6 \times 12 = 72 \text{ cm}^2
$

Perimeter:
Top horizontal part: $6 + 6 + 6 = 18$ cm
Side vertical parts: $2 + 6 = 8$ cm (left side) and $2 + 6 = 8$ cm (right side)
Lower vertical segments: Three segments of 2 cm each on both left and right, totaling $2 + 2 + 2 = 6$ cm on each side.
Bottom horizontal part: $6$ cm
$
18 + 8 + 8 + 6 + 6 + 6 = 52 \text{ cm}
$

(b)
$
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
$
Base = 16 cm
Height = 8.4 cm

$
\text{Area} = \frac{1}{2} \times 16 \times 8.4
$
$
= \frac{16 \times 8.4}{2} = \frac{134.4}{2} = 67.2 \text{ cm}^2
$

(c)
circumference of a circle
$
C = 2\pi r
$
$
28 = 2\pi r
$
$
r = \frac{28}{2\pi} = \frac{14}{\pi}
$
$
r \approx \frac{14}{3.1416} \approx 4.46 \text{ cm}
$

(d)
The volume of a cube
$
V = s^3
$
$ V = 125 $
$
s^3 = 125
$
$
s = \sqrt[3]{125} = 5 \text{ m}
$

The surface area of a cube
$
\text{Surface Area} = 6s^2
$
$
= 6(5^2) = 6(25) = 150 \text{ m}^2
$

Question6

6(a)(i): C4.5
6(a)(ii): C4.5
6(a)(iii): C4.5
6(b)(i)(a): C7.1
6(b)(i)(b): C7.1
6(b)(ii): C7.1

(a) For each quadrilateral, draw any lines of symmetry and write down its mathematical name.

(b) The diagram shows three triangles A, B, and C, on a grid.

(i) Describe fully the single transformation that maps

(a) triangle A onto triangle B

(b) triangle A onto triangle C.

(ii) On the grid, reflect triangle A in the line x = -1.

▶️Answer/Explanation

<p(a)(i) Two correct diagonals
rhombus
(a)(ii) 1 correct diagonal
kite
(b)(i)(a) Enlargement
(sf=) 3 oe
(centre) (4, 6) oe
(b)(i)(b) Rotation
90° clockwise oe
(centre) (0, 0) oe
(b)(ii) Correct reflection, points (-3, 2) (-6, 2) (-3, 4)

Detailed Solution

(a)(i)

(a)(ii)

(b)(i)(a)

(b)(i)(b)

(b)(ii)

Triangle A:
$
(1,4), (1,2), (4,2)
$
$
x’ = 2k – x, \quad y’ = y
$
Point (1,4):
$
x’ = 2(-1) – 1 = -2 – 1 = -3, \quad y’ = 4
$
Reflection: (-3,4)

Point (1,2):
$
x’ = 2(-1) – 1 = -3, \quad y’ = 2
$
Reflection: (-3,2)

Point (4,2):
$
x’ = 2(-1) – 4 = -2 – 4 = -6, \quad y’ = 2
$
Reflection: (-6,2)

Question7

7(a)(i): C3.5
7(a)(ii): C3.2
7(a)(iii): C3.1
7(b)(i): C2.10
7(b)(ii): C2.10
7(b)(iii): C2.10
7(b)(iv): C2.5

(a) (i) Find the equation of line L.

Give your answer in the form y = mx + c.

(ii) On the grid, draw the line y = 1.

(iii) Write down the coordinates of the point where the two lines intersect.

(b) (i) Complete the table of values for y = x² + x – 8.

(ii) On the grid, draw the graph of y = x² + x – 8 for -4 ≤ x ≤ 4.

(iii) Write down the equation of the line of symmetry of the graph.

(iv) Use your graph to solve the equation x² + x – 8 = 0.

▶️Answer/Explanation

<p(a)(i) [y =] 1.5x – 2 final answer
(a)(ii) Correct ruled line
(a)(iii) (2, 1)
(b)(i) -6 -6 12
(b)(ii) Correct and accurate curve
(b)(iii) x = -½ oe
(b)(iv) −3.5 to −3.3 2.3 to 2.5

Detailed Solution

(i)
$
y = mx + c
$

the line $ L $ crosses the y-axis at $ (0, -2) $,
$
c = -2
$

$
m = \frac{\text{change in } y}{\text{change in } x}
$
$
(0, -2) \quad \text{and} \quad (2, 1)
$
$
m = \frac{1 – (-2)}{2 – 0}
$
$
m = \frac{1 + 2}{2} = \frac{3}{2}
$
$
y = \frac{3}{2}x – 2
$

(ii)

(iii)
$
\frac{3}{2}x – 2 = 1
$
$
\frac{3}{2}x = 3
$
$
x = 2
$

$
\mathrm{(2,1)}
$

(b) (i)
$ y = x^2 + x – 8 $.

For $ x = -2 $
$
y = (-2)^2 + (-2) – 8
$
$
= 4 – 2 – 8 = -6
$

For $ x = 1 $
$
y = (1)^2 + (1) – 8
$
$
= 1 + 1 – 8 = -6
$

For $ x = 4 $
$
y = (4)^2 + (4) – 8
$
$
= 16 + 4 – 8 = 12
$
(ii)

(iii)

The line of symmetry of a quadratic function $ y = ax^2 + bx + c $
$
x = \frac{-b}{2a}
$
For $ y = x^2 + x – 8 $,
$ a = 1 $, $ b = 1 $
$
x = \frac{-1}{2(1)} = \frac{-1}{2} = -0.5
$
$
x = -\frac{1}{2}
$

(iv)

$ x^2 + x – 8 = 0 $, points where the graph intersects the x-axis (i.e., where $ y = 0 $).

Question8

8(a): C2.1
8(b): C2.2
8(c): C2.2
8(d): C2.5
8(e): C2.2
8(f): C2.5

(a) T = P + 5 – 3Q

Find the value of T when P = 6 and Q = 8.

(b) Simplify.

3a – b + 7a + 2b – 4a

5(2x – 3y)

(d) Solve.

5x – 1 = 3x + 1 + 9

(e) Make t the subject of the formula p = (5t – 3)

(f) Entry to a castle costs $\$x$ for an adult and $\$y$ for a child.

Entry for 2 adults and 3 children costs $\$15.00$.
Entry for 3 adults and 5 children costs $\$23.50$.

Write down a pair of simultaneous equations to show this information and solve them to find the value of x and the value of y.

You must show all your working.

▶️Answer/Explanation

<p(a) 54
(b) 5a – 3b final answer
(c) 10x – 15y final answer
(d) 10
(e) [t=] (p + 3) / 5 oe final answer
(f) 2x + 3y = 15 and 3x + 5y = 23.5
correctly equating one set of coefficients
correct method to eliminate one variable
[x =] 4.5
[y =] 2

Detailed Solution

(a)
$
T = P + 5 – 3Q
$
$ P = 6 $ and $ Q = 8 $
$
T = 6 + 5 – 3(8)
$
$
T = 6 + 5 – 24
$
$
T = -13
$

(b)
$
3a – b + 7a + 2b – 4a
$

like term
$
(3a + 7a – 4a) + (-b + 2b)
$
$
6a + b
$

(d)
$ 5x – 1 = 3x + 1 + 9 $
$
5x – 1 = 3x + 10
$
$
5x – 3x = 10 + 1
$
$
2x = 11
$
$
x = \frac{11}{2}
$

(e)
$
p = 5t – 3
$
$
p + 3 = 5t
$
$
t = \frac{p + 3}{5}
$

(f)
Entry for 2 adults and 3 children costs $\$15.00$
$
2x + 3y = 15
$

Entry for 3 adults and 5 children costs $\$23.50$
$
3x + 5y = 23.50
$

$
(3)(2x + 3y) = 3(15)
$
$
6x + 9y = 45
$
$
(2)(3x + 5y) = 2(23.50)
$
$
6x + 10y = 47
$
$
(6x + 10y) – (6x + 9y) = 47 – 45
$
$
6x + 10y – 6x – 9y = 2
$
$
y = 2
$

$ y = 2 $ into the first equation
$
2x + 3(2) = 15
$
$
2x + 6 = 15
$
$
2x = 9
$
$
x = 4.50
$
$
\mathrm{x = 4.50, \quad y = 2}
$

Question9

9(a)(i): C2.7
9(a)(ii): C2.7
9(a)(iii): C2.7
9(b)(i): C2.7
9(b)(ii): C2.7

(a) These are the first four terms of a sequence.

2 8 14 20

(i) Write down the next term.

(ii) Write down the term to term rule for continuing the sequence.

(iii) Find an expression for the nth term.

(b) (i) Find the first three terms of the sequence with nth term n² + 5.

(ii) These are the first four terms of another sequence.

7 10 15 22

Find an expression for the nth term of this sequence.

▶️Answer/Explanation

<p(a)(i) 26
(a)(ii) add 6 oe
(a)(iii) 6n – 4 oe final answer
(b)(i) 6 9 14
(b)(ii) n² + 6 oe final answer

(a) (i)
sequence: 2, 8, 14, 20
$ 8 – 2 = 6 $
$ 14 – 8 = 6 $
$ 20 – 14 = 6 $
The sequence increases by 6 each time.
$
20 + 6 = 26
$

(ii)
To continue the sequence, add 6 to the previous term.
So, the term-to-term rule is:
Add 6 to the previous term

(iii)
first term $ a = 2 $ and common difference $ d = 6 $.

$
T_n = a + (n – 1) d
$
$
T_n = 2 + (n – 1) \times 6
$
$
T_n = 2 + 6n – 6
$
$
T_n = 6n – 4

(b)(i)
nth term $ n^2 + 5 $
When $ n = 1 $:
$
1^2 + 5 = 1 + 5 = 6
$
When $ n = 2 $:
$
2^2 + 5 = 4 + 5 = 9
$
When $ n = 3 $:
$
3^2 + 5 = 9 + 5 = 14
$
the first three terms
$
6, 9, 14
$

(ii)
Assuming $ T_n = an^2 + bn + c $.

the first three term
1. $ a + b + c = 7 $
2. $ 4a + 2b + c = 10 $
3. $ 9a + 3b + c = 15 $

Solving
$ 3a + b = 3 $
$ 5a + b = 5 $
$ a = 1 $, $ b = 0 $, $ c = 6 $

Thus, the nth term
$
T_n = n^2 + 6
$

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