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Question 1 Report
A designer is preparing a card template for a stationery order. The template is an L-shape, formed from a rectangle \(20\) cm by \(15\) cm with a smaller rectangle \(8\) cm by \(6\) cm removed from one corner, as shown.
An L-shape area is found by treating it as a large rectangle with a smaller rectangle cut away from one corner.
(a) The overall bounding rectangle is \( 20 \times 15 = 300 \ \text{cm}^2 \) [1 mark]. The rectangle removed from the corner is \( 8 \times 6 = 48 \ \text{cm}^2 \) [1 mark]. Subtracting the removed piece from the full rectangle gives the template's area: \( 300 - 48 = 252 \ \text{cm}^2 \) [1 mark].
Splitting an L-shape into "big rectangle minus small rectangle" is usually quicker and less error-prone than splitting it into two separate rectangles that must be added, since the given side lengths already match the two rectangles in this subtraction.
An L-shape area is found by treating it as a large rectangle with a smaller rectangle cut away from one corner.
(a) The overall bounding rectangle is \( 20 \times 15 = 300 \ \text{cm}^2 \) [1 mark]. The rectangle removed from the corner is \( 8 \times 6 = 48 \ \text{cm}^2 \) [1 mark]. Subtracting the removed piece from the full rectangle gives the template's area: \( 300 - 48 = 252 \ \text{cm}^2 \) [1 mark].
Splitting an L-shape into "big rectangle minus small rectangle" is usually quicker and less error-prone than splitting it into two separate rectangles that must be added, since the given side lengths already match the two rectangles in this subtraction.
Question 2 Report
Two similar triangular set-square rulers, sold together as a matching pair by a stationery shop for use in geometry lessons, have corresponding sides in the ratio \(2 : 5\). The smaller set-square has a perimeter of 24 cm. Work out the perimeter of the larger set-square. (3)
For similar shapes, perimeter is a length (a 1-dimensional quantity), so it scales by exactly the same linear scale factor as any individual side.
The ratio of corresponding sides, \(2:5\), gives a scale factor from the smaller to the larger set-square of
\[5\div2=2.5\] [1 mark]Since perimeter is built entirely from lengths, it scales by this same linear factor, not by its square (which is only for area) or its cube (which is only for volume). [1 mark]
Applying the factor to the smaller perimeter:
\[24\times2.5=60 \text{ cm}\] [1 mark]Recognising that perimeter behaves like a length, not an area, is the key idea: multiplying by \(2.5\) directly (rather than \(2.5^{2}=6.25\)) is what similarity rules require for any 1-dimensional measurement.
For similar shapes, perimeter is a length (a 1-dimensional quantity), so it scales by exactly the same linear scale factor as any individual side.
The ratio of corresponding sides, \(2:5\), gives a scale factor from the smaller to the larger set-square of
\[5\div2=2.5\] [1 mark]Since perimeter is built entirely from lengths, it scales by this same linear factor, not by its square (which is only for area) or its cube (which is only for volume). [1 mark]
Applying the factor to the smaller perimeter:
\[24\times2.5=60 \text{ cm}\] [1 mark]Recognising that perimeter behaves like a length, not an area, is the key idea: multiplying by \(2.5\) directly (rather than \(2.5^{2}=6.25\)) is what similarity rules require for any 1-dimensional measurement.
Question 3 Report
A bakery is planning exactly how much dough its overnight shift should prepare each morning, and models the number of loaves, \(L\), it can bake from one batch of dough using \(L = 3n^2 - 5\), where \(n\) is the number of trays used. One batch produced \(43\) loaves.
Find the value of \(n\). (3)
Setting the given model equal to the known number of loaves produces a quadratic equation in \(n\); solving it by isolating \(n^2\) and taking a square root gives two possible roots, of which only the positive one makes sense as a number of trays.
\(3n^2-5=43\), so \(3n^2=48\) [1 mark]
\(n^2=16\) [1 mark]
\(n=4\), rejecting \(n=-4\) since \(n\) is a number of trays and cannot be negative. [1 mark]
Setting the given model equal to the known number of loaves produces a quadratic equation in \(n\); solving it by isolating \(n^2\) and taking a square root gives two possible roots, of which only the positive one makes sense as a number of trays.
\(3n^2-5=43\), so \(3n^2=48\) [1 mark]
\(n^2=16\) [1 mark]
\(n=4\), rejecting \(n=-4\) since \(n\) is a number of trays and cannot be negative. [1 mark]
Question 4 Report
A household's annual electricity bill is \(\pounds C\), found from \(C=4.5\times10^2 \times n\), where \(n\) is the number of people in the household. A family is billed \(\pounds2.7\times10^3\).
A formula linking a bill to the number of people can be rearranged to find an unknown quantity, then reused for other household sizes, and comparing two suppliers means evaluating each one's total charge for the same scenario.
(a) Rearranging \(C=4.5\times10^{2}\times n\) for \(n\) by dividing both sides by \(4.5\times10^{2}\):
\[n=2.7\times10^{3}\div4.5\times10^{2}=2700\div450=6\] [2 marks](b) Substituting \(n=4\) into the original formula:
\[C=4\times4.5\times10^{2}=1800=1.8\times10^{3}\] [1 mark](c) The rival supplier's cost for 6 people is a fixed charge plus a per-person charge:
\[1.6\times10^{3}+6\times1\times10^{2}=1600+600=2200=2.2\times10^{3}\]Comparing this to the original supplier's charge for 6 people, \(\pounds2.7\times10^{3}=\pounds2700\): since \(\pounds2200 \lt \pounds2700\), the rival supplier is cheaper, by \(2700-2200=\pounds500=\pounds5\times10^{2}\). [2 marks]
The original supplier charges purely per person, while the rival charges a fixed amount plus a smaller per-person rate; comparing the two total costs for the same household size, rather than comparing the per-person rates alone, is what correctly identifies which is cheaper.
A formula linking a bill to the number of people can be rearranged to find an unknown quantity, then reused for other household sizes, and comparing two suppliers means evaluating each one's total charge for the same scenario.
(a) Rearranging \(C=4.5\times10^{2}\times n\) for \(n\) by dividing both sides by \(4.5\times10^{2}\):
\[n=2.7\times10^{3}\div4.5\times10^{2}=2700\div450=6\] [2 marks](b) Substituting \(n=4\) into the original formula:
\[C=4\times4.5\times10^{2}=1800=1.8\times10^{3}\] [1 mark](c) The rival supplier's cost for 6 people is a fixed charge plus a per-person charge:
\[1.6\times10^{3}+6\times1\times10^{2}=1600+600=2200=2.2\times10^{3}\]Comparing this to the original supplier's charge for 6 people, \(\pounds2.7\times10^{3}=\pounds2700\): since \(\pounds2200 \lt \pounds2700\), the rival supplier is cheaper, by \(2700-2200=\pounds500=\pounds5\times10^{2}\). [2 marks]
The original supplier charges purely per person, while the rival charges a fixed amount plus a smaller per-person rate; comparing the two total costs for the same household size, rather than comparing the per-person rates alone, is what correctly identifies which is cheaper.
Question 5 Report
A community library receives a donation of \(\pounds T\), split between the children's section and the adult section in the ratio \(5:7\). The children's section then receives a further \(\pounds 48\) from a literacy grant, changing the ratio of children's section funding to adult section funding to \(3:4\).
A total split in a given ratio gives each section's original amount as a fraction of the total, and adding a fixed sum to only one section then creates a new ratio equation that can be solved for the original total.
(a) The children's section originally holds \( \dfrac{5}{12}T = \dfrac{5T}{12} \). After the £48 grant, its new amount is \( \dfrac{5T}{12}+48 \) [1 mark].
(b) The adult section is unaffected, staying at \( \dfrac{7T}{12} \). Since the new ratio of children's to adult funding is \(3:4\): \( \dfrac{\frac{5T}{12}+48}{\frac{7T}{12}}=\dfrac{3}{4} \) [1 mark]. Cross-multiplying gives \( 4\left(\dfrac{5T}{12}+48\right)=3\left(\dfrac{7T}{12}\right) \), i.e. \( \dfrac{20T}{12}+192=\dfrac{21T}{12} \) [1 mark]. Subtracting \( \dfrac{20T}{12} \) from both sides gives \( 192=\dfrac{T}{12} \) [1 mark], so \( T=192\times12=2304 \) [1 mark].
(c) The original amount allocated to the adult section is \( \dfrac{7(2304)}{12}=7\times192=£1344 \) [2 marks].
Only the children's section receives the £48 grant, which is why the adult section's expression, \(\dfrac{7T}{12}\), appears unchanged in the new ratio equation while the children's expression gains the extra "+48" term.
A total split in a given ratio gives each section's original amount as a fraction of the total, and adding a fixed sum to only one section then creates a new ratio equation that can be solved for the original total.
(a) The children's section originally holds \( \dfrac{5}{12}T = \dfrac{5T}{12} \). After the £48 grant, its new amount is \( \dfrac{5T}{12}+48 \) [1 mark].
(b) The adult section is unaffected, staying at \( \dfrac{7T}{12} \). Since the new ratio of children's to adult funding is \(3:4\): \( \dfrac{\frac{5T}{12}+48}{\frac{7T}{12}}=\dfrac{3}{4} \) [1 mark]. Cross-multiplying gives \( 4\left(\dfrac{5T}{12}+48\right)=3\left(\dfrac{7T}{12}\right) \), i.e. \( \dfrac{20T}{12}+192=\dfrac{21T}{12} \) [1 mark]. Subtracting \( \dfrac{20T}{12} \) from both sides gives \( 192=\dfrac{T}{12} \) [1 mark], so \( T=192\times12=2304 \) [1 mark].
(c) The original amount allocated to the adult section is \( \dfrac{7(2304)}{12}=7\times192=£1344 \) [2 marks].
Only the children's section receives the £48 grant, which is why the adult section's expression, \(\dfrac{7T}{12}\), appears unchanged in the new ratio equation while the children's expression gains the extra "+48" term.
Question 6 Report
While revising equations that involve fractions, a student is given the equation \(\frac{2}{x} + \frac{1}{3} = 1\), where \(x\) is not zero. Solve the equation to find the value of \(x\), showing full working at every stage. (4)
This question solves an equation containing two different algebraic fractions by clearing both denominators at once with a common multiple.
The two denominators are \(x\) and \(3\), so multiplying every term by their lowest common multiple, \(3x\), clears both fractions simultaneously:
\[3x \times \dfrac{2}{x} + 3x \times \dfrac{1}{3} = 3x \times 1\] \[6 + x = 3x\][1 mark]
Collecting the \(x\) terms on one side:
\[6 = 3x - x\][1 mark]
\[6 = 2x\][1 mark]
\[x = 3\][1 mark]
Checking in the original equation: \(\dfrac{2}{3} + \dfrac{1}{3} = 1\), which is correct. Multiplying every single term, including the \(1\) on the right-hand side, by \(3x\) is what allows both fractions to disappear in one step.
This question solves an equation containing two different algebraic fractions by clearing both denominators at once with a common multiple.
The two denominators are \(x\) and \(3\), so multiplying every term by their lowest common multiple, \(3x\), clears both fractions simultaneously:
\[3x \times \dfrac{2}{x} + 3x \times \dfrac{1}{3} = 3x \times 1\] \[6 + x = 3x\][1 mark]
Collecting the \(x\) terms on one side:
\[6 = 3x - x\][1 mark]
\[6 = 2x\][1 mark]
\[x = 3\][1 mark]
Checking in the original equation: \(\dfrac{2}{3} + \dfrac{1}{3} = 1\), which is correct. Multiplying every single term, including the \(1\) on the right-hand side, by \(3x\) is what allows both fractions to disappear in one step.
Question 7 Report
A corner shop prices a delivery item using \(f(x) = 1.2x + 3\), where \(x\) is the wholesale cost in pounds. A loyalty discount is then applied using \(g(x) = x - 4\).
An inverse function \( f^{-1}(x) \) undoes what \( f(x) \) does, found by writing \( y=f(x) \) and rearranging to make \( x \) the subject; a composite function \( gf(x) \) means applying \( f \) first and then \( g \) to the result.
(a) Writing \( y = 1.2x+3 \) and rearranging: subtracting 3 from both sides gives \( y-3=1.2x \), then dividing by 1.2 gives \( x = \dfrac{y-3}{1.2} \). Swapping \(x\) and \(y\) to express the inverse in the usual function notation gives \( f^{-1}(x) = \dfrac{x-3}{1.2} \) [2 marks].
(b) \( gf(x) \) means apply \( f \) first, then apply \( g \) to that result: \( gf(x) = g(f(x)) = f(x) - 4 = (1.2x+3) - 4 = 1.2x - 1 \) [2 marks].
(c) Setting \( gf(x) = 32 \) gives \( 1.2x - 1 = 32 \), so adding 1 to both sides gives \( 1.2x = 33 \), and dividing by 1.2 gives \( x = 27.5 \) [1 mark].
The order in \( gf(x) \) always means "\(f\) then \(g\)", reading right to left, which is why the calculation applies \(f\)'s rule first and then subtracts 4 from that result, not the other way round.
An inverse function \( f^{-1}(x) \) undoes what \( f(x) \) does, found by writing \( y=f(x) \) and rearranging to make \( x \) the subject; a composite function \( gf(x) \) means applying \( f \) first and then \( g \) to the result.
(a) Writing \( y = 1.2x+3 \) and rearranging: subtracting 3 from both sides gives \( y-3=1.2x \), then dividing by 1.2 gives \( x = \dfrac{y-3}{1.2} \). Swapping \(x\) and \(y\) to express the inverse in the usual function notation gives \( f^{-1}(x) = \dfrac{x-3}{1.2} \) [2 marks].
(b) \( gf(x) \) means apply \( f \) first, then apply \( g \) to that result: \( gf(x) = g(f(x)) = f(x) - 4 = (1.2x+3) - 4 = 1.2x - 1 \) [2 marks].
(c) Setting \( gf(x) = 32 \) gives \( 1.2x - 1 = 32 \), so adding 1 to both sides gives \( 1.2x = 33 \), and dividing by 1.2 gives \( x = 27.5 \) [1 mark].
The order in \( gf(x) \) always means "\(f\) then \(g\)", reading right to left, which is why the calculation applies \(f\)'s rule first and then subtracts 4 from that result, not the other way round.
Question 8 Report
A delivery van's odometer shows that a journey covers \(186\) km, correct to the nearest km. The journey took \(3.4\) hours, correct to 1 decimal place.
The upper bound of a compound quantity like speed, formed by dividing distance by time, comes from dividing the largest possible distance by the smallest possible time, since making the numerator bigger and the denominator smaller both push the result upward.
(a) Rounded to the nearest km, half of 1 km is 0.5 km, so the upper bound of the distance is
\[186+0.5=186.5 \text{ km}\] [1 mark](b) Rounded to 1 decimal place, half of 0.1 hours is 0.05 hours, so the lower bound of the time is
\[3.4-0.05=3.35 \text{ hours}\] [1 mark](c) The upper bound of average speed uses the upper bound of distance divided by the lower bound of time:
\[186.5 \div 3.35 = 55.6716\ldots\]Rounded to 3 significant figures, this is \(55.7\) km/h. [2 marks]
Speed is distance divided by time, so its bounds behave oppositely for numerator and denominator: to maximise the result, the distance must be pushed to its upper bound while the time is pushed to its lower bound, and mixing this up is the most common error in bounds questions involving division.
The upper bound of a compound quantity like speed, formed by dividing distance by time, comes from dividing the largest possible distance by the smallest possible time, since making the numerator bigger and the denominator smaller both push the result upward.
(a) Rounded to the nearest km, half of 1 km is 0.5 km, so the upper bound of the distance is
\[186+0.5=186.5 \text{ km}\] [1 mark](b) Rounded to 1 decimal place, half of 0.1 hours is 0.05 hours, so the lower bound of the time is
\[3.4-0.05=3.35 \text{ hours}\] [1 mark](c) The upper bound of average speed uses the upper bound of distance divided by the lower bound of time:
\[186.5 \div 3.35 = 55.6716\ldots\]Rounded to 3 significant figures, this is \(55.7\) km/h. [2 marks]
Speed is distance divided by time, so its bounds behave oppositely for numerator and denominator: to maximise the result, the distance must be pushed to its upper bound while the time is pushed to its lower bound, and mixing this up is the most common error in bounds questions involving division.
Question 9 Report
A household is trying to bring its finances back under control after an expensive winter, and its monthly budget shortfall, in pounds, satisfies \(x^2 - 6x - 2 = 0\), where \(x\) is the number of streaming subscriptions the family cancels.
Solve \(x^2 - 6x - 2 = 0\) by completing the square, giving your answers in the form \(a \pm \sqrt{b}\). (4)
Completing the square rewrites a quadratic \( x^2+bx+c \) as \( (x+\frac{b}{2})^2 - (\frac{b}{2})^2 + c \), which turns the equation into one that can be solved by taking a square root directly, giving exact surd answers.
For \( x^2-6x-2=0 \), halving the coefficient of \(x\) gives \(-3\), so \( x^2-6x \) can be written as \( (x-3)^2-9 \) (since \( (x-3)^2 = x^2-6x+9 \), the \(-9\) corrects for the extra \(+9\) introduced). Substituting this back gives \( (x-3)^2-9-2=0 \) [1 mark], which simplifies to \( (x-3)^2 = 11 \) [1 mark].
Taking the square root of both sides, and keeping both the positive and negative root since squaring loses the sign, gives \( x-3 = \pm\sqrt{11} \) [1 mark]. Adding 3 to both sides gives \( x = 3 \pm \sqrt{11} \) [1 mark].
Leaving the answer as \( 3 \pm \sqrt{11} \), rather than converting to a decimal, gives the exact solutions requested; \( \sqrt{11} \) is irrational, so no exact decimal exists, and rounding it early would lose the precision the question asks for.
Completing the square rewrites a quadratic \( x^2+bx+c \) as \( (x+\frac{b}{2})^2 - (\frac{b}{2})^2 + c \), which turns the equation into one that can be solved by taking a square root directly, giving exact surd answers.
For \( x^2-6x-2=0 \), halving the coefficient of \(x\) gives \(-3\), so \( x^2-6x \) can be written as \( (x-3)^2-9 \) (since \( (x-3)^2 = x^2-6x+9 \), the \(-9\) corrects for the extra \(+9\) introduced). Substituting this back gives \( (x-3)^2-9-2=0 \) [1 mark], which simplifies to \( (x-3)^2 = 11 \) [1 mark].
Taking the square root of both sides, and keeping both the positive and negative root since squaring loses the sign, gives \( x-3 = \pm\sqrt{11} \) [1 mark]. Adding 3 to both sides gives \( x = 3 \pm \sqrt{11} \) [1 mark].
Leaving the answer as \( 3 \pm \sqrt{11} \), rather than converting to a decimal, gives the exact solutions requested; \( \sqrt{11} \) is irrational, so no exact decimal exists, and rounding it early would lose the precision the question asks for.
Question 10 Report
The time, \(T\) hours, taken for a bus to complete a fixed route is inversely proportional to its average speed, \(s\) km/h. At an average speed of \(40\) km/h the route takes \(3\) hours.
Inverse proportion between two quantities means their product is always the same constant, \(k\), so \( T = \dfrac{k}{s} \), meaning time falls as speed rises.
(a) Substituting the known pair, \( s = 40 \) km/h giving \( T = 3 \) hours, the constant is \( k = T \times s = 3 \times 40 = 120 \) [1 mark].
(b) Using this constant with the new speed, \( T = \dfrac{120}{60} = 2 \) hours [2 marks].
Because \( T \) and \( s \) are inversely proportional, increasing the speed from 40 km/h to 60 km/h (a factor of \( 1.5 \)) reduces the time by the same factor, from 3 hours to \( 3 \div 1.5 = 2 \) hours, which is a useful check on the answer.
Inverse proportion between two quantities means their product is always the same constant, \(k\), so \( T = \dfrac{k}{s} \), meaning time falls as speed rises.
(a) Substituting the known pair, \( s = 40 \) km/h giving \( T = 3 \) hours, the constant is \( k = T \times s = 3 \times 40 = 120 \) [1 mark].
(b) Using this constant with the new speed, \( T = \dfrac{120}{60} = 2 \) hours [2 marks].
Because \( T \) and \( s \) are inversely proportional, increasing the speed from 40 km/h to 60 km/h (a factor of \( 1.5 \)) reduces the time by the same factor, from 3 hours to \( 3 \div 1.5 = 2 \) hours, which is a useful check on the answer.
Question 11 Report
Route planners write the ratio between the express and local bus fares for a shared journey as \(64^{\frac{2}{3}}\) hundred pounds per year, before rounding the figure for the new public timetable that goes on display at every stop along the corridor.
Work out the value of \(64^{\frac{2}{3}}\). (3)
A fractional power \( a^{\frac{m}{n}} \) means "take the \(n\)th root of \(a\), then raise the result to the power \(m\)"; working with the root first usually keeps the numbers smaller.
The denominator of the fractional power, 3, means taking the cube root of 64 first: \( 64^{\frac{1}{3}} = 4 \), since \( 4^3 = 64 \) [1 mark]. The numerator, 2, means then squaring this result: \( 4^2 = 16 \) [1 mark]. Combining these two steps, \( 64^{\frac{2}{3}} = \left(64^{\frac{1}{3}}\right)^2 = 4^2 = 16 \) [1 mark].
Taking the cube root before squaring (rather than squaring 64 first and then trying to cube-root a much larger number) is the practical reason this order is always used when evaluating a fractional index by hand.
A fractional power \( a^{\frac{m}{n}} \) means "take the \(n\)th root of \(a\), then raise the result to the power \(m\)"; working with the root first usually keeps the numbers smaller.
The denominator of the fractional power, 3, means taking the cube root of 64 first: \( 64^{\frac{1}{3}} = 4 \), since \( 4^3 = 64 \) [1 mark]. The numerator, 2, means then squaring this result: \( 4^2 = 16 \) [1 mark]. Combining these two steps, \( 64^{\frac{2}{3}} = \left(64^{\frac{1}{3}}\right)^2 = 4^2 = 16 \) [1 mark].
Taking the cube root before squaring (rather than squaring 64 first and then trying to cube-root a much larger number) is the practical reason this order is always used when evaluating a fractional index by hand.
Question 12 Report
The table shows how a football club's income is split between two sources, and the percentage increase in each source this year.
| Source | Share of income | Increase this year |
|---|---|---|
| Ticket sales | 70% | 15% |
| Merchandise sales | 30% | 25% |
This question tests the weighted-average method for combining percentage increases in different proportions of a total, which is more general than the compound-multiplier method used when a single quantity changes twice.
Ticket sales make up \(70\%\) of income and rose by \(15\%\); merchandise makes up \(30\%\) of income and rose by \(25\%\). The overall percentage increase in total income is the weighted average of the two increases, weighted by their share of income:
\[(0.7 \times 15) + (0.3 \times 25) = 10.5 + 7.5 = 18\%\][4 marks]
The overall increase of \(18\%\) is closer to the merchandise sales increase of \(25\%\) than a simple average of \(15\%\) and \(25\%\) (which would be \(20\%\)) might suggest at first glance, but comparing \(18\%\) directly to each source, it sits closer to the ticket sales increase of \(15\%\), because ticket sales make up the larger share (\(70\%\)) of the club's income [1 mark].
The overall percentage change is pulled towards whichever source contributes the greater share of the total, which is why the weights in part (a) must match the shares in the table exactly.
This question tests the weighted-average method for combining percentage increases in different proportions of a total, which is more general than the compound-multiplier method used when a single quantity changes twice.
Ticket sales make up \(70\%\) of income and rose by \(15\%\); merchandise makes up \(30\%\) of income and rose by \(25\%\). The overall percentage increase in total income is the weighted average of the two increases, weighted by their share of income:
\[(0.7 \times 15) + (0.3 \times 25) = 10.5 + 7.5 = 18\%\][4 marks]
The overall increase of \(18\%\) is closer to the merchandise sales increase of \(25\%\) than a simple average of \(15\%\) and \(25\%\) (which would be \(20\%\)) might suggest at first glance, but comparing \(18\%\) directly to each source, it sits closer to the ticket sales increase of \(15\%\), because ticket sales make up the larger share (\(70\%\)) of the club's income [1 mark].
The overall percentage change is pulled towards whichever source contributes the greater share of the total, which is why the weights in part (a) must match the shares in the table exactly.
Question 13 Report
On sports day, flags \(S\) and \(F\) are fixed 80 m apart along a sideline. Tent \(T\) sits on the perpendicular bisector of \(SF\), 30 m from midpoint \(M\), shown in the diagram.
Every point on the perpendicular bisector of a line segment is the same distance from both endpoints, and this shared distance can be found using Pythagoras' theorem once the perpendicular offset and the half-length of the segment are known.
(a) By definition, a point on the perpendicular bisector of \(SF\) is equidistant from \(S\) and \(F\). [1 mark]
(b) Since \(SF=80\) m, the midpoint distance is \(MS=80\div2=40\) m. \(T\) sits 30 m from \(M\) along the perpendicular, so triangle \(TMS\) is right-angled at \(M\), giving:
\[TS=\sqrt{TM^{2}+MS^{2}}=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50 \text{ m}\] [3 marks](c) An arc of radius equal to \(TS\), drawn centred at \(S\), would pass through \(T\), so the radius that fixes \(T\) by construction, centred at \(S\), is 50 m. [1 mark]
(d) By the perpendicular bisector property from part (a), \(TF=TS=50\) m as well. Since the requirement is at least 45 m from each flag, and \(50 \geq 45\), the tent's position satisfies the rule. [2 marks]
Because \(T\) lies exactly on the perpendicular bisector, finding its distance to just one flag, \(S\), using Pythagoras' theorem automatically gives its distance to the other flag, \(F\), too, without a second calculation.
Every point on the perpendicular bisector of a line segment is the same distance from both endpoints, and this shared distance can be found using Pythagoras' theorem once the perpendicular offset and the half-length of the segment are known.
(a) By definition, a point on the perpendicular bisector of \(SF\) is equidistant from \(S\) and \(F\). [1 mark]
(b) Since \(SF=80\) m, the midpoint distance is \(MS=80\div2=40\) m. \(T\) sits 30 m from \(M\) along the perpendicular, so triangle \(TMS\) is right-angled at \(M\), giving:
\[TS=\sqrt{TM^{2}+MS^{2}}=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50 \text{ m}\] [3 marks](c) An arc of radius equal to \(TS\), drawn centred at \(S\), would pass through \(T\), so the radius that fixes \(T\) by construction, centred at \(S\), is 50 m. [1 mark]
(d) By the perpendicular bisector property from part (a), \(TF=TS=50\) m as well. Since the requirement is at least 45 m from each flag, and \(50 \geq 45\), the tent's position satisfies the rule. [2 marks]
Because \(T\) lies exactly on the perpendicular bisector, finding its distance to just one flag, \(S\), using Pythagoras' theorem automatically gives its distance to the other flag, \(F\), too, without a second calculation.
Question 14 Report
Two courier firms quote prices for a Saturday delivery round. Fastlane's price is \(C = 4n + 6\) and Roadrunner's price is \(C = 5.5n\), with \(C\) in pounds for \(n\) parcels carried on the round. Find the value of \(n\) for which the two firms charge exactly the same amount. (3)
Two pricing formulae charge the same amount exactly when they are set equal to each other; collecting the \(n\) terms on one side and the constants on the other then isolates \(n\).
\(4n+6=5.5n\) [1 mark]
Rearranging: \(6=1.5n\) [1 mark]
\(n=4\) [1 mark]
Two pricing formulae charge the same amount exactly when they are set equal to each other; collecting the \(n\) terms on one side and the constants on the other then isolates \(n\).
\(4n+6=5.5n\) [1 mark]
Rearranging: \(6=1.5n\) [1 mark]
\(n=4\) [1 mark]
Question 15 Report
As part of household budgeting, a family models their heating cost, \(C\) pounds per month, as directly proportional to the square of the thermostat setting, \(t\) degrees above a fixed baseline. At \(t = 4\), \(C = £19.20\).
"Directly proportional to the square of \(t\)" means \( C = kt^2 \) for a fixed constant \(k\), so a given percentage increase in the thermostat setting causes a larger percentage increase in cost.
(a) Substituting \( t = 4 \), \( C = 19.20 \) gives \( 19.20 = k \times 4^2 = 16k \), so \( k = 19.20 \div 16 = 1.2 \) [2 marks].
(b) Using this constant with \( t = 6 \) gives \( C = 1.2 \times 6^2 = 1.2 \times 36 = £43.20 \) [2 marks].
(c) Raising the thermostat setting by 50% multiplies \( t \) by \( 1.5 \). Since \( C \) is proportional to \( t^2 \), the cost is multiplied by \( 1.5^2 = 2.25 \), regardless of the actual value of \( t \) [1 mark]. Applying this to the current cost of £28.80 gives a new cost of \( 28.80 \times 2.25 = £64.80 \), which is greater than £60, as required [1 mark].
This method avoids ever finding the actual thermostat setting: because cost is proportional to the square of the setting, any percentage change in the setting can be converted straight into the corresponding multiplier for cost, here \( 1.5^2 = 2.25 \).
"Directly proportional to the square of \(t\)" means \( C = kt^2 \) for a fixed constant \(k\), so a given percentage increase in the thermostat setting causes a larger percentage increase in cost.
(a) Substituting \( t = 4 \), \( C = 19.20 \) gives \( 19.20 = k \times 4^2 = 16k \), so \( k = 19.20 \div 16 = 1.2 \) [2 marks].
(b) Using this constant with \( t = 6 \) gives \( C = 1.2 \times 6^2 = 1.2 \times 36 = £43.20 \) [2 marks].
(c) Raising the thermostat setting by 50% multiplies \( t \) by \( 1.5 \). Since \( C \) is proportional to \( t^2 \), the cost is multiplied by \( 1.5^2 = 2.25 \), regardless of the actual value of \( t \) [1 mark]. Applying this to the current cost of £28.80 gives a new cost of \( 28.80 \times 2.25 = £64.80 \), which is greater than £60, as required [1 mark].
This method avoids ever finding the actual thermostat setting: because cost is proportional to the square of the setting, any percentage change in the setting can be converted straight into the corresponding multiplier for cost, here \( 1.5^2 = 2.25 \).
Question 16 Report
A taxi driver's satnav map is drawn to scale, so the triangular route between three towns, X, Y and Z, on the map is similar to the actual route triangle.
On the map, \(XY = 4.5\) cm, \(YZ = 6\) cm and \(XZ = 7.5\) cm. In real life the actual distance \(XY\) is \(27\) km.
A map scale expresses the ratio of a distance on the map to the corresponding real-life distance, and once found, the same ratio converts every other length on the map to real life.
(a) Converting the real distance to centimetres to match the map's units: \(27\) km \(=27\times1000\times100=2700000\) cm. The scale factor, map to real life, is
\[XY_{\text{map}}:XY_{\text{real}}=4.5:2700000\]Dividing both sides by 4.5 to get the ratio into the form \(1:n\):
\[1:600000\] [2 marks](b) Multiplying each map length by the scale factor 600000 (then converting back to km):
\[YZ=6\times600000=3600000 \text{ cm}=36 \text{ km}\] \[XZ=7.5\times600000=4500000 \text{ cm}=45 \text{ km}\] [2 marks](c) The full round trip covers \(XY+YZ+ZX=27+36+45=108\) km. At £1.20 per km:
\[108\times1.20=\pounds129.60\] [2 marks]Converting the real-life distance into the map's units (cm) before dividing is essential; comparing 4.5 cm directly with 27 km without a unit conversion would give a scale factor with the wrong number of zeros.
A map scale expresses the ratio of a distance on the map to the corresponding real-life distance, and once found, the same ratio converts every other length on the map to real life.
(a) Converting the real distance to centimetres to match the map's units: \(27\) km \(=27\times1000\times100=2700000\) cm. The scale factor, map to real life, is
\[XY_{\text{map}}:XY_{\text{real}}=4.5:2700000\]Dividing both sides by 4.5 to get the ratio into the form \(1:n\):
\[1:600000\] [2 marks](b) Multiplying each map length by the scale factor 600000 (then converting back to km):
\[YZ=6\times600000=3600000 \text{ cm}=36 \text{ km}\] \[XZ=7.5\times600000=4500000 \text{ cm}=45 \text{ km}\] [2 marks](c) The full round trip covers \(XY+YZ+ZX=27+36+45=108\) km. At £1.20 per km:
\[108\times1.20=\pounds129.60\] [2 marks]Converting the real-life distance into the map's units (cm) before dividing is essential; comparing 4.5 cm directly with 27 km without a unit conversion would give a scale factor with the wrong number of zeros.
Question 17 Report
Trying to save more each month, a household carefully tracks the weekly surplus in its food budget, in pounds, using the expression \(3x^2 + 5x - 2\), where \(x\) is the number of weeks since the family started meal-planning.
Factorising a quadratic where the \( x^2 \) coefficient is not 1 means finding two numbers that multiply to give (leading coefficient) \(\times\) (constant term) and add to give the middle coefficient, then splitting the middle term using those numbers.
(a) For \( 3x^2+5x-2 \), the product \( 3 \times (-2) = -6 \) and the required sum is \( 5 \); the numbers \( 6 \) and \( -1 \) satisfy \( 6 \times (-1) = -6 \) and \( 6+(-1)=5 \). Splitting the middle term: \( 3x^2+6x-x-2 = 3x(x+2) -1(x+2) = (3x-1)(x+2) \) [2 marks].
(b) The surplus is zero when either bracket is zero: \( 3x-1=0 \) gives \( x=\dfrac{1}{3} \); \( x+2=0 \) gives \( x=-2 \) [1 mark].
Both roots are mathematically valid answers to "when is the surplus zero", even though \( x=-2 \) (two weeks before meal-planning started) may not make sense in the household's actual timeline; the question only asks for the values, not for a real-world interpretation of each.
Factorising a quadratic where the \( x^2 \) coefficient is not 1 means finding two numbers that multiply to give (leading coefficient) \(\times\) (constant term) and add to give the middle coefficient, then splitting the middle term using those numbers.
(a) For \( 3x^2+5x-2 \), the product \( 3 \times (-2) = -6 \) and the required sum is \( 5 \); the numbers \( 6 \) and \( -1 \) satisfy \( 6 \times (-1) = -6 \) and \( 6+(-1)=5 \). Splitting the middle term: \( 3x^2+6x-x-2 = 3x(x+2) -1(x+2) = (3x-1)(x+2) \) [2 marks].
(b) The surplus is zero when either bracket is zero: \( 3x-1=0 \) gives \( x=\dfrac{1}{3} \); \( x+2=0 \) gives \( x=-2 \) [1 mark].
Both roots are mathematically valid answers to "when is the surplus zero", even though \( x=-2 \) (two weeks before meal-planning started) may not make sense in the household's actual timeline; the question only asks for the values, not for a real-world interpretation of each.
Question 18 Report
Wheelworks offers two repair-warranty schemes for wheel customers. Under Scheme 1, the probability a wheel needs a free repair within a year is \(0.15\). Under Scheme 2, the probability a wheel fails an initial inspection is \(0.4\); if it fails, the probability it then needs a free repair is \(0.5\), and if it passes, that probability is \(0.05\).
The overall probability of an event that can happen along two different paths of a tree diagram is the sum of the probabilities of each path, and multiplying by the number of wheels serviced then converts a probability into an expected count.
(a) Following the "fail, then repair" branch: \( P(\text{fail, repair}) = 0.4 \times 0.5 = 0.2 \) [1 mark]. Following the "pass, then repair" branch: \( P(\text{pass, repair}) = 0.6 \times 0.05 = 0.03 \) [1 mark]. Since a wheel needing repair happens via exactly one of these two paths, the overall probability is their sum: \( 0.2+0.03=0.23 \) [1 mark].
(b) Multiplying each scheme's probability of needing a repair by the 200 wheels serviced gives the expected number of repairs: Scheme 1, \( 200 \times 0.15 = 30 \); Scheme 2, \( 200 \times 0.23 = 46 \) [2 marks].
(c) Since \( 46 > 30 \), Scheme 2 is expected to need more free repairs over the year, so it is likely to cost the mechanic more [2 marks].
Even though most wheels pass Scheme 2's inspection (60% of them), the 40% that fail have a relatively high 50% repair rate, which is enough to push Scheme 2's overall repair probability above Scheme 1's flat 15% rate.
The overall probability of an event that can happen along two different paths of a tree diagram is the sum of the probabilities of each path, and multiplying by the number of wheels serviced then converts a probability into an expected count.
(a) Following the "fail, then repair" branch: \( P(\text{fail, repair}) = 0.4 \times 0.5 = 0.2 \) [1 mark]. Following the "pass, then repair" branch: \( P(\text{pass, repair}) = 0.6 \times 0.05 = 0.03 \) [1 mark]. Since a wheel needing repair happens via exactly one of these two paths, the overall probability is their sum: \( 0.2+0.03=0.23 \) [1 mark].
(b) Multiplying each scheme's probability of needing a repair by the 200 wheels serviced gives the expected number of repairs: Scheme 1, \( 200 \times 0.15 = 30 \); Scheme 2, \( 200 \times 0.23 = 46 \) [2 marks].
(c) Since \( 46 > 30 \), Scheme 2 is expected to need more free repairs over the year, so it is likely to cost the mechanic more [2 marks].
Even though most wheels pass Scheme 2's inspection (60% of them), the 40% that fail have a relatively high 50% repair rate, which is enough to push Scheme 2's overall repair probability above Scheme 1's flat 15% rate.
Question 19 Report
A community library charges £18 a year, including 5 free book loans; each further loan costs £1.20.
| Fee | £18 |
|---|---|
| Free loans | 5 |
| Extra loan cost | £1.20 |
This question builds a piecewise linear cost model from a fee-plus-per-unit description, evaluates it, forms and solves an equation from a known total, and checks a numerical claim.
Borrowing \(9\) books means \(9-5=4\) loans beyond the \(5\) free ones, each costing \(£1.20\):
\[T = £18 + (9-5) \times £1.20 = £18 + £4.80 = £22.80\][2 marks]
For a general number of books \(b \ge 5\):
\[T = 18 + 1.2(b-5)\][1 mark]
Setting \(T=£32.40\):
\[18+1.2(b-5) = 32.40\] \[1.2(b-5) = 14.40\] \[b-5 = 12\] \[b = 17 \text{ books}\][2 marks]
For \(b=20\):
\[T = 18+1.2(20-5) = 18+18 = £36\]Since \(£36 \lt £40\), the third member's claim is correct [1 mark].
The formula in part (b) only applies once at least \(5\) books have been borrowed (since \((b-5)\) must not be negative), which matches the "\(b \ge 5\)" condition stated in the question.
This question builds a piecewise linear cost model from a fee-plus-per-unit description, evaluates it, forms and solves an equation from a known total, and checks a numerical claim.
Borrowing \(9\) books means \(9-5=4\) loans beyond the \(5\) free ones, each costing \(£1.20\):
\[T = £18 + (9-5) \times £1.20 = £18 + £4.80 = £22.80\][2 marks]
For a general number of books \(b \ge 5\):
\[T = 18 + 1.2(b-5)\][1 mark]
Setting \(T=£32.40\):
\[18+1.2(b-5) = 32.40\] \[1.2(b-5) = 14.40\] \[b-5 = 12\] \[b = 17 \text{ books}\][2 marks]
For \(b=20\):
\[T = 18+1.2(20-5) = 18+18 = £36\]Since \(£36 \lt £40\), the third member's claim is correct [1 mark].
The formula in part (b) only applies once at least \(5\) books have been borrowed (since \((b-5)\) must not be negative), which matches the "\(b \ge 5\)" condition stated in the question.
Question 20 Report
A market trader designs a triangular canopy, shape \(A\), with vertices \((1, 1)\), \((4, 1)\) and \((1, 3)\), to be printed on a stall banner drawn on a coordinate grid.
Combining two reflections in intersecting lines produces a single rotation about the point where the lines cross, through twice the angle between the lines; here \(y=x\) and the \(x\)-axis meet at the origin at \(45^\circ\), so the combined effect should be a rotation of \(90^\circ\) about \((0,0)\).
Combining two reflections in intersecting lines produces a single rotation about the point where the lines cross, through twice the angle between the lines; here \(y=x\) and the \(x\)-axis meet at the origin at \(45^\circ\), so the combined effect should be a rotation of \(90^\circ\) about \((0,0)\).
Question 21 Report
A bicycle repair shop offers a voucher that reduces a repair bill to three-quarters of its original value, after which a fixed \(\pounds5\) collection fee is added, giving a final bill of \(\pounds41\). Taking the original repair cost as \(y\) pounds, form an equation and solve it to find \(y\). (3)
This question translates a worded description involving a fraction of an unknown and a fixed fee into a linear equation, and solves it.
Reducing the bill to three-quarters gives \(\dfrac{3}{4}y\); adding the fixed \(£5\) collection fee gives the final bill:
\[\dfrac{3}{4}y + 5 = 41\][1 mark]
Subtracting \(5\) from both sides:
\[\dfrac{3}{4}y = 36\][1 mark]
Multiplying both sides by \(\dfrac{4}{3}\):
\[y = £48\][1 mark]
The original repair cost was \(£48\). Checking: \(\dfrac{3}{4}(48)+5 = 36+5=41\), which matches.
This question translates a worded description involving a fraction of an unknown and a fixed fee into a linear equation, and solves it.
Reducing the bill to three-quarters gives \(\dfrac{3}{4}y\); adding the fixed \(£5\) collection fee gives the final bill:
\[\dfrac{3}{4}y + 5 = 41\][1 mark]
Subtracting \(5\) from both sides:
\[\dfrac{3}{4}y = 36\][1 mark]
Multiplying both sides by \(\dfrac{4}{3}\):
\[y = £48\][1 mark]
The original repair cost was \(£48\). Checking: \(\dfrac{3}{4}(48)+5 = 36+5=41\), which matches.
Question 22 Report
A bicycle repair shop sets up a savings fund to replace its diagnostic tools in future years. The fund grows by compound interest at a fixed rate of 4% per year, compounded annually, and after 2 years the fund is worth £1352.
This question writes a general compound-interest expression and then uses a given final value to solve for the original principal.
Growing at \(4\%\) per year, compounded annually, for \(2\) years, the fund's value is:
\[P(1.04)^2\][2 marks]
Setting this equal to \(£1352\):
\[P(1.04)^2 = 1352\] \[P \times 1.0816 = 1352\] \[P = 1352 \div 1.0816 = £1250\][2 marks]
The original amount invested was \(£1250\). Dividing by the full two-year multiplier \(1.0816\) (rather than by \(1.08\), which would only reverse one year of interest) correctly undoes both years of compounding at once.
This question writes a general compound-interest expression and then uses a given final value to solve for the original principal.
Growing at \(4\%\) per year, compounded annually, for \(2\) years, the fund's value is:
\[P(1.04)^2\][2 marks]
Setting this equal to \(£1352\):
\[P(1.04)^2 = 1352\] \[P \times 1.0816 = 1352\] \[P = 1352 \div 1.0816 = £1250\][2 marks]
The original amount invested was \(£1250\). Dividing by the full two-year multiplier \(1.0816\) (rather than by \(1.08\), which would only reverse one year of interest) correctly undoes both years of compounding at once.
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