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Question 1 Report
The distances driven on 100 delivery rounds are grouped like this: 40 rounds under 5 km, 36 rounds from 5 km up to 8 km, and 24 rounds from 8 km up to 20 km.
The three classes have different widths, so a histogram needs frequency density, and an estimated mean needs each class represented by its midpoint.
(a)
(b)
Notice how the densities rank differently from the frequencies: the widest class has the most spread-out data and so the shortest bar, even though it contains 24 rounds. Drawing bars of height 40, 36 and 24 would badly misrepresent the distribution.
The mean of 6.7 km is higher than the median class because the long 8 km to 20 km class pulls it up; with grouped data the answer is always an estimate, since the individual distances within each class are unknown.
The three classes have different widths, so a histogram needs frequency density, and an estimated mean needs each class represented by its midpoint.
(a)
(b)
Notice how the densities rank differently from the frequencies: the widest class has the most spread-out data and so the shortest bar, even though it contains 24 rounds. Drawing bars of height 40, 36 and 24 would badly misrepresent the distribution.
The mean of 6.7 km is higher than the median class because the long 8 km to 20 km class pulls it up; with grouped data the answer is always an estimate, since the individual distances within each class are unknown.
Question 2 Report
Bramley Farm harvested 640 tonnes of wheat last season. This season the harvest rose 25 per cent. Next season it is expected to fall 12 per cent.
Percentage changes are handled with multipliers, and each change is applied to the amount immediately before it, not to the original figure.
(a) A rise of \(25\) per cent means this season is \(125\) per cent of last season, a multiplier of \(1.25\):
\[640 \times 1.25\][1]
\[= 800 \text{ tonnes}\][1]
(b) A fall of \(12\) per cent leaves \(100 - 12 = 88\) per cent, a multiplier of \(0.88\). The question says the fall applies to next season relative to this season, so it acts on the \(800\) tonnes:
\[800 \times 0.88\][1]
\[= 704 \text{ tonnes}\][1]
Applying the \(12\) per cent fall to the original \(640\) tonnes is the error to avoid; a percentage change always refers to the figure it is measured from.
The two changes do not cancel out even though \(25\) and \(12\) might look close to balancing. The combined multiplier is \(1.25 \times 0.88 = 1.1\), so the expected harvest is \(10\) per cent above the \(640\) tonnes of last season, because the \(12\) per cent fall is taken from the larger figure of \(800\).
Percentage changes are handled with multipliers, and each change is applied to the amount immediately before it, not to the original figure.
(a) A rise of \(25\) per cent means this season is \(125\) per cent of last season, a multiplier of \(1.25\):
\[640 \times 1.25\][1]
\[= 800 \text{ tonnes}\][1]
(b) A fall of \(12\) per cent leaves \(100 - 12 = 88\) per cent, a multiplier of \(0.88\). The question says the fall applies to next season relative to this season, so it acts on the \(800\) tonnes:
\[800 \times 0.88\][1]
\[= 704 \text{ tonnes}\][1]
Applying the \(12\) per cent fall to the original \(640\) tonnes is the error to avoid; a percentage change always refers to the figure it is measured from.
The two changes do not cancel out even though \(25\) and \(12\) might look close to balancing. The combined multiplier is \(1.25 \times 0.88 = 1.1\), so the expected harvest is \(10\) per cent above the \(640\) tonnes of last season, because the \(12\) per cent fall is taken from the larger figure of \(800\).
Question 3 Report
A triangular table top in the school canteen has two edges meeting at 58°. One of them is 1.4 m long and the top covers 0.85 m².
(a) The area formula \(\frac{1}{2}ab\sin C\) uses the two sides that meet at the known angle, so it can be rearranged to find the missing side.
(b) Two sides and the angle between them are now known, so the cosine rule gives the third edge.
Use the unrounded 1.43185 in part (b) rather than 1.43, since the value is squared and doubled in the formula. The third edge is shorter than either of the other two, which is consistent with the 58° angle being the smallest angle in the triangle.
(a) The area formula \(\frac{1}{2}ab\sin C\) uses the two sides that meet at the known angle, so it can be rearranged to find the missing side.
(b) Two sides and the angle between them are now known, so the cosine rule gives the third edge.
Use the unrounded 1.43185 in part (b) rather than 1.43, since the value is squared and doubled in the formula. The third edge is shorter than either of the other two, which is consistent with the 58° angle being the smallest angle in the triangle.
Question 4 Report
A triangular reading table in the community library is the triangle ABC. \(AB = 1.8\) m, \(AC = 2.4\) m and angle BAC is 74°.
Two sides and the angle between them are known, so the cosine rule is the tool for the third side. The included angle \(BAC\) sits between \(AB\) and \(AC\) and faces the side \(BC\).
(a)
\[BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC)\] \[= 1.8^2 + 2.4^2 - 2 \times 1.8 \times 2.4 \times \cos 74^{\circ} = 3.24 + 5.76 - 8.64 \times 0.275637...\] \[= 9 - 2.3815... = 6.6185...\][1]
\[BC = \sqrt{6.6185...} = 2.5726...\]so \(BC = 2.57\) m correct to 3 significant figures. [1]
The cosine rule generalises Pythagoras: if the angle were \(90\)° then \(\cos 90^{\circ} = 0\) and the last term would vanish. Because \(74\)° is acute the term is subtracted, making \(BC\) shorter than the \(3\) m that Pythagoras would give.
(b) With a full side and its opposite angle now available, the sine rule is quicker than a second cosine rule. Angle \(ABC\) is opposite \(AC = 2.4\) m, and angle \(BAC = 74\)° is opposite \(BC\):
\[\frac{\sin(ABC)}{2.4} = \frac{\sin 74^{\circ}}{2.5726...} \implies \sin(ABC) = \frac{2.4 \times \sin 74^{\circ}}{2.5726...} = 0.89683...\][1]
\[\text{angle } ABC = \sin^{-1}(0.89683...) = 63.734...\]so angle \(ABC = 63.7\)° correct to 1 decimal place. [1]
The obtuse alternative \(116.3\)° is rejected because it would push the angle sum past \(180\)° when combined with the \(74\)° already known.
(c) Two sides and the included angle again, this time for area:
\[\text{Area} = \frac{1}{2} \times 1.8 \times 2.4 \times \sin 74^{\circ} = 2.16 \times 0.961262... = 2.0763...\]so the area is \(2.08\) m\(^2\) correct to 3 significant figures. [1]
(d) The edging runs round the perimeter:
\[1.8 + 2.4 + 2.5726... = 6.7726... \text{ m}\]Since \(6.77\) m is less than \(7\) m, one length is enough. [1]
Use the unrounded \(BC\) in the perimeter. The margin is only about \(0.23\) m, so this is a genuinely close comparison rather than an obvious one.
Two sides and the angle between them are known, so the cosine rule is the tool for the third side. The included angle \(BAC\) sits between \(AB\) and \(AC\) and faces the side \(BC\).
(a)
\[BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC)\] \[= 1.8^2 + 2.4^2 - 2 \times 1.8 \times 2.4 \times \cos 74^{\circ} = 3.24 + 5.76 - 8.64 \times 0.275637...\] \[= 9 - 2.3815... = 6.6185...\][1]
\[BC = \sqrt{6.6185...} = 2.5726...\]so \(BC = 2.57\) m correct to 3 significant figures. [1]
The cosine rule generalises Pythagoras: if the angle were \(90\)° then \(\cos 90^{\circ} = 0\) and the last term would vanish. Because \(74\)° is acute the term is subtracted, making \(BC\) shorter than the \(3\) m that Pythagoras would give.
(b) With a full side and its opposite angle now available, the sine rule is quicker than a second cosine rule. Angle \(ABC\) is opposite \(AC = 2.4\) m, and angle \(BAC = 74\)° is opposite \(BC\):
\[\frac{\sin(ABC)}{2.4} = \frac{\sin 74^{\circ}}{2.5726...} \implies \sin(ABC) = \frac{2.4 \times \sin 74^{\circ}}{2.5726...} = 0.89683...\][1]
\[\text{angle } ABC = \sin^{-1}(0.89683...) = 63.734...\]so angle \(ABC = 63.7\)° correct to 1 decimal place. [1]
The obtuse alternative \(116.3\)° is rejected because it would push the angle sum past \(180\)° when combined with the \(74\)° already known.
(c) Two sides and the included angle again, this time for area:
\[\text{Area} = \frac{1}{2} \times 1.8 \times 2.4 \times \sin 74^{\circ} = 2.16 \times 0.961262... = 2.0763...\]so the area is \(2.08\) m\(^2\) correct to 3 significant figures. [1]
(d) The edging runs round the perimeter:
\[1.8 + 2.4 + 2.5726... = 6.7726... \text{ m}\]Since \(6.77\) m is less than \(7\) m, one length is enough. [1]
Use the unrounded \(BC\) in the perimeter. The margin is only about \(0.23\) m, so this is a genuinely close comparison rather than an obvious one.
Question 5 Report
A corner shop has \(12\frac{1}{2}\) kg of flour. Each customer order uses \(1\frac{3}{4}\) kg.
Dividing by a mixed number is much safer once both quantities are improper fractions, because the division rule applies directly to those.
(a) Converting:
\[12\frac{1}{2} = \frac{25}{2} \qquad \text{and} \qquad 1\frac{3}{4} = \frac{7}{4}\][1]
Dividing by a fraction is the same as multiplying by its reciprocal, so turn the second fraction upside down:
\[\frac{25}{2} \div \frac{7}{4} = \frac{25}{2} \times \frac{4}{7} = \frac{100}{14} = \frac{50}{7}\][1]
\[\frac{50}{7} = 7\frac{1}{7}\]Only whole orders can be filled, so the answer is rounded down: \(7\) complete orders. [1]
Rounding down rather than to the nearest whole number is essential here, since the eighth order could not be completed with the flour that remains.
(b) The flour used by \(7\) orders is
\[7 \times \frac{7}{4} = \frac{49}{4} \text{ kg}\]Subtracting from the stock, over a common denominator of \(4\):
\[\frac{25}{2} - \frac{49}{4} = \frac{50}{4} - \frac{49}{4} = \frac{1}{4} \text{ kg}\][1]
The leftover of \(\frac{1}{4}\) kg agrees with the \(\frac{1}{7}\) of an order left in part (a), since \(\frac{1}{7} \times \frac{7}{4} = \frac{1}{4}\) kg. Note that \(\frac{1}{7}\) is a fraction of an order, not a mass in kilograms, which is why part (b) needs its own calculation.
Dividing by a mixed number is much safer once both quantities are improper fractions, because the division rule applies directly to those.
(a) Converting:
\[12\frac{1}{2} = \frac{25}{2} \qquad \text{and} \qquad 1\frac{3}{4} = \frac{7}{4}\][1]
Dividing by a fraction is the same as multiplying by its reciprocal, so turn the second fraction upside down:
\[\frac{25}{2} \div \frac{7}{4} = \frac{25}{2} \times \frac{4}{7} = \frac{100}{14} = \frac{50}{7}\][1]
\[\frac{50}{7} = 7\frac{1}{7}\]Only whole orders can be filled, so the answer is rounded down: \(7\) complete orders. [1]
Rounding down rather than to the nearest whole number is essential here, since the eighth order could not be completed with the flour that remains.
(b) The flour used by \(7\) orders is
\[7 \times \frac{7}{4} = \frac{49}{4} \text{ kg}\]Subtracting from the stock, over a common denominator of \(4\):
\[\frac{25}{2} - \frac{49}{4} = \frac{50}{4} - \frac{49}{4} = \frac{1}{4} \text{ kg}\][1]
The leftover of \(\frac{1}{4}\) kg agrees with the \(\frac{1}{7}\) of an order left in part (a), since \(\frac{1}{7} \times \frac{7}{4} = \frac{1}{4}\) kg. Note that \(\frac{1}{7}\) is a fraction of an order, not a mass in kilograms, which is why part (b) needs its own calculation.
Question 6 Report
Two firms price a taxi ride. Ace charges \(y = 2x + 6\) pounds for \(x\) miles. Bell charges \(y = 3x + 2\) pounds. The grid shows both lines.
Each fare is a fixed charge plus a rate per mile, so on the graph the intercept is the fixed charge and the gradient is the charge per mile.
(a) The Ace line meets the vertical axis at 6, which is the fare for zero miles, so the fixed charge is £6. This matches the \(+6\) in \(y = 2x + 6\). [1]
(b) The two lines cross at a distance of 4 miles, so that is where the fares are equal. [1]
(c)
(d) Substituting into either formula: \(2\times 4 + 6 = 14\), and \(3\times 4 + 2 = 14\), so both firms charge £14 for that journey. [1]
(e) For 9 miles, Ace charges \(2\times 9 + 6 = 24\) pounds and Bell charges \(3\times 9 + 2 = 29\) pounds, a difference of £5. [1]
(f) For a journey such as the 9 mile ride, choose Ace: its gradient of 2 pounds per mile is smaller than Bell's 3 pounds per mile, so beyond the 4 mile crossing point Ace is always cheaper, and the gap widens by £1 for every further mile. [1]
The advice depends on the distance. Below 4 miles Bell is the cheaper firm, because its smaller fixed charge of £2 outweighs its higher rate; the crossing point is exactly where the two effects balance.
Each fare is a fixed charge plus a rate per mile, so on the graph the intercept is the fixed charge and the gradient is the charge per mile.
(a) The Ace line meets the vertical axis at 6, which is the fare for zero miles, so the fixed charge is £6. This matches the \(+6\) in \(y = 2x + 6\). [1]
(b) The two lines cross at a distance of 4 miles, so that is where the fares are equal. [1]
(c)
(d) Substituting into either formula: \(2\times 4 + 6 = 14\), and \(3\times 4 + 2 = 14\), so both firms charge £14 for that journey. [1]
(e) For 9 miles, Ace charges \(2\times 9 + 6 = 24\) pounds and Bell charges \(3\times 9 + 2 = 29\) pounds, a difference of £5. [1]
(f) For a journey such as the 9 mile ride, choose Ace: its gradient of 2 pounds per mile is smaller than Bell's 3 pounds per mile, so beyond the 4 mile crossing point Ace is always cheaper, and the gap widens by £1 for every further mile. [1]
The advice depends on the distance. Below 4 miles Bell is the cheaper firm, because its smaller fixed charge of £2 outweighs its higher rate; the crossing point is exactly where the two effects balance.
Question 7 Report
A trader at the local market loses money in any week when \(x^2 - 7x + 10\) takes a negative value. Here \(x\) is the number of hours the stall opens each day. Solve the inequality \(x^2 - 7x + 10 < 0\) to find the hours she should avoid. (3)
A quadratic inequality is solved in two stages: find where the expression equals zero, then decide which side of those values makes it negative.
Test a value in each region to confirm: at \(x = 0\) the expression is \(+10\), at \(x = 3\) it is \(9 - 21 + 10 = -2\), and at \(x = 6\) it is \(36 - 42 + 10 = +4\). Only the middle region is negative.
The classic error is writing \(x < 2\) or \(x > 5\), which is the answer to the opposite inequality. Deciding the shape of the parabola first, rather than guessing, prevents that.
A quadratic inequality is solved in two stages: find where the expression equals zero, then decide which side of those values makes it negative.
Test a value in each region to confirm: at \(x = 0\) the expression is \(+10\), at \(x = 3\) it is \(9 - 21 + 10 = -2\), and at \(x = 6\) it is \(36 - 42 + 10 = +4\). Only the middle region is negative.
The classic error is writing \(x < 2\) or \(x > 5\), which is the answer to the opposite inequality. Deciding the shape of the parabola first, rather than guessing, prevents that.
Question 8 Report
Each of 90 students at a sports day plays football, netball, both or neither. In the Venn diagram \(x\) play both and \(2x\) play neither.
The four regions of the diagram do not overlap and account for all \(90\) students, so their expressions add to \(90\). Reading the diagram, \(38\) play football only, \(x\) play both, \(25\) play netball only, and \(2x\) play neither.
(a)
\[38 + x + 25 + 2x = 90\][1]
Collecting terms, the constants give \(38 + 25 = 63\) and the \(x\) terms give \(x + 2x = 3x\):
\[63 + 3x = 90 \implies 3x = 27 \implies x = 9 \text{ as required.}\][1]
(b) The netball set covers two regions, netball only and the overlap:
\[25 + 9 = 34 \text{ netball players}\] \[P(\text{netball}) = \frac{34}{90} = \frac{17}{45}\][1]
(c) This is a conditional probability. The student is chosen from the netball players only, so the denominator is \(34\) rather than \(90\). Of those \(34\), the \(9\) in the overlap also play football. [1]
\[P(\text{football given netball}) = \frac{9}{34}\][1]
The change of denominator between parts (b) and (c) is the point being tested. In part (b) the student comes from the whole year group; in part (c) the pool has already been narrowed to netball players. Using \(\frac{9}{90}\) in part (c) would throw away the information the question supplies.
Substituting \(x = 9\) checks the diagram: the regions hold \(38\), \(9\), \(25\) and \(18\) students, totalling \(90\).
The four regions of the diagram do not overlap and account for all \(90\) students, so their expressions add to \(90\). Reading the diagram, \(38\) play football only, \(x\) play both, \(25\) play netball only, and \(2x\) play neither.
(a)
\[38 + x + 25 + 2x = 90\][1]
Collecting terms, the constants give \(38 + 25 = 63\) and the \(x\) terms give \(x + 2x = 3x\):
\[63 + 3x = 90 \implies 3x = 27 \implies x = 9 \text{ as required.}\][1]
(b) The netball set covers two regions, netball only and the overlap:
\[25 + 9 = 34 \text{ netball players}\] \[P(\text{netball}) = \frac{34}{90} = \frac{17}{45}\][1]
(c) This is a conditional probability. The student is chosen from the netball players only, so the denominator is \(34\) rather than \(90\). Of those \(34\), the \(9\) in the overlap also play football. [1]
\[P(\text{football given netball}) = \frac{9}{34}\][1]
The change of denominator between parts (b) and (c) is the point being tested. In part (b) the student comes from the whole year group; in part (c) the pool has already been narrowed to netball players. Using \(\frac{9}{90}\) in part (c) would throw away the information the question supplies.
Substituting \(x = 9\) checks the diagram: the regions hold \(38\), \(9\), \(25\) and \(18\) students, totalling \(90\).
Question 9 Report
Two similar conical heaps of grain stand in a barn. The mass of a heap is proportional to the cube of its height. The smaller heap is 1.2 m high and has mass 540 kg.
The figure shows two similar cones, one 1.2 m high and one 2 m high. Mass proportional to the cube of the height gives \(M = kh^3\).
Find the constant from the smaller heap: \(1.2^3 = 1.728\), so \(k = \frac{540}{1.728} = 312.5\) and \(M = 312.5h^3\).
(a)
(b)
The cube law follows from similarity: because the heaps are the same shape, every length scales by the same factor, so the volume, and with constant density the mass, scales by the cube of that factor. Part (a) can be checked that way: \(\left(\frac{2}{1.2}\right)^3 = 4.6296\ldots\) and \(540 \times 4.6296\ldots = 2500\) kg.
Cube the height before multiplying by \(k\), and take the cube root, not the square root, when reversing the process in part (b).
The figure shows two similar cones, one 1.2 m high and one 2 m high. Mass proportional to the cube of the height gives \(M = kh^3\).
Find the constant from the smaller heap: \(1.2^3 = 1.728\), so \(k = \frac{540}{1.728} = 312.5\) and \(M = 312.5h^3\).
(a)
(b)
The cube law follows from similarity: because the heaps are the same shape, every length scales by the same factor, so the volume, and with constant density the mass, scales by the cube of that factor. Part (a) can be checked that way: \(\left(\frac{2}{1.2}\right)^3 = 4.6296\ldots\) and \(540 \times 4.6296\ldots = 2500\) kg.
Cube the height before multiplying by \(k\), and take the cube root, not the square root, when reversing the process in part (b).
Question 10 Report
A farm keeps 90 animals. Set \(C\) is the animals kept in the north field and set \(D\) is the animals that are milked. Each region of the Venn diagram is written in terms of \(y\).
The four regions of the diagram do not overlap and between them account for every animal, so their expressions add to the total.
(a)
\[2y + (y - 4) + (y + 6) + 12\]Collecting the \(y\) terms gives \(2y + y + y = 4y\), and the constants give \(-4 + 6 + 12 = 14\):
\[= 4y + 14\][1]
The \(-4\) must be subtracted, not added; treating it as \(+4\) is the usual slip.
(b) The farm keeps \(90\) animals, so
\[4y + 14 = 90 \implies 4y = 76\][1]
\[y = 19\][1]
(c) \(C \cap D'\) means in \(C\) but not in \(D\), which is the part of the \(C\) circle outside the overlap:
\[n(C \cap D') = 2y = 38\][1]
(d) \((C \cup D)'\) is everything outside the union, so it is the region beyond both circles:
\[n((C \cup D)') = 12\][1]
Substituting \(y = 19\) checks the whole diagram: the regions hold \(38\), \(15\), \(25\) and \(12\) animals, totalling \(90\).
Note that \((C \cup D)'\) and \(C' \cap D'\) describe the same region, a result known as one of De Morgan's laws: being outside both circles is the same as being outside their union.
The four regions of the diagram do not overlap and between them account for every animal, so their expressions add to the total.
(a)
\[2y + (y - 4) + (y + 6) + 12\]Collecting the \(y\) terms gives \(2y + y + y = 4y\), and the constants give \(-4 + 6 + 12 = 14\):
\[= 4y + 14\][1]
The \(-4\) must be subtracted, not added; treating it as \(+4\) is the usual slip.
(b) The farm keeps \(90\) animals, so
\[4y + 14 = 90 \implies 4y = 76\][1]
\[y = 19\][1]
(c) \(C \cap D'\) means in \(C\) but not in \(D\), which is the part of the \(C\) circle outside the overlap:
\[n(C \cap D') = 2y = 38\][1]
(d) \((C \cup D)'\) is everything outside the union, so it is the region beyond both circles:
\[n((C \cup D)') = 12\][1]
Substituting \(y = 19\) checks the whole diagram: the regions hold \(38\), \(15\), \(25\) and \(12\) animals, totalling \(90\).
Note that \((C \cup D)'\) and \(C' \cap D'\) describe the same region, a result known as one of De Morgan's laws: being outside both circles is the same as being outside their union.
Question 11 Report
A school canteen models the cost of ingredients by the curve \(y = 3x^{2} + 2\). The region \(R\) is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).
(a) Substituting \(x = 3\) into \(y = 3x^{2} + 2\):
\[y = 3(9) + 2 = 29\][1]
(b) The area between a curve and the \(x\)-axis, from \(x = a\) to \(x = b\), is found by integrating the equation of the curve and evaluating between those limits. The region \(R\) shown runs from \(x = 1\) to \(x = 3\), and the curve lies above the \(x\)-axis throughout, so the integral gives the area directly.
Integrating raises each index by one and divides by the new index:
\[\int (3x^{2} + 2)\,dx = \frac{3x^{3}}{3} + 2x = x^{3} + 2x\][1]
Evaluating at the upper limit and then at the lower limit:
\[\text{at } x = 3: \quad 27 + 6 = 33 \qquad \text{at } x = 1: \quad 1 + 2 = 3\][1]
The area is the upper value minus the lower value:
\[\text{Area} = 33 - 3 = 30\][1]
The constant of integration is not needed for a definite integral, because it appears in both evaluations and cancels in the subtraction.
A rough check supports the answer: the region is \(2\) units wide and the curve rises from \(y = 5\) at \(x = 1\) to \(y = 29\) at \(x = 3\), so an area of \(30\) square units is the right order of size.
(a) Substituting \(x = 3\) into \(y = 3x^{2} + 2\):
\[y = 3(9) + 2 = 29\][1]
(b) The area between a curve and the \(x\)-axis, from \(x = a\) to \(x = b\), is found by integrating the equation of the curve and evaluating between those limits. The region \(R\) shown runs from \(x = 1\) to \(x = 3\), and the curve lies above the \(x\)-axis throughout, so the integral gives the area directly.
Integrating raises each index by one and divides by the new index:
\[\int (3x^{2} + 2)\,dx = \frac{3x^{3}}{3} + 2x = x^{3} + 2x\][1]
Evaluating at the upper limit and then at the lower limit:
\[\text{at } x = 3: \quad 27 + 6 = 33 \qquad \text{at } x = 1: \quad 1 + 2 = 3\][1]
The area is the upper value minus the lower value:
\[\text{Area} = 33 - 3 = 30\][1]
The constant of integration is not needed for a definite integral, because it appears in both evaluations and cancels in the subtraction.
A rough check supports the answer: the region is \(2\) units wide and the curve rises from \(y = 5\) at \(x = 1\) to \(y = 29\) at \(x = 3\), so an area of \(30\) square units is the right order of size.
Question 12 Report
Rain falling on a shed roof runs into a 150 litre water butt. The roof is the rectangle shown, each length correct to 1 decimal place. One storm drops 18 mm of rain, to the nearest millimetre.
The figure shows a rectangular roof measuring 3.2 m by 2.5 m. Each length is to 1 decimal place, so the true length lies between 3.15 m and 3.25 m and the true width between 2.45 m and 2.55 m.
(a) Area is a product, so it is greatest when both sides take their upper bounds.
(b) Lower bound \(= 3.15 \times 2.45 = 7.7175\) m\(^2\). [1]
(c) The rain forms a thin layer over the roof, so the volume is area times depth, with the depth converted to metres.
(d) The butt holds 150 litres, but as much as about 153 litres could arrive, and \(153 > 150\), so the butt could overflow. [1]
The rainfall of 18 mm to the nearest millimetre has a half-unit of 0.5 mm, giving an upper bound of 18.5 mm. Converting millimetres to metres by dividing by 1000, rather than by 100, is essential: using 0.185 m would give a volume ten times too large and a completely different conclusion.
The figure shows a rectangular roof measuring 3.2 m by 2.5 m. Each length is to 1 decimal place, so the true length lies between 3.15 m and 3.25 m and the true width between 2.45 m and 2.55 m.
(a) Area is a product, so it is greatest when both sides take their upper bounds.
(b) Lower bound \(= 3.15 \times 2.45 = 7.7175\) m\(^2\). [1]
(c) The rain forms a thin layer over the roof, so the volume is area times depth, with the depth converted to metres.
(d) The butt holds 150 litres, but as much as about 153 litres could arrive, and \(153 > 150\), so the butt could overflow. [1]
The rainfall of 18 mm to the nearest millimetre has a half-unit of 0.5 mm, giving an upper bound of 18.5 mm. Converting millimetres to metres by dividing by 1000, rather than by 100, is essential: using 0.185 m would give a volume ten times too large and a completely different conclusion.
Question 13 Report
A workshop makes closed cylindrical drums for a transport depot. Each drum has radius \(6\) cm and height \(14\) cm. Calculate the total surface area of one drum, correct to 3 significant figures. (3)
A closed cylinder has three surfaces: two circular ends and one curved surface. Imagine unrolling the curved surface: it becomes a rectangle whose width is the circumference \(2\pi r\) and whose height is \(h\), which is why its area is \(2\pi r h\).
The word "closed" is doing real work in this question. An open drum, such as a bucket, would have only one circular end, giving \(36\pi + 168\pi = 204\pi\). Adding the ends is also the step most often forgotten, leaving only the curved surface.
Collecting the terms as multiples of \(\pi\) and rounding only once at the end keeps the third significant figure reliable.
A closed cylinder has three surfaces: two circular ends and one curved surface. Imagine unrolling the curved surface: it becomes a rectangle whose width is the circumference \(2\pi r\) and whose height is \(h\), which is why its area is \(2\pi r h\).
The word "closed" is doing real work in this question. An open drum, such as a bucket, would have only one circular end, giving \(36\pi + 168\pi = 204\pi\). Adding the ends is also the step most often forgotten, leaving only the curved surface.
Collecting the terms as multiples of \(\pi\) and rounding only once at the end keeps the third significant figure reliable.
Question 14 Report
At a festival, a rope of length 34 metres runs round a rectangular first aid tent covering 60 square metres. The tent measures \(a\) metres by \(b\) metres.
Two facts are given about the same rectangle: the rope round it is its perimeter, and the ground it covers is its area.
(a)
(b) Divide the perimeter equation by 2 to get \(a + b = 17\), so \(b = 17 - a\). Substituting into the area equation gives \(a(17 - a) = 60\), that is \(17a - a^2 = 60\). Multiplying through by \(-1\) and rearranging gives \(a^2 - 17a + 60 = 0\) as required. [1]
(c)
Check both conditions: \(2(5 + 12) = 34\) metres of rope and \(5\times 12 = 60\) square metres of ground.
Here both roots are positive, so neither is rejected. Recognising that the pair of roots represents the pair of sides is exactly what the final mark rewards.
Two facts are given about the same rectangle: the rope round it is its perimeter, and the ground it covers is its area.
(a)
(b) Divide the perimeter equation by 2 to get \(a + b = 17\), so \(b = 17 - a\). Substituting into the area equation gives \(a(17 - a) = 60\), that is \(17a - a^2 = 60\). Multiplying through by \(-1\) and rearranging gives \(a^2 - 17a + 60 = 0\) as required. [1]
(c)
Check both conditions: \(2(5 + 12) = 34\) metres of rope and \(5\times 12 = 60\) square metres of ground.
Here both roots are positive, so neither is rejected. Recognising that the pair of roots represents the pair of sides is exactly what the final mark rewards.
Question 15 Report
Cold drinks at the club pool come in two similar cone shaped cups, 9 cm and 12 cm tall.
Between similar solids, lengths scale by \(k\), surface areas by \(k^2\) and volumes by \(k^3\).
(a) The scale factor from the smaller cup to the larger is \(\frac{12}{9} = \frac{4}{3}\). [1]
(b)
(c)
Part (c) runs in the opposite direction to parts (a) and (b), so the scale factor must be inverted before cubing; using \(\left(\frac{4}{3}\right)^3\) would enlarge rather than reduce and give 1517 ml, more than the larger cup itself.
Working with the fractions rather than decimals keeps the arithmetic exact, since \(\frac{16}{9}\) and \(\frac{27}{64}\) both cancel neatly against the given figures.
Between similar solids, lengths scale by \(k\), surface areas by \(k^2\) and volumes by \(k^3\).
(a) The scale factor from the smaller cup to the larger is \(\frac{12}{9} = \frac{4}{3}\). [1]
(b)
(c)
Part (c) runs in the opposite direction to parts (a) and (b), so the scale factor must be inverted before cubing; using \(\left(\frac{4}{3}\right)^3\) would enlarge rather than reduce and give 1517 ml, more than the larger cup itself.
Working with the fractions rather than decimals keeps the arithmetic exact, since \(\frac{16}{9}\) and \(\frac{27}{64}\) both cancel neatly against the given figures.
Question 16 Report
The diagram shows the floor of a new changing room. A square store cupboard fills one corner. The rest is used for changing. All lengths are in metres.
The figure shows the whole floor as a rectangle \((3x - 2)\) m by \((x + 1)\) m, with a square store cupboard of side \((x - 1)\) m filling the bottom-right corner. The changing area is what remains after the cupboard is taken out.
(a) Multiplying each term of the first bracket by each term of the second:
\[(3x - 2)(x + 1) = 3x^2 + 3x - 2x - 2 = 3x^2 + x - 2\][2]
(b) A square bracket means the bracket multiplied by itself, not each term squared:
\[(x - 1)^2 = (x - 1)(x - 1) = x^2 - x - x + 1 = x^2 - 2x + 1\][1]
Writing \(x^2 - 1\) here is the classic error; the middle term \(-2x\) comes from the two cross products.
(c) The changing floor is the whole rectangle minus the square cupboard. Keep the second expression bracketed so every term changes sign:
\[(3x^2 + x - 2) - (x^2 - 2x + 1) = 3x^2 + x - 2 - x^2 + 2x - 1 = 2x^2 + 3x - 3 \text{ m}^2 \text{ as required.}\][2]
(d) Setting the expression equal to the given area and collecting on one side:
\[2x^2 + 3x - 3 = 24 \implies 2x^2 + 3x - 27 = 0 \text{ as required.}\][1]
(e) Two numbers are needed that multiply to \(2 \times (-27) = -54\) and add to \(3\), namely \(9\) and \(-6\), which leads to
\[(x - 3)(2x + 9) = 0\]so \(x = 3\) or \(x = -4.5\). [1]
The negative root is rejected because it would make every side length negative. The root that fits the room is \(x = 3\). [1]
Checking: the whole floor is \(7\) m by \(4\) m, giving \(28\) m\(^2\); the cupboard is \(2\) m by \(2\) m, giving \(4\) m\(^2\); and \(28 - 4 = 24\) m\(^2\) as stated.
The figure shows the whole floor as a rectangle \((3x - 2)\) m by \((x + 1)\) m, with a square store cupboard of side \((x - 1)\) m filling the bottom-right corner. The changing area is what remains after the cupboard is taken out.
(a) Multiplying each term of the first bracket by each term of the second:
\[(3x - 2)(x + 1) = 3x^2 + 3x - 2x - 2 = 3x^2 + x - 2\][2]
(b) A square bracket means the bracket multiplied by itself, not each term squared:
\[(x - 1)^2 = (x - 1)(x - 1) = x^2 - x - x + 1 = x^2 - 2x + 1\][1]
Writing \(x^2 - 1\) here is the classic error; the middle term \(-2x\) comes from the two cross products.
(c) The changing floor is the whole rectangle minus the square cupboard. Keep the second expression bracketed so every term changes sign:
\[(3x^2 + x - 2) - (x^2 - 2x + 1) = 3x^2 + x - 2 - x^2 + 2x - 1 = 2x^2 + 3x - 3 \text{ m}^2 \text{ as required.}\][2]
(d) Setting the expression equal to the given area and collecting on one side:
\[2x^2 + 3x - 3 = 24 \implies 2x^2 + 3x - 27 = 0 \text{ as required.}\][1]
(e) Two numbers are needed that multiply to \(2 \times (-27) = -54\) and add to \(3\), namely \(9\) and \(-6\), which leads to
\[(x - 3)(2x + 9) = 0\]so \(x = 3\) or \(x = -4.5\). [1]
The negative root is rejected because it would make every side length negative. The root that fits the room is \(x = 3\). [1]
Checking: the whole floor is \(7\) m by \(4\) m, giving \(28\) m\(^2\); the cupboard is \(2\) m by \(2\) m, giving \(4\) m\(^2\); and \(28 - 4 = 24\) m\(^2\) as stated.
Question 17 Report
A community library is designing a floor pattern. The grid shows triangles \(P\), \(Q\) and \(R\).
Reading the vertices off the grid first makes every part straightforward. From the diagram, \(P\) has vertices \((1, 1)\), \((3, 1)\) and \((1, 4)\); \(Q\) has \((5, 1)\), \((7, 1)\) and \((5, 4)\); and \(R\) has \((-1, 1)\), \((-3, 1)\) and \((-1, 4)\).
(a) Comparing corresponding vertices, \((1, 1) \to (5, 1)\), \((3, 1) \to (7, 1)\) and \((1, 4) \to (5, 4)\). Every point moves \(4\) to the right and \(0\) up, and the triangle keeps the same orientation and size. That is a translation. [1]
The translation vector is \(\begin{pmatrix} 4 \\ 0 \end{pmatrix}\). [1]
A translation is described fully by its vector alone; no centre or line is needed.
(b) Here \((1, 1) \to (-1, 1)\), \((3, 1) \to (-3, 1)\) and \((1, 4) \to (-1, 4)\). The \(y\)-coordinates are unchanged while the \(x\)-coordinates change sign, and the triangle is turned over rather than slid. That is a reflection. [1]
The mirror line is the \(y\)-axis, the line \(x = 0\). [1]
A reflection is described fully by naming the mirror line, so both the word and the line are needed for the two marks.
(c) Reflecting in the \(x\)-axis leaves \(x\) unchanged and reverses the sign of \(y\), so \((x, y) \to (x, -y)\). Applying this to \(R\):
\[(-1, 1) \to (-1, -1), \quad (-3, 1) \to (-3, -1), \quad (-1, 4) \to (-1, -4)\]Triangle \(T\) has vertices \((-1, -1)\), \((-3, -1)\) and \((-1, -4)\). [1]
(d) She is correct. A rotation of \(180\)° about the origin sends \((x, y)\) to \((-x, -y)\). Applying that to \(P\):
\[(1, 1) \to (-1, -1), \quad (3, 1) \to (-3, -1), \quad (1, 4) \to (-1, -4)\]These are exactly the vertices of \(T\) found in part (c), so the rotation does map \(P\) onto \(T\). [1]
This is worth noticing as a general result: a reflection in the \(y\)-axis followed by a reflection in the \(x\)-axis is equivalent to a single \(180\)° rotation about the origin, which is precisely the route \(P \to R \to T\) taken here. A rotation of \(180\)° needs no direction stated, since clockwise and anticlockwise give the same image.
Reading the vertices off the grid first makes every part straightforward. From the diagram, \(P\) has vertices \((1, 1)\), \((3, 1)\) and \((1, 4)\); \(Q\) has \((5, 1)\), \((7, 1)\) and \((5, 4)\); and \(R\) has \((-1, 1)\), \((-3, 1)\) and \((-1, 4)\).
(a) Comparing corresponding vertices, \((1, 1) \to (5, 1)\), \((3, 1) \to (7, 1)\) and \((1, 4) \to (5, 4)\). Every point moves \(4\) to the right and \(0\) up, and the triangle keeps the same orientation and size. That is a translation. [1]
The translation vector is \(\begin{pmatrix} 4 \\ 0 \end{pmatrix}\). [1]
A translation is described fully by its vector alone; no centre or line is needed.
(b) Here \((1, 1) \to (-1, 1)\), \((3, 1) \to (-3, 1)\) and \((1, 4) \to (-1, 4)\). The \(y\)-coordinates are unchanged while the \(x\)-coordinates change sign, and the triangle is turned over rather than slid. That is a reflection. [1]
The mirror line is the \(y\)-axis, the line \(x = 0\). [1]
A reflection is described fully by naming the mirror line, so both the word and the line are needed for the two marks.
(c) Reflecting in the \(x\)-axis leaves \(x\) unchanged and reverses the sign of \(y\), so \((x, y) \to (x, -y)\). Applying this to \(R\):
\[(-1, 1) \to (-1, -1), \quad (-3, 1) \to (-3, -1), \quad (-1, 4) \to (-1, -4)\]Triangle \(T\) has vertices \((-1, -1)\), \((-3, -1)\) and \((-1, -4)\). [1]
(d) She is correct. A rotation of \(180\)° about the origin sends \((x, y)\) to \((-x, -y)\). Applying that to \(P\):
\[(1, 1) \to (-1, -1), \quad (3, 1) \to (-3, -1), \quad (1, 4) \to (-1, -4)\]These are exactly the vertices of \(T\) found in part (c), so the rotation does map \(P\) onto \(T\). [1]
This is worth noticing as a general result: a reflection in the \(y\)-axis followed by a reflection in the \(x\)-axis is equivalent to a single \(180\)° rotation about the origin, which is precisely the route \(P \to R \to T\) taken here. A rotation of \(180\)° needs no direction stated, since clockwise and anticlockwise give the same image.
Question 18 Report
A bakery mixes flour and butter in the ratio \(5 : 2\) by mass. One batch of pastry uses \(m\) grams of flour.
The ratio \(5 : 2\) means that for every \(5\) parts of flour there are \(2\) parts of butter, so the butter is \(\frac{2}{5}\) of the flour by mass.
(a)
\[\text{butter} = \frac{2m}{5} \text{ grams}\][1]
(b) The batch is flour plus butter, and the two must be written over a common denominator before adding. Since \(m = \frac{5m}{5}\):
\[m + \frac{2m}{5} = \frac{5m + 2m}{5} = \frac{7m}{5} = 2100\][1]
Multiplying both sides by \(5\) and dividing by \(7\):
\[7m = 10500 \implies m = 1500 \text{ grams}\][1]
(c) The butter is what remains of the batch once the flour is accounted for:
\[2100 - 1500 = 600 \text{ grams}\][1]
This agrees with the expression from part (a), since \(\frac{2 \times 1500}{5} = 600\) grams.
The whole batch is \(5 + 2 = 7\) parts, so each part is \(2100 \div 7 = 300\) grams, giving \(5 \times 300 = 1500\) g of flour and \(2 \times 300 = 600\) g of butter. That parts method is a quick independent check. The error to avoid is treating \(m\) as the total mass rather than the mass of flour, which the question defines explicitly.
The ratio \(5 : 2\) means that for every \(5\) parts of flour there are \(2\) parts of butter, so the butter is \(\frac{2}{5}\) of the flour by mass.
(a)
\[\text{butter} = \frac{2m}{5} \text{ grams}\][1]
(b) The batch is flour plus butter, and the two must be written over a common denominator before adding. Since \(m = \frac{5m}{5}\):
\[m + \frac{2m}{5} = \frac{5m + 2m}{5} = \frac{7m}{5} = 2100\][1]
Multiplying both sides by \(5\) and dividing by \(7\):
\[7m = 10500 \implies m = 1500 \text{ grams}\][1]
(c) The butter is what remains of the batch once the flour is accounted for:
\[2100 - 1500 = 600 \text{ grams}\][1]
This agrees with the expression from part (a), since \(\frac{2 \times 1500}{5} = 600\) grams.
The whole batch is \(5 + 2 = 7\) parts, so each part is \(2100 \div 7 = 300\) grams, giving \(5 \times 300 = 1500\) g of flour and \(2 \times 300 = 600\) g of butter. That parts method is a quick independent check. The error to avoid is treating \(m\) as the total mass rather than the mass of flour, which the question defines explicitly.
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