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Question 1 Report
A school canteen is fitting a triangular serving hatch \(ABC\) into an awkward corner of the kitchen, so that trays can be passed through to the dining hall. Angle \(A = 58^{\circ}\), angle \(B = 71^{\circ}\) and \(AB = 2.4\) m, shown below.
Work out the length \(BC\), giving your answer correct to 3 significant figures. (4)
Once two angles of a triangle are known, the third follows from the angle sum of a triangle; the sine rule \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) then finds an unknown side from a known side and the angles opposite each of them.
Angle \(C=180^\circ-58^\circ-71^\circ=51^\circ\) [1 mark]
\(BC\) is opposite angle \(A\) and \(AB\) is opposite angle \(C\), so \(\dfrac{BC}{\sin A}=\dfrac{AB}{\sin C}\) [1 mark]
\(BC=\dfrac{2.4\times\sin58^\circ}{\sin51^\circ}\) [1 mark]
\(BC=2.62\) m (3 s.f.) [1 mark]
Once two angles of a triangle are known, the third follows from the angle sum of a triangle; the sine rule \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) then finds an unknown side from a known side and the angles opposite each of them.
Angle \(C=180^\circ-58^\circ-71^\circ=51^\circ\) [1 mark]
\(BC\) is opposite angle \(A\) and \(AB\) is opposite angle \(C\), so \(\dfrac{BC}{\sin A}=\dfrac{AB}{\sin C}\) [1 mark]
\(BC=\dfrac{2.4\times\sin58^\circ}{\sin51^\circ}\) [1 mark]
\(BC=2.62\) m (3 s.f.) [1 mark]
Question 2 Report
A farmer fences a triangular paddock \(ABC\) so that the two fences \(AB\) and \(AC\) are equal in length. The apex angle at \(A\) is \((4x)^\circ\) and each base angle is \((x + 15)^\circ\), as shown.
This question uses the angle sum of a triangle, together with the fact that two base angles of an isosceles triangle are equal, to form and solve an equation.
Since \(AB = AC\), the base angles at \(B\) and \(C\) are equal, each \((x+15)^\circ\). The three angles of triangle \(ABC\) sum to \(180^\circ\):
\[4x + (x+15) + (x+15) = 180\] \[6x + 30 = 180\] \[6x = 150\] \[x = 25\][2 marks]
Substituting \(x=25\) into the apex angle expression:
\[4(25) = 100^\circ\][1 mark]
Recognising that the two fences of equal length force the two base angles to be equal is the key step; without that fact there would be three unknown angles and only one equation.
This question uses the angle sum of a triangle, together with the fact that two base angles of an isosceles triangle are equal, to form and solve an equation.
Since \(AB = AC\), the base angles at \(B\) and \(C\) are equal, each \((x+15)^\circ\). The three angles of triangle \(ABC\) sum to \(180^\circ\):
\[4x + (x+15) + (x+15) = 180\] \[6x + 30 = 180\] \[6x = 150\] \[x = 25\][2 marks]
Substituting \(x=25\) into the apex angle expression:
\[4(25) = 100^\circ\][1 mark]
Recognising that the two fences of equal length force the two base angles to be equal is the key step; without that fact there would be three unknown angles and only one equation.
Question 3 Report
A family attends a music festival. Adult tickets cost \(£24.75\) and child tickets cost \(£14.50\).
The total ticket cost is the number of each ticket type multiplied by its price, added together; adding the programme gives the total spend, and the change from a payment is the amount paid minus the total spend.
The total ticket cost is the number of each ticket type multiplied by its price, added together; adding the programme gives the total spend, and the change from a payment is the amount paid minus the total spend.
Question 4 Report
A bakery surveys how long, in minutes, customers wait in the queue during its Saturday morning rush. The histogram shows the frequency density for each waiting-time interval.
On a histogram, frequency is frequency density multiplied by class width; adding every class's frequency gives the total surveyed, and a "10 minutes or more" percentage is found by summing the frequencies of every class at or beyond that boundary, dividing by the total, and converting to a percentage.
On a histogram, frequency is frequency density multiplied by class width; adding every class's frequency gives the total surveyed, and a "10 minutes or more" percentage is found by summing the frequencies of every class at or beyond that boundary, dividing by the total, and converting to a percentage.
Question 5 Report
At a school sports day, the height of a thrown shot above the ground, \(h\) metres, \(t\) seconds after release, is modelled by \(h = -5t^2 + 8t + 1.8\), shown below.
This question reads an initial value from a projectile model, uses the quadratic formula to find when it lands, and uses completed-square form to find and check its maximum height against a fixed limit.
At release, \(t=0\), so:
\[h = -5(0)^2+8(0)+1.8 = 1.8 \text{ m}\][1 mark]
The shot lands when \(h=0\):
\[-5t^2+8t+1.8=0\]Multiplying by \(-10\) to clear the decimal and make the leading coefficient positive:
\[50t^2-80t-18=0 \Rightarrow 25t^2-40t-9=0\][1 mark]
Using the quadratic formula with \(a=25\), \(b=-40\), \(c=-9\):
\[t = \dfrac{40 \pm \sqrt{(-40)^2-4(25)(-9)}}{2(25)} = \dfrac{40 \pm \sqrt{2500}}{50} = \dfrac{40\pm50}{50}\]giving \(t=1.8\) or \(t=-0.2\) [1 mark]. Time cannot be negative, so \(t=1.8\) seconds [1 mark].
Completing the square:
\[h = -5(t^2-1.6t)+1.8 = -5\left[(t-0.8)^2-0.64\right]+1.8 = -5(t-0.8)^2+5\]The maximum height is \(5\) m, occurring at \(t=0.8\) s [2 marks].
Since the maximum height reached, \(5\) m, is less than \(5.2\) m, the shot never reaches the height of the crossbar [1 mark].
Because part (c)'s completed-square form gives an exact maximum of \(5\) m, no further checking of intermediate values of \(t\) is needed to answer part (d): the entire path never exceeds \(5\) m.
This question reads an initial value from a projectile model, uses the quadratic formula to find when it lands, and uses completed-square form to find and check its maximum height against a fixed limit.
At release, \(t=0\), so:
\[h = -5(0)^2+8(0)+1.8 = 1.8 \text{ m}\][1 mark]
The shot lands when \(h=0\):
\[-5t^2+8t+1.8=0\]Multiplying by \(-10\) to clear the decimal and make the leading coefficient positive:
\[50t^2-80t-18=0 \Rightarrow 25t^2-40t-9=0\][1 mark]
Using the quadratic formula with \(a=25\), \(b=-40\), \(c=-9\):
\[t = \dfrac{40 \pm \sqrt{(-40)^2-4(25)(-9)}}{2(25)} = \dfrac{40 \pm \sqrt{2500}}{50} = \dfrac{40\pm50}{50}\]giving \(t=1.8\) or \(t=-0.2\) [1 mark]. Time cannot be negative, so \(t=1.8\) seconds [1 mark].
Completing the square:
\[h = -5(t^2-1.6t)+1.8 = -5\left[(t-0.8)^2-0.64\right]+1.8 = -5(t-0.8)^2+5\]The maximum height is \(5\) m, occurring at \(t=0.8\) s [2 marks].
Since the maximum height reached, \(5\) m, is less than \(5.2\) m, the shot never reaches the height of the crossbar [1 mark].
Because part (c)'s completed-square form gives an exact maximum of \(5\) m, no further checking of intermediate values of \(t\) is needed to answer part (d): the entire path never exceeds \(5\) m.
Question 6 Report
A phone company has \(180\) customers split between three tariffs, shown in the pie chart below. The Pay-as-you-go sector has an angle of \(90^\circ\) and the Basic sector has an angle of \(120^\circ\); the Premium sector's angle is not marked. (a) Work out the angle of the Premium sector. (1) (b) Work out the number of customers on the Premium tariff. (2)
A pie chart divides a full turn of \( 360^\circ \) between all the categories, so an unmarked sector's angle is found by subtracting the known angles from \( 360^\circ \), and converting an angle to a quantity uses the same fraction of the total that the angle is of the full circle.
(a) Since the three sectors must add up to a full turn, the Premium sector's angle is \( 360-90-120=150^\circ \) [1 mark].
(b) The Premium sector represents the fraction \( \dfrac{150}{360} \) of all 180 customers: \( \dfrac{150}{360}\times 180=75 \) customers [2 marks].
As a check, the same method applied to the other two sectors gives \( \dfrac{90}{360}\times180=45 \) Pay-as-you-go and \( \dfrac{120}{360}\times180=60 \) Basic customers; adding all three, \( 45+60+75=180 \), confirms every customer is accounted for.
A pie chart divides a full turn of \( 360^\circ \) between all the categories, so an unmarked sector's angle is found by subtracting the known angles from \( 360^\circ \), and converting an angle to a quantity uses the same fraction of the total that the angle is of the full circle.
(a) Since the three sectors must add up to a full turn, the Premium sector's angle is \( 360-90-120=150^\circ \) [1 mark].
(b) The Premium sector represents the fraction \( \dfrac{150}{360} \) of all 180 customers: \( \dfrac{150}{360}\times 180=75 \) customers [2 marks].
As a check, the same method applied to the other two sectors gives \( \dfrac{90}{360}\times180=45 \) Pay-as-you-go and \( \dfrac{120}{360}\times180=60 \) Basic customers; adding all three, \( 45+60+75=180 \), confirms every customer is accounted for.
Question 7 Report
Two triangular flags used at a music festival are mathematically similar, shown in the diagram. Flag A has an area of \(72\) cm\(^2\) and a base of \(12\) cm. Flag B is an enlargement of Flag A with an area of \(162\) cm\(^2\).
For similar triangles, the standard area formula \(\text{area}=\frac{1}{2}\times\text{base}\times\text{height}\) finds an unknown height, and the ratio of areas of similar shapes equals the square of the ratio of their corresponding lengths.
(a) Substituting the known area and base of Flag A into the area formula:
\[72=\dfrac{1}{2}\times12\times h\]so \(72=6h\), giving \(h=12\) cm. [2 marks]
(b) The area scale factor from Flag A to Flag B is the ratio of their areas:
\[\dfrac{162}{72}=2.25\]Since area scales with the square of the length scale factor, the length scale factor is the square root of this:
\[\sqrt{2.25}=1.5\] [2 marks](c) Multiplying Flag A's base by the length scale factor:
\[12\times1.5=18 \text{ cm}\] [1 mark]Finding the area scale factor first and only then taking its square root is essential; applying \(2.25\) directly to a length (rather than \(1.5\)) would badly overstate Flag B's base.
For similar triangles, the standard area formula \(\text{area}=\frac{1}{2}\times\text{base}\times\text{height}\) finds an unknown height, and the ratio of areas of similar shapes equals the square of the ratio of their corresponding lengths.
(a) Substituting the known area and base of Flag A into the area formula:
\[72=\dfrac{1}{2}\times12\times h\]so \(72=6h\), giving \(h=12\) cm. [2 marks]
(b) The area scale factor from Flag A to Flag B is the ratio of their areas:
\[\dfrac{162}{72}=2.25\]Since area scales with the square of the length scale factor, the length scale factor is the square root of this:
\[\sqrt{2.25}=1.5\] [2 marks](c) Multiplying Flag A's base by the length scale factor:
\[12\times1.5=18 \text{ cm}\] [1 mark]Finding the area scale factor first and only then taking its square root is essential; applying \(2.25\) directly to a length (rather than \(1.5\)) would badly overstate Flag B's base.
Question 8 Report
Amara receives a monthly allowance. She spends \(\frac{2}{5}\) of it on transport, then spends \(\frac{1}{4}\) of what remains on lunches, and saves the rest.
This question applies a fraction of a fraction to a whole, tracking how much of an original amount is used at each stage before finding what remains.
After spending \(\dfrac{2}{5}\) on transport, the remaining fraction is:
\[1 - \dfrac{2}{5} = \dfrac{3}{5}\][1 mark]
Lunches cost \(\dfrac{1}{4}\) of what remains, which is \(\dfrac{1}{4}\) of the original allowance's \(\dfrac{3}{5}\):
\[\dfrac{1}{4} \times \dfrac{3}{5} = \dfrac{3}{20}\][1 mark]
The fraction saved is what remains after transport, minus what is spent on lunches:
\[\dfrac{3}{5} - \dfrac{3}{20} = \dfrac{12}{20} - \dfrac{3}{20} = \dfrac{9}{20}\]as required [1 mark].
[1 mark]
The lunch fraction in part (b) is taken of the remaining \(\dfrac{3}{5}\), not of the whole original allowance, which is why "\(\dfrac{1}{4}\) of what remains" must be multiplied by \(\dfrac{3}{5}\) rather than used on its own.
This question applies a fraction of a fraction to a whole, tracking how much of an original amount is used at each stage before finding what remains.
After spending \(\dfrac{2}{5}\) on transport, the remaining fraction is:
\[1 - \dfrac{2}{5} = \dfrac{3}{5}\][1 mark]
Lunches cost \(\dfrac{1}{4}\) of what remains, which is \(\dfrac{1}{4}\) of the original allowance's \(\dfrac{3}{5}\):
\[\dfrac{1}{4} \times \dfrac{3}{5} = \dfrac{3}{20}\][1 mark]
The fraction saved is what remains after transport, minus what is spent on lunches:
\[\dfrac{3}{5} - \dfrac{3}{20} = \dfrac{12}{20} - \dfrac{3}{20} = \dfrac{9}{20}\]as required [1 mark].
[1 mark]
The lunch fraction in part (b) is taken of the remaining \(\dfrac{3}{5}\), not of the whole original allowance, which is why "\(\dfrac{1}{4}\) of what remains" must be multiplied by \(\dfrac{3}{5}\) rather than used on its own.
Question 9 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 10 Report
A car park operator charges permit holders an annual fee that rises by the same percentage, \(r\%\), at the start of each year. In Year 1 the fee is \(£300\).
This question builds a general compound-growth expression, uses given data to confirm the growth rate, then projects forward and checks the projection against a stated limit.
With the fee rising by \(r\%\) at the start of each year, the fee in Year 3 has grown for \(2\) years from the Year 1 fee of \(£300\):
\[300\left(1+\dfrac{r}{100}\right)^2\][2 marks]
Setting this equal to \(£349.92\):
\[300\left(1+\dfrac{r}{100}\right)^2 = 349.92\] \[\left(1+\dfrac{r}{100}\right)^2 = 1.1664\] \[1+\dfrac{r}{100} = \sqrt{1.1664} = 1.08\]so \(r = 8\), as required [2 marks].
The Year 5 fee has grown for \(4\) years from Year 1:
\[£300 \times 1.08^4 = £300 \times 1.36048896 = £408.15 \text{ (nearest penny)}\][2 marks]
The Year 6 fee has grown for \(5\) years:
\[£300 \times 1.08^5 = £440.80 \text{ (nearest penny)}\]Since \(£440.80 \lt £450\), continuing to raise the fee by \(8\%\) each year does keep the Year 6 fee within the operator's limit [2 marks].
Counting years of growth carefully is essential: the Year 1 fee has had no rises yet, so the Year \(n\) fee has grown for \((n-1)\) years, not \(n\) years.
This question builds a general compound-growth expression, uses given data to confirm the growth rate, then projects forward and checks the projection against a stated limit.
With the fee rising by \(r\%\) at the start of each year, the fee in Year 3 has grown for \(2\) years from the Year 1 fee of \(£300\):
\[300\left(1+\dfrac{r}{100}\right)^2\][2 marks]
Setting this equal to \(£349.92\):
\[300\left(1+\dfrac{r}{100}\right)^2 = 349.92\] \[\left(1+\dfrac{r}{100}\right)^2 = 1.1664\] \[1+\dfrac{r}{100} = \sqrt{1.1664} = 1.08\]so \(r = 8\), as required [2 marks].
The Year 5 fee has grown for \(4\) years from Year 1:
\[£300 \times 1.08^4 = £300 \times 1.36048896 = £408.15 \text{ (nearest penny)}\][2 marks]
The Year 6 fee has grown for \(5\) years:
\[£300 \times 1.08^5 = £440.80 \text{ (nearest penny)}\]Since \(£440.80 \lt £450\), continuing to raise the fee by \(8\%\) each year does keep the Year 6 fee within the operator's limit [2 marks].
Counting years of growth carefully is essential: the Year 1 fee has had no rises yet, so the Year \(n\) fee has grown for \((n-1)\) years, not \(n\) years.
Question 11 Report
Two bus stops \(P\) and \(Q\) lie 40 m apart along a straight road, shown in the diagram. A new stop must be nearer to \(P\) than to \(Q\). Two candidate sites are marked: site 1, 15 m from \(P\), and site 2, 28 m from \(P\), both measured towards \(Q\).
The set of points equally distant from two fixed points is the perpendicular bisector of the segment joining them; any point closer to one of the two points than the other lies on that point's side of this boundary line.
(a) The boundary separating points nearer to \(P\) from points nearer to \(Q\) is the perpendicular bisector of \(PQ\), which lies at the midpoint of the 40 m distance between them:
\[40\div2=20 \text{ m from } P\] [1 mark](b) Site 1 is 15 m from \(P\). Since \(15 \lt 20\), site 1 lies on \(P\)'s side of the boundary, so it is nearer to \(P\) than to \(Q\), and it satisfies the requirement. Site 2 is 28 m from \(P\). Since \(28 \gt 20\), site 2 lies on \(Q\)'s side of the boundary, so it is nearer to \(Q\), and it does not satisfy the requirement. [2 marks]
Comparing each site's distance from \(P\) directly with the 20 m boundary found in part (a), rather than measuring from \(Q\) as well, is enough to decide which side of the perpendicular bisector each site falls on.
The set of points equally distant from two fixed points is the perpendicular bisector of the segment joining them; any point closer to one of the two points than the other lies on that point's side of this boundary line.
(a) The boundary separating points nearer to \(P\) from points nearer to \(Q\) is the perpendicular bisector of \(PQ\), which lies at the midpoint of the 40 m distance between them:
\[40\div2=20 \text{ m from } P\] [1 mark](b) Site 1 is 15 m from \(P\). Since \(15 \lt 20\), site 1 lies on \(P\)'s side of the boundary, so it is nearer to \(P\) than to \(Q\), and it satisfies the requirement. Site 2 is 28 m from \(P\). Since \(28 \gt 20\), site 2 lies on \(Q\)'s side of the boundary, so it is nearer to \(Q\), and it does not satisfy the requirement. [2 marks]
Comparing each site's distance from \(P\) directly with the 20 m boundary found in part (a), rather than measuring from \(Q\) as well, is enough to decide which side of the perpendicular bisector each site falls on.
Question 12 Report
The school canteen records the lunch choices of Year 10 and Year 11 students on one Tuesday, split by jacket potato or panini, in the table below. The total number of students recorded that day was 96.
| Jacket potato | Panini | |
|---|---|---|
| Year 10 | 2x | 20 |
| Year 11 | 16 | 3x |
Every entry in the table must add up to the given total, which turns the table into a linear equation in \(x\); once \(x\) is known, any single cell divided by the grand total gives a probability.
(a) Adding all four entries in the table and setting the total equal to 96 gives \( 2x+20+16+3x=96 \) [1 mark]. Collecting like terms gives \( 5x+36=96 \), so subtracting 36 from both sides gives \( 5x=60 \), and dividing by 5 gives \( x=12 \) [1 mark].
(b) The Year 11 panini entry is \( 3x = 3(12) = 36 \) students. The probability that a randomly selected student is a Year 11 panini student is this count out of the total 96 recorded: \( \dfrac{36}{96} \) [1 mark], which simplifies (dividing top and bottom by 12) to \( \dfrac{3}{8} \) [1 mark].
Substituting \( x=12 \) back into every cell (\( 2x=24 \), \( 3x=36 \)) and checking that \( 24+20+16+36=96 \) confirms the value of \( x \) before it is used to answer part (b).
Every entry in the table must add up to the given total, which turns the table into a linear equation in \(x\); once \(x\) is known, any single cell divided by the grand total gives a probability.
(a) Adding all four entries in the table and setting the total equal to 96 gives \( 2x+20+16+3x=96 \) [1 mark]. Collecting like terms gives \( 5x+36=96 \), so subtracting 36 from both sides gives \( 5x=60 \), and dividing by 5 gives \( x=12 \) [1 mark].
(b) The Year 11 panini entry is \( 3x = 3(12) = 36 \) students. The probability that a randomly selected student is a Year 11 panini student is this count out of the total 96 recorded: \( \dfrac{36}{96} \) [1 mark], which simplifies (dividing top and bottom by 12) to \( \dfrac{3}{8} \) [1 mark].
Substituting \( x=12 \) back into every cell (\( 2x=24 \), \( 3x=36 \)) and checking that \( 24+20+16+36=96 \) confirms the value of \( x \) before it is used to answer part (b).
Question 13 Report
A student is investigating a relationship between two quantities. The table below shows recorded values of \(x\) and \(y\), where \(y\) is directly proportional to \(x^2\).
| x | 5 | 8 |
|---|---|---|
| y | 75 | ? |
"Directly proportional to \(x^2\)" means \( y = kx^2 \) for a fixed constant \(k\), which must first be found from the known pair of values before it can be used elsewhere in the table or to solve for \(x\).
(a) Substituting \( x = 5 \), \( y = 75 \) gives \( 75 = k \times 5^2 = 25k \), so \( k = 75 \div 25 = 3 \) [1 mark].
(b) The completed table below uses \( y = 3x^2 \) for \( x = 8 \): \( y = 3 \times 8^2 = 3 \times 64 = 192 \) [2 marks].
| x | 5 | 8 |
|---|---|---|
| y | 75 | 192 |
(c) Setting \( y = 48 \) in the formula gives \( 48 = 3x^2 \), so \( x^2 = 16 \), and taking the positive square root (as requested) gives \( x = 4 \) [1 mark].
Squaring \(x\) means the formula \( y = 3x^2 \) always has two solutions for \(x\) given a value of \(y\), \( x = 4 \) or \( x = -4 \); asking specifically for the positive value avoids any ambiguity.
"Directly proportional to \(x^2\)" means \( y = kx^2 \) for a fixed constant \(k\), which must first be found from the known pair of values before it can be used elsewhere in the table or to solve for \(x\).
(a) Substituting \( x = 5 \), \( y = 75 \) gives \( 75 = k \times 5^2 = 25k \), so \( k = 75 \div 25 = 3 \) [1 mark].
(b) The completed table below uses \( y = 3x^2 \) for \( x = 8 \): \( y = 3 \times 8^2 = 3 \times 64 = 192 \) [2 marks].
| x | 5 | 8 |
|---|---|---|
| y | 75 | 192 |
(c) Setting \( y = 48 \) in the formula gives \( 48 = 3x^2 \), so \( x^2 = 16 \), and taking the positive square root (as requested) gives \( x = 4 \) [1 mark].
Squaring \(x\) means the formula \( y = 3x^2 \) always has two solutions for \(x\) given a value of \(y\), \( x = 4 \) or \( x = -4 \); asking specifically for the positive value avoids any ambiguity.
Question 14 Report
A stationery company packs a set of gel pens into a small cardboard cuboid box, ready for an online order. The box measures 8 cm by 5 cm by 3 cm, as shown below.
The volume of a cuboid is length times width times height.
\(V=8\times5\times3\) [1 mark]
\(=120\) cm\(^3\) [1 mark]
The volume of a cuboid is length times width times height.
\(V=8\times5\times3\) [1 mark]
\(=120\) cm\(^3\) [1 mark]
Question 15 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 16 Report
A science lab runs the same reaction-time experiment once in the morning and once in the afternoon, each with \(30\) trials. The box plots show the reaction times, in seconds, recorded in each session.
On a box plot, the interquartile range is the width of the box; comparing two sessions uses the median for typical speed (a lower reaction time is faster) and the interquartile range for consistency (a smaller IQR means less spread in the middle half of the results).
On a box plot, the interquartile range is the width of the box; comparing two sessions uses the median for typical speed (a lower reaction time is faster) and the interquartile range for consistency (a smaller IQR means less spread in the middle half of the results).
Question 17 Report
A taxi firm paints its logo as a coloured sector of a circle on each rear car door. The sector has radius \(9\) cm and angle \(120^{\circ}\), measured from the centre of the circle.
The arc length of a sector is the fraction of the full circle's circumference that the sector's angle represents: \( \text{arc length} = \dfrac{\theta}{360} \times 2\pi r \), where \( \theta \) is the sector angle at the centre.
(a) Substituting \( \theta = 120^\circ \) and \( r = 9 \) cm: \( \text{arc length} = \dfrac{120}{360} \times 2\pi \times 9 \) [1 mark]. Since \( \dfrac{120}{360} = \dfrac{1}{3} \), this simplifies to \( \dfrac{1}{3} \times 18\pi = 6\pi \) cm, left as a multiple of \( \pi \) as requested [1 mark].
The fraction \( \dfrac{\theta}{360} \) always compares the sector's angle with a full turn; forgetting to convert the angle into this fraction (and instead using \( \theta \) directly) is the usual mistake with sector questions.
The arc length of a sector is the fraction of the full circle's circumference that the sector's angle represents: \( \text{arc length} = \dfrac{\theta}{360} \times 2\pi r \), where \( \theta \) is the sector angle at the centre.
(a) Substituting \( \theta = 120^\circ \) and \( r = 9 \) cm: \( \text{arc length} = \dfrac{120}{360} \times 2\pi \times 9 \) [1 mark]. Since \( \dfrac{120}{360} = \dfrac{1}{3} \), this simplifies to \( \dfrac{1}{3} \times 18\pi = 6\pi \) cm, left as a multiple of \( \pi \) as requested [1 mark].
The fraction \( \dfrac{\theta}{360} \) always compares the sector's angle with a full turn; forgetting to convert the angle into this fraction (and instead using \( \theta \) directly) is the usual mistake with sector questions.
Question 18 Report
Two mathematically similar conical flasks, kept side by side on a shelf in a school science lab preparation room, have heights 8 cm and 20 cm. The smaller flask has a capacity of 32 ml.
Work out the capacity of the larger flask. (3)
For similar solids, capacity (a volume) scales with the cube of the linear scale factor between corresponding lengths such as height.
Using the two heights, which are corresponding lengths:
\[\text{scale factor}=20\div8=2.5\] [1 mark]Capacity is 3-dimensional, so it scales with the cube of this linear factor:
\[\text{volume scale factor}=2.5^{3}=15.625\] [1 mark]Applying this to the smaller flask's capacity:
\[32\times15.625=500 \text{ ml}\] [1 mark]The height ratio, 2.5, describes how the flasks compare in one dimension only; because capacity depends on all three dimensions at once, it grows much faster than height, which is why the larger flask holds more than 15 times as much rather than only 2.5 times as much.
For similar solids, capacity (a volume) scales with the cube of the linear scale factor between corresponding lengths such as height.
Using the two heights, which are corresponding lengths:
\[\text{scale factor}=20\div8=2.5\] [1 mark]Capacity is 3-dimensional, so it scales with the cube of this linear factor:
\[\text{volume scale factor}=2.5^{3}=15.625\] [1 mark]Applying this to the smaller flask's capacity:
\[32\times15.625=500 \text{ ml}\] [1 mark]The height ratio, 2.5, describes how the flasks compare in one dimension only; because capacity depends on all three dimensions at once, it grows much faster than height, which is why the larger flask holds more than 15 times as much rather than only 2.5 times as much.
Question 19 Report
A household is laying a new patio, shaped as a rectangle measuring \(6\) m by \(4\) m with a quarter-circle flower bed of radius \(2\) m cut from one corner, shown below.
The paved area is the full rectangle with the quarter-circle flower bed removed, so it is found by subtracting the two areas.
(a) A quarter circle of radius 2 m has area \( \dfrac{1}{4} \times \pi \times 2^2 = \dfrac{1}{4} \times 4\pi = \pi \ \text{m}^2 \), left as a multiple of \( \pi \) as requested [2 marks].
(b) The full rectangle has area \( 6 \times 4 = 24 \ \text{m}^2 \). Removing the flower bed gives the paved area \( 24 - \pi = 20.858\ldots \), which rounds to \( 20.9 \ \text{m}^2 \) to 3 significant figures [2 marks].
(c) Using the unrounded paved area, the cost is \( 38 \times (24 - \pi) = 912 - 38\pi = 792.619\ldots \), which rounds to \( £792.62 \) to the nearest penny [2 marks].
(d) Since \( £792.62 \) is more than the \( £750 \) budget, the budget is not enough. The shortfall is \( £792.62 - £750 = £42.62 \) [1 mark].
Working with the exact multiple of \( \pi \) throughout, rather than a rounded decimal from part (b), is what keeps the cost in part (c) accurate to the nearest penny; rounding too early can shift the final pence figure.
The paved area is the full rectangle with the quarter-circle flower bed removed, so it is found by subtracting the two areas.
(a) A quarter circle of radius 2 m has area \( \dfrac{1}{4} \times \pi \times 2^2 = \dfrac{1}{4} \times 4\pi = \pi \ \text{m}^2 \), left as a multiple of \( \pi \) as requested [2 marks].
(b) The full rectangle has area \( 6 \times 4 = 24 \ \text{m}^2 \). Removing the flower bed gives the paved area \( 24 - \pi = 20.858\ldots \), which rounds to \( 20.9 \ \text{m}^2 \) to 3 significant figures [2 marks].
(c) Using the unrounded paved area, the cost is \( 38 \times (24 - \pi) = 912 - 38\pi = 792.619\ldots \), which rounds to \( £792.62 \) to the nearest penny [2 marks].
(d) Since \( £792.62 \) is more than the \( £750 \) budget, the budget is not enough. The shortfall is \( £792.62 - £750 = £42.62 \) [1 mark].
Working with the exact multiple of \( \pi \) throughout, rather than a rounded decimal from part (b), is what keeps the cost in part (c) accurate to the nearest penny; rounding too early can shift the final pence figure.
Question 20 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 21 Report
A phone company's monthly profit, in thousands of pounds, from adjusting its tariff price by \(x\) pounds is modelled by \(P=x^3-12x^2+36x+5\), valid for \(0 \leq x \leq 8\).
Stationary points of a cubic are found by setting its derivative to zero, and the second derivative distinguishes a local maximum from a local minimum at each one, which is exactly what is needed to choose the price change that maximises profit.
(a) Differentiating \(P=x^{3}-12x^{2}+36x+5\) term by term:
\[\dfrac{dP}{dx}=3x^{2}-24x+36\] [1 mark](b) Setting the derivative to zero:
\[3x^{2}-24x+36=0\]Dividing every term by 3:
\[x^{2}-8x+12=0\]Factorising:
\[(x-2)(x-6)=0\]giving \(x=2\) or \(x=6\). [2 marks]
(c) Differentiating again:
\[\dfrac{d^{2}P}{dx^{2}}=6x-24\]At \(x=2\): \(6(2)-24=-12 \lt 0\), so this is a maximum. At \(x=6\): \(6(6)-24=12 \gt 0\), so this is a minimum. [2 marks]
(d) Since maximising profit requires the maximum, the company should choose \(x=2\). Substituting into the original profit formula:
\[P(2)=2^{3}-12(2)^{2}+36(2)+5=8-48+72+5=37\]So the resulting profit is \(\pounds37000\) (since \(P\) is measured in thousands of pounds). [2 marks]
A cubic profit model typically has one local maximum and one local minimum; the second derivative test in part (c) is what tells the company which of the two candidate price changes, \(x=2\) or \(x=6\), is actually the profitable one to choose rather than the one that minimises profit.
Stationary points of a cubic are found by setting its derivative to zero, and the second derivative distinguishes a local maximum from a local minimum at each one, which is exactly what is needed to choose the price change that maximises profit.
(a) Differentiating \(P=x^{3}-12x^{2}+36x+5\) term by term:
\[\dfrac{dP}{dx}=3x^{2}-24x+36\] [1 mark](b) Setting the derivative to zero:
\[3x^{2}-24x+36=0\]Dividing every term by 3:
\[x^{2}-8x+12=0\]Factorising:
\[(x-2)(x-6)=0\]giving \(x=2\) or \(x=6\). [2 marks]
(c) Differentiating again:
\[\dfrac{d^{2}P}{dx^{2}}=6x-24\]At \(x=2\): \(6(2)-24=-12 \lt 0\), so this is a maximum. At \(x=6\): \(6(6)-24=12 \gt 0\), so this is a minimum. [2 marks]
(d) Since maximising profit requires the maximum, the company should choose \(x=2\). Substituting into the original profit formula:
\[P(2)=2^{3}-12(2)^{2}+36(2)+5=8-48+72+5=37\]So the resulting profit is \(\pounds37000\) (since \(P\) is measured in thousands of pounds). [2 marks]
A cubic profit model typically has one local maximum and one local minimum; the second derivative test in part (c) is what tells the company which of the two candidate price changes, \(x=2\) or \(x=6\), is actually the profitable one to choose rather than the one that minimises profit.
Question 22 Report
A skewer used to test loaves in a bakery oven touches a circular baking tray, centre O, at point T, meeting the worktop at point P outside the tray. Angle TPO is \(x^{\circ}\) and angle TOP is \((2x + 15)^{\circ}\).
A tangent is always perpendicular to the radius at the point where it touches the circle, which fixes one angle of triangle OTP so the triangle's angle sum gives an equation in \(x\).
(a) With angle OTP \(=90^\circ\) (tangent perpendicular to radius), the angles of triangle OTP sum to \(180^\circ\): \( 90+x+(2x+15)=180 \) [1 mark]. Combining like terms gives \( 3x+105=180 \), so subtracting 105 and dividing by 3 gives \( x=25 \) [1 mark].
(b) Substituting \( x=25 \): angle TOP \(=2(25)+15=65^\circ\) [1 mark].
As a check, all three angles of triangle OTP now sum correctly: \( 90+25+65=180 \), confirming both the right angle assumption and the value of \(x\).
A tangent is always perpendicular to the radius at the point where it touches the circle, which fixes one angle of triangle OTP so the triangle's angle sum gives an equation in \(x\).
(a) With angle OTP \(=90^\circ\) (tangent perpendicular to radius), the angles of triangle OTP sum to \(180^\circ\): \( 90+x+(2x+15)=180 \) [1 mark]. Combining like terms gives \( 3x+105=180 \), so subtracting 105 and dividing by 3 gives \( x=25 \) [1 mark].
(b) Substituting \( x=25 \): angle TOP \(=2(25)+15=65^\circ\) [1 mark].
As a check, all three angles of triangle OTP now sum correctly: \( 90+25+65=180 \), confirming both the right angle assumption and the value of \(x\).
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