Loading....
|
Press & Hold to Drag Around |
|||
|
Click Here to Close |
|||
Question 1 Report
Priya's bus leaves the depot on time. Between stops it runs 3 minutes early, then 7 minutes late, then a further 2 minutes early, each change measured against the timetable.
This question tests writing directed numbers from a word description and combining them to find a net result.
(a) Running early is treated as negative and running late as positive, so the three changes in order are
\[-3, \; +7, \; -2\][2] (1 for the correct signs, 1 for all three values correct)
(b) Adding the three changes gives the overall lateness or earliness at the final stop:
\[-3 + 7 - 2 = 2\]Since the result is positive, the bus is \(2\) minutes late overall. [2] (1 for a correct running total, 1 for correctly stating the bus is 2 minutes late)
The sign of the final total tells you the direction of the overall result: here a positive total means late, but a negative total would have meant the bus was early overall.
This question tests writing directed numbers from a word description and combining them to find a net result.
(a) Running early is treated as negative and running late as positive, so the three changes in order are
\[-3, \; +7, \; -2\][2] (1 for the correct signs, 1 for all three values correct)
(b) Adding the three changes gives the overall lateness or earliness at the final stop:
\[-3 + 7 - 2 = 2\]Since the result is positive, the bus is \(2\) minutes late overall. [2] (1 for a correct running total, 1 for correctly stating the bus is 2 minutes late)
The sign of the final total tells you the direction of the overall result: here a positive total means late, but a negative total would have meant the bus was early overall.
Question 2 Report
Two building sites owned by the same developer are being compared for their progress this quarter. Site A has completed 156 out of 200 planned housing units. Site B has completed 133 out of 175 planned housing units.
This question tests converting two fractions with different totals into percentages so that progress can be compared fairly.
(a) Each site's percentage progress is the number of units completed divided by the number planned, multiplied by 100:
\[\text{Site A: } \frac{156}{200}\times 100 = 78\%\] \[\text{Site B: } \frac{133}{175}\times 100 = 76\% \text{ <b>[3]</b>}\](b) Comparing the percentages, Site A has made the greater percentage progress. [1]
Even though Site B has completed fewer housing units in total, 133 against 156, converting to a common scale of 100 shows that Site A is proportionally further ahead of its own target, which is what "percentage progress" measures.
This question tests converting two fractions with different totals into percentages so that progress can be compared fairly.
(a) Each site's percentage progress is the number of units completed divided by the number planned, multiplied by 100:
\[\text{Site A: } \frac{156}{200}\times 100 = 78\%\] \[\text{Site B: } \frac{133}{175}\times 100 = 76\% \text{ <b>[3]</b>}\](b) Comparing the percentages, Site A has made the greater percentage progress. [1]
Even though Site B has completed fewer housing units in total, 133 against 156, converting to a common scale of 100 shows that Site A is proportionally further ahead of its own target, which is what "percentage progress" measures.
Question 3 Report
The diagram shows a rectangular field on a farm, with length \((x + 9)\) metres and width \((x - 2)\) metres. Its area is 312 m\(^2\).
This question tests forming a quadratic equation from the area of a rectangle shown in a diagram, solving it, and using the result to state a missing dimension.
(a) The area is length times width; expanding the product and setting it equal to 312:
\[(x+9)(x-2)=312 \Rightarrow x^2+7x-18=312 \Rightarrow x^2+7x-330=0\ \text{<b>[3]</b>}\](b) Look for two numbers that multiply to \(-330\) and add to \(7\): these are \(22\) and \(-15\), giving the factorisation:
\[(x+22)(x-15)=0 \Rightarrow x=-22 \text{ or } x=15\]Since \(x\) must be positive for the dimensions to make sense, \(x=15\). [3]
(c) Substituting \(x=15\) into the width expression:
\[\text{width}=x-2=13 \text{ metres}\ \text{<b>[1]</b>}\]Both dimensions, \((x+9)\) and \((x-2)\), must give positive lengths once \(x\) is found; checking \(x-2>0\) confirms that \(x=15\) is the physically sensible root, since \(x=-22\) would make both sides of the rectangle negative.
This question tests forming a quadratic equation from the area of a rectangle shown in a diagram, solving it, and using the result to state a missing dimension.
(a) The area is length times width; expanding the product and setting it equal to 312:
\[(x+9)(x-2)=312 \Rightarrow x^2+7x-18=312 \Rightarrow x^2+7x-330=0\ \text{<b>[3]</b>}\](b) Look for two numbers that multiply to \(-330\) and add to \(7\): these are \(22\) and \(-15\), giving the factorisation:
\[(x+22)(x-15)=0 \Rightarrow x=-22 \text{ or } x=15\]Since \(x\) must be positive for the dimensions to make sense, \(x=15\). [3]
(c) Substituting \(x=15\) into the width expression:
\[\text{width}=x-2=13 \text{ metres}\ \text{<b>[1]</b>}\]Both dimensions, \((x+9)\) and \((x-2)\), must give positive lengths once \(x\) is found; checking \(x-2>0\) confirms that \(x=15\) is the physically sensible root, since \(x=-22\) would make both sides of the rectangle negative.
Question 4 Report
Nnamdi prices sweets in glass jars along the counter of his corner shop, from 20p to 29p, a penny apart. These ten prices are the universal set \(\mathscr{E} = \{20, 21, 22, 23, 24, 25, 26, 27, 28, 29\}\). Set \(Y\) is the even prices, \(Y = \{x : x \text{ is even}, x \in \mathscr{E}\}\).
Set-builder notation defines membership by a rule applied to every element of \(\mathscr{E}\); here "even" is checked against each price from 20p to 29p.
\(n(Y) + n(Y')\) always equals \(n(\mathscr{E})\) for any set \(Y\): together a set and its complement account for every element of the universal set exactly once.
Set-builder notation defines membership by a rule applied to every element of \(\mathscr{E}\); here "even" is checked against each price from 20p to 29p.
\(n(Y) + n(Y')\) always equals \(n(\mathscr{E})\) for any set \(Y\): together a set and its complement account for every element of the universal set exactly once.
Question 5 Report
At the school sports day, ticket income is calculated using the formula \(R = 3.5a + 1.5c\), where \(R\) is the total income in pounds, \(a\) is the number of adult tickets sold and \(c\) is the number of child tickets sold. On Friday, the school sold 42 adult tickets and 65 child tickets.
This question tests substituting into a formula with two variables, then using one known total to find the other variable.
Part (b) is testing that the known variable, \(a\), is substituted and its contribution subtracted from the total income first, leaving a simple one-step equation in \(c\) to solve, rather than trying to rearrange the two-variable formula directly.
This question tests substituting into a formula with two variables, then using one known total to find the other variable.
Part (b) is testing that the known variable, \(a\), is substituted and its contribution subtracted from the total income first, leaving a simple one-step equation in \(c\) to solve, rather than trying to rearrange the two-variable formula directly.
Question 6 Report
Mr Okafor is teaching his class about combining sets. He draws a Venn diagram showing two sets, \(P\) and \(Q\), inside the universal set \(\mathscr{E}\), and shades one region in grey for the class to describe.
A Venn diagram represents sets as regions inside a rectangle (the universal set \(\mathscr{E}\)). Shading identifies a specific combination of set membership, which can be written using set notation such as intersection (\(\cap\)), union (\(\cup\)), or complement (\(\,'\)).
Whenever two circles overlap on a Venn diagram, the region belonging to one circle only must have the overlapping part subtracted out first; simply reading off \(n(P) = 12\) for the shaded region would wrongly include the 4 elements shared with \(Q\).
A Venn diagram represents sets as regions inside a rectangle (the universal set \(\mathscr{E}\)). Shading identifies a specific combination of set membership, which can be written using set notation such as intersection (\(\cap\)), union (\(\cup\)), or complement (\(\,'\)).
Whenever two circles overlap on a Venn diagram, the region belonging to one circle only must have the overlapping part subtracted out first; simply reading off \(n(P) = 12\) for the shaded region would wrongly include the 4 elements shared with \(Q\).
Question 7 Report
Elmswood bus depot recorded the number of passengers carried by three routes last Saturday, shown in the table below. Each total happens to be a perfect square. The depot manager wants the square root of each total worked out by hand before the figures go into next month's report.
| Route | Passengers |
|---|---|
| A | 324 |
| B | 441 |
| C | 576 |
A square root undoes squaring, so finding \(\sqrt{324}\), \(\sqrt{441}\) and \(\sqrt{576}\) means finding the number that, when multiplied by itself, gives each total.
A square root undoes squaring, so finding \(\sqrt{324}\), \(\sqrt{441}\) and \(\sqrt{576}\) means finding the number that, when multiplied by itself, gives each total.
Question 8 Report
A corner shop owner tracks the shop's weekly takings in a notebook kept beside the till, hoping to spot which weeks are consistently quieter than others before deciding whether to change the opening hours. Over five weeks, the takings, in pounds, were 1250, 1380, 1190, 1420 and 1310.
This question tests calculating a median and a range from a raw list of weekly takings.
The median (£1310) is not pulled towards either extreme, since it depends only on the middle-ranked value, whereas the range of £230 shows the actual size of the swing between the best and worst weeks.
This question tests calculating a median and a range from a raw list of weekly takings.
The median (£1310) is not pulled towards either extreme, since it depends only on the middle-ranked value, whereas the range of £230 shows the actual size of the swing between the best and worst weeks.
Question 9 Report
A school runs a quiz night to raise money for charity. In the quiz, a correct answer scores +5 points, a wrong answer scores \(-2\) points, and an unanswered question scores 0 points. One team answers all 12 questions in the final round: 7 correct, 3 wrong, and 2 unanswered.
This question tests applying a scoring rule to directed numbers and then comparing two totals.
(a) Each correct answer scores \(+5\) and each wrong answer scores \(-2\), while unanswered questions score 0. With 7 correct, 3 wrong and 2 unanswered:
\[7 \times 5 = 35, \qquad 3 \times (-2) = -6\] \[35 + (-6) + 0 = 29\]The team's total score is \(29\). [3] (1 for \(7\times5=35\), 1 for \(3\times(-2)=-6\), 1 for the total 29)
(b) The difference between the two teams' scores is found by subtracting the second team's score from the first:
\[29 - (-4) = 29 + 4 = 33\]The difference between the two teams' scores is \(33\) points. [2] (1 for setting up \(29-(-4)\), 1 for the value 33)
When a scoring rule mixes positive and negative points, calculate each category's contribution separately before adding them together, so a sign is not lost partway through a longer sum.
This question tests applying a scoring rule to directed numbers and then comparing two totals.
(a) Each correct answer scores \(+5\) and each wrong answer scores \(-2\), while unanswered questions score 0. With 7 correct, 3 wrong and 2 unanswered:
\[7 \times 5 = 35, \qquad 3 \times (-2) = -6\] \[35 + (-6) + 0 = 29\]The team's total score is \(29\). [3] (1 for \(7\times5=35\), 1 for \(3\times(-2)=-6\), 1 for the total 29)
(b) The difference between the two teams' scores is found by subtracting the second team's score from the first:
\[29 - (-4) = 29 + 4 = 33\]The difference between the two teams' scores is \(33\) points. [2] (1 for setting up \(29-(-4)\), 1 for the value 33)
When a scoring rule mixes positive and negative points, calculate each category's contribution separately before adding them together, so a sign is not lost partway through a longer sum.
Question 10 Report
As part of planning her garden budget for the coming summer, Priya is marking out a rectangular flower bed at the end of the lawn, near where the old greenhouse used to stand before it was taken down. The diagram shows its dimensions in metres, and the finished bed will cover 36 square metres.
This question tests forming a quadratic equation from the area of a rectangle described in an algebraic diagram, then solving it by factorisation and rejecting the solution that does not make physical sense.
(a) The diagram shows a rectangle of length \((x+5)\) and width \(x\), with area \(36\) square metres, so:
\[x(x+5)=36 \quad\Rightarrow\quad x^2+5x=36 \quad\Rightarrow\quad x^2+5x-36=0\]Moving the \(36\) across to form an equation equal to zero is what turns the area statement into the quadratic that has to be shown. [2]
(b) Find two numbers multiplying to \(-36\) and adding to \(5\): these are \(9\) and \(-4\), since \(9\times(-4)=-36\) and \(9+(-4)=5\):
\[(x+9)(x-4)=0 \quad\Rightarrow\quad x=-9 \text{ or } x=4\]Since \(x\) is a physical width, it cannot be negative, so the negative solution \(x=-9\) is rejected, and the width of the flower bed is \(x=4\) metres. [3] (1 for the correct factor pair, 1 for both roots, 1 for selecting \(x=4\) with a reason)
This "show that" and "solve" pairing is a common exam structure: first turn a real-world area or perimeter statement into a quadratic equal to zero, then solve by factorising and discard whichever root cannot represent a length, since a negative width is not physically possible.
This question tests forming a quadratic equation from the area of a rectangle described in an algebraic diagram, then solving it by factorisation and rejecting the solution that does not make physical sense.
(a) The diagram shows a rectangle of length \((x+5)\) and width \(x\), with area \(36\) square metres, so:
\[x(x+5)=36 \quad\Rightarrow\quad x^2+5x=36 \quad\Rightarrow\quad x^2+5x-36=0\]Moving the \(36\) across to form an equation equal to zero is what turns the area statement into the quadratic that has to be shown. [2]
(b) Find two numbers multiplying to \(-36\) and adding to \(5\): these are \(9\) and \(-4\), since \(9\times(-4)=-36\) and \(9+(-4)=5\):
\[(x+9)(x-4)=0 \quad\Rightarrow\quad x=-9 \text{ or } x=4\]Since \(x\) is a physical width, it cannot be negative, so the negative solution \(x=-9\) is rejected, and the width of the flower bed is \(x=4\) metres. [3] (1 for the correct factor pair, 1 for both roots, 1 for selecting \(x=4\) with a reason)
This "show that" and "solve" pairing is a common exam structure: first turn a real-world area or perimeter statement into a quadratic equal to zero, then solve by factorising and discard whichever root cannot represent a length, since a negative width is not physically possible.
Question 11 Report
A delivery driver leaves the depot at 08:15 and drives 96 km to a warehouse, arriving at 09:45. After unloading for 20 minutes, she drives back along a different route of 108 km, arriving at the depot at 11:35.
This question tests calculating times and average speeds for the two stages of a round trip, combining them for an overall average speed, and comparing that result with a target.
The unloading time is deliberately excluded from part (d); an average driving speed is calculated only from time spent moving, not from the total time the driver was away from the depot.
This question tests calculating times and average speeds for the two stages of a round trip, combining them for an overall average speed, and comparing that result with a target.
The unloading time is deliberately excluded from part (d); an average driving speed is calculated only from time spent moving, not from the total time the driver was away from the depot.
Question 12 Report
A company invests £5000 in a maintenance fund that earns compound interest at 3.5% per year.
This question again compares compound and simple interest, this time asking why the type of growth normally described as faster, compound interest, actually gives the smaller return over this particular period.
(a) Compound interest after 4 years at 3.5%:
\[ £5000 \times 1.035^4 = £5000 \times 1.147523 = £5737.62 \](correct to the nearest penny) [3]
(b) Interest earned:
\[ £5737.62 - £5000 = £737.62 \][1]
(c) Simple interest at 4% over 4 years:
\[ £5000 \times 0.04 \times 4 = £800, \quad \text{value} = £5000 + £800 = £5800 \][2]
(d) Comparing the two schemes:
\[ £5800 - £5737.62 = £62.38 \]The simple interest scheme gives the greater return after 4 years, by £62.38. [2]
(e) Compound interest only overtakes a given simple interest rate once the compound rate and the elapsed time are large enough for the repeated compounding to make up the gap between the two rates. Here the compound rate, 3.5%, is already lower than the simple rate, 4%, so compound interest starts behind; four years is not long enough for compounding on the smaller rate to close that gap and pull ahead. [4]
This shows that "compound interest grows faster" is only true in the long run and only once its own rate is being compared fairly; a lower compound rate can lose to a higher simple rate for many years, exactly as it does here.
This question again compares compound and simple interest, this time asking why the type of growth normally described as faster, compound interest, actually gives the smaller return over this particular period.
(a) Compound interest after 4 years at 3.5%:
\[ £5000 \times 1.035^4 = £5000 \times 1.147523 = £5737.62 \](correct to the nearest penny) [3]
(b) Interest earned:
\[ £5737.62 - £5000 = £737.62 \][1]
(c) Simple interest at 4% over 4 years:
\[ £5000 \times 0.04 \times 4 = £800, \quad \text{value} = £5000 + £800 = £5800 \][2]
(d) Comparing the two schemes:
\[ £5800 - £5737.62 = £62.38 \]The simple interest scheme gives the greater return after 4 years, by £62.38. [2]
(e) Compound interest only overtakes a given simple interest rate once the compound rate and the elapsed time are large enough for the repeated compounding to make up the gap between the two rates. Here the compound rate, 3.5%, is already lower than the simple rate, 4%, so compound interest starts behind; four years is not long enough for compounding on the smaller rate to close that gap and pull ahead. [4]
This shows that "compound interest grows faster" is only true in the long run and only once its own rate is being compared fairly; a lower compound rate can lose to a higher simple rate for many years, exactly as it does here.
Question 13 Report
A school athletics squad trains for sports day by running relay laps. The table shows the distance, in metres, run by four squad members during one training week.
| Member | Distance (m) |
|---|---|
| Priya | \(4.2 \times 10^3\) |
| Sam | \(3.8 \times 10^3\) |
| Leo | \(5.1 \times 10^3\) |
| Aisha | \(4.6 \times 10^3\) |
Comparing four training distances and then combining two of them uses the same standard-form rules used throughout this topic.
All four distances here share the power \(10^3\), so this question really tests whether the leading numbers can be compared and added correctly once the shared power of 10 is recognised.
Comparing four training distances and then combining two of them uses the same standard-form rules used throughout this topic.
All four distances here share the power \(10^3\), so this question really tests whether the leading numbers can be compared and added correctly once the shared power of 10 is recognised.
Question 14 Report
The owner of a corner shop borrows £1500 from a bank to buy a new fridge for the shop after the old one breaks down, and agrees to repay the loan over 3 years at a simple interest rate of 5% per year.
Simple interest is calculated only on the original amount borrowed, using \(\text{interest} = \text{principal} \times \text{rate} \times \text{time}\), and the total repayment adds this interest to the loan.
(a) Interest over 3 years:
\[ £1500 \times 0.05 \times 3 = £225 \][2]
(b) Total repaid:
\[ £1500 + £225 = £1725 \][2]
Because this is simple rather than compound interest, the £225 could also be found by working out one year's interest, £75, and multiplying by 3, since the interest does not itself earn further interest.
Simple interest is calculated only on the original amount borrowed, using \(\text{interest} = \text{principal} \times \text{rate} \times \text{time}\), and the total repayment adds this interest to the loan.
(a) Interest over 3 years:
\[ £1500 \times 0.05 \times 3 = £225 \][2]
(b) Total repaid:
\[ £1500 + £225 = £1725 \][2]
Because this is simple rather than compound interest, the £225 could also be found by working out one year's interest, £75, and multiplying by 3, since the interest does not itself earn further interest.
Question 15 Report
The sports day committee has a list of 5 girls and 3 boys who volunteered to be reserve runners. Two reserves are picked at random from the list, and the name picked first is not put back before the second is drawn.
This is a "without replacement" probability question: once the first reserve is picked, the list has one fewer name, which changes the probabilities for the second pick. The probabilities of consecutive events on one branch are multiplied together.
This is a "without replacement" probability question: once the first reserve is picked, the list has one fewer name, which changes the probabilities for the second pick. The probabilities of consecutive events on one branch are multiplied together.
Question 16 Report
A land surveyor measured the lengths, in metres, of 9 fence sections around a paddock, shown in the table below. The fencing contractor uses these figures to plan the next repair visit. Sections shorter than the paddock's minimum boundary length will be replaced first.
| Section | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| Length (m) | 18 | 22 | 15 | 25 | 19 | 21 | 17 | 23 | 20 |
This question tests calculating a median and an interquartile range from a raw list of fence lengths.
With 9 values, each quartile is itself the average of two values once the median has been set aside, since the remaining four values in each half have no single middle value of their own.
This question tests calculating a median and an interquartile range from a raw list of fence lengths.
With 9 values, each quartile is itself the average of two values once the median has been set aside, since the remaining four values in each half have no single middle value of their own.
Question 17 Report
A straight hedge runs parallel to a straight fence at the edge of a field. A footpath crosses both in a single straight line, and where the footpath crosses the fence it makes an angle of 132 degrees with the fence.
This question tests co-interior (allied) angles, which occur between two parallel lines cut by a transversal.
The hedge and fence are parallel, and the footpath is the transversal crossing both. The marked \(132^\circ\) angle where the footpath meets the fence, and the required angle where the footpath meets the hedge on the same side of the transversal, are co-interior angles, meaning they lie between the two parallel lines on the same side of the transversal. Co-interior angles between parallel lines always sum to \(180^\circ\):
\[132^\circ + \text{required angle} = 180^\circ\]So the required angle is:
\[180^\circ - 132^\circ = 48^\circ\][3] (identifying the co-interior angle relationship, stating the sum is \(180^\circ\), and subtracting to find \(48^\circ\))
Co-interior angles are the one pair of parallel-line angles that are supplementary (add to \(180^\circ\)) rather than equal; the other pairs, corresponding and alternate angles, are always equal. Checking whether the two marked angles are on the same side of the transversal, and between or outside the parallel lines, tells you which rule to apply.
This question tests co-interior (allied) angles, which occur between two parallel lines cut by a transversal.
The hedge and fence are parallel, and the footpath is the transversal crossing both. The marked \(132^\circ\) angle where the footpath meets the fence, and the required angle where the footpath meets the hedge on the same side of the transversal, are co-interior angles, meaning they lie between the two parallel lines on the same side of the transversal. Co-interior angles between parallel lines always sum to \(180^\circ\):
\[132^\circ + \text{required angle} = 180^\circ\]So the required angle is:
\[180^\circ - 132^\circ = 48^\circ\][3] (identifying the co-interior angle relationship, stating the sum is \(180^\circ\), and subtracting to find \(48^\circ\))
Co-interior angles are the one pair of parallel-line angles that are supplementary (add to \(180^\circ\)) rather than equal; the other pairs, corresponding and alternate angles, are always equal. Checking whether the two marked angles are on the same side of the transversal, and between or outside the parallel lines, tells you which rule to apply.
Question 18 Report
At a car workshop, the probability that a car booked in has an engine fault is 0.18. Independently, the probability that the same car has a battery fault is 0.09.
The two faults are independent, so the probability of both occurring, or of neither occurring, is found by multiplying the individual probabilities.
The two faults are independent, so the probability of both occurring, or of neither occurring, is found by multiplying the individual probabilities.
Question 19 Report
The diagram shows a cuboid with length 8 cm, width 6 cm and height 4 cm. The vertices are labelled \(A\) to \(H\).
Calculate the length of the space diagonal \(AG\). Give your answer correct to 3 significant figures. (5)
This question tests finding the space diagonal of a cuboid by applying Pythagoras' theorem twice: once across the base, then again up through the solid.
First find the base diagonal \(AC\), using the length \(AB = 8\) cm and width \(BC = 6\) cm: [2]
\[AC^2 = 8^2 + 6^2 = 64 + 36 = 100\] \[AC = \sqrt{100} = 10\text{ cm}\]The space diagonal \(AG\) is the hypotenuse of a second right-angled triangle formed by the base diagonal \(AC\) and the vertical edge \(CG = 4\) cm: [3]
\[AG^2 = AC^2 + CG^2 = 10^2 + 4^2 = 100 + 16 = 116\] \[AG = \sqrt{116} = 10.8\text{ cm (3 s.f.)}\]Finding the space diagonal of a cuboid always works this way: find a face (or base) diagonal first with Pythagoras, then use that diagonal as one side of a second right-angled triangle with the remaining edge.
This question tests finding the space diagonal of a cuboid by applying Pythagoras' theorem twice: once across the base, then again up through the solid.
First find the base diagonal \(AC\), using the length \(AB = 8\) cm and width \(BC = 6\) cm: [2]
\[AC^2 = 8^2 + 6^2 = 64 + 36 = 100\] \[AC = \sqrt{100} = 10\text{ cm}\]The space diagonal \(AG\) is the hypotenuse of a second right-angled triangle formed by the base diagonal \(AC\) and the vertical edge \(CG = 4\) cm: [3]
\[AG^2 = AC^2 + CG^2 = 10^2 + 4^2 = 100 + 16 = 116\] \[AG = \sqrt{116} = 10.8\text{ cm (3 s.f.)}\]Finding the space diagonal of a cuboid always works this way: find a face (or base) diagonal first with Pythagoras, then use that diagonal as one side of a second right-angled triangle with the remaining edge.
Question 20 Report
Two coffees and three pastries cost £9.50. One coffee and two pastries cost £5.50.
Find the cost of one coffee and the cost of one pastry. (3)
This problem needs to be translated into two simultaneous equations before it can be solved: let \(c\) be the cost of one coffee and \(p\) the cost of one pastry, in pounds.
Multiplying an equation by a constant so that one variable's coefficient matches in both equations is the standard first step whenever the coefficients are not already equal, as here with the coffee terms \(2c\) and \(c\).
This problem needs to be translated into two simultaneous equations before it can be solved: let \(c\) be the cost of one coffee and \(p\) the cost of one pastry, in pounds.
Multiplying an equation by a constant so that one variable's coefficient matches in both equations is the standard first step whenever the coefficients are not already equal, as here with the coffee terms \(2c\) and \(c\).
Question 21 Report
A train travels away from a station. Its distance, \(d\) km, from the station after \(t\) minutes is given by \(d = 1.5t\) for \(0 \leq t \leq 20\). The table below shows some values.
| \(t\) | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| \(d\) | 0 | 15 | 30 |
The relationship \(d = 1.5t\) is directly proportional: the graph is a straight line through the origin with gradient \(1.5\). To complete the table, substitute each given value of \(t\); to draw the graph, plot the table's points and join them with a straight line since the relationship is linear.
| \(t\) | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| \(d\) | 0 | 7.5 | 15 | 22.5 | 30 |
A directly proportional graph always passes through the origin; once you have one other point, or the gradient, you can extend the line accurately without needing every table value plotted first.
The relationship \(d = 1.5t\) is directly proportional: the graph is a straight line through the origin with gradient \(1.5\). To complete the table, substitute each given value of \(t\); to draw the graph, plot the table's points and join them with a straight line since the relationship is linear.
| \(t\) | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| \(d\) | 0 | 7.5 | 15 | 22.5 | 30 |
A directly proportional graph always passes through the origin; once you have one other point, or the gradient, you can extend the line accurately without needing every table value plotted first.
Would you like to proceed with this action?