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Question 1 Report
A corner shop buys tins of soup from a supplier. The supplier raises the wholesale price by 8%, so the shop now pays 54p for each tin.
When the increased price is known, the original price is found by dividing by the increase multiplier, since multiplying the unknown original by that same multiplier is what produced the increased price.
(a) Let the original wholesale price be \(p\). Since an 8% increase means \(p \times 1.08 = 54\):
\[p = 54 \div 1.08\] [1 mark]
\[= 50\text{p}\] [1 mark]
Checking: \(50 \times 1.08 = 54\)p, which matches the new price [1 mark].
(b) Profit is the selling price minus the (new) wholesale cost:
\[75 - 54 = 21\text{p}\] [1 mark]
The percentage profit is this profit as a fraction of the wholesale price actually paid, times 100:
\[\frac{21}{54} \times 100 = 38.888\ldots\% = 38.9\%\] (1 d.p.) [1 mark]
Percentage profit is always measured against the cost price (54p), not the selling price (75p), because profit is defined relative to what was spent.
When the increased price is known, the original price is found by dividing by the increase multiplier, since multiplying the unknown original by that same multiplier is what produced the increased price.
(a) Let the original wholesale price be \(p\). Since an 8% increase means \(p \times 1.08 = 54\):
\[p = 54 \div 1.08\] [1 mark]
\[= 50\text{p}\] [1 mark]
Checking: \(50 \times 1.08 = 54\)p, which matches the new price [1 mark].
(b) Profit is the selling price minus the (new) wholesale cost:
\[75 - 54 = 21\text{p}\] [1 mark]
The percentage profit is this profit as a fraction of the wholesale price actually paid, times 100:
\[\frac{21}{54} \times 100 = 38.888\ldots\% = 38.9\%\] (1 d.p.) [1 mark]
Percentage profit is always measured against the cost price (54p), not the selling price (75p), because profit is defined relative to what was spent.
Question 2 Report
A workshop moves a template for a medal-ribbon logo, used in school sports day, across a planning grid by the vector \(\binom{2}{-5}\). Triangle \(E\), with vertices \((1, 0)\), \((2, 4)\), \((3, 0)\), is set out below.
A translation by the column vector \(\binom{2}{-5}\) means every point of the shape moves 2 units to the right and 5 units down, so \(2\) is added to every \(x\)-coordinate and \(5\) is subtracted from every \(y\)-coordinate. Applying this to each vertex of triangle \(E\): \((1, 0) \to (1+2, 0-5) = (3, -5)\), \((2, 4) \to (2+2, 4-5) = (4, -1)\), and \((3, 0) \to (3+2, 0-5) = (5, -5)\), giving the image vertices \((3, -5)\), \((4, -1)\), \((5, -5)\) [3 marks]. Since a translation preserves both shape and size, the image triangle is congruent to \(E\), simply repositioned lower and to the right.
A translation by the column vector \(\binom{2}{-5}\) means every point of the shape moves 2 units to the right and 5 units down, so \(2\) is added to every \(x\)-coordinate and \(5\) is subtracted from every \(y\)-coordinate. Applying this to each vertex of triangle \(E\): \((1, 0) \to (1+2, 0-5) = (3, -5)\), \((2, 4) \to (2+2, 4-5) = (4, -1)\), and \((3, 0) \to (3+2, 0-5) = (5, -5)\), giving the image vertices \((3, -5)\), \((4, -1)\), \((5, -5)\) [3 marks]. Since a translation preserves both shape and size, the image triangle is congruent to \(E\), simply repositioned lower and to the right.
Question 3 Report
A water treatment plant recorded the volume of water used each day over one month, in thousands of litres. The box plot summarises the results.
The box plot's five marked values, \(120, 150, 175, 210, 260\) thousand litres, give the minimum, lower quartile, median, upper quartile and maximum daily volume used.
The box plot's five marked values, \(120, 150, 175, 210, 260\) thousand litres, give the minimum, lower quartile, median, upper quartile and maximum daily volume used.
Question 4 Report
A farmer fixes three straight fence rails to one post, all on the same side of a horizontal ground rail. The three angles the rails make with the ground rail and with each other are in the ratio \(2:3:4\), written as \(2k^{\circ}\), \(3k^{\circ}\) and \(4k^{\circ}\), as marked.
This question shares a straight-line total according to a given ratio.
(a) Since all three angles lie on the same side of the ground rail, they sum to \(180^\circ\): \(2k + 3k + 4k = 180\). [1 mark]
Simplifying gives \(9k = 180\), so \(k = 20\). [1 mark]
(b) The largest angle is \(4k = 4(20)\). [1 mark]
This gives \(80^\circ\). [1 mark]
Writing the ratio as \(2k, 3k, 4k\) turns three unknown angles into a single unknown \(k\), which the straight-line total can then be used to solve for.
This question shares a straight-line total according to a given ratio.
(a) Since all three angles lie on the same side of the ground rail, they sum to \(180^\circ\): \(2k + 3k + 4k = 180\). [1 mark]
Simplifying gives \(9k = 180\), so \(k = 20\). [1 mark]
(b) The largest angle is \(4k = 4(20)\). [1 mark]
This gives \(80^\circ\). [1 mark]
Writing the ratio as \(2k, 3k, 4k\) turns three unknown angles into a single unknown \(k\), which the straight-line total can then be used to solve for.
Question 5 Report
Two steel props cross at a single bolted joint on a building site, one prop running level and the other running diagonally. The angle between the level prop and the diagonal prop is \(115^{\circ}\), as shown.
Work out the size of angle \(x\). (2)
This question uses the vertically opposite angle rule where two straight props cross.
Angle \(x\) and the marked \(115^\circ\) angle are formed on opposite sides of the crossing point by the same two straight props, making them vertically opposite angles. [1 mark]
Vertically opposite angles are always equal, so \(x = 115^\circ\). [1 mark]
Vertically opposite angles arise whenever two straight lines cross: the two angle pairs facing each other across the intersection point are always equal, regardless of how steeply the lines cross.
This question uses the vertically opposite angle rule where two straight props cross.
Angle \(x\) and the marked \(115^\circ\) angle are formed on opposite sides of the crossing point by the same two straight props, making them vertically opposite angles. [1 mark]
Vertically opposite angles are always equal, so \(x = 115^\circ\). [1 mark]
Vertically opposite angles arise whenever two straight lines cross: the two angle pairs facing each other across the intersection point are always equal, regardless of how steeply the lines cross.
Question 6 Report
The diagram shows a weekly household chores chart divided into 12 equal boxes, one for each chore on the list. The shaded boxes show the chores that have already been completed this week.
Work out the fraction of the chores, in its simplest form, that still need to be completed. (3)
This question uses a diagram to test reading data before simplifying a fraction. The chart is divided into 12 equal boxes, one for each chore, and the shaded boxes represent chores already completed. Counting the diagram, 8 of the 12 boxes are shaded, so \(12 - 8 = 4\) boxes remain unshaded [1 mark].
The fraction of chores still to be completed is therefore the unshaded boxes over the total, \(\frac{4}{12}\) [1 mark]. Simplifying by dividing the numerator and denominator by their highest common factor, 4, gives \(\frac{4}{12} = \frac{1}{3}\) [1 mark]. So one third of the chores still need to be completed this week.
This question uses a diagram to test reading data before simplifying a fraction. The chart is divided into 12 equal boxes, one for each chore, and the shaded boxes represent chores already completed. Counting the diagram, 8 of the 12 boxes are shaded, so \(12 - 8 = 4\) boxes remain unshaded [1 mark].
The fraction of chores still to be completed is therefore the unshaded boxes over the total, \(\frac{4}{12}\) [1 mark]. Simplifying by dividing the numerator and denominator by their highest common factor, 4, gives \(\frac{4}{12} = \frac{1}{3}\) [1 mark]. So one third of the chores still need to be completed this week.
Question 7 Report
A delivery drone icon for a freezer-label icon, tracked as part of a corner shop, is translated twice on a coordinate grid. Point \(A\) starts at \((2, 1)\) and is translated by the vector \(\binom{x}{3}\) to reach the point \((6, 4)\).
(a) A translation by the vector \(\binom{x}{3}\) adds \(x\) to the starting \(x\)-coordinate and adds 3 to the starting \(y\)-coordinate. Point \(A\) starts at \((2, 1)\) and reaches \((6, 4)\), so looking only at the \(x\)-coordinates, \(2 + x = 6\), which gives \(x = 4\) [2 marks]. This can be checked against the \(y\)-coordinates too: \(1 + 3 = 4\), which matches the given image point.
(b) The image point after the first translation is \((6, 4)\). Applying the second translation \(\binom{-1}{y}\) takes this to \((5, 10)\), so looking at the \(y\)-coordinates, \(4 + y = 10\), which gives \(y = 6\) [2 marks]. Checking the \(x\)-coordinates: \(6 + (-1) = 5\), which matches the final point given.
(a) A translation by the vector \(\binom{x}{3}\) adds \(x\) to the starting \(x\)-coordinate and adds 3 to the starting \(y\)-coordinate. Point \(A\) starts at \((2, 1)\) and reaches \((6, 4)\), so looking only at the \(x\)-coordinates, \(2 + x = 6\), which gives \(x = 4\) [2 marks]. This can be checked against the \(y\)-coordinates too: \(1 + 3 = 4\), which matches the given image point.
(b) The image point after the first translation is \((6, 4)\). Applying the second translation \(\binom{-1}{y}\) takes this to \((5, 10)\), so looking at the \(y\)-coordinates, \(4 + y = 10\), which gives \(y = 6\) [2 marks]. Checking the \(x\)-coordinates: \(6 + (-1) = 5\), which matches the final point given.
Question 8 Report
A water company records the fraction of capacity currently held in three reservoirs, shown in the table.
| Reservoir | Fraction full | Full capacity (ML) |
|---|---|---|
| A | \(\frac{5}{8}\) | 4000 |
| B | \(\frac{3}{10}\) | 6000 |
| C | \(\frac{7}{12}\) | 3600 |
(a) This part tests finding a fraction of an amount from a table. Reservoir A's fraction full is \(\frac{5}{8}\), so the volume currently held is \(\frac{5}{8} \times 4000 = \frac{5 \times 4000}{8} = 2500\) megalitres [2 marks].
(b) Reservoir B holds \(\frac{3}{10} \times 6000 = 1800\) megalitres, and Reservoir C holds \(\frac{7}{12} \times 3600 = 2100\) megalitres [2 marks]. Adding all three reservoirs together, \(2500 + 1800 + 2100 = 6400\) megalitres are currently held in total [1 mark].
(a) This part tests finding a fraction of an amount from a table. Reservoir A's fraction full is \(\frac{5}{8}\), so the volume currently held is \(\frac{5}{8} \times 4000 = \frac{5 \times 4000}{8} = 2500\) megalitres [2 marks].
(b) Reservoir B holds \(\frac{3}{10} \times 6000 = 1800\) megalitres, and Reservoir C holds \(\frac{7}{12} \times 3600 = 2100\) megalitres [2 marks]. Adding all three reservoirs together, \(2500 + 1800 + 2100 = 6400\) megalitres are currently held in total [1 mark].
Question 9 Report
At the school sports day, a stall sells 260 bottles of water in total during the afternoon. 45% of these bottles are sold to spectators watching from the stands, and the rest are sold to competitors.
Work out how many bottles are sold to competitors. (3)
Since every bottle sold goes to either a spectator or a competitor, the number bought by competitors is found by subtracting the spectators' share from the total sold.
The number of bottles sold to spectators is 45% of the 260 total:
\[260 \times 0.45 = 117\] [1 mark]
The remaining bottles are sold to competitors:
\[260 - 117\] [1 mark]
\[= 143 \text{ bottles}\] [1 mark]
As a check, \(143 \div 260 \times 100 = 55\%\), which is exactly \(100\% - 45\%\), confirming the two groups of buyers make up all 260 bottles.
Since every bottle sold goes to either a spectator or a competitor, the number bought by competitors is found by subtracting the spectators' share from the total sold.
The number of bottles sold to spectators is 45% of the 260 total:
\[260 \times 0.45 = 117\] [1 mark]
The remaining bottles are sold to competitors:
\[260 - 117\] [1 mark]
\[= 143 \text{ bottles}\] [1 mark]
As a check, \(143 \div 260 \times 100 = 55\%\), which is exactly \(100\% - 45\%\), confirming the two groups of buyers make up all 260 bottles.
Question 10 Report
Households on a street are surveyed by a local researcher about their weekly grocery spend, rounded to the nearest £10, as part of a study into the cost of living. The table shows the amount spent and the number of households reporting it.
| Amount spent (£) | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| Number of households | 3 | 7 | 9 | 5 | 2 |
The mode is read straight from the highest frequency, the range compares the extreme amounts spent, and the mean uses every household's spending weighted by how many households reported it.
The mode is read straight from the highest frequency, the range compares the extreme amounts spent, and the mean uses every household's spending weighted by how many households reported it.
Question 11 Report
A tailor charges a fixed fee for cutting a pattern plus an amount for every metre of fabric bought. The total cost, \(y\) pounds, for \(x\) metres of fabric follows a straight-line rule that passes through the points \((3, 26)\) and \((7, 42)\).
Find the equation of this line in the form \(y = mx + c\). (3)
The tailor's total cost is a straight-line rule, so knowing two points on that line is enough to find both the cost per metre (the gradient) and the fixed cutting fee (the intercept).
Using the two given points, \((3, 26)\) and \((7, 42)\), the gradient is:
\[\frac{42 - 26}{7 - 3} = \frac{16}{4} = 4 \quad \textbf{[1 mark]}\]
Substituting the point \((3, 26)\) and gradient 4 into \(y = mx + c\):
\[26 = 4(3) + c \implies 26 = 12 + c \implies c = 14 \quad \textbf{[1 mark]}\]
So the equation of the line is:
\[y = 4x + 14 \quad \textbf{[1 mark]}\]
Here the gradient, 4, is the cost per metre of fabric, and the intercept, 14, is the fixed fee for cutting the pattern.
Exam tip: when a straight-line rule is described only through two points, always find the gradient first using both points, then substitute one point back in to find the intercept.
The tailor's total cost is a straight-line rule, so knowing two points on that line is enough to find both the cost per metre (the gradient) and the fixed cutting fee (the intercept).
Using the two given points, \((3, 26)\) and \((7, 42)\), the gradient is:
\[\frac{42 - 26}{7 - 3} = \frac{16}{4} = 4 \quad \textbf{[1 mark]}\]
Substituting the point \((3, 26)\) and gradient 4 into \(y = mx + c\):
\[26 = 4(3) + c \implies 26 = 12 + c \implies c = 14 \quad \textbf{[1 mark]}\]
So the equation of the line is:
\[y = 4x + 14 \quad \textbf{[1 mark]}\]
Here the gradient, 4, is the cost per metre of fabric, and the intercept, 14, is the fixed fee for cutting the pattern.
Exam tip: when a straight-line rule is described only through two points, always find the gradient first using both points, then substitute one point back in to find the intercept.
Question 12 Report
A building firm mixes concrete in equal batches. Each batch uses 62.4 kg of cement, and 3 batches are needed for a job. Cement costs £0.20 per kg.
This question tests multiplying a mass per batch by the number of batches, finding the cost of that mass, and comparing the cost with a fixed budget.
Finding the total mass first, before applying the price per kilogram, avoids having to multiply three numbers together in a single step.
This question tests multiplying a mass per batch by the number of batches, finding the cost of that mass, and comparing the cost with a fixed budget.
Finding the total mass first, before applying the price per kilogram, avoids having to multiply three numbers together in a single step.
Question 13 Report
The total time, \(T\) minutes, for an athlete to complete a sports day circuit is given by \(T = 4s + 2h\), where \(s\) is the number of running stations and \(h\) is the number of hurdles.
(a) Substituting \(s = 6\) and \(h = 3\) into \(T = 4s + 2h\) gives \(T = 4(6) + 2(3) = 24 + 6 = 30\) minutes [2 marks].
(b) In the formula, \(2h\) is the part of the total time contributed by the hurdles: it represents the total time added by the hurdles, at 2 minutes for each of the \(h\) hurdles [1 mark].
(c) Subtracting \(4s\) from both sides of \(T = 4s + 2h\) gives \(T - 4s = 2h\) [1 mark], and dividing both sides by 2 gives \(h = \frac{T - 4s}{2}\) [1 mark].
(a) Substituting \(s = 6\) and \(h = 3\) into \(T = 4s + 2h\) gives \(T = 4(6) + 2(3) = 24 + 6 = 30\) minutes [2 marks].
(b) In the formula, \(2h\) is the part of the total time contributed by the hurdles: it represents the total time added by the hurdles, at 2 minutes for each of the \(h\) hurdles [1 mark].
(c) Subtracting \(4s\) from both sides of \(T = 4s + 2h\) gives \(T - 4s = 2h\) [1 mark], and dividing both sides by 2 gives \(h = \frac{T - 4s}{2}\) [1 mark].
Question 14 Report
A corner shop orders 80 punnets of strawberries at the start of the week and sells them for £1.20 each. By Thursday, 65% of the punnets have sold at the full price. On Friday, the shop reduces the remaining punnets by 40% to clear the stock, and all of these sell.
This question tracks stock sold at two different prices across two days, before combining both days' income and comparing the week's total with what the shop paid its supplier.
(a) Punnets sold at full price by Thursday:
\[80 \times 0.65 = 52 \text{ punnets}\] [1 mark]
(b) The 40% reduction on Friday multiplies the £1.20 price by \(0.60\):
\[1.20 \times 0.60\] [1 mark]
\[= \pounds 0.72\] [1 mark]
(c) The remaining \(80 - 52 = 28\) punnets all sell at this reduced price on Friday:
\[28 \times 0.72\] [1 mark]
\[= \pounds 20.16\]
(d) Thursday's full-price income is \(52 \times 1.20 = \pounds 62.40\). Adding Friday's £20.16 gives the week's total income:
\[62.40 + 20.16 = \pounds 82.56\] [1 mark]
Since \(\pounds 82.56 > \pounds 60\), this is more than the £60 paid to the supplier, so the shop makes a profit on the batch [1 mark]. The full week's income must combine both the full-price sales up to Thursday and the reduced-price clearance on Friday, since neither day's takings alone represent the whole week.
This question tracks stock sold at two different prices across two days, before combining both days' income and comparing the week's total with what the shop paid its supplier.
(a) Punnets sold at full price by Thursday:
\[80 \times 0.65 = 52 \text{ punnets}\] [1 mark]
(b) The 40% reduction on Friday multiplies the £1.20 price by \(0.60\):
\[1.20 \times 0.60\] [1 mark]
\[= \pounds 0.72\] [1 mark]
(c) The remaining \(80 - 52 = 28\) punnets all sell at this reduced price on Friday:
\[28 \times 0.72\] [1 mark]
\[= \pounds 20.16\]
(d) Thursday's full-price income is \(52 \times 1.20 = \pounds 62.40\). Adding Friday's £20.16 gives the week's total income:
\[62.40 + 20.16 = \pounds 82.56\] [1 mark]
Since \(\pounds 82.56 > \pounds 60\), this is more than the £60 paid to the supplier, so the shop makes a profit on the batch [1 mark]. The full week's income must combine both the full-price sales up to Thursday and the reduced-price clearance on Friday, since neither day's takings alone represent the whole week.
Question 15 Report
The organisers arrange energy drink crates into a perfect cube stack for the sports day refreshment stall, using 125 crates in total.
A cube number is a whole number multiplied by itself three times; finding a cube root means finding the whole number that produces a given cube.
(a) Testing whole numbers, \(5 \times 5 \times 5 = 125\), so the cube root of 125 is 5. [1 mark]
(b) Squaring this value: \(5^2 = 25\). [1 mark]
(c) Using one fewer crate along each edge gives a cube of side \(5 - 1 = 4\), so the second stack contains \(4^3 = 4 \times 4 \times 4 = 64\) crates. [2 marks]
Reducing the edge length by just 1 crate does not reduce the total by a small fixed amount: because volume scales with the cube of the edge length, the total drops from 125 crates to 64, a loss of 61 crates from removing only one layer off each edge.
A cube number is a whole number multiplied by itself three times; finding a cube root means finding the whole number that produces a given cube.
(a) Testing whole numbers, \(5 \times 5 \times 5 = 125\), so the cube root of 125 is 5. [1 mark]
(b) Squaring this value: \(5^2 = 25\). [1 mark]
(c) Using one fewer crate along each edge gives a cube of side \(5 - 1 = 4\), so the second stack contains \(4^3 = 4 \times 4 \times 4 = 64\) crates. [2 marks]
Reducing the edge length by just 1 crate does not reduce the total by a small fixed amount: because volume scales with the cube of the edge length, the total drops from 125 crates to 64, a loss of 61 crates from removing only one layer off each edge.
Question 16 Report
A corner shop sells rice in bags. The cost of a bag of rice is directly proportional to its mass. A bag with a mass of 2.4 kg costs £4.32.
Because the cost of the rice is directly proportional to its mass, the price per kilogram is a fixed rate found by dividing the given cost by the given mass; that rate then applies to any other mass.
(a) The cost per kilogram is \(4.32 \div 2.4 = £1.80\). [1 mark]
(b) Multiplying the rate by the new mass: \(1.80 \times 3.5 = £6.30\). [1 mark]
This is the same "find the rate, then scale" method used throughout direct proportion problems: once £1.80 per kg is established, it applies to a bag of any size.
Because the cost of the rice is directly proportional to its mass, the price per kilogram is a fixed rate found by dividing the given cost by the given mass; that rate then applies to any other mass.
(a) The cost per kilogram is \(4.32 \div 2.4 = £1.80\). [1 mark]
(b) Multiplying the rate by the new mass: \(1.80 \times 3.5 = £6.30\). [1 mark]
This is the same "find the rate, then scale" method used throughout direct proportion problems: once £1.80 per kg is established, it applies to a bag of any size.
Question 17 Report
At the school sports day, the winning relay team is presented with a small wooden trophy in the shape of a cube. Each edge of the trophy measures 5 cm, as shown in the diagram.
Work out the volume of the trophy. (2)
A cube has all three dimensions equal, so its volume is simply the edge length multiplied by itself three times.
With edge length 5 cm:
\[5 \times 5 \times 5 = 125 \text{ cm}^3 \quad \textbf{[2 marks]}\]
Exam tip: a cube's volume formula, \(a^3\), is a special case of the cuboid formula \(l \times w \times h\) where all three lengths happen to be equal.
A cube has all three dimensions equal, so its volume is simply the edge length multiplied by itself three times.
With edge length 5 cm:
\[5 \times 5 \times 5 = 125 \text{ cm}^3 \quad \textbf{[2 marks]}\]
Exam tip: a cube's volume formula, \(a^3\), is a special case of the cuboid formula \(l \times w \times h\) where all three lengths happen to be equal.
Question 18 Report
Two bus routes run past a transport hub near a corner shop. Route X departs every \(2^1\) minutes, while Route Y departs every \(2^{-1}\) hours. The hub manager compares both intervals before adjusting the printed timetable.
This question compares two bus intervals given in different units, so a unit conversion is needed before the intervals can be compared directly.
(a) Route X's interval is \(2^1\) minutes: \[ 2^1 = 2 \text{ minutes} \] [1 mark]
(b) Route Y's interval is \(2^{-1}\) hours; evaluate the power first, then convert to minutes: \[ 2^{-1} \text{ hours} = \frac{1}{2} \text{ hour} = 0.5 \times 60 = 30 \text{ minutes} \] [2 marks]
(c) With both intervals in the same unit, divide the longer interval by the shorter one to count how many Route X buses fit between two Route Y buses: \[ 30 \div 2 = 15 \text{ buses} \] [2 marks]
The question is a reminder that index laws only combine values with a common base and common units; converting hours to minutes first is essential before any comparison or division makes sense.
This question compares two bus intervals given in different units, so a unit conversion is needed before the intervals can be compared directly.
(a) Route X's interval is \(2^1\) minutes: \[ 2^1 = 2 \text{ minutes} \] [1 mark]
(b) Route Y's interval is \(2^{-1}\) hours; evaluate the power first, then convert to minutes: \[ 2^{-1} \text{ hours} = \frac{1}{2} \text{ hour} = 0.5 \times 60 = 30 \text{ minutes} \] [2 marks]
(c) With both intervals in the same unit, divide the longer interval by the shorter one to count how many Route X buses fit between two Route Y buses: \[ 30 \div 2 = 15 \text{ buses} \] [2 marks]
The question is a reminder that index laws only combine values with a common base and common units; converting hours to minutes first is essential before any comparison or division makes sense.
Question 19 Report
For school sports day, a hexagonal post cap holds a triangular banner. Five of the post cap's interior angles are recorded in the table below.
| Corner | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Interior angle | 125 | 110 | 135 | 118 | 122 | x |
This question combines a polygon's interior angle sum, the exterior/interior angle relationship, and a triangle's angle sum with an isosceles check.
(a) The interior angles of a hexagon sum to \((6-2) \times 180 = 720^\circ\). [1 mark]
(b) Setting the five known angles plus \(x\) equal to this total: \(125+110+135+118+122+x = 720\), so \(610 + x = 720\), giving \(x = 110\). [2 marks]
(c) The exterior angle at that corner and the interior angle \(x\) lie on a straight line, so the exterior angle is \(180 - 110 = 70^\circ\). [1 mark]
(d) Using this \(70^\circ\) as one angle of the banner, with a second angle of \(55^\circ\), the banner's third angle is \(180 - 70 - 55 = 55^\circ\). [2 marks]
(e) Since two of the banner's angles are both \(55^\circ\), yes, the banner is isosceles (equal angles mean the sides opposite them are also equal). [1 mark]
This question links three separate angle facts in sequence: a hexagon's fixed interior-angle total, the straight-line relationship between interior and exterior angles, and the triangle angle sum, each feeding the next calculation.
This question combines a polygon's interior angle sum, the exterior/interior angle relationship, and a triangle's angle sum with an isosceles check.
(a) The interior angles of a hexagon sum to \((6-2) \times 180 = 720^\circ\). [1 mark]
(b) Setting the five known angles plus \(x\) equal to this total: \(125+110+135+118+122+x = 720\), so \(610 + x = 720\), giving \(x = 110\). [2 marks]
(c) The exterior angle at that corner and the interior angle \(x\) lie on a straight line, so the exterior angle is \(180 - 110 = 70^\circ\). [1 mark]
(d) Using this \(70^\circ\) as one angle of the banner, with a second angle of \(55^\circ\), the banner's third angle is \(180 - 70 - 55 = 55^\circ\). [2 marks]
(e) Since two of the banner's angles are both \(55^\circ\), yes, the banner is isosceles (equal angles mean the sides opposite them are also equal). [1 mark]
This question links three separate angle facts in sequence: a hexagon's fixed interior-angle total, the straight-line relationship between interior and exterior angles, and the triangle angle sum, each feeding the next calculation.
Question 20 Report
A quality inspector at a building site tested the mass, in kilograms, of 12 concrete blocks from one batch, marking each result on a dot plot before the load could be signed off.
Reading a dot plot means counting the stacked dots above each mass to rebuild the twelve individual readings: \(18, 19, 19, 20, 20, 20, 21, 21, 22, 23, 23, 24\) kg.
Reading a dot plot means counting the stacked dots above each mass to rebuild the twelve individual readings: \(18, 19, 19, 20, 20, 20, 21, 21, 22, 23, 23, 24\) kg.
Question 21 Report
A builder mixes a batch of concrete and pours it into a rectangular foundation strip. The strip is 2.5 m long, 1.2 m wide and 0.4 m deep. He needs the volume before ordering the ready-mixed delivery, since the supplier charges by the cubic metre.
The volume of a rectangular concrete foundation strip is found by multiplying its three dimensions together, exactly as for any cuboid.
Substituting the three dimensions:
\[V = 2.5 \times 1.2 \times 0.4\]
\[V = 1.2 \text{ m}^3\] [2 marks]
Exam tip: check that all three dimensions are already given in the same unit (here, metres) before multiplying, since mixing units would give a volume in the wrong scale entirely.
The volume of a rectangular concrete foundation strip is found by multiplying its three dimensions together, exactly as for any cuboid.
Substituting the three dimensions:
\[V = 2.5 \times 1.2 \times 0.4\]
\[V = 1.2 \text{ m}^3\] [2 marks]
Exam tip: check that all three dimensions are already given in the same unit (here, metres) before multiplying, since mixing units would give a volume in the wrong scale entirely.
Question 22 Report
A school displays trophies from sports day in a growing block arrangement; the first three patterns below hold 9, 13 and 17 trophies.
An arithmetic sequence with a formula \(a_1 + (n-1)d\) uses the first term \(a_1\) and the common difference \(d\); a maximum capacity limits how large the pattern number can be, since beyond a certain point the pattern needs more trophies than the cabinet can hold.
The trophies rise from 9 by 4 each pattern, so the common difference is \[d = 4\] Using the first term 9 and this common difference, the \(n\)th term is \(9+(n-1)(4)\), which expands to \[4n + 5\] [2 marks]
Substituting \(n=8\) gives \[4(8) + 5 = 32 + 5 = 37 \text{ trophies}\] [1 mark]
Setting \(4n+5 \le 100\) and rearranging gives \(4n \le 95\), so \[n \le 23.75\] The greatest whole number satisfying this is \(n=23\), so Pattern 23 is the largest pattern that fits in the cabinet. [2 marks]
Pattern 25 needs \(4(25)+5=105\) trophies, which is more than the cabinet's limit of 100 trophies, so Pattern 25 could not be displayed. [1 mark]
An arithmetic sequence with a formula \(a_1 + (n-1)d\) uses the first term \(a_1\) and the common difference \(d\); a maximum capacity limits how large the pattern number can be, since beyond a certain point the pattern needs more trophies than the cabinet can hold.
The trophies rise from 9 by 4 each pattern, so the common difference is \[d = 4\] Using the first term 9 and this common difference, the \(n\)th term is \(9+(n-1)(4)\), which expands to \[4n + 5\] [2 marks]
Substituting \(n=8\) gives \[4(8) + 5 = 32 + 5 = 37 \text{ trophies}\] [1 mark]
Setting \(4n+5 \le 100\) and rearranging gives \(4n \le 95\), so \[n \le 23.75\] The greatest whole number satisfying this is \(n=23\), so Pattern 23 is the largest pattern that fits in the cabinet. [2 marks]
Pattern 25 needs \(4(25)+5=105\) trophies, which is more than the cabinet's limit of 100 trophies, so Pattern 25 could not be displayed. [1 mark]
Question 23 Report
A tool-hire depot stocks drills that are either cordless or corded. A drill is chosen at random from the depot. The probability that it is cordless is 0.58, and there are 87 cordless drills at the depot.
This question tests reversing "probability = frequency / total" to find a total, then using the complement to find the remaining count.
(a) Since \( P(\text{cordless}) = \dfrac{\text{number cordless}}{\text{total}} \), the total number of drills at the depot is:
\[ \text{Total} = 87 \div 0.58 = 150 \] [2 marks]
(b) Every drill is either cordless or corded, so the number of corded drills is the total minus the number of cordless drills:
\[ \text{Corded drills} = 150 - 87 = 63 \] [1 mark]
This can be checked using the complement probability: \( (1 - 0.58) \times 150 = 0.42 \times 150 = 63 \).
This question tests reversing "probability = frequency / total" to find a total, then using the complement to find the remaining count.
(a) Since \( P(\text{cordless}) = \dfrac{\text{number cordless}}{\text{total}} \), the total number of drills at the depot is:
\[ \text{Total} = 87 \div 0.58 = 150 \] [2 marks]
(b) Every drill is either cordless or corded, so the number of corded drills is the total minus the number of cordless drills:
\[ \text{Corded drills} = 150 - 87 = 63 \] [1 mark]
This can be checked using the complement probability: \( (1 - 0.58) \times 150 = 0.42 \times 150 = 63 \).
Question 24 Report
Two building site teams each had 10 workers. The back-to-back stem-and-leaf diagram shows the hours each worker logged in one week.
| Team A | Stem | Team B |
|---|---|---|
| 8 5 4 2 | 2 | 0 3 6 9 |
| 8 5 3 1 | 3 | 0 2 4 7 9 |
| 4 1 | 4 | 2 |
| Key: Team A 2 | 2 means 22 hours; Team B 2 | 0 means 20 hours | ||
This question tests reading two sets of data from a back-to-back stem-and-leaf diagram, finding each team's median, comparing them against a bonus threshold, and correcting a total for omitted workers.
(a) Team A's ten hours, read from the diagram and placed in order, are 22, 24, 25, 28, 31, 33, 35, 38, 41, 44. With 10 values, the median is the mean of the 5th and 6th values:
\[ \frac{31+33}{2} = 32 \text{ hours} \] [1 mark]
(b) Team B's ten hours in order are 20, 23, 26, 29, 30, 32, 34, 37, 39, 42, giving a median of
\[ \frac{30+32}{2} = 31 \text{ hours} \] [1 mark]
(c) The bonus requires a median exceeding 31 hours. Team A's median of 32 hours exceeds 31, but Team B's median of exactly 31 hours does not, so only Team A qualifies for the bonus. [1 mark]
(d) Adding all ten of Team A's logged hours from the diagram gives 321 hours. Including the 2 extra workers who each logged 20 hours but were omitted from the report,
\[ 321 + (2 \times 20) = 361 \text{ hours} \] [1 mark]
The reported total of 320 hours in part (d)'s scenario is close to, but not exactly, the diagram's own total of 321 hours; either way, the two extra workers' 40 hours must still be added on top of whichever total is being corrected.
This question tests reading two sets of data from a back-to-back stem-and-leaf diagram, finding each team's median, comparing them against a bonus threshold, and correcting a total for omitted workers.
(a) Team A's ten hours, read from the diagram and placed in order, are 22, 24, 25, 28, 31, 33, 35, 38, 41, 44. With 10 values, the median is the mean of the 5th and 6th values:
\[ \frac{31+33}{2} = 32 \text{ hours} \] [1 mark]
(b) Team B's ten hours in order are 20, 23, 26, 29, 30, 32, 34, 37, 39, 42, giving a median of
\[ \frac{30+32}{2} = 31 \text{ hours} \] [1 mark]
(c) The bonus requires a median exceeding 31 hours. Team A's median of 32 hours exceeds 31, but Team B's median of exactly 31 hours does not, so only Team A qualifies for the bonus. [1 mark]
(d) Adding all ten of Team A's logged hours from the diagram gives 321 hours. Including the 2 extra workers who each logged 20 hours but were omitted from the report,
\[ 321 + (2 \times 20) = 361 \text{ hours} \] [1 mark]
The reported total of 320 hours in part (d)'s scenario is close to, but not exactly, the diagram's own total of 321 hours; either way, the two extra workers' 40 hours must still be added on top of whichever total is being corrected.
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