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Question 1 Report
A corner shop owner wants to decide whether to expand the range of energy drinks, set \(E\), or the range of magazines, set \(M\). In a survey of 80 customers, \(n(E) = 34\), \(n(M) = 29\), and \(n(E \cap M) = 11\).
The inclusion-exclusion rule, \(n(E\cup M)=n(E)+n(M)-n(E\cap M)\), finds how many customers buy either product; the region outside both circles is then whatever remains of the 80 customers surveyed.
(a) \(n(E\cup M) = 34+29-11 = 52\). [2 marks]
(b) Customers buying neither product are those outside both circles: \(80-52 = 28\). [2 marks]
(c) Customers buying only energy drinks: \(34-11=23\). Customers buying only magazines: \(29-11=18\). Since \(23\gt18\), more customers buy energy drinks exclusively, so the owner should expand the energy drinks range. [2 marks]
The inclusion-exclusion rule, \(n(E\cup M)=n(E)+n(M)-n(E\cap M)\), finds how many customers buy either product; the region outside both circles is then whatever remains of the 80 customers surveyed.
(a) \(n(E\cup M) = 34+29-11 = 52\). [2 marks]
(b) Customers buying neither product are those outside both circles: \(80-52 = 28\). [2 marks]
(c) Customers buying only energy drinks: \(34-11=23\). Customers buying only magazines: \(29-11=18\). Since \(23\gt18\), more customers buy energy drinks exclusively, so the owner should expand the energy drinks range. [2 marks]
Question 2 Report
A music festival's rectangular main stage, shown in the diagram, has length \((3x-2)\) m and width \(x\) m. The perimeter of the whole stage is exactly 36 m.
Work out the value of \(x\). (3)
The stage is a rectangle with length \((3x-2)\) m and width \(x\) m, so its perimeter is twice the sum of one length and one width; setting this expression equal to the given perimeter of 36 m gives an equation to solve for \(x\).
Perimeter \(=2[(3x-2)+x]=36\). [1 mark]
Expanding the bracket and simplifying: \(2(4x-2)=36\), so \(8x-4=36\). [1 mark]
Adding 4 to both sides gives \(8x=40\), so \(x=5\). [1 mark]
Checking: with \(x=5\), the length is \(3(5)-2=13\) m and the width is \(5\) m, giving a perimeter of \(2(13+5)=36\) m, which matches.
The stage is a rectangle with length \((3x-2)\) m and width \(x\) m, so its perimeter is twice the sum of one length and one width; setting this expression equal to the given perimeter of 36 m gives an equation to solve for \(x\).
Perimeter \(=2[(3x-2)+x]=36\). [1 mark]
Expanding the bracket and simplifying: \(2(4x-2)=36\), so \(8x-4=36\). [1 mark]
Adding 4 to both sides gives \(8x=40\), so \(x=5\). [1 mark]
Checking: with \(x=5\), the length is \(3(5)-2=13\) m and the width is \(5\) m, giving a perimeter of \(2(13+5)=36\) m, which matches.
Question 3 Report
In a laboratory, four sensor mounting points A, B, C and D have position vectors from a fixed origin O given by \(\overrightarrow{OA}=2\mathbf{a}\), \(\overrightarrow{OB}=2\mathbf{a}+3\mathbf{b}\), \(\overrightarrow{OC}=5\mathbf{a}+5\mathbf{b}\) and \(\overrightarrow{OD}=5\mathbf{a}+2\mathbf{b}\).
Every side of quadrilateral ABCD can be written in terms of \(\mathbf{a}\) and \(\mathbf{b}\) by subtracting position vectors; showing that opposite sides \(\overrightarrow{AB}\) and \(\overrightarrow{DC}\) are identical vectors proves ABCD is a parallelogram, since equal vectors mean equal length and the same direction.
(a) \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(2\mathbf{a}+3\mathbf{b})-2\mathbf{a}=3\mathbf{b}\). [2 marks]
(b) \(\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=(5\mathbf{a}+5\mathbf{b})-(5\mathbf{a}+2\mathbf{b})=3\mathbf{b}\). [2 marks]
(c) Since \(\overrightarrow{AB}=\overrightarrow{DC}=3\mathbf{b}\), side AB is parallel to side DC and equal in length to it, which is exactly the condition needed for ABCD to be a parallelogram. [1 mark]
(d) The midpoint of AC has position vector \(\tfrac12(\overrightarrow{OA}+\overrightarrow{OC})=\tfrac12(2\mathbf{a}+5\mathbf{a}+5\mathbf{b})=\tfrac12(7\mathbf{a}+5\mathbf{b})=3.5\mathbf{a}+2.5\mathbf{b}\). [1 mark]
A quadrilateral is proved to be a parallelogram from vectors by finding just one pair of opposite sides and showing they are equal as vectors, not merely equal in length; being equal vectors automatically guarantees they are also parallel.
Every side of quadrilateral ABCD can be written in terms of \(\mathbf{a}\) and \(\mathbf{b}\) by subtracting position vectors; showing that opposite sides \(\overrightarrow{AB}\) and \(\overrightarrow{DC}\) are identical vectors proves ABCD is a parallelogram, since equal vectors mean equal length and the same direction.
(a) \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(2\mathbf{a}+3\mathbf{b})-2\mathbf{a}=3\mathbf{b}\). [2 marks]
(b) \(\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}=(5\mathbf{a}+5\mathbf{b})-(5\mathbf{a}+2\mathbf{b})=3\mathbf{b}\). [2 marks]
(c) Since \(\overrightarrow{AB}=\overrightarrow{DC}=3\mathbf{b}\), side AB is parallel to side DC and equal in length to it, which is exactly the condition needed for ABCD to be a parallelogram. [1 mark]
(d) The midpoint of AC has position vector \(\tfrac12(\overrightarrow{OA}+\overrightarrow{OC})=\tfrac12(2\mathbf{a}+5\mathbf{a}+5\mathbf{b})=\tfrac12(7\mathbf{a}+5\mathbf{b})=3.5\mathbf{a}+2.5\mathbf{b}\). [1 mark]
A quadrilateral is proved to be a parallelogram from vectors by finding just one pair of opposite sides and showing they are equal as vectors, not merely equal in length; being equal vectors automatically guarantees they are also parallel.
Question 4 Report
A household budgeting app models monthly savings, in pounds, as \(S = \dfrac{t^{2}-9}{t-3}\), for \(t \gt 3\), where \(t\) is the number of months since the account opened.
(a) The numerator, \(t^{2}-9\), is a difference of two squares, since \(9=3^{2}\), so it factorises as \((t-3)(t+3)\). Cancelling the common factor \((t-3)\) with the denominator (valid since \(t>3\) means \(t-3\) is never zero) gives \(S=t+3\). [2 marks]
(b) Substituting \(t=9\) into the simplified form: \(S=9+3=12\). [1 mark]
(c) Adding the fixed \(\pounds15\) joining bonus to \(S\): \(T(t)=(t+3)+15=t+18\). [1 mark]
(d) Setting \(T(t)=50\): \(t+18=50\), so \(t=32\). [2 marks]
(e) Since \(t=32\) months is longer than the \(24\)-month savings plan being considered, this value of \(t\) is not realistic within that plan. [1 mark]
The restriction \(t>3\) given in the question is exactly what makes the cancellation in part (a) valid; without it, \(t=3\) would need special treatment, since the original fraction would be \(\dfrac{0}{0}\) there.
(a) The numerator, \(t^{2}-9\), is a difference of two squares, since \(9=3^{2}\), so it factorises as \((t-3)(t+3)\). Cancelling the common factor \((t-3)\) with the denominator (valid since \(t>3\) means \(t-3\) is never zero) gives \(S=t+3\). [2 marks]
(b) Substituting \(t=9\) into the simplified form: \(S=9+3=12\). [1 mark]
(c) Adding the fixed \(\pounds15\) joining bonus to \(S\): \(T(t)=(t+3)+15=t+18\). [1 mark]
(d) Setting \(T(t)=50\): \(t+18=50\), so \(t=32\). [2 marks]
(e) Since \(t=32\) months is longer than the \(24\)-month savings plan being considered, this value of \(t\) is not realistic within that plan. [1 mark]
The restriction \(t>3\) given in the question is exactly what makes the cancellation in part (a) valid; without it, \(t=3\) would need special treatment, since the original fraction would be \(\dfrac{0}{0}\) there.
Question 5 Report
A bicycle repair shop manager models the maximum number of punctures her new apprentice can realistically fix in a single working day, \(x\), using \(ax - 7 \leq 9\), where \(a\) is a positive integer. She knows the solution to this inequality is \(x \leq 4\).
Solving the inequality \(ax-7\leqslant9\) in terms of \(a\) and matching the result to the known solution \(x\leqslant4\) finds \(a\); that same value of \(a\) is then used to solve a second, related inequality.
(a) Adding 7: \(ax\leqslant16\). Dividing by the positive constant \(a\): \(x\leqslant\dfrac{16}{a}\). Since this must match the given solution \(x\leqslant4\), \(\dfrac{16}{a}=4\), giving \(a=4\). [2 marks]
(b) Substituting \(a=4\) into \(4x-7\lt25\): adding 7 gives \(4x\lt32\), and dividing by 4 gives \(x\lt8\). The greatest integer value satisfying this strict inequality is \(x=7\). [2 marks]
Solving the inequality \(ax-7\leqslant9\) in terms of \(a\) and matching the result to the known solution \(x\leqslant4\) finds \(a\); that same value of \(a\) is then used to solve a second, related inequality.
(a) Adding 7: \(ax\leqslant16\). Dividing by the positive constant \(a\): \(x\leqslant\dfrac{16}{a}\). Since this must match the given solution \(x\leqslant4\), \(\dfrac{16}{a}=4\), giving \(a=4\). [2 marks]
(b) Substituting \(a=4\) into \(4x-7\lt25\): adding 7 gives \(4x\lt32\), and dividing by 4 gives \(x\lt8\). The greatest integer value satisfying this strict inequality is \(x=7\). [2 marks]
Question 6 Report
A household's electricity bill has a fixed standing charge of \(\pounds 18\) plus a rate of \(\pounds 0.72\) per unit of electricity used. The total bill for one month is \(\pounds 86.40\). Letting \(u\) be the number of units used, form and solve an equation to find \(u\). (4)
The total bill is made up of a fixed standing charge plus a rate for every unit of electricity used, so setting up an equation with \(u\) as the number of units used and solving it finds how many units were used that month.
Setting up the equation for the total bill: \(18+0.72u=86.40\). [1 mark]
Subtracting the standing charge from both sides: \(0.72u=86.40-18=68.40\). [1 mark]
Dividing both sides by \(0.72\) isolates \(u\): \(u=68.40\div0.72=95\) units. [2 marks]
Checking the answer: \(18+0.72\times95=18+68.40=\pounds86.40\), which matches the given total bill exactly.
The total bill is made up of a fixed standing charge plus a rate for every unit of electricity used, so setting up an equation with \(u\) as the number of units used and solving it finds how many units were used that month.
Setting up the equation for the total bill: \(18+0.72u=86.40\). [1 mark]
Subtracting the standing charge from both sides: \(0.72u=86.40-18=68.40\). [1 mark]
Dividing both sides by \(0.72\) isolates \(u\): \(u=68.40\div0.72=95\) units. [2 marks]
Checking the answer: \(18+0.72\times95=18+68.40=\pounds86.40\), which matches the given total bill exactly.
Question 7 Report
A local market trader hangs a rectangular stall sign, shown in the diagram, with sides \(\sqrt{50}\) m and \(\sqrt{18}\) m.
Work out the area of the sign. Give your answer as an integer. (2)
The diagram gives the two side lengths of the rectangular sign as surds, \(\sqrt{50}\) m and \(\sqrt{18}\) m; the area of a rectangle is the product of its two sides, and surds multiply using \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\).
Multiplying the two labelled sides: \(\sqrt{50}\times\sqrt{18} = \sqrt{50\times18} = \sqrt{900}\). [1 mark]
Since \(900=30^{2}\), \(\sqrt{900}=30\), so the area of the sign is \(30\) m\(^{2}\). [1 mark]
Exam tip: multiplying the numbers under the roots first (\(50\times18=900\)) before taking the square root avoids having to simplify two separate surds.
The diagram gives the two side lengths of the rectangular sign as surds, \(\sqrt{50}\) m and \(\sqrt{18}\) m; the area of a rectangle is the product of its two sides, and surds multiply using \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\).
Multiplying the two labelled sides: \(\sqrt{50}\times\sqrt{18} = \sqrt{50\times18} = \sqrt{900}\). [1 mark]
Since \(900=30^{2}\), \(\sqrt{900}=30\), so the area of the sign is \(30\) m\(^{2}\). [1 mark]
Exam tip: multiplying the numbers under the roots first (\(50\times18=900\)) before taking the square root avoids having to simplify two separate surds.
Question 8 Report
The graph shows the distance, \(s\) metres, of a car from the car park barrier, \(t\) seconds after moving off, modelled by \(s=6t^{2}-t^{3}\) for \(0 \le t \le 6\), as shown.
(a) Reading from the graph, the curve returns to \(s=0\) at \(t=6\) seconds (confirmed algebraically, since \(s=6t^{2}-t^{3}=t^{2}(6-t)\) is zero at \(t=0\) and \(t=6\)). [1 mark]
(b) Differentiating \(s=6t^{2}-t^{3}\) term by term gives \(\dfrac{ds}{dt}=12t-3t^{2}\). [1 mark]
(c) The velocity is the value of \(\dfrac{ds}{dt}\) at the given time. Substituting \(t=2\): \(12(2)-3(2)^{2}=24-12=12\) m/s. [2 marks]
(d) The car is momentarily at rest when its velocity is zero: \(12t-3t^{2}=0\), which factorises as \(3t(4-t)=0\), giving \(t=0\) or \(t=4\). Other than the start, this occurs at \(t=4\) s. The distance travelled by then is \(s=6(4)^{2}-(4)^{3}=96-64=32\) m. [2 marks]
The derivative gives the car's velocity at any instant; where that derivative is zero, the distance-time graph has a turning point, here a local maximum distance from the barrier before the car starts moving back towards it.
(a) Reading from the graph, the curve returns to \(s=0\) at \(t=6\) seconds (confirmed algebraically, since \(s=6t^{2}-t^{3}=t^{2}(6-t)\) is zero at \(t=0\) and \(t=6\)). [1 mark]
(b) Differentiating \(s=6t^{2}-t^{3}\) term by term gives \(\dfrac{ds}{dt}=12t-3t^{2}\). [1 mark]
(c) The velocity is the value of \(\dfrac{ds}{dt}\) at the given time. Substituting \(t=2\): \(12(2)-3(2)^{2}=24-12=12\) m/s. [2 marks]
(d) The car is momentarily at rest when its velocity is zero: \(12t-3t^{2}=0\), which factorises as \(3t(4-t)=0\), giving \(t=0\) or \(t=4\). Other than the start, this occurs at \(t=4\) s. The distance travelled by then is \(s=6(4)^{2}-(4)^{3}=96-64=32\) m. [2 marks]
The derivative gives the car's velocity at any instant; where that derivative is zero, the distance-time graph has a turning point, here a local maximum distance from the barrier before the car starts moving back towards it.
Question 9 Report
An allotment holder's crop weight, \(x\) kg, increases by 25% one year, then decreases by 20% the following year.
(a) A 25% increase corresponds to a multiplier of \(1.25\), and the following 20% decrease corresponds to a multiplier of \(1-0.20=0.8\), applied to the already-increased weight. So the weight after both changes is \(1.25\times0.8x\), which simplifies to \(1.0x\), since \(1.25\times0.8=1.0\). [2 marks]
(b) Since the combined multiplier is exactly \(1.0\), the final weight equals the starting weight exactly, so there is no overall change in the crop weight over the two years, it is neither an increase nor a decrease. [3 marks]
It is a common misconception that a 25% increase followed by a 20% decrease would cancel to leave a net 5% increase; because each percentage is calculated on a different amount (the second on the already-larger weight), the two multipliers instead combine to exactly \(1.0\) here.
(a) A 25% increase corresponds to a multiplier of \(1.25\), and the following 20% decrease corresponds to a multiplier of \(1-0.20=0.8\), applied to the already-increased weight. So the weight after both changes is \(1.25\times0.8x\), which simplifies to \(1.0x\), since \(1.25\times0.8=1.0\). [2 marks]
(b) Since the combined multiplier is exactly \(1.0\), the final weight equals the starting weight exactly, so there is no overall change in the crop weight over the two years, it is neither an increase nor a decrease. [3 marks]
It is a common misconception that a 25% increase followed by a 20% decrease would cancel to leave a net 5% increase; because each percentage is calculated on a different amount (the second on the already-larger weight), the two multipliers instead combine to exactly \(1.0\) here.
Question 10 Report
A festival car park attendant records the number of vehicles parked on Friday and on Saturday, shown in the table.
| Day | Vehicles |
|---|---|
| Friday | 240 |
| Saturday | 300 |
Work out the percentage increase in the number of vehicles from Friday to Saturday. (2)
A percentage increase compares the change in the number of vehicles to the original (Friday's) number: the increase is \(300-240=60\) vehicles, so the percentage increase is \(\dfrac{60}{240}\times100\). [1 mark]
This gives \(25\%\). [1 mark]
Dividing the increase by Friday's total, not Saturday's, is essential; a percentage change is always calculated relative to the starting value, here the smaller number of vehicles.
A percentage increase compares the change in the number of vehicles to the original (Friday's) number: the increase is \(300-240=60\) vehicles, so the percentage increase is \(\dfrac{60}{240}\times100\). [1 mark]
This gives \(25\%\). [1 mark]
Dividing the increase by Friday's total, not Saturday's, is essential; a percentage change is always calculated relative to the starting value, here the smaller number of vehicles.
Question 11 Report
A gardener orders three separate sections of decorative edging with lengths \(\sqrt{45}\), \(\sqrt{20}\) and \(\sqrt5\) metres to surround a newly planted flower bed on the allotment. Simplify \(\sqrt{45} - \sqrt{20} + \sqrt5\) fully, giving your answer in the form \(k\sqrt5\). (3)
To add or subtract surds, each one must first be written in terms of the same surd, by taking out the largest perfect square factor from underneath each root.
Simplify each edging length: \(\sqrt{45} = \sqrt{9\times5} = 3\sqrt5\), and \(\sqrt{20} = \sqrt{4\times5} = 2\sqrt5\); \(\sqrt5\) is already in simplest form. [1 mark] Now every term is a multiple of \(\sqrt5\), so they can be combined like ordinary algebraic terms: \(3\sqrt5 - 2\sqrt5 + \sqrt5 = (3-2+1)\sqrt5 = 2\sqrt5\). [2 marks] Comparing with the form \(k\sqrt5\) gives \(k=2\).
Exam tip: surds can only be combined once they share the same number under the root sign - always simplify every surd in an expression before adding or subtracting.
To add or subtract surds, each one must first be written in terms of the same surd, by taking out the largest perfect square factor from underneath each root.
Simplify each edging length: \(\sqrt{45} = \sqrt{9\times5} = 3\sqrt5\), and \(\sqrt{20} = \sqrt{4\times5} = 2\sqrt5\); \(\sqrt5\) is already in simplest form. [1 mark] Now every term is a multiple of \(\sqrt5\), so they can be combined like ordinary algebraic terms: \(3\sqrt5 - 2\sqrt5 + \sqrt5 = (3-2+1)\sqrt5 = 2\sqrt5\). [2 marks] Comparing with the form \(k\sqrt5\) gives \(k=2\).
Exam tip: surds can only be combined once they share the same number under the root sign - always simplify every surd in an expression before adding or subtracting.
Question 12 Report
A bicycle repair shop buys chains in bulk. Each chain has a breaking strength of \(8.4\times10^{3}\) newtons. The shop tests a batch of \(2.5\times10^{2}\) chains together, adding their individual breaking strengths, to check against the supplier\'s claimed batch strength of \(2\times10^{6}\) newtons.
(a) To multiply the two standard-form quantities, multiply the decimal parts and the powers of \(10\) separately. The decimal parts give \(8.4\times2.5=21\). [1 mark] The powers of \(10\) combine by adding their indices: \(10^{3}\times10^{2}=10^{5}\). [1 mark] Combining, the total is \(21\times10^{5}\); since \(21\) is not between \(1\) and \(10\), this must be adjusted to standard form: \(21\times10^{5}=2.1\times10^{6}\) newtons. [1 mark]
(b) Comparing the actual total, \(2.1\times10^{6}\) newtons, with the supplier's claimed \(2\times10^{6}\) newtons: since \(2.1\times10^{6} > 2\times10^{6}\), the batch meets, and slightly exceeds, the supplier's claim. [2 marks]
Writing \(21\times10^{5}\) as \(2.1\times10^{6}\) is not just presentation; a coefficient outside the range \(1\) to \(10\) is not valid standard form, and the shift from \(10^{5}\) to \(10^{6}\) compensates exactly for moving the decimal point in \(21\).
(a) To multiply the two standard-form quantities, multiply the decimal parts and the powers of \(10\) separately. The decimal parts give \(8.4\times2.5=21\). [1 mark] The powers of \(10\) combine by adding their indices: \(10^{3}\times10^{2}=10^{5}\). [1 mark] Combining, the total is \(21\times10^{5}\); since \(21\) is not between \(1\) and \(10\), this must be adjusted to standard form: \(21\times10^{5}=2.1\times10^{6}\) newtons. [1 mark]
(b) Comparing the actual total, \(2.1\times10^{6}\) newtons, with the supplier's claimed \(2\times10^{6}\) newtons: since \(2.1\times10^{6} > 2\times10^{6}\), the batch meets, and slightly exceeds, the supplier's claim. [2 marks]
Writing \(21\times10^{5}\) as \(2.1\times10^{6}\) is not just presentation; a coefficient outside the range \(1\) to \(10\) is not valid standard form, and the shift from \(10^{5}\) to \(10^{6}\) compensates exactly for moving the decimal point in \(21\).
Question 13 Report
A bike shop calculates the recommended tyre pressure, \(p\) psi, for a given tyre size as directly proportional to the rider's weight, \(w\) kg. A rider weighing 70 kg needs a pressure of 63 psi.
Direct proportion between pressure and weight means pressure equals a constant, \(k\), multiplied by weight; \(k\) is found from the known rider before being used to predict the pressure for a different rider and check it against the safety limit.
Because pressure is directly, not inversely, proportional to weight here, a heavier rider needs a higher recommended pressure, not a lower one, which is why the pressure rises from \(63\) psi to \(81\) psi as the weight increases from \(70\) kg to \(90\) kg.
Direct proportion between pressure and weight means pressure equals a constant, \(k\), multiplied by weight; \(k\) is found from the known rider before being used to predict the pressure for a different rider and check it against the safety limit.
Because pressure is directly, not inversely, proportional to weight here, a heavier rider needs a higher recommended pressure, not a lower one, which is why the pressure rises from \(63\) psi to \(81\) psi as the weight increases from \(70\) kg to \(90\) kg.
Question 14 Report
The times, in minutes after 6 a.m., that trains depart from a station during the morning are given by the set \(T = \{x : 0 \leqslant x \leqslant 90, x \text{ is a multiple of } 15\}\). Work out the value of \(n(T)\), showing the members of \(T\) you have used. (3)
Listing every value that satisfies the set's condition, then counting them, gives \(n(T)\); here the condition restricts \(T\) to multiples of 15 within a given range.
The members of \(T\) satisfying \(0\leqslant x\leqslant90\) and "\(x\) is a multiple of 15" are \(0, 15, 30, 45, 60, 75, 90\). [1 mark]
This is every multiple of 15 from 0 up to 90 inclusive. [1 mark]
Counting them gives \(n(T) = 7\). [1 mark]
Listing every value that satisfies the set's condition, then counting them, gives \(n(T)\); here the condition restricts \(T\) to multiples of 15 within a given range.
The members of \(T\) satisfying \(0\leqslant x\leqslant90\) and "\(x\) is a multiple of 15" are \(0, 15, 30, 45, 60, 75, 90\). [1 mark]
This is every multiple of 15 from 0 up to 90 inclusive. [1 mark]
Counting them gives \(n(T) = 7\). [1 mark]
Question 15 Report
A cycle-hire scheme used by public transport commuters charges a fixed cost plus an hourly rate. The table below shows the cost, \(C\) pounds, of hiring a bike for \(t\) whole hours.
| t (hours) | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| C (£) | 3.20 | 5.40 | 7.60 | 9.80 |
(a) The cost rises by the same amount, \(\pounds2.20\), for every extra hour (\(5.40-3.20=2.20\), \(7.60-5.40=2.20\), and \(9.80-7.60=2.20\)), so the gradient of the formula \(C=at+b\) is \(a=2.2\). Substituting the point \(t=1\), \(C=3.20\): \(2.2(1)+b=3.2\), so \(b=1\). [2 marks]
(b) With \(C=2.2t+1\), substituting \(t=9\): \(C=2.2(9)+1=19.8+1=\pounds20.80\). [1 mark]
The constant \(b=1\) is what the formula would predict as a fixed base cost even before any hourly charge is added, while \(a=2.2\) is the cost of each additional hour beyond that; extending the pattern in the table by \(5\) more hours from \(t=4\) (\(\pounds9.80+5\times\pounds2.20=\pounds20.80\)) gives the same answer as a useful check.
(a) The cost rises by the same amount, \(\pounds2.20\), for every extra hour (\(5.40-3.20=2.20\), \(7.60-5.40=2.20\), and \(9.80-7.60=2.20\)), so the gradient of the formula \(C=at+b\) is \(a=2.2\). Substituting the point \(t=1\), \(C=3.20\): \(2.2(1)+b=3.2\), so \(b=1\). [2 marks]
(b) With \(C=2.2t+1\), substituting \(t=9\): \(C=2.2(9)+1=19.8+1=\pounds20.80\). [1 mark]
The constant \(b=1\) is what the formula would predict as a fixed base cost even before any hourly charge is added, while \(a=2.2\) is the cost of each additional hour beyond that; extending the pattern in the table by \(5\) more hours from \(t=4\) (\(\pounds9.80+5\times\pounds2.20=\pounds20.80\)) gives the same answer as a useful check.
Question 16 Report
A bicycle repair shop hires part-time staff alongside 3 full-time staff, each paid \(\pounds 340\) a week. Each part-time worker is paid \(\pounds w\) a week.
The shop's weekly cost is built from a fixed part (the full-time wages) and a variable part (the part-time wages), so each part of this question either forms that total, substitutes a known value, or solves for an unknown rate.
(a) Three full-time staff at \(\pounds340\) each cost \(3\times340=1020\); adding \(p\) part-time staff at \(\pounds w\) each gives the expression \(1020+pw\). [1 mark]
(b) With \(p=4\), the budget equation is \(1020+4w=1500\). Subtracting 1020 gives \(4w=480\), so \(w=120\); each part-time worker earns \(\pounds120\) a week. [3 marks]
(c) Substituting \(w=120\) into the general expression from part (a) gives the simplified cost formula \(1020+120p\). [2 marks]
(d) Setting this expression equal to \(1980\): \(1020+120p=1980\), so \(120p=960\) and \(p=8\). [2 marks]
(e) Since \(p=8\) is a positive whole number, the shop really could hire exactly 8 part-time staff, so this value is realistic. [1 mark]
This question rewards keeping the formula in terms of \(p\) and \(w\) as long as possible: once \(w\) is pinned down in part (b), every later part is a straightforward substitution rather than a fresh setup.
The shop's weekly cost is built from a fixed part (the full-time wages) and a variable part (the part-time wages), so each part of this question either forms that total, substitutes a known value, or solves for an unknown rate.
(a) Three full-time staff at \(\pounds340\) each cost \(3\times340=1020\); adding \(p\) part-time staff at \(\pounds w\) each gives the expression \(1020+pw\). [1 mark]
(b) With \(p=4\), the budget equation is \(1020+4w=1500\). Subtracting 1020 gives \(4w=480\), so \(w=120\); each part-time worker earns \(\pounds120\) a week. [3 marks]
(c) Substituting \(w=120\) into the general expression from part (a) gives the simplified cost formula \(1020+120p\). [2 marks]
(d) Setting this expression equal to \(1980\): \(1020+120p=1980\), so \(120p=960\) and \(p=8\). [2 marks]
(e) Since \(p=8\) is a positive whole number, the shop really could hire exactly 8 part-time staff, so this value is realistic. [1 mark]
This question rewards keeping the formula in terms of \(p\) and \(w\) as long as possible: once \(w\) is pinned down in part (b), every later part is a straightforward substitution rather than a fresh setup.
Question 17 Report
A box for a school stationery order contains 30 pens, of which 4 are defective and do not write. Two pens are selected at random from the box, without replacement, for a quality check. Work out the probability that both selected pens work correctly. (3)
Choosing two pens without replacement means the second probability depends on what was removed first; multiplying the two conditional probabilities along the "both work" branch gives the required probability.
26 of the 30 pens work correctly, so \(P(\text{1st works}) = \dfrac{26}{30}\). [1 mark]
Given the first worked, 25 working pens remain out of 29, so \(P(\text{2nd works}\mid\text{1st works}) = \dfrac{25}{29}\). [1 mark]
Multiplying along the branch: \(P(\text{both work}) = \dfrac{26}{30}\times\dfrac{25}{29} = \dfrac{650}{870} = \dfrac{65}{87}\). [1 mark]
Choosing two pens without replacement means the second probability depends on what was removed first; multiplying the two conditional probabilities along the "both work" branch gives the required probability.
26 of the 30 pens work correctly, so \(P(\text{1st works}) = \dfrac{26}{30}\). [1 mark]
Given the first worked, 25 working pens remain out of 29, so \(P(\text{2nd works}\mid\text{1st works}) = \dfrac{25}{29}\). [1 mark]
Multiplying along the branch: \(P(\text{both work}) = \dfrac{26}{30}\times\dfrac{25}{29} = \dfrac{650}{870} = \dfrac{65}{87}\). [1 mark]
Question 18 Report
A traffic cone at the entrance to a car park has a base radius of 6 cm and a height of 14 cm. Work out the volume of the cone, giving your answer as a multiple of \(\pi\). (2)
The traffic cone's volume follows the standard formula for a cone, \(V=\dfrac13\pi r^{2}h\), using the given base radius and height.
Substituting the radius \(6\) cm and height \(14\) cm into the cone volume formula: \(V=\dfrac13\pi(6)^{2}(14)\). [1 mark]
Evaluating: \(\dfrac13\pi(36)(14)=\dfrac13\pi(504)=168\pi\) cm\(^{3}\). [1 mark]
Leaving the answer as a multiple of \(\pi\), as requested, avoids any rounding and keeps the value exact; the \(\dfrac13\) divides evenly into \(504\) here, since \(504=3\times168\).
The traffic cone's volume follows the standard formula for a cone, \(V=\dfrac13\pi r^{2}h\), using the given base radius and height.
Substituting the radius \(6\) cm and height \(14\) cm into the cone volume formula: \(V=\dfrac13\pi(6)^{2}(14)\). [1 mark]
Evaluating: \(\dfrac13\pi(36)(14)=\dfrac13\pi(504)=168\pi\) cm\(^{3}\). [1 mark]
Leaving the answer as a multiple of \(\pi\), as requested, avoids any rounding and keeps the value exact; the \(\dfrac13\) divides evenly into \(504\) here, since \(504=3\times168\).
Question 19 Report
At the repair shop, the distance, \(d\) metres, a bike coasts after its brakes are released varies directly with the square of the speed, \(s\) mph, it had when the brakes were applied. Coasting from 12 mph covers 7.2 m.
Direct proportion to the square of the speed means coasting distance equals a constant, \(k\), multiplied by speed squared; \(k\) is found from the known test before predicting the distance at a different speed and comparing it with the length of the bay.
The speed increased from \(12\) mph to \(20\) mph, a factor of \(\frac{5}{3}\), but because coasting distance depends on the square of speed, the distance increased by a factor of \(\left(\frac{5}{3}\right)^2\approx2.78\) (from \(7.2\) m to \(20\) m); this faster-than-linear growth is exactly why higher release speeds are disproportionately harder to stop safely within a fixed space.
Direct proportion to the square of the speed means coasting distance equals a constant, \(k\), multiplied by speed squared; \(k\) is found from the known test before predicting the distance at a different speed and comparing it with the length of the bay.
The speed increased from \(12\) mph to \(20\) mph, a factor of \(\frac{5}{3}\), but because coasting distance depends on the square of speed, the distance increased by a factor of \(\left(\frac{5}{3}\right)^2\approx2.78\) (from \(7.2\) m to \(20\) m); this faster-than-linear growth is exactly why higher release speeds are disproportionately harder to stop safely within a fixed space.
Question 20 Report
A farmer extends an orchard by adding a rectangular section measuring \(x\) m by \((x+5)\) m next to a triangular section with base \(x\) m and height \(8\) m. The total extra area is \(220\) m\(^2\).
(a) The rectangular section has area \(x(x+5) = x^{2}+5x\), and the triangular section has area \(\dfrac{1}{2}(x)(8) = 4x\). Adding these and setting the total equal to \(220\) m\(^{2}\) gives \(x^{2}+5x+4x = x^{2}+9x = 220\). [2 marks]
(b) Moving every term to one side gives \(x^{2}+9x-220=0\). Factorising requires two numbers that multiply to \(-220\) and add to \(9\): these are \(20\) and \(-11\), so \((x+20)(x-11)=0\), giving \(x=-20\) or \(x=11\). [3 marks]
(c) Since \(x\) is a length, it cannot be negative, so \(x=-20\) is rejected and \(x=11\) m. [1 mark]
Quadratic equations arising from area or length problems almost always produce one negative and one positive root; only the positive root has physical meaning, which is why every solution of this type should end with a check against the context.
(a) The rectangular section has area \(x(x+5) = x^{2}+5x\), and the triangular section has area \(\dfrac{1}{2}(x)(8) = 4x\). Adding these and setting the total equal to \(220\) m\(^{2}\) gives \(x^{2}+5x+4x = x^{2}+9x = 220\). [2 marks]
(b) Moving every term to one side gives \(x^{2}+9x-220=0\). Factorising requires two numbers that multiply to \(-220\) and add to \(9\): these are \(20\) and \(-11\), so \((x+20)(x-11)=0\), giving \(x=-20\) or \(x=11\). [3 marks]
(c) Since \(x\) is a length, it cannot be negative, so \(x=-20\) is rejected and \(x=11\) m. [1 mark]
Quadratic equations arising from area or length problems almost always produce one negative and one positive root; only the positive root has physical meaning, which is why every solution of this type should end with a check against the context.
Question 21 Report
When the expression \((x+p)(x+5)\) is expanded and simplified, the coefficient of x in the result is 12. This information is used to find the missing constant p.
Find the value of p. (3)
Expanding \((x+p)(x+5)\) using the standard method of multiplying each term in the first bracket by each term in the second: \(x^{2}+5x+px+5p\), which collects to \(x^{2}+(p+5)x+5p\). [1 mark]
The coefficient of \(x\) in this expansion is \(p+5\). Since this is given as \(12\): \(p+5=12\). [1 mark]
Subtracting \(5\) from both sides gives \(p=7\). [1 mark]
Comparing coefficients like this, matching the \(x\) term in the general expanded form to the given numerical value, works because the expansion is an identity that holds for every value of \(x\), so the coefficients themselves must match exactly.
Expanding \((x+p)(x+5)\) using the standard method of multiplying each term in the first bracket by each term in the second: \(x^{2}+5x+px+5p\), which collects to \(x^{2}+(p+5)x+5p\). [1 mark]
The coefficient of \(x\) in this expansion is \(p+5\). Since this is given as \(12\): \(p+5=12\). [1 mark]
Subtracting \(5\) from both sides gives \(p=7\). [1 mark]
Comparing coefficients like this, matching the \(x\) term in the general expanded form to the given numerical value, works because the expansion is an identity that holds for every value of \(x\), so the coefficients themselves must match exactly.
Question 22 Report
Two school sports day fundraising stalls model their profit, in pounds, as \(P_{1} = 3x^{2}-2x\) and \(P_{2} = 3x^{2}+4x-30\), where \(x\) is the number of raffle tickets sold, in tens.
Subtracting one profit expression from the other cancels the matching \(3x^{2}\) terms, leaving a simple linear expression that can be set to zero (to find equal profit) or evaluated directly (to compare a specific case).
(a) \(P_{2}-P_{1} = (3x^{2}+4x-30)-(3x^{2}-2x)\). The \(3x^{2}\) terms cancel, leaving \(4x-(-2x)-30 = 4x+2x-30 = 6x-30\). [2 marks]
(b) Setting the difference to zero: \(6x-30=0\), so \(x=5\) (tens of tickets). [2 marks]
(c) At \(x=3\): \(6(3)-30 = -12\), which is negative, meaning \(P_{2}\lt P_{1}\). So Stall 1 is the more profitable stall when \(x=3\). [2 marks]
Subtracting one profit expression from the other cancels the matching \(3x^{2}\) terms, leaving a simple linear expression that can be set to zero (to find equal profit) or evaluated directly (to compare a specific case).
(a) \(P_{2}-P_{1} = (3x^{2}+4x-30)-(3x^{2}-2x)\). The \(3x^{2}\) terms cancel, leaving \(4x-(-2x)-30 = 4x+2x-30 = 6x-30\). [2 marks]
(b) Setting the difference to zero: \(6x-30=0\), so \(x=5\) (tens of tickets). [2 marks]
(c) At \(x=3\): \(6(3)-30 = -12\), which is negative, meaning \(P_{2}\lt P_{1}\). So Stall 1 is the more profitable stall when \(x=3\). [2 marks]
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