Question1
1(a)(i): C9.2, C9.4
1(a)(ii): C1.4
1(a)(iii): C1.11
1(a)(iv): C1.1
1(a)(v): C1.11
1(b)(i): C8.1
1(b)(ii): C1.2
1(b)(iii): C8.1
1(b)(iv): C1.2
(a) Claudia asks some students to choose their favourite science from biology, chemistry and physics. The pie chart shows the results.
(i) Find the percentage of students who choose chemistry.
(ii) Find the fraction of students who choose physics. Give your answer in its simplest form.
(iii) For the number of students choosing each subject, find the ratio biology : chemistry : physics. Give your answer in its simplest form.
(iv) Marcus says:
‘I do not know how many people choose chemistry, but I do know it is an even number.’
Explain how Marcus knows this.
(v) Claudia now tells Marcus that 26 students choose chemistry.
Work out how many students choose physics.
(b) The Venn diagram shows information about the number of students in a class who study geography (G) and history (H).
(i) Work out the number of students in the class.
(ii) Find n(G).
(iii) One of the students is chosen at random.
Find the probability that this student studies geography and history.
(iv) One of the students who studies geography and history stops studying history.
Complete this Venn diagram to show this change.
▶️Answer/Explanation
<p(a)(i) 25
(a)(ii) 5/8 cao
(a)(iii) 1:2:5
(a)(iv) Correct explanation e.g., if chemistry has an odd number then biology would not be a whole number of students
(a)(v) 65
(b)(i) 28
(b)(ii) 14
(b)(iii) 9/28 oe
(b)(iv) 6, 8, 10, 4
Detailed Solution
(a)(i)
Total angle in a pie chart = 360°
Angle for chemistry = 90°
$
\frac{90^\circ}{360^\circ} = \frac{1}{4}
$
$
\frac{1}{4} \times 100\% = \rm{25\%}
$
(a)ii)
Angle for physics = 225°
Total angle = 360°
$
\frac{225^\circ}{360^\circ}
$
GCD of 225 and 360 = 45
$
\frac{225 \div 45}{360 \div 45} = \frac{5}{8}
$
(a)(iii)
Biology angle = 45°
Chemistry angle = 90°
Physics angle = 225°
$
\text{Ratio} = 45 : 90 : 225
$
Divide each by 45
$
\frac{45}{45} : \frac{90}{45} : \frac{225}{45} = 1 : 2 : 5
$
(a)(iv)
From part (iii),
Biology : Chemistry : Physics = 1 : 2 : 5
Let the number of students be \( x \), then:
Chemistry = \( 2x \)
\( 2x \) is always even (2 multiplied by any integer), the number of students choosing chemistry must be even.
(a)(v)
From the ratio Chemistry = 2 parts = 26 students
$
1 \text{ part} = \frac{26}{2} = 13
$
Physics = 5 parts
$
5 \times 13 = \rm{65 \text{ students}}
$
(b)(i)
Only Geography = 5
Both Geography and History = 9
Only History = 10
Neither subject = 4
$
\text{Total} = 5 + 9 + 10 + 4 = \rm{28}
$
28 students in the class.
(b)(ii)
\( n(G) \) represents the number of students who study Geography, including those who study both subjects.
$
n(G) = \text{Only Geography} + \text{Both Geography and History} = 5 + 9 = 14
$
(b)(iii)
Students who study both: 9
Total students: 28
$
P(\text{Geography and History}) = \frac{9}{28}
$
(b)(iv)
One of the 9 students studying both subjects now studies only Geography.
Only Geography: \( 5 + 1 = 6 \)
Both Geography and History: \( 9 – 1 = 8 \)
Only History: remains 10
Neither: remains 4
Question2
2(a): C1.6
2(b): C1.13
2(c): C1.13
2(d)(i): C2.5
2(d)(ii): C2.9
2(e): C1.6
2(f): C1.1
2(g): C1.11
(a) Bananas cost $\$1.20 per$ kilogram and apples cost $\$2.25$ per bag.
Work out the total cost of 3.5 kg of bananas and 2 bags of apples.
(b) Students receive a 10% discount on their shopping.
Before the discount, the cost of a student’s shopping is $16.80.
Work out the amount of the discount.
(c) The cost of a cabbage increases by 15%.
Calculate the new price if the original price is $1.80.
(d) Some customers have their shopping delivered to their home.
The cost is $\$5$ plus $\$1.50$ for each kilometre travelled from the shop to their home.
(i) Show that the cost for a customer living 10km from the shop is $\$20$.
(e) A bottle of water costs $1.55.
Suki has $20.
Work out the maximum number of bottles Suki can buy and the change she receives.
(f) A farmer delivers eggs to the shop in trays of 50.
The eggs are then put into boxes of 12.
There are no eggs left in the trays and all of the egg boxes are full.
Work out the smallest possible number of eggs that the farmer delivers.
(g) The shop sells bottles of orange juice in three different sizes.
Work out which bottle is the best value.
Show how you decide.
▶️Answer/Explanation
<p(a) 8.7[0]
(b) 1.68 cao
(c) 2.07 cao
(d)(i) 5 + 10 × 1.5 [= 20]
(d)(ii) Ruled line from (0, 5) to (10, 20)
(e) 12 1.4[0]
(f) 300
(g) A
With correct comparisons made of the 3 bottles with suitable accuracy shown
Detailed Solution
(a) Total cost of 3.5 kg of bananas and 2 bags of apples
Bananas: $\$1.20$ per kg
Apples: $\$2.25$ per bag
$
\text{Bananas cost} = 3.5 \times 1.20 = 4.20
$
$
\text{Apples cost} = 2 \times 2.25 = 4.50
$
$
\text{Total cost} = 4.20 + 4.50 = \mathrm{\$8.70}
$
(b) Student receives a $10\%$ discount on $\$16.80$
$
\text{Discount} = 10\% \text{ of } 16.80 = 0.10 \times 16.80 = \mathrm{\$1.68}
$
(c)
Price increase of $15\% $on $\$1.80$
$
\text{Increase} = 15\% \text{ of } 1.80 = 0.15 \times 1.80 = 0.27
$
$
\text{New price} = 1.80 + 0.27 = \mathrm{\$2.07}
$
(d)(i)
Cost for 10 km delivery
Fixed cost = $\$5$
Cost per km = $\$1.50$
Distance = 10 km
$
\text{Total cost} = 5 + (1.50 \times 10) = 5 + 15 = \mathrm{\$20}
$
(d)(ii)
The equation for the delivery cost is
$
\text{Cost} = 5 + 1.5x \quad \text{where } x = \text{distance in km}
$
At \( x = 0 \), cost = $\$5$
At \( x = 10 \), cost = $\$20 $
(e)
Suki buys bottles of water
Cost per bottle = $\$1.55$
Total money = $\$20$
$
\text{Max number of bottles} = \left\lfloor \frac{20}{1.55} \right\rfloor = 12
$
$
\text{Cost of 12 bottles} = 12 \times 1.55 = 18.60
$
$
\text{Change} = 20 – 18.60 = \mathrm{\$1.40}
$
(f)
LCM (Least Common Multiple) of 50 and 12
Prime factorization
50 = 2 × 5²
12 = 2² × 3
$
\text{LCM} = 2² \times 3 \times 5² = 4 \times 3 \times 25 = \mathrm{300}
$
the smallest number of eggs that fit evenly in both trays and boxes is 300 eggs.
(g)
Bottle A
Volume: 0.5 litres
Price: $\$1.30$
$
\text{Price per litre} = \frac{1.30}{0.5} = \mathrm{2.60 \text{ per litre}}
$
Bottle B
Volume: 1.2 litres
Price: $\$3.20$
$
\text{Price per litre} = \frac{3.20}{1.2} \approx \mathrm{2.67 \text{ per litre}}
$
Bottle C
Volume: 2 litres
Price: $\$5.25$
$
\text{Price per litre} = \frac{5.25}{2} = \mathrm{2.625 \text{ per litre}}
$
Bottle A: $\$2.60$ per litre
Bottle B: $\$2.67$ per litre
Bottle C: $\$2.625$ per litre
The cheapest price per litre is from Bottle A.
Question3
3(a): C5.2
3(b): C5.2
3(c): C5.2
3(d): C5.4
3(e): C5.3
3(f)(i): C4.6
3(f)(ii): C4.6
(a) The diagram shows a shape on 1 cm² grid.
Work out the area of the shape.
(b)
The diagram shows a rectangle.
Work out the perimeter of the rectangle.
(c) A square has an area of 841 cm².
Work out the length of one side of the square.
(d) The diagram shows a cuboid made from 1 cm³ cubes.
(i) Work out the volume of the cuboid.
(ii) Write down the dimensions of a different cuboid that can be made using all of the cubes.
(e) The diagram shows three small circles and one large circle. The large circle has radius 20 cm. The small circles each have radius 4 cm.
Work out the shaded area.
Give your answer in terms of π.
(f) The exterior angle of a 9-sided regular polygon is 40°.
(i) Work out the size of the interior angle of this polygon.
(ii) The diagram shows a regular pentagon inside part of a regular 9-sided polygon.
Work out the value of x.
▶️Answer/Explanation
<p(a) 9.5
(b) 38
(c) 29
(d)(i) 48
(d)(ii) Correct dimensions
(e) 352π cao nfww
(f)(i) 140
(f)(ii) 16
Detailed Solution
(a)
1. Rectangular part (bottom):
Length: 6 cm
Height: 1 cm
$
\text{Area}_{\text{rectangle}} = 6 \times 1 = 6 \text{ cm}^2
$
2. Triangular part (top):
2 full triangles
1 half triangle
Each triangle
Base: 2 cm
Height: 1 cm
$
\text{Area of one full triangle} = \frac{1}{2} \times 2 \times 1 = 1 \text{ cm}^2
$
$
\text{2 full triangles} = 2 \times 1 = 2 \text{ cm}^2
$
$
\text{1 half triangle} = \frac{1}{2} \times 1 = 0.5 \text{ cm}^2
$
$
\text{Total triangle area} = 2 + 0.5 = 2.5 \text{ cm}^2
$
3.Extra square (top-right corner)
It’s 1 cm × 1 cm =
$
1 \text{ cm}^2
$
Total area of the shape
$
6 \text{ (rectangle)} + 2.5 \text{ (triangles)} + 1 \text{ (extra square)} = \rm{9.5 \text{ cm}^2}
$
(b)
Length = 12 cm
Width = 7 cm
$
P = 2 \times (\text{length} + \text{width})
$
$
P = 2 \times (12 + 7) = 2 \times 19 = 38 \text{ cm}
$
(c)
$
\text{Area} = \text{side}^2
$
$
s^2 = 841
$
$
s = \sqrt{841} = 29 \text{ cm}
$
(d) (i)
$
V = \text{length} \times \text{width} \times \text{height}
$
$
V = \text{8} \times \text{2} \times \text{3}
$
$
V = 48 \rm{ cm^3}
$
(d) (ii)
A different cuboid using all the same cubes
The total number of 1 cm³ cubes is 48.
(length × width × height) that multiplies to 48.
$
4 \times 4 \times 3 = 48 \text{ cm}^3
$
(e)
Radius of large circle = 20 cm
Radius of each small circle = 4 cm
Number of small circles = 3
$
\text{Area}_{\text{large}} = \pi r^2 = \pi (20)^2 = 400\pi
$
$
\text{Area}_{\text{small}} = \pi (4)^2 = 16\pi
$
Total area of 3 small circles
$
3 \times 16\pi = 48\pi
$
$
\text{Shaded area} = \text{Area}_{\text{large}} – \text{Area}_{\text{3 small}} = 400\pi – 48\pi = \rm{352\pi \text{ cm}^2}
$
(f) (i)
exterior angle = 40°
Interior angle =
$
180^\circ – \text{exterior angle} = 180^\circ – 40^\circ = \rm{140^\circ}
$
(ii)
A regular pentagon inside a 9-sided polygon
Two identical angles labeled \( x \), formed between the outer polygon and the inner pentagon
regular pentagon
$
\text{Interior angle} = \frac{(5 – 2) \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ
$
Interior angle of the 9-sided polygon
$
\text{Interior angle} = 140^\circ
$
$
x + 108^\circ + x = 140^\circ
$
$
2x + 108^\circ = 140^\circ
$
$
2x = 32^\circ \Rightarrow x = \rm{16^\circ}
$
Question4
4(a)(i): C1.15
4(a)(ii): C1.15
4(a)(iii): C2.9
4(a)(iv): C1.12
4(b)(i): C3.1
4(b)(ii): C3.1
4(b)(iii): C3.1
(a) A boat sails from A to B.
The travel graph shows this journey.
(i) Write down the time that the boat leaves A.
(ii) Work out how long, in minutes, it takes the boat to sail from A to B.
(iii) The boat stays at B for 20 minutes. The boat then sails to C at a constant speed of 8 km/h.
Complete the travel graph.
(iv) Work out the average speed, in km/h, for the whole journey from A to C.
(b) The scale drawing shows the positions of two ports, X and Y. The scale is 1 cm represents 8 km.
(i) Measure the bearing of Y from X.
(ii) A boat, B, is 52 km from X and 80 km from Y.
On the scale drawing, mark the position of B.
(iii) A ship, S, is on a bearing of 284° from X.
Work out the bearing of X from S.
▶️Answer/Explanation
<p(a)(i) 10 20
(a)(ii) 40
(a)(iii) Ruled lines from (11 00, 6) to (11 20, 6) and (11 20, 6) to (12 50, 18)
(a)(iv) 7.2
b)(i) [0]65
(b)(ii) B correctly marked 6.5 cm from X and 10 cm from Y
(b)(iii) 104
Detailed Solution
(a)
(i)
From the graph, the boat starts at A (0 km) at 10:20.
(ii)
The boat reaches B (6 km) at 11:00.
The boat leaves A at 10:20.
The time taken to travel from A to B
$
11:00 – 10:20 = 40 \text{ minutes}
$
(iii)
The boat stays at B for 20 minutes, meaning it remains at 6 km from 11:00 to 11:20.
The boat then sails to C at a constant speed of 8 km/h.
The distance from B to C
$
18 – 6 = 12 \text{ km}
$
$
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{12}{8} = 1.5 \text{ hours} = 90 \text{ minutes}
$
The boat leaves B at 11:20.
Adding 90 minutes → it reaches C at 12:50.
horizontal line from (11:00, 6 km) to (11:20, 6 km) represent the stop at B.
straight line from (11:20, 6 km) to (12:50, 18 km) represent the journey to C.
(iv)
$
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}
$
Total Distance = \( 18 \) km (A to C).
Total Time
A to B: 40 minutes
Waiting at B: 20 minutes
B to C: 90 minutes
Total: \( 40 + 20 + 90 = 150 \) minutes = \( \frac{150}{60} = 2.5 \) hours.
$
\text{Average Speed} = \frac{18}{2.5} = 7.2 \text{ km/h}
$
(b)(i)
A bearing is measured clockwise from North at the starting point (X) to the direction of the destination (Y).
Place a protractor at point X.
the 0° mark with the North direction.
Measure the clockwise angle to Y.
(ii)
B is 52 km from X
B is 80 km from Y
Scale: 1 cm = 8 km
$
\frac{52}{8} = 6.5 \text{ cm (from X)}
$
$
\frac{80}{8} = 10 \text{ cm (from Y)}
$
a compass radius to 6.5 cm and draw a circle around X.
compass radius to 10 cm and draw a circle around Y.
(iii)
S is on a bearing of 284° from X.
Bearings are always measured clockwise from North.
$
\text{Bearing of X from S} = 284° – 180° = 104°
$
Question5
5(a): C2.7
5(b): C2.7
5(c)(i): C2.7
5(c)(ii): C2.7
5(d)(i): C2.7
5(d)(ii): C2.7
5(d)(iii): C2.7
The grid shows the first three diagrams in a sequence. Each diagram is made using sticks.
(a) On the grid, draw Diagram 4.
(b) Complete the table.
(c) (i) Find an expression, in terms of n, for the number of sticks in Diagram n.
(ii) One of the diagrams has 73 sticks.
Work out its Diagram number.
(d) (i) Show that the total number of sticks needed to make the first 3 diagrams is 27.
(ii) The total number of sticks needed to make the first k diagrams is 2k² + 3k.
Show that this expression gives the correct total number of sticks needed to make the first 3 diagrams.
(iii) Tobias wants to make the first 10 diagrams. He has already made the first 3 diagrams. He has 240 sticks left to make the remaining 7 diagrams.
Work out how many sticks he has left when all 10 diagrams are made.
▶️Answer/Explanation
<p(a) Correct diagram drawn
(b) 17, 21
c)(i) 4n + 1 oe final answer
(c)(ii) 18 nfw
(d)(i) 5 + 9 + 13 [= 27]
d)(ii) 2 × 3² + 3 × 3 leading to 27 as final answer
(d)(iii) 37
Detailed Solution
(a)
(b)
(c)(i)
The difference between consecutive terms
$
9 – 5 = 4, \quad 13 – 9 = 4
$
difference is constant, this follows an arithmetic sequence
$
S_n = a + (n – 1)d
$
\( a = 5 \) (first term),
\( d = 4 \) (common difference).
$
S_n = 5 + (n – 1) \times 4
$
$
S_n = 4n + 1
$
(c)(ii)
$
4n + 1 = 73
$
$
4n = 72
$
$
n = 18
$
(d)(i)
The total number of sticks for the first 3 diagrams
$
S_1 + S_2 + S_3
$
\( S_n = 4n + 1 \)
$
S_1 = 4(1) + 1 = 5
$
$
S_2 = 4(2) + 1 = 9
$
$
S_3 = 4(3) + 1 = 13
$
$
5 + 9 + 13 = 27
$
(d)(ii)
\( T_k = 2k^2 + 3k \) for \( k = 3 \)
$
T_3 = 2(3)^2 + 3(3)
$
$
= 2(9) + 9
$
$
= 18 + 9 = 27
$
(d)(iii)
The total sticks for the first 10 diagrams
$
T_{10} = 2(10)^2 + 3(10)
$
$
= 2(100) + 30
$
$
= 200 + 30 = 230
$
He has already made the first 3 diagrams, which required 27 sticks.
sticks needed for the remaining 7 diagrams
$
T_{10} – T_{3} = 230 – 27 = 203
$
He has 240 sticks left.
Sticks remaining after making the remaining 7 diagrams
$
240 – 203 = 37
$
Question6
6(b)(i): C2.10
6(b)(ii): C2.10
6(c)(i): C2.10
6(c)(ii): C2.10
6(c)(iii): C2.10
6(d): C2.5
(a) Write down the equation of line L in the form y = mx + c.
(b) (i) Complete the table of values for y = x2 – 3x – 3.
(ii) On the grid, draw the graph of y = x2 – 3x – 3 for -2 ≤ x ≤ 5.
(c) (i) Write down the coordinates of the lowest point of the graph of y = x2 – 3x – 3.
(ii) On the grid, draw the line of symmetry of the graph of y = x2 – 3x – 3.
(iii) Write down the equation of the line of symmetry.
(d) Write down the coordinates of the point where line L intersects the graph of y = x2 – 3x – 3 for x > 0.
▶️Answer/Explanation
(a) [y =] -2x + 3 final answer
(b)(i) 7, -3, -3, 7
(b)(ii) Correct and accurate curve
(c)(i) (1.5, k) where -5.5 ≤ k < -5
(c)(ii) Line x = 1.5 drawn accurately
(c)(iii) x = 1.5 oe
(d) (3, -3)
Detailed Solution
(a)
$m = \frac{y_2 – y_1}{x_2 – x_1}$
(x1, y1) = (0, 3) and (x2, y2) = (1.5, 0).
$m = \frac{0 – 3}{1.5 – 0}$
$m = \frac{-3}{1.5}$
$m = -2$
The y-intercept is the point where the line crosses the y-axis, which is (0, c).
The point (0, 3)
$c = 3$
$y = mx + c$
$y = -2x + 3$
(b)(i)
$
y = x^2 – 3x – 3
$
\(x= -2 \)
\( y=(-2)^2 – 3(-2) – 3 = 4 + 6 – 3 = 7 \)
\( x= 0 \)
\( y=(0)^2 – 3(0) – 3 = -3 \)
\( x= 3 \)
\( y=(3)^2 – 3(3) – 3 = 9 – 9 – 3 = -3 \)
\( x= 5 \)
\(y= (5)^2 – 3(5) – 3 = 25 – 15 – 3 = 7 \)
(b)(ii)
(c)(i) Lowest point of the graph (Vertex)
$
y = ax^2 + bx + c
$
$
x = \frac{-b}{2a}
$
$
x = \frac{-(-3)}{2(1)} = \frac{3}{2} = 1.5
$
$
y = (1.5)^2 – 3(1.5) – 3
$
$
y = 2.25 – 4.5 – 3 = -5.25
$
(c)(ii)
(c)(iii)
The vertex is at \( x = 1.5 \), the equation of the line of symmetry
$
x = 1.5
$
(d)
$
L : y = -2x + 3
$
$
x^2 – 3x – 3 = -2x + 3
$
$
x^2 – 3x – 3 + 2x – 3 = 0
$
$
x^2 – x – 6 = 0
$
$
(x – 3)(x + 2) = 0
$
$
x – 3 = 0 \quad \Rightarrow \quad x = 3
$
$
x + 2 = 0 \quad \Rightarrow \quad x = -2
$
$
y = -2(3) + 3 = -6 + 3 = -3
$
$
(3, -3)
$
$
y = -2(-2) + 3 = 4 + 3 = 7
$
$
(-2, 7)
$
Question7
7(a)(i): C4.6
7(a)(ii): C4.6
7(a)(iii): C4.6
7(b): C4.6
7(c): C6.2
7(d)(i): C4.4
7(d)(ii): C4.4
7(e): C5.2
(a) The diagram shows a triangle ABC and a straight line BCD.
(i) Angle ACB = 108°.
Write down the mathematical name for this type of angle.
(ii) Work out the value of x.
(iii) Work out the value of y.
(b) Show that the mean of the angles in any triangle is 60°.
(c) The diagram shows a right-angled triangle.
Calculate the value of h.
(d) Triangle ABC is similar to triangle PQR.
(i) Calculate PR.
(ii) Calculate BC.
(e) The diagram shows a right-angled triangle.
Calculate the perimeter of this triangle.
▶️Answer/Explanation
<p(a)(i) Obtuse
(a)(ii) 72
(a)(iii) 26
(b) 180 / 3
(c) 9.77 or 9.766…
(d)(i) 10.8
(d)(ii) 1.76
(e) 60
Detailed Solution
(a)(i)
An angle between \( 90^\circ \) and \( 180^\circ \) is called an obtuse angle.
(a)(ii)
\( x^\circ \) and \( 108^\circ \) form a straight line, they are supplementary, meaning their sum is \( 180^\circ \)
$
x + 108 = 180
$
$
x = 180 – 108 = 72^\circ
$
a)(iii)
the sum of the angles in \( \triangle ABC \) is:
$
46^\circ + y^\circ + 108^\circ = 180^\circ
$
$
y + 152 = 180
$
$
y = 180 – 152= 28^\circ
$
(b)
The sum of the angles in any triangle
$
180^\circ
$
Since a triangle has 3 angles
$
\frac{180}{3} = 60^\circ
$
(c)
cosine rule:
$
\cos(35^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}}
$
$
\cos(35^\circ) = \frac{8}{h}
$
$
h = \frac{8}{\cos(35^\circ)}
$
$
h \approx \frac{8}{0.8192} = 9.77 \text{ cm}
$
(d) (i)
the triangles are similar, their sides are proportional:
$
\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}
$
\( AB = 1.36 \) cm, \( PQ = 6.12 \) cm
\( AC = 2.4 \) cm, \( PR = ? \)
Using the proportion
$
\frac{AC}{PR} = \frac{AB}{PQ}
$
$
PR = \frac{2.4 \times 6.12}{1.36}
$
$
PR = \frac{14.688}{1.36} = 10.8 \text{ cm}
$
(ii)
Using the proportion
$
BC = \frac{QR \times AB}{PQ}
$
$
BC = \frac{7.92 \times 1.36}{6.12}
$
$
BC = \frac{10.7712}{6.12} = 1.76 \text{ cm}
$
(e)
One leg = 24 cm
Hypotenuse = 26 cm
$
a^2 + b^2 = c^2
$
$
24^2 + b^2 = 26^2
$
$
576 + b^2 = 676
$
$
b^2 = 100
$
$
b = 10 \text{ cm}
$
$
\text{Perimeter} = 24 + 10 + 26 = 60 \text{ cm}
$
Question8
8(a): C1.10
8(b): C1.11
8(c): C1.8
8(d): C1.13
(a) The length, l m, of a piece of wire is 18.7 metres, correct to the nearest 10 centimetres.
Complete the statement about the value of l.
(b) 850 metres of wire has a mass of 130.5 kilograms.
Work out the length of wire, in metres, that has a mass of 900 grams.
(c) Aluminium is used to make the wire. The mass of 1 cm³ of aluminium is 2.7 grams.
Work out the mass, in grams, of 6000 cm³ of aluminium.
Give your answer in standard form.
(d) A 12 metre length of wire increases in length to 12.017 metres when its temperature rises.
Calculate the percentage increase in the length of the wire.
▶️Answer/Explanation
<p(a) 18.65 18.75
(b) 5.86 or 5.862…
(c) 1.62 × 10⁴ cao
(d) 0.142 or 0.1416 to 0.1417
Detailed Solution
(a)
Nearest 10 cm = Nearest 0.1 m
lower bound
$
18.7 – 0.05 = 18.65 \text{ m}
$
upper bound
$
18.7 + 0.05 = 18.75 \text{ m}
$
(b)
$
900 \text{ g} = 0.9 \text{ kg}
$
$
\text{Mass per metre} = \frac{130.5}{850} = 0.1535 \text{ kg/m}
$
$
\text{Length} = \frac{0.9}{0.1535} \approx 5.864 \text{ m}
$
(c)
$
\text{Mass} = 6000 \times 2.7 = 16200 \text{ g}
$
standard form
$
\rm{1.62 \times 10^4 \text{ g}}
$
(d)
$
\text{Increase} = 12.017 – 12 = 0.017 \text{ m}
$
$
\text{Percentage increase} = \left(\frac{0.017}{12}\right) \times 100 \approx 0.1417\%
$