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More questions on this topic: IGCSE Algebra: Substitution (118)
7 Here is a rule to work out the cost of buying a number of floorboards.
Cost = £55 \(\times\) number of floorboards
Toby buys a number of floorboards.
The cost is £330
Work out the number of floorboards Toby buys. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 6 | M1 | for \(330 \div 55\) |
| A1 | cao |
Build up methods are allowed
Condone one error in build up method
\(55 \times 6 = 330\), 330 or 55 on answer line scores M1A0
ignore use of £ eg £6
29 \(w = 40 - t^2\)
\[p = \frac{h - 5}{3}\]
| Answer | Mark | Mark scheme |
|---|---|---|
| 15 | M1 | for correct substitution, eg \(40 - (-5)^2\) |
| A1 | cao |
Condone missing brackets
| Answer | Mark | Mark scheme |
|---|---|---|
| \(h = 3p + 5\) | M1 | for a correct first step, eg \(3p = h - 5\) or for isolating the \(\dfrac{h}{3}\) term, eg \(p + \dfrac{5}{3} = \dfrac{h}{3}\) |
| A1 | for \(h = 3p + 5\) oe eg \(h = 3\left(p + \dfrac{5}{3}\right)\) |
Award M1 for \(3p + 5\) without seeing \(h = 3p + 5\)
22 The number of hours, \(H\), that some machines take to make 5000 bottles is given by
\[H = \frac{72}{n} \qquad \text{where } n \text{ is the number of machines.}\]
On Monday, 6 machines made 5000 bottles.
On Tuesday, 9 machines made 5000 bottles.
The machines took more time to make the bottles on Monday than on Tuesday.
How much more time? (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 4 | M1 | for method to use formula, eg \(72 \div 6\ (= 12)\) or \(72 \div 9\ (= 8)\) |
| A1 | cao |
Can be implied by \(\dfrac{72}{6}\) or \(\dfrac{72}{9}\)
13
\(T = x + 2y\)
\(x = 3\) and \(y = -4\)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(6cd\) | B1 | for \(6cd\) or \(6dc\) or \(cd6\) oe |
Inclusion of \(\times\) scores B0, eg \(6 \times cd\) oe
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-5\) | M1 | for \(3 + 2 \times -4\ (= 3 - 8)\) |
| A1 | cao |
3 The number of hours, \(H\), that some machines take to make 5000 bottles is given by
\(H = \dfrac{72}{n}\) where \(n\) is the number of machines.
On Monday, 6 machines made 5000 bottles.
On Tuesday, 9 machines made 5000 bottles.
The machines took more time to make the bottles on Monday than on Tuesday.
How much more time? (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 4 | M1 | for method to use formula, eg \(72 \div 6\ (= 12)\) or \(72 \div 9\ (= 8)\) |
| A1 | cao |
Can be implied by \(\dfrac{72}{6}\) or \(\dfrac{72}{9}\)
12 \(P = 2g + 4h\)
\(V = 3r - q\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 26 | M1 | for substitution eg \(2 \times 3 + 4 \times 5\) or \(6 + 20\) |
| A1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 13 | M1 | for substitution eg \(38 = 2g + 4 \times 3\) or a complete numerical method eg \((38 - 4 \times 3) \div 2\) or for a correct first step to rearrange eg \(P - 4h = 2g\) or \(\dfrac{P}{2} = g + \dfrac{4h}{2}\) oe |
| A1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-11\) | M1 | for \(3 \times -3 = -9\) oe or a full substitution eg \((3 \times -3) - 2\) |
| A1 | cao |
Condone absence of brackets
18 Naomi has \(b\) bags of apples and \(c\) crates of apples.
There are 5 apples in each bag.
There are 28 apples in each crate.
Naomi has a total of \(T\) apples.
Write a formula for \(T\) in terms of \(b\) and \(c\). (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(T = 5b + 28c\) | M1 | for \(5b\) or \(28c\) or \(T =\) a linear expression in \(b\) and/or \(c\) |
| M1 | for \(5b + 28c\) or partially correct formula, eg \(T = 5b\ (+ kc)\) oe or \(T = 28c\ (+ kb)\) oe | |
| A1 | for \(T = 5b + 28c\) oe |
Allow \(5 \times b\) and \(28 \times c\) throughout
8 Safiya wants to hire a van.
She uses this rule to work out the cost of hiring a van for a number of days.
| Cost = £45 × number of days |
Safiya is going to hire the van for 7 days.
Work out the cost. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 315 | M1 | for \(45 \times 7\) |
| A1 | cao |
23 \(T = 4m^2 - 11\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 25 | M1 | for (\(T =\)) \(4 \times (-3)^2 - 11\) or \(4 \times (-3)^2 = 36\) |
| A1 | cao |
Can accept missing brackets
| Answer | Mark | Mark scheme |
|---|---|---|
| \(p = \dfrac{d - 4}{3}\) oe | M1 | for a correct first step, eg \(d - 4 = 3p\) or \(\dfrac{d}{3} = p + \dfrac{4}{3}\) or for \(\dfrac{d - 4}{3}\) as answer |
| A1 | for \(p = \dfrac{d - 4}{3}\) oe |
May be in unsimplified form,
eg \(d - 4 = 3p + 4 - 4\)
14 \(y = 6x - 5\)
Work out the value of \(y\) when \(x = 4\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 19 | M1 | for a correct substitution, eg \((y =)\ 6 \times 4 - 5\) |
| A1 | cao |
2 \(T = 4m^2 - 11\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 25 | M1 | for (\(T =\)) \(4 \times (-3)^2 - 11\) or \(4 \times (-3)^2 = 36\) |
| A1 | cao |
Can accept missing brackets.
| Answer | Mark | Mark scheme |
|---|---|---|
| \(p = \dfrac{d - 4}{3}\) oe | M1 | for a correct first step, eg \(d - 4 = 3p\) or \(\dfrac{d}{3} = p + \dfrac{4}{3}\) or for \(\dfrac{d - 4}{3}\) as answer |
| A1 | for \(p = \dfrac{d - 4}{3}\) oe |
May be in unsimplified form,
eg \(d - 4 = 3p + 4 - 4\)
28 The number of days, \(d\), that it will take to build a house is given by
\[d = \frac{720}{n}\]where \(n\) is the number of workers used each day.
Ali’s company will take 40 days to build the house.
Hayley’s company will take 30 days to build the house.
Hayley’s company will have to use more workers each day than Ali’s company.
How many more? (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 6 | M1 | for \(720 \div 40\ (= 18)\) or \(720 \div 30\ (= 24)\) |
| M1 | for a complete process eg \((720 \div 30) - (720 \div 40)\) or \(\text{``}18\text{''} \times 4/3 - \text{``}18\text{''}\) or \(\text{``}24\text{''} - \text{``}24\text{''} \times 3/4\) | |
| A1 | cao |
17 \(x - 1 = 2\)
Work out the value of \(2x^2\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 18 | P1 | for process to solve \(x - 1 = 2\), eg. \(x = 2 + 1\ (= 3)\) or for \(2x = 6\) |
| P1 | for \(2 \times 9\) | |
| A1 | cao |
Can award mark for \(3 - 1 = 2\)
7 \(w = 4u + 3\)
Find the value of \(w\) when \(u = 8\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 35 | M1 | for \(4 \times 8\ (=32)\) |
| A1 | cao |
Award this mark if used ambiguously
eg \(4 \times 8 + 3 = 4 \times 11\) as long as \(4 \times 8\) is stated
15 You can use this rule to work out the total hire charge, in pounds (£), for hiring a 3D printer for a number of weeks.
Total hire charge (£) \(=\) number of weeks \(\times\ 70 + 50\)
Mia wants to hire a 3D printer for 4 weeks.
Zahir hires a 3D printer.
The total hire charge is £680
| Answer | Mark | Mark scheme |
|---|---|---|
| 330 | M1 | for \(4 \times 70 + 50\) oe |
| A1 | cao |
May be seen as sum of four 70s and a 50
\(n \times (70 + 50)\) or ambiguous working gets 0 marks
| Answer | Mark | Mark scheme |
|---|---|---|
| 9 | M1 | for use of inverse operations eg \((680 - 50) \div 70\) OR rearranges an equation to solve eg \(70x + 50 = 680\) rearranged to isolate \(x\) term. OR ft (a) eg \(((680 - \text{``}330\text{''}) \div 70) + 4\) |
| A1 | cao or ft their (a) |
Need not have brackets; can be written in an incorrect order if the intention is clear
A correct but embedded answer gets 1 mark
11 \(P = 7r + 3q\)
Work out the value of \(P\) when \(r = 5\) and \(q = -4\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 23 | M1 | for substitution eg. \(7 \times 5\) and \(3 \times -4\) or \(7(5) + 3(-4)\) |
| A1 | cao |
\(7 \times 5\ (= 35)\) and \(3 \times -4\ (= -12)\) may be seen separately but both must be seen for the award of M1
21 \(v^2 = u^2 + 2as\)
\(u = 12 \qquad a = -3 \qquad s = 18\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 6 or \(-6\) | M1 | for \(12^2 + 2 \times -3 \times 18\ (= 36)\) |
| A1 | for 6 or \(-6\), accept \(\pm 6\) |
Terms may be partially evaluated.
Only one value is required for full marks
| Answer | Mark | Mark scheme |
|---|---|---|
| \(s = \dfrac{v^2 - u^2}{2a}\) | M1 | for subtracting \(u^2\) from both sides or dividing all terms by \(2a\) as the first step |
| A1 | \(s = \dfrac{v^2 - u^2}{2a}\) oe |
Must see this step carried out, not just the intention shown
9 \(g = 9\)
\(h = 4\)
Work out the value of \(2g + 3h\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 30 | M1 | \(2 \times 9 + 3 \times 4\) |
| A1 | cao |
May be shown in stages but an intention to add \(2 \times 9\) and \(3 \times 4\) must be clear
2 \(v^2 = u^2 + 2as\)
\(u = 12 \qquad a = -3 \qquad s = 18\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 6 or \(-6\) | M1 | for \(12^2 + 2 \times -3 \times 18\ (= 36)\) |
| A1 | for 6 or \(-6\), accept \(\pm 6\) |
Terms may be partially evaluated.
Only one value is required for full marks
| Answer | Mark | Mark scheme |
|---|---|---|
| \(s = \dfrac{v^2 - u^2}{2a}\) | M1 | for subtracting \(u^2\) from both sides or dividing all terms by \(2a\) as the first step |
| A1 | \(s = \dfrac{v^2 - u^2}{2a}\) oe |
Must see this step carried out, not just the intention shown
16 \(P = 4x + 3y\)
\(x = 5\)
\(y = -2\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 14 | M1 | for \(4 \times 5\) and \(3 \times -2\), the substitution may be seen in two separate calculations, eg \(4 \times 5\ (= 20)\) and \(3 \times -2\ (= -6)\) |
| A1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(4e^2 + 8e\) | B2 | for \(4e^2 + 8e\) |
| (B1 | for \(4e^2\) or \(8e\)) |
Note: \(4e^2 + 8e = 12e^3\) for example gets B1 only
| Answer | Mark | Mark scheme |
|---|---|---|
| 11 | M1 | for a correct first step eg \(3 \times m - 3 \times 4 = 21\) oe or \(m - 4 = 21 \div 3\ (= 7)\) oe |
| A1 | cao |
Showing \(\div 3\) by each side of equation is sufficient
16 \(v = u + at\)
\(u = 1 \qquad a = -3 \qquad t = \dfrac{1}{2}\)
Work out the value of \(v\). (2)
| Answer | Mark | Notes |
|---|---|---|
| \(-\dfrac{1}{2}\) | M1 | for substitution with operations shown e.g. \(1 + {-3} \times \dfrac{1}{2}\) or \(1 - \dfrac{3}{2}\) or \(1\frac{1}{2}\) or \(-1\frac{1}{2}\) |
| A1 | oe |
11 \(T = 4v + 3\)
| Answer | Mark | Notes |
|---|---|---|
| 11 | M1 | substitutes \(v = 2\) eg \(4 \times 2 + 3\) or \(8 + 3\) |
| A1 | cao |
| Answer | Mark | Notes |
|---|---|---|
| \(v = \dfrac{T - 3}{4}\) | M1 | correct first step to rearrange by isolating \(4v\) or dividing each term by 4, eg \(T - 3 = 4v\) |
| A1 | fully correct answer |
13 \(y = 4x - 3t\)
\(x = 2\)
\(t = 5\)
\(y = 4x - 3t\)
\(y = 30\)
\(t = 2\)
| Answer | Mark | Notes |
|---|---|---|
| \(-7\) | 2 | M1 for substitution eg \(4 \times 2 - 3 \times 5\) A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 9 | 2 | M1 for substitution eg \(30 = 4x - 3 \times 2\) or \(\dfrac{30 + 6}{4}\) or \(y + 3t = 4x\) oe A1 cao |
1
\(v = 2t^2\)
\(t = 3\)
| Answer | Mark | Notes |
|---|---|---|
| \(x^{12}\) | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 2 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 18 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| example given | 1 | B1 for stating a value of \(n\) for which \(6n + 1\) is not prime eg 4, 8, 9, 14, 19, ... ,1000 etc |
14 This rule is used to work out the cost of hiring a village hall.
Total cost = £8 for each hour plus a fixed charge of £5
Jackie hires the hall for 7 hours.
James pays £29 to hire the hall.
| Answer | Mark | Notes |
|---|---|---|
| 61 | 2 | M1 for a complete method eg \(7 \times 8 + 5\) A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 3 | 3 | M1 for \(29 - 5\) (= 24) or for \(8x + 5 = 29\) M1 for \(\text{``}24\text{''} \div 8\) or for \(8x = 24\) A1 cao |
13 \(h = 3t^2\)
\(h = 3t^2\)
| Answer | Mark | Notes |
|---|---|---|
| 75 | 2 | M1 for \(3 \times 5 \times 5\) oe A1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(108 = 3t^2\) \(t^2 = \tfrac{108}{3} = 36\) | \(\pm 6\) | 2 | M1 for \((t^2 =) \dfrac{108}{3}\) (= 36) A1 for 6 or \(-6\) or both |
| Answer | Mark | Notes |
|---|---|---|
| \(a = \dfrac{v - u}{t}\) | 2 | M1 for subtracting \(u\) from both sides or dividing both sides by \(t\) as a first step A1 for \(a = \dfrac{v - u}{t}\) oe |
11
\(k = 5\)
\(m = 3\)
| Answer | Mark | Notes |
|---|---|---|
| \(3h\) | 1 | B1 |
| Answer | Mark | Notes |
|---|---|---|
| \(3pr\) | 1 | B1 |
| Answer | Mark | Notes |
|---|---|---|
| \(2x + 7y\) | 2 | M1 for \(2x\) or \(7y\) A1 for \(2x + 7y\) |
| Answer | Mark | Notes |
|---|---|---|
| 22 | 2 | M1 for correct substitution e.g. \(2 \times 5 + 4 \times 3\) A1 cao |
23 The body mass index, \(B\), for a person of mass \(m\) kg and height \(h\) metres is given by the formula
\[B = \frac{m}{h^2}\]Usman has a mass of 50 kg.
He has a height of 1.57 m.
Tom’s height is 1.80 m.
He wants his body mass index to be 21
Tom is a ski jumper.
The maximum length of skis he can use is 145% of his height.
Tom’s height is 1.80 m.
| Answer | Mark | Notes |
|---|---|---|
| 20.3 | 2 | M1 for \(\dfrac{50}{1.57^2}\) oe A1 for answer in range 20.2 to 20.3 |
| Answer | Mark | Notes |
|---|---|---|
| 68.04 | 2 | M1 for \((m =)\ 1.8^2 \times 21\) oe A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 2.61 | 3 | M2 for a complete method to find 145% of 1.8, eg. \(\frac{145}{100} \times 1.80\) oe (M1 for a method to find 45% of 1.8, eg. \(\frac{45}{100} \times 1.80\) (= 0.81) or for a multiplication factor of 1.45) A1 cao |
6 The body mass index, \(B\), for a person of mass \(m\) kg and height \(h\) metres is given by the formula \[B = \frac{m}{h^2}\]
Usman has a mass of 50 kg.
He has a height of 1.57 m.
Tom’s height is 1.80 m.
He wants his body mass index to be 21
Tom is a ski jumper.
The maximum length of skis he can use is 145% of his height.
Tom’s height is 1.80 m.
| Answer | Mark | Notes |
|---|---|---|
| 20.3 | 2 | M1 for \(\dfrac{50}{1.57^2}\) oe A1 for answer in range 20.2 to 20.3 |
| Answer | Mark | Notes |
|---|---|---|
| 68.04 | 2 | M1 for (\(m\) =) \(1.8^2 \times 21\) oe A1 for 68.04 |
| Answer | Mark | Notes |
|---|---|---|
| 2.61 | 3 | M2 for a complete method to find 145% of 1.8, eg. \(\dfrac{145}{100} \times 1.80\) oe (M1 for a method to find 45% of 1.8, eg. \(\dfrac{45}{100} \times 1.80\) (= 0.81) or for a multiplication factor of 1.45) A1 cao |
15 \(p = 2e + 3g\)
\(e = 6.5\)
\(g = 2\)
Work out the value of \(p\). (2)
| Answer | Mark | Notes |
|---|---|---|
| 19 | 2 | M1 for \(2 \times 6.5 + 3 \times 2\) or 13 and 6 A1 cao |
12 You can use this rule to work out the total hire charge, in pounds (£), for hiring a satellite phone.
Total hire charge = number of weeks \(\times\) 90 + 50
Ismail wants to hire a satellite phone for 4 weeks.
Dominik hires a satellite phone.
His total hire charge is £860
| Answer | Mark | Notes |
|---|---|---|
| 410 | 2 | M1 for \(4 \times 90 + 50\) (= 410) A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 9 | 3 | M1 for one inverse operation eg \(-50\) or \(\div 90\) M1 for complete inverse operations, eg \((860 - 50) \div 90\) accept \(860 - 50 \div 90\) A1 cao OR M1 ft for finding the difference to part (a), ie \(860 - \text{``}410\text{''}\) (= 450) M1 for \(\text{``}450\text{''} \div 90\) A1 cao |
17 \(a = 4\)
\(b = -5\)
\(x = 3\)
| Answer | Mark | Notes |
|---|---|---|
| \(-7\) | 2 | M1 for \(2 \times 4 + 3 \times -5\) or \(8 - 15\) oe A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 36 | 1 | B1 cao |
2 \(x = 3\)
| Answer | Mark | Notes |
|---|---|---|
| 36 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 2.5 oe | 2 | M1 for collecting the terms in \(x\) or the number terms in an equation, eg. \(5x - x + 4 = 14\) or \(5x = 14 - 4 + x\) A1 for 2.5 oe (accept \(\tfrac{10}{4}\)) |
7
| Answer | Mark | Notes |
|---|---|---|
| 31 | 2 | M1 for \(3 \times 5 + 2 \times 8\) or 15 and 16 A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(6x + 8y\) | 2 | M1 for \(6x\) or \(8y\) A1 for \(6x + 8y\) oe as final answer |
3 \(f = 3g + 7h\)
| Answer | Mark | Notes |
|---|---|---|
| \(-1\) | 2 | M1 for \(3 \times -5 + 7 \times 2\) A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(3(x + 2)\) | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(7y - 16\) | 2 | M1 for intention to expand a bracket eg \(5y - 10\) or \(2y - 6\) A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(m^8\) | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(p^4\) | 1 | B1 cao |
18 \(x = -5\)
\(y = 2\)
Janet buys \(p\) packets of sweets.
There are 10 sweets in each packet.
Janet eats 7 of the sweets.
| Answer | Mark | Notes |
|---|---|---|
| \(-7\) | 2 | M1 for \(3 \times -5\) (\(= -15\)) and \(4 \times 2\) (\(=8\)) A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| (i) \(10p\) | 2 | B1 for \(10p\) oe |
| (ii) \(10p - 7\) | B1ft for “\(10p\)” \(- 7\) Note “\(10p\)” MUST be an algebraic expression |
12 You can change temperatures from °F to °C by using the formula \[C = \frac{5(F - 32)}{9}\]
\(F\) is the temperature in °F.
\(C\) is the temperature in °C.
The minimum temperature in an elderly person’s home should be 20°C.
Mrs Smith is an elderly person.
The temperature in Mrs Smith’s home is 77°F.
| Answer | Mark | Notes |
|---|---|---|
| No, temp is 25°C | 3 | M1 for substitution of 77 into the RHS of the formula A1 for 25 cao or for 225/9 and 180/9 cao C1 (dep on M1) for conclusion (ft) following from working shown OR M1 for substitution of 20 into LHS of formula and correct process to find F A1 for 68 cao C1 (dep on M1) for conclusion (ft) following from working shown |
| Answer | Mark | Notes |
|---|---|---|
| \(F = \dfrac{9}{5}C + 32\) | 3 | M1 for expansion of the brackets (eg \(5 \times F - 5 \times 32\)) or an attempt to multiply both sides by 9, or divide both sides by 5 as the first step. M1 (dep) for a correct second step A1 for \(F = \dfrac{9}{5}C + 32\) oe |
10 You can use these rules to change temperatures from \(^\circ\)C to \(^\circ\)F.
| approximate rule |
|---|
| Multiply the \(^\circ\)C temperature by 2 and then add 30 |
| exact rule |
|---|
| Multiply the \(^\circ\)C temperature by 1.8 and then add 32 |
Amy uses the approximate rule to change 20\(^\circ\)C to \(^\circ\)F.
Dan uses the exact rule to change 20\(^\circ\)C to \(^\circ\)F.
Jade uses the approximate rule to change a temperature from \(^\circ\)C to \(^\circ\)F.
The result is 110\(^\circ\)F.
| Answer | Mark | Notes |
|---|---|---|
| 2 | 4 | M1 for 20 × 2 + 30 (=70) M1 for 20 × 1.8 + 32 (=68) M1 (dep on M1) for “70” – “68” A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 40 | 3 | M1 for 110 – 30 ÷ 2 or 110 = ? × 2 + 30 or 110 – 30 or ÷ 2 seen as second operation M1 for “(110 – 30)” ÷ 2 A1 cao NB accept reverse flowcharts for inverse operations SC if exact rule used: B2 for “(110 – 32)” ÷ 1.8 |
10 Here is a number machine.
input → ÷ 4 → +7 → output
Complete this table for the number machine.
| Input | Output |
|---|---|
| 8 | 9 |
| 12 | |
| 27 |
(2)
| Answer | Mark | Notes |
|---|---|---|
| (12) 10 | 1 | B1 cao |
| 80 (27) | 1 | B1 cao |
4 \(a = 5\)
\(b = 3\)
Work out the value of \(4a + 2b\) (2)
| Answer | Mark | Notes |
|---|---|---|
| 26 | 2 | M1 for correct substitution into expression \(4 \times 5 + 2 \times 3\) A1 cao |
15 Here is a number machine.
input → ×6 → + 3 → output
Complete this table for the number machine.
| Input | Output |
|---|---|
| 0.5 | |
| 2 | 15 |
| 3 | |
| 33 |
(3)
| Answer | Mark | Notes |
|---|---|---|
| (a) 6 | 1 | B1 cao |
| (b) 21 | 1 | B1 cao |
| (c) 5 | 1 | B1 cao |
*14 Angela and Michelle both work as waitresses at the same restaurant.
This formula is used to work out the total amount of money each waitress gets.
Total amount = £6.50 × number of hours worked + tips
The table shows the number of hours Angela and Michelle each worked last Saturday.
It also shows the tips they got.
| Number of hours worked | Tips | |
|---|---|---|
| Angela | 8 | £12 |
| Michelle | 7 | £15 |
Who got the higher total amount of money last Saturday?
You must show clearly how you got your answer. (4)
| Answer | Mark | Notes |
|---|---|---|
| Correct statement | 4 | M1 for 6.50 × 8 + 12 or 6.50 × 7 + 15 M1 for 6.50 × 8 + 12 and 6.50 × 7 + 15 A1 for 64 and 60.5(0) C1 (dep on first M1) for correct statement ft their figures OR M1 for 6.50 × (8−7) or 15−12 M1 for 6.50 × (8−7) and 15−12 A1 for 6.5(0) and 3 C1 (dep on first M1) for correct statement ft their figures [SC If no working shown B1 for 64 and 60.5(0) or B1 for 6.5(0) and 3] |
8 \(a = 4b\)
\(P = 4d - 3\)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| 4 × 3 | 12 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 5 | 2 | M1 for 4 × 2 – 3 A1 cao |
18 \(f = 8\)
\(T = 3g + 5h\)
\(g = -2\)
\(h = 4\)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(2 \times 8 = 16\) \(16 + 7\) | 23 | 2 | M1 for correct substitution of 8 into the expression \(2f\) A1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(3 \times -2 + 5 \times 4\) | 14 | 2 | M1 for correct substitution into the formula A1 cao |
14 \(A = 4bc\)
\(A = 100\)
\(b = 2\)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(100 = 4 \times 2 \times c\) | 12.5 | 2 | M1 for correct substitution into formula A1 for 12.5 oe |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(m^2 = \dfrac{k + 1}{4}\) \(4m^2 = k + 1\) \(k = 4m^2 - 1\) or \(2m = \sqrt{(k + 1)}\) \(4m^2 = k + 1\) \(k = 4m^2 - 1\) | \(k = 4m^2 - 1\) | 3 | M1 for correct method to clear fraction or remove square root sign M1 (dep) for a fully correct method to both clear fraction and remove square root sign A1 for \(k = 4m^2 - 1\) or \(k = (2m + 1)(2m - 1)\) |
13 You can use this rule to work out the total cost of hiring a car.
Total cost = £4 per hour plus £12
Arun hires a car for 5 hours.
Raj hires a car.
The total cost is £40
| Answer | Mark | Notes |
|---|---|---|
| 32 | 2 | M1 for \(4 \times 5 + 12\) oe A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 7 | 3 | M1 for \(40 - 12\) or 28 seen M1 (dep) for \('28' \div 4\) A1 cao OR M1 for \(12 + 4 + 4 + \ldots\) M1 for \(12 + 7 \times 4\) oe A1 cao OR M1 for \(12 + 4x = 40\) oe M1 for \(4x = 40 - 12\) oe A1 cao NOTE: A correct embedded answer scores M2 A0 OR M1 ft for \('32' + 4\ldots\) or \(40 - '32'\) M1 ft for \(5 + 1 \ldots\) oe A1 ft Note: Do not follow through from part a an answer of 40 |
18 \(y = 4x + c\)
\(x = 7.5\)
\(c = 5.4\)
\(y = 4x + c\)
\(y = 18.8\)
\(c = -2.4\)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(y = 4 \times 7.5 + 5.4\) | 35.4 | 2 | M1 for \(4 \times 7.5 + 5.4\) A1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(18.8 = 4x - 2.4\) \(x = \dfrac{18.8 + 2.4}{4}\) | 5.3 | 2 | M1 for intention to add 2.4 to 18.8 or to subtract \(-2.4\) from 18.8 or to divide 18.8 and \((-)2.4\) by 4 A1 cao |
10 You can use this rule to work out the total charge for hiring a concrete mixer.
Total charge = £30 plus £8 each day
Esme hired a concrete mixer for 4 days.
William also hired a concrete mixer.
The total charge was £110
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(30 + 8 \times 4\) | 62 | 2 | M1 for \(30 + 8 \times 4\) or attempt to add four 8s to 30 (allow one error in addition) A1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(110 - 30 = 80\) \(80 \div 8 = 10\) OR \(110 - 62 = 48\) \(48 \div 8 = 6\) \(4 + 6 = 10\) | 10 | 3 | M1 for \(110 - 30\) (=80) M1 (dep) for \(\text{`}80\text{'} \div 8\) or A1 cao OR M1 for \(110 - 62\) (= 48) M1(dep) for \(\text{`}48\text{'} \div 8 = 6\) A1 cao |
4 Here is a two-stage number machine.
It multiplies by 5 and then subtracts 3
| Input | Output |
|---|---|
| 1 | 2 |
| 2 | 7 |
| 5 | 22 |
| 7 | |
| 47 |
Here is a different two-stage number machine.
When the input is 10, the output is 60
| Answer | Mark | Notes |
|---|---|---|
| 32 and 10 | 2 | B1 for 32 in the correct place B1 for 10 in the correct place |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(10 \times 3 \times 2 = 60\) or \(10 \times 3 + 30 = 60\) | \(\times 2\) or \(+30\) | 1 | B1 for \(\times 2\) or \(+30\) |
15 The table shows the minimum distance and the maximum distance people should sit from different sized TV screens when watching TV.
| TV screen size (inches) | Minimum distance (feet) | Maximum distance (feet) |
|---|---|---|
| 28 | 3.5 | 7 |
| 30 | 3.75 | 7.5 |
| 32 | 4 | 8 |
| 36 | 4.5 | 9 |
| 38 | 4.75 | 9.5 |
| 40 | 5 | 10 |
| 42 | 5.25 | 10.5 |
James has a TV with a screen size of 32 inches.
He is going to watch his TV.
Sheraz has a TV with a screen size of 38 inches.
He is going to watch his TV.
You can use this rule to work out the maximum distance to sit from a TV screen.
Maximum distance in feet = screen size in inches \(\div\) 4
Clare has a TV with a screen size of 24 inches.
She is going to watch her TV.
Richard should sit a maximum distance of 12 feet from his screen.
| Answer | Mark | Notes |
|---|---|---|
| 4 | 1 | B1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(9.5 - 4.75 =\) OR \(9.5 \div 2 =\) | 4.75 | 2 | M1 for \(9.5 - 4.75\) or \(9.5 \div 2\) or \(4.75 - 9.5\) A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 6 | 1 | B1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(12 \times 4 =\) | 48 | 2 | M1 for \(\times 4\) seen or identifying \(+ 0.5\) for every 2 inches or \(12 + 12 + 12 + 12\) oe or build up method eg 12, 24, 36, 48 allow one error A1 cao |
14 \(P = 3.5x - y\)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(3.5 \times 12 - 5\) | 37 | 2 | M1 for \(3.5 \times 12 - 5\) or \(42 - 5\) A1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(3.5 \times -9 - -6\) | \(-25.5\) | 2 | M1 for \(3.5 \times -9 - -6\) or \(3.5 \times -9 + 6\) or sight of \(-31.5\) A1 for \(-25.5\) or \(-\frac{51}{2}\) or \(-25\frac{1}{2}\) |
5
You can work out the amount of medicine, \(c\) m\(l\), to give to a child by using the formula
\[c = \frac{ma}{150}\]\(m\) is the age of the child, in months.
\(a\) is an adult dose, in m\(l\).
A child is 30 months old.
An adult’s dose is 40 m\(l\).
Work out the amount of medicine you can give to the child. (2)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(c = \dfrac{30 \times 40}{150}\) | 8 | 2 | M1 for \(\dfrac{30 \times 40}{150}\) or 1200 seen A1 cao |
16 \(x\) is a positive even number.
\(y = (x - 3)(x - 5)(x + 6)\)
Without expanding brackets,
explain why there is only one value of \(x\) for which \(y\) is negative. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| When \(x = 2\), \((y =)\ 24\) or when \(x = 2\), negative (\(\times\)) negative (\(\times\)) positive = positive | B1 | oe |
| When \(x = 4\), \((y =)\ {-10}\) or when \(x = 4\), positive (\(\times\)) negative (\(\times\)) positive = negative | B1 | oe |
| When \(x = 6\), \((y =)\ 36\) or when \(x = 6\), positive (\(\times\)) positive (\(\times\)) positive = positive | B1 | oe SC1 correct value of \(y\) for any positive even value of \(x\) greater than 6 |
| SC1 should only be considered when the response scores B0B0B0 | |
| Ignore any other evaluation attempts or written statements alongside correct responses | |
| A value of \(x\) may be implied by consistent substitution into the equation eg \((2 - 3)(2 - 5)(2 + 6) = 24\) implies \(x = 2\) | B1 |
| For any response using negative/positive, \(x = 2\) can be implied by stating ‘below 4’ and \(x = 6\) can be implied by stating ‘above 4’, or equivalent | |
| For any response using negative/positive, multiplication must be shown, referred to or implied, and the sign of the result must be stated |
16
Assume the job needs completing in 20 days.
How many workers are needed? [3 marks]
What does this mean about the number of workers that are needed?
Tick one box. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(20 = \dfrac{800}{w}\) or \((w =)\ \dfrac{800}{d}\) | M1 | oe |
| \(800 \div 20\) | M1dep | oe |
| 40 | A1 |
| Answer | Mark | Comments |
|---|---|---|
| It is greater than the answer to part (a) | B1 |
10 Work out the value of \(\quad 2(a^2 + 3a) \quad\) when \(\quad a = 4\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: substitutes first | ||
| \(4^2\) or 16 or 32 or \(3 \times 4\) or 12 or 24 | M1 | oe implied by 28 |
| \(2 \times (4 \times 4 + 3 \times 4)\) | M1dep | oe |
| 56 | A1 | |
| Alternative method 2: multiplies out first | ||
| \(2 \times a^2\) or \(2 \times 3a\) | M1 | oe |
| \(2 \times 4 \times 4 + 2 \times 3 \times 4\) | M1dep | oe |
| 56 | A1 | |
| Alternative method 3: factorises first | ||
| \(2a(a + 3)\) or \(2 \times a\) or \(a + 3\) | M1 | oe |
| \(2 \times 4 \times (4 + 3)\) | M1dep | oe |
| 56 | A1 | |
| Inclusion of \(a\) can score a maximum of one mark eg1 \(4^2 = 16 \quad 16 \times 2 = 32a\) eg2 \(56a\) | M1 only M1 only |
| \(2 \times 4^2 + 2 \times 3a\) is not enough for the second mark of Alt 2 | M1M0A0 |
23 The total cost of a taxi ride is calculated by adding
a fixed charge of £4
and
a charge of £2 per mile.
Write a formula to work out the total cost, £\(C\), of a journey of \(m\) miles. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2m + 4\) or \(2 \times m + 4\) | B2 | oe B1 \(2m\) or \(m2\) or \(200m + 400\) or \(km + 4\) with k an integer, \(k \ne 0\) or 2 |
| Ignore any attempt to substitute a value for \(m\) after \(2m\) seen for B1 | |
| Condone \(m2 + 4\) (not testing simplification) | B2 |
| Accept use of another letter for \(m\) other than \(c\) eg1 \(2x + 4\) eg2 \(2c + 4\) | B2 B0 |
| Condone inclusion of £ sign for B1 eg1 £\(2m + 4\) eg2 £\(2m\) | B1 B1 |
12
| Answer | Mark | Comments |
|---|---|---|
| \((-4)^2 + 7 \times -4\) or \(-4(-4 + 7)\) or 16 or \(-28\) | M1 | oe eg \((-4)^2 + 7(-4)\) |
| \(-12\) | A1 | SC1 \(-44\) |
| SC is for \(-4^2 + 7 \times -4 = -16 - 28 = -44\) | |
| Embedded 16 or \(-28\) seen eg \(16 + 7x\) without correct answer | M1A0 |
| Values may be implied eg1 \((-4)^2 + 7 \times 4 = 44\) 16 is implied eg2 only answer 44 | M1A0 M0A0 |
| Further correct work eg \(16 - 28 = -12\), Answer \(3x\) | M1A1 |
| Further incorrect work eg \(16 - 28 = -12\), Answer \(-12x\) | M1A0 |
| \(+-28\) is the same as \(-28\) | M1 |
| Only \(-4^2 + 7 \times -4\) | M0A0 |
| \(-4^2 + 7 \times -4 = -16 + 28 = 12\) | M0A0 |
| \(16x\) does not imply 16 | M0A0 |
| Answer | Mark | Comments |
|---|---|---|
| \(y + 1\) or \(1 + y\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \(4a + 8\) or \(8 + 4a\) or \(5a\) | M1 | |
| \(5a + 8\) or \(8 + 5a\) | A1 |
| Further incorrect work or simplification eg \(5a + 8\), Answer \(13a\) | M1A0 |
5 The total cost of a taxi ride is calculated by adding
a fixed charge of £4
and
a charge of £2 per mile.
Write a formula to work out the total cost, £\(C\), of a journey of \(m\) miles. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2m + 4\) or \(2 \times m + 4\) | B2 | oe B1 \(2m\) or \(m2\) or \(200m + 400\) or \(km + 4\) with k an integer, \(k \ne 0\) or 2 |
| Ignore any attempt to substitute a value for \(m\) after \(2m\) seen for B1 | |
| Condone \(m2 + 4\) (not testing simplification) | B2 |
| Accept use of another letter for \(m\) other than \(c\) eg1 \(2x + 4\) eg2 \(2c + 4\) | B2 B0 |
| Condone inclusion of £ sign for B1 eg1 £\(2m + 4\) eg2 £\(2m\) | B1 B1 |
14 A graph passes through the points \((3, 15)\) and \((7, w)\)
Work out the value of \(w\) that this would give. [3 marks]
What does this mean about the actual value of \(w\)?
Tick one box. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(15 = 3^2 + c\) or \((c =)\ 6\) | M1 | oe |
| \(7^2 +\) their 6 | M1dep | oe |
| 55 | A1 |
| Answer | Mark | Comments |
|---|---|---|
| It is impossible to tell | B1 |
11 A window cleaner uses this formula.
| \(C = 2W + 5\) |
\(C\) = cost, in £, for the customer
\(W\) = number of windows to be cleaned
Show why this cost must be incorrect. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(2 \times 6\) or 12 or 1700 | M1 | oe |
| 17(.00) | A1 | accept 1700p |
| Answer | Mark | Comments |
|---|---|---|
| 9.5 or \(9\frac{1}{2}\) or valid reason | B1 | eg shows that 9 windows cost £23 and 10 windows cost £25 or you can’t have half a window |
| Calculated values must be correct | |
| (19, 21,) 23, 25 (, 27…) | B1 |
| Even number \(+\) 5 can’t be even / can’t end in a 4 | B1 |
| There’s no whole number you can times by 2 to get to 19 There’s no number you can times by 2 to get to 19 | B1 B0 |
| Even \(+\) 5 \(=\) odd, 24 is even Even \(+\) 5 \(=\) odd | B1 B0 |
| Costs are all always odd numbers, 24 is even Costs are all always odd numbers | B1 B0 |
| 19 is a prime number and so doesn’t divide exactly by 2 19 is a prime number | B1 B0 |
5
Work out the value of \(d\) when \(\quad g = 15 \quad\) and \(\quad h = 63\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(15^2 - 2 \times 63\) or 225 or 126 | M1 | \(-126\) implies M1 |
| 99 | A1 |
| Answer | Mark | Comments |
|---|---|---|
| \(m - k\) or \(-k + m\) | B1 |
8
\[T = 5P - W\]| Answer | Mark | Comments |
|---|---|---|
| \(5 \times 4\) or 20 | M1 | oe |
| 18 | A1 |
| Answer | Mark | Comments |
|---|---|---|
| \(-40 + 10\) or \(-30\) or \(-40 = 5P - 10\) or \(P = \dfrac{T + W}{5}\) | M1 | |
| their \(-30 \div 5\) | M1dep | |
| \(-6\) | A1 | SC2 \(-10\) with \(-50\) seen |
| Embedded answer of \(-6\) | M1M1A0 |
| SC2 \(-10\) with \(-50\) seen for \(-40 + 10 = -50\) and then \(\div 5\) | |
| \(-40 = 5P - 10\) may use a different letter or symbol for \(P\) but not \(T\) or \(W\) |
16 A graph has the equation \(\quad y = x^2 + px + r \quad\) where \(p\) and \(r\) are constants.
The graph passes through the points (0, 4), (1, 3) and \((8, w)\)
Work out the value of \(w\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4 = 0^2 + p \times 0 + r\) or \(r = 4\) | M1 | oe equation may be implied |
| \(1^2 + p\ (\times 1) +\) their \(4 = 3\) or \(p = -2\) | M1 | oe equation allow their 4 to be \(r\) |
| \(8^2 + (\text{their } {-2}) \times 8 +\) their 4 or \(64 - 16 + 4\) | M1dep | oe dep on M1M1 do not allow their 4 to be \(r\) |
| 52 | A1 |
14 To travel to a festival, a group of people will hire a minibus.
This formula has all costs in £
| Cost per person \(= \dfrac{165 + \text{cost of the minibus}}{\text{number of people in the group}}\) |
Work out the cost per person. [2 marks]
The cost per person will be £50
Work out the cost of this minibus. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{165 + 567}{12}\) or \(\dfrac{732}{12}\) | M1 | oe |
| 61 | A1 | SC1 212.25 |
| Only \(165 + 567 \div 12\) with brackets missing | M0A0 |
| 61.00 | M1A1 |
| 61.0 | M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(50 = \dfrac{165 + x}{15}\) or \(50 \times 15\) or 750 seen | M1 | oe eg \(750 = 165 +\) cost of minibus any letter or symbol or word(s) |
| \(50 \times 15 - 165\) | M1dep | oe |
| 585 | A1 | SC1 915 |
| Alternative method 2 | ||
| \(165 \div 15\) or 11 | M1 | oe |
| \((50 -\) their \(11) \times 15\) or \(39 \times 15\) | M1dep | oe |
| 585 | A1 | SC1 915 |
Up to M2 may be awarded for correct work with no answer, or incorrect answer, even if this is seen amongst multiple attempts
\((165 +\) any value\() \div 15\) does not imply M1 unless set up as an equation for the first mark of Alt 1
Allow 12 as a misread for 15
8 Work out the value of \(\quad 3r + 4t \quad\) when \(\quad r = 13\) and \(t = -2\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3 \times 13 + 4 \times -2\) or \((3r =)\ 39\) or \((4t =)\ {-8}\) | M1 | oe |
| 31 | A1 |
| \(39 + 8\) | M1A0 |
| 39 or −8 may be implied by a calculation eg \(3 \times 13 + 4 \times 2 = 47\) | M1A0 |
| 47 only does not imply 39 | M0A0 |
| Values are not implied by incorrect expressions eg only \(39r\) | M0 |
| Incorrect further work | A0 |
23 Here are two simultaneous equations.
\(y = x^2 + 7x - c\)
and
\(y = 3x + d\)
There is a solution when \(\quad x = 5\)
Work out the value of \(\quad c + d\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(5^2 + 7 \times 5 - c\) or \(60 - c\) and \(3 \times 5 + d\) or \(15 + d\) | M1 | oe |
| \(25 + 35 - c = 15 + d\) or \(60 - c = 15 + d\) or \(c = 60 - y\) and \(d = y - 15\) and \(c + d = 60 - y + y - 15\) | M1dep | oe equation with squaring and multiplications correctly completed |
| 45 | A1 | |
| Alternative method 2 | ||
| \(x^2 + 7x - c = 3x + d\) or \(x^2 + 7x - c - (3x + d) = 0\) or \(x^2 + 7x - c - 3x - d = 0\) or \(3x + d - (x^2 + 7x - c) = 0\) or \(3x + d - x^2 - 7x + c = 0\) | M1 | oe |
| (\(c + d\) =) \(x^2 + 7x - 3x\) or (\(c + d\) =) \(x^2 + 4x\) and substitutes \(x = 5\) | M1dep | oe |
| 45 | A1 | |
| Once \(c + d = 45\) is seen, ignore further attempts to find values for \(c\) or \(d\) | |
| 45 on answer line with no working or no incorrect working | M1M1A1 |
12
| \(8a\) | \(a^2 - b\) |
When \(\quad a = 25 \quad\) the expressions have the same value.
Work out the value of \(b\). [3 marks]
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(3x + 2y\) | B2 | either order B1 \(3x\) or \(2y\) |
| Further incorrect work after a B2 response is B1 | |
| eg \(3x + 2y\) followed by \(5xy\) | B1 |
| Further incorrect work after a B1 response is B1 | |
| eg \(15x + 2y\) followed by \(30xy\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \(8 \times 25\) or 200 or \(25^2\) or 625 | M1 | oe |
| \(8 \times 25\) or 200 and \(25^2 - b\) or \(625 - b\) or \(25^2 - 8 \times 25\) or \(625 - 200\) | M1dep | oe may be seen in an equation |
| 425 | A1 |
| Embedded answer | M1M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| \(3w + 5\) | B1 |
20 \(\dfrac{a}{b} = 3c\)
\(\dfrac{b}{c} = 2\)
Work out the value of \(a\) when \(\quad c = 8\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(b = 2c\) or \(b = 16\) or \(\dfrac{a}{b} \times \dfrac{b}{c} = 3c \times 2\) | M1 | oe eg \(\dfrac{a}{2c} = 3c\) or \(\dfrac{a}{16} = 3c\) |
| \((a =)\ 6c^2\) or \(3 \times 8 \times 2 \times 8\) or \(24 \times 16\) or \(6 \times 8^2\) or \(6 \times 64\) | M1dep | oe |
| 384 | A1 |
| \(\dfrac{b}{8} = 2\) | M0 |
| \(\dfrac{a}{b} = 24\) | M0 |
15
| \(x\) | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
| \(y\) | 2 | −1 | −2 | −1 |
| Answer | Mark | Comments | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| B1 |
| Answer | Mark | Comments |
|---|---|---|
| Plots at least four points correctly | M1 | ft their points from part (a) \(\pm\,\dfrac{1}{2}\) small square |
| Correct graph drawn through the seven correct points | A1 | \(\pm\,\dfrac{1}{2}\) small square smooth quadratic curve |
| Correct graph drawn without plotting the correct points | M1A1 |
5 Work out the value of \(\quad a^2 - 4a \quad\) when \(\quad a = 10\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 60 | B2 | B1 \(10^2 - 4 \times 10\) or \(10^2 - 4(10)\) or 100 or 40 or \(-40\) seen or implied |
| Embedded correct value eg \(100 - 4a = 96a\) | B1 |
| Further correct work eg \(100 - 40 = 60\), Answer \(6a\) | B2 |
| Further incorrect work eg \(100 - 40 = 60\), Answer \(60a\) | B1 |
| Values may be implied eg1 \(10^2 - 4 = 96\) implies 100 eg2 96 only does not imply 100 | B1 B0 |
| Incorrect calculations cannot imply a value eg \(10^2 - 4a = 96a\) | B0 |
| \(100a\) does not imply 100 | B0 |
22 This formula converts temperature in degrees Fahrenheit (\(F\)) to kelvin (\(K\))
| \(K = \dfrac{5}{9}(F - 32) + 273\) |
A pottery oven is heated to 2192 degrees Fahrenheit.
Work out this temperature in kelvin. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 2160 | B1 | may be implied by 240 or 10 800 |
| \(\dfrac{5 \times \text{their } 2160}{9}\) or \(5 \times 240\) or \(10\,800 \div 9\) or 1200 | M1 | oe |
| 1473 | A1 |
| Accept 0.55 or 0.56 or better for \(\dfrac{5}{9}\) | |
| eg \(\dfrac{5}{9}(2160) + 273\) (no indication that they know to multiply by \(\dfrac{5}{9}\)) eg \(\dfrac{5}{9} \times (2160) + 273\) eg 2130, \(5 \times 2130 \div 9\) | B1M0A0 B1M1A0 B0M1A0 |
20
\(a \leqslant 12 \qquad b \lt 9\)
Work out the largest possible value of \(\quad 2a + b\) [2 marks]
Show that \(\dfrac{y}{x}\) could equal 4 [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \((24 + 8 =)\ 32\) | B2 | B1 \((2a =)\ 2 \times 12\) or \((2a =)\ 24\) or \((b =)\ 8\) |
| 32 with no incorrect working | B2 |
| 32 from incorrect working eg \(22 + 10 = 32\) | B0 |
| \(24 + 9 = 33\) | B1 |
| \(22 + 8 = 30\) | B1 |
| \(24a\) without a B1 response | B0 |
| \(8b\) without a B1 response | B0 |
| \(24a + 8b\) without a B1 response | B0 |
| Use of inequalities in answer without a B1 response | B0 |
| Answer | Mark | Comments |
|---|---|---|
| An example where \(x\) and \(y\) are both negative and \(\dfrac{y}{x} = 4\) | B1 | eg \(x = -1\) and \(y = -4\) values of \(x\) and \(y\) can be implied eg \(\dfrac{-12}{-3}\ (= 4)\) |
| Correct use of \(\div\) instead of fractions is allowed eg \(-12 \div -3\) | B1 |
| Must show the fraction or division or state which is \(x\) and which is \(y\) eg \(-1\) and \(-4\) | B0 |
| Decimals and / or fractions may be used eg \(\dfrac{-6.4}{-1.6}\) or \(\dfrac{-2}{-\frac{1}{2}}\) | B1 |
| One correct example among several attempts | B1 |
3 Circle the expression that has the smallest value when \(\quad x = 4\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(x - 4\) | B1 |
26 Given that \(\quad a \times 60 = b \quad\) work out the value of \(\quad \dfrac{4b}{a}\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(60 \times 4\) or \(4(a \times 60)\) or \(4a \times 60\) or \(\dfrac{b}{a} = 60\) or \(\dfrac{4b}{b/60}\) or \(4b = 240a\) or \(\dfrac{240a}{a}\) | M1 | accept any multiplication signs |
| 240 | A1 | Condone \(\dfrac{240}{1}\) |
| Correct answer found by substituting appropriate values for \(a\) and \(b\) | M1A1 |
| Incorrect answer found by substituting appropriate values for \(a\) and \(b\) | M0A0 |
| Award M1 for \(60 \times 4\) or 240 in working, either as individual expressions or as part of longer expressions eg \(4 \times 60 = 240\), answer \(240b\) eg \(\dfrac{4 \times 60 \times a}{4b}\) | M1A0 M1A0 |
| Do not award M1 for 240 within a list of multiples of 60 that continues beyond 240 |
8 The cost of a taxi journey is
£3 plus £2 per mile.
Circle the cost of a journey of 6 miles. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| £15 | B1 |
7 Given that \(\quad a \times 60 = b \quad\) work out the value of \(\quad \dfrac{4b}{a}\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(60 \times 4\) or \(4(a \times 60)\) or \(4a \times 60\) or \(\dfrac{b}{a} = 60\) or \(\dfrac{4b}{b/60}\) or \(4b = 240a\) or \(\dfrac{240a}{a}\) | M1 | accept any multiplication signs |
| 240 | A1 | Condone \(\dfrac{240}{1}\) |
| Correct answer found by substituting appropriate values for \(a\) and \(b\) | M1A1 |
| Incorrect answer found by substituting appropriate values for \(a\) and \(b\) | M0A0 |
| Award M1 for \(60 \times 4\) or 240 in working, either as individual expressions or as part of longer expressions eg \(4 \times 60 = 240\), answer \(240b\) eg \(\dfrac{4 \times 60 \times a}{4b}\) | M1A0 M1A0 |
| Do not award M1 for 240 within a list of multiples of 60 that continues beyond 240 |
7 Choose one of the following to make a correct statement each time. [4 marks]
is less than is equal to is greater than
| When \(a = 3\) | \(4a\) | …………………………… | \(a + 7\) |
| When \(b = 8\) | \(2b - 6\) | …………………………… | \(18 - b\) |
| When \(c = 0.5\) | \(3c\) | …………………………… | \(c + 1\) |
| When \(d = -1\) | \(d\) | …………………………… | \(d^2\) |
| Answer | Mark | Comments |
|---|---|---|
| is greater than | B1 | allow > |
| is equal to | B1 | allow = |
| is equal to | B1 | allow = |
| is less than | B1 | allow < |
Do not allow \(\geqslant\) or \(\leqslant\) or \(\equiv\)
Do not allow contradictions eg < is greater than
22 Here is a formula.
\[T = n^2 - \frac{12}{n}\]| Answer | Mark | Comments |
|---|---|---|
| 22.6 or \(\dfrac{113}{5}\) or \(22\dfrac{3}{5}\) | B1 |
| Condone \(22\dfrac{6}{10}\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(n^2\) will be positive and \(\dfrac{12}{n}\) will be negative and positive – negative = positive | B2 | oe B1 \(n^2\) will be positive or \(\dfrac{12}{n}\) will be negative |
| Alternative method 2 | ||
| \(n^2\) will be positive and \(-\dfrac{12}{n}\) will be positive and positive + positive = positive | B2 | oe B1 \(n^2\) will be positive or \(-\dfrac{12}{n}\) will be positive |
For ‘\(n^2\) will be positive’ accept the square of a negative number is a positive
For ‘\(n^2\) will be positive’ condone square or squared numbers are positive
For ‘positive – negative = positive’ condone +(ve) \(-\) \(-\)(ve) = +(ve)
19 You are given that \(\quad 4a - 2b = 10\)
Write an expression in \(a\), \(b\) and \(c\) that is equal to 23
Give your answer in its simplest form.
You must show your working. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 5 | B1 |
| Condone \(10 - 5 = 5\) | B1 |
| Condone \(x = 5\) | B1 |
| \(\dfrac{10}{2}\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(-10\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| Unsimplified expression in \(a\), \(b\) and \(c\) which would evaluate to 23 | M1 | eg \(2(4a - 2b) + a + c\) or \(8a - 4b + a + c\) or \(11(a + c) - (4a - 2b)\) or \(11a + 11c - 4a + 2b\) |
| Simplified expression in \(a\), \(b\) and \(c\) which would evaluate to 23 | A1 | eg \(9a - 4b + c\) \(7a + 2b + 11c\) SC2 Values assigned to \(a\), \(b\) and \(c\) which satisfy original equations and expression given which has value 23 eg \(a = 3, b = 1, c = 0\) and \(7a + 2b + c\) |
| There are infinitely many correct solutions. Allow solutions where the coefficients are not integers if initial working is shown. eg \(3(4a - 2b) - \dfrac{7}{3}(a + c) = \dfrac{29}{3}a - 6b - \dfrac{7}{3}c\) | M1A1 |
| \(5a - 2b + c + 10 = 23\) | M1A1 |
| Condone ‘= 23’ after the expression | |
| Answer using only two variables eg \(2.3(4a - 2b)\) | M0A0 |
1 Circle the point that lies on the curve \(\quad y = x^2 - 4x + 1\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \((-1, 6)\) | B1 |
19 \(a = 7\) and \(b = 2\)
Work out the value of \(\quad \dfrac{a}{b} - a^b\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 3.5 or \(3\dfrac{1}{2}\) or 49 or \((49 =)\ \dfrac{98}{2}\) | M1 | |
| \(3.5 - 49\) or \(49 - 3.5\) or \(3\dfrac{1}{2} - 49\) or \(49 - 3\dfrac{1}{2}\) or \(\dfrac{7}{2} - \dfrac{98}{2}\) or \(\dfrac{98}{2} - \dfrac{7}{2}\) | M1dep | 45.5 (oe) implies M2 |
| \(-45.5\) or \(-45\dfrac{1}{2}\) or \(-\dfrac{91}{2}\) | A1 |
| \(\dfrac{7}{2}\) without \(\dfrac{98}{2}\) | M0 |
| \(7^2\) without 49 | M0 |
| \(\dfrac{7}{2} - 7^2\) (no further correct work) | M0 |
| \(7^2 = 14\), \(3.5 - 14 = -10.5\) | M1M0A0 |
| \(\dfrac{7}{2} - 49\) | M1 |
| \(3.5 - 7^2\) | M1 |
15 A company uses this formula to work out the cost, £\(A\), of a taxi ride.
\[A = 4 + 1.8m + b\]£4 is a fixed charge
\(m\) is the number of miles travelled
£\(b\) is a charge for booking online
She pays £20 altogether.
How much is the charge for booking online? [3 marks]
has a fixed charge of £3
charges £1.90 per mile
has no charge for booking online.
Write a formula for the cost, £\(C\), of a taxi ride with this company. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(1.8(0) \times 8\) or 14.4(0) | M1 | implied by 5.6(0) or 18.4(0) |
| \(20 -\) their \(14.4(0) - 4\) or \(20 - 18.4(0)\) or 1.6 | M1dep | |
| 1.60 | A1 | condone £1.60p |
| Alternative method 2 | ||
| \(b = A - 4 - 1.8m\) | M1 | oe correct formula with \(b\) as the subject |
| \(20 - 4 - 1.8(0) \times 8\) or 1.6 | M1dep | |
| 1.60 | A1 | condone £1.60p |
| \(1.8(0) \times 8\) may be within an incorrect calculation eg \(4 + 1.8(0) \times 8 + 20\) | M1 |
| Answer | Mark | Comments |
|---|---|---|
| \(C = 3 + 1.9(0)m\) | B1 | oe formula with \(C\) as subject accept \(C = 3 + 1.9(0) \times m\) condone \(+ 0\) or \(+ 0b\) |
| \(3 + 1.9m\) | B0 |
| Do not accept eg \(A = \ldots\) for \(C = \ldots\) | B0 |
| Allow \(m\) to be \(\times\) mile(s) but not a different letter unless defined eg1 \(C = 3 + 1.9(0) \times\) miles eg2 \(C = 3 + 1.9(0)\) miles eg3 \(C = 3 + 1.9(0)\) per mile or \(C = 3 + 1.9(0)\)pm eg4 \(C = 3 + 1.9(0)x\) | B1 B0 B0 B0 |
| Ignore £ inserted in part or all of equation eg \(C = 3 + £1.90m\) | B1 |
| Correct formula followed by substitution (and evaluation) | B1 |
20 Here is an inflated swimming ring with dimensions in centimetres.
The volume of the ring, \(V\) cm\(^3\), is given by
| \(V = 0.25\pi^2(b - a)^2(b + a)\) |
Work out the volume when \(\quad a = 20 \quad\) and \(\quad b = 30\)
Give your answer to 3 significant figures. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(0.25\pi^2(30 - 20)^2(30 + 20)\) or \(0.25\pi^2 \times 10^2 \times 50\) | M1 | oe allow use of \(\pi\) as [3.14, 3.142] |
| [12 320, 12 340.21] | A1 | may be implied |
| 12 300 or \(1.23 \times 10^4\) with no value outside [12 320, 12 340.21] seen | A1 |
| \(0.25\pi^2(30 - 20)^2(30 + 20)\) 12 300 | M1 A1(implied)A1 |
| 12 300 with no incorrect working | M1A1A1 |
| 12 300.0 is not to 3 significant figures | |
| M1 gained followed by answer 12 300.0 | M1A0A0 |
| Do not allow misreads eg \(0.25\pi^2(30 + 20)^2(30 + 20)\) | M0A0A0 |
| Brackets expanded correctly and values substituted | M1 |
13 Here is a formula for the amount of water needed to cook rice.
\[w = 1.5r + 0.5\]\(w\) is the number of cups of water needed
\(r\) is the number of cups of rice to be cooked
| Answer | Mark | Comments |
|---|---|---|
| \(1.5 \times 7 + 0.5\) or \(10.5 + 0.5\) | M1 | oe |
| 11 | A1 |
| \(1.5 \times 7 = 10.5\) and \(0.5 \times 7 = 3.5\) and \(10.5 + 3.5 = 14\) | M0A0 |
| \(7 \times 1.5r + 0.5\) | M0A0 |
| \(7 \times 1.5r + 0.5\) and answer \(11r\) | M0A0 |
| \(7 \times 1.5r + 0.5\) and answer 11 (has recovered) | M1A1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(20 - 0.5\) or 19.5 or \(r = \dfrac{w - 0.5}{1.5}\) | M1 | oe |
| their \(19.5 \div 1.5\) | M1dep | oe \((20 - 0.5) \div 1.5\) is M2 |
| 13 | A1 | |
| Alternative method 2 | ||
| 20 – their 11 from part (a) or 9 | M1 | implied by ‘6 extra cups (of rice)’ |
| \(7 +\) (their \(9 \div 1.5\)) or \(7 + 6\) | M1dep | |
| 13 | A1 | |
| 13 from incorrect working eg rounding \(20 \div 1.5 = 13\) eg scaling 11 and rounding ie \(20 \div 11 \times 7 = 13\) | M0M0A0 |
| Brackets omitted ie \(20 - 0.5 \div 1.5\), unless recovered | M0M0A0 |
| \(1.5 \times 13 + 0.5 = 20\), unless 13 selected | M1M1A0 |
| \(1.5 \times 13 = 19.5\), unless 13 selected | M1M1A0 |
| Trial and improvement, unless answer 13 | M0M0A0 |
17 Here is a formula to convert degrees Celsius (°C) to degrees Fahrenheit (°F).
\[F = 1.8C + 32\]\(F\) is the number of degrees Fahrenheit
\(C\) is the number of degrees Celsius
Nick says,
“Because the temperature is negative in Celsius, it must be negative in Fahrenheit.”
Is he correct?
You must show your working. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(1.8 \times -40 + 32\) or –72 | M1 | oe eg \(1.8(-40) + 32\) |
| \(1.8 \times -40 + 32 = -40\) or \(1.8 \times -40 = -72\) and \(-72 + 32 = -40\) | A1 | oe eg \(1.8(-40) + 32 = -40\) Full working must be seen oe eg \(1.8 \times -40 = -72\) and \(-40 - 32 = -72\) |
| Alternative method 2 | ||
| \(\dfrac{-40 - 32}{1.8}\) or –72 | M1 | |
| \(\dfrac{-40 - 32}{1.8} = -40\) or \(-40 - 32 = -72\) and \(-72 \div 1.8 = -40\) | A1 | Full working must be seen oe eg \(-40 - 32 = -72\) and \(-40 \times 1.8 = -72\) |
| Alternative method 3 | ||
| \(F = 1.8F + 32\) and \(F - 1.8F = 32\) or \(0.8F = -32\) | M1 | Forms equation in one variable and collects terms correctly using any letter oe eg \(1.8F - F = -32\) or \(-0.8F = 32\) |
| (\(F =\)) \(-32 \div 0.8\) and \(F = -40\) | A1 | Full working must be seen oe eg (\(F =\)) \(32 \div -0.8\) and \(F = -40\) |
| Ignore units | |
| 72 does not imply M1 | |
| Only \(-72 + 32 = -40\) | M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| No and 5 or No and correctly evaluated counter example | B1 |
| No, anything between –17°C and 0°C is positive in Fahrenheit | B1 |
| No, anything between 0°F and 32°F is negative in Celsius | B1 |
| Unless the range from –17°C to 0°C is given, then the counter example must be evaluated correctly | |
| No because \(1.8 \times -15\) is –27, and 32 – 27 = 4 | B0 |
| Any temperature in Celsius between \(-17\dfrac{7}{9}\)°C and 0°C can be used as a counter-example | |
| eg1 \(1.8 \times -10 + 32 = 14\) so No | B1 |
| eg2 \(1.8 \times -1 + 32 = 30.2\) so No | B1 |
| No because 14°F is –10°C | B1 |
| Accept No because –10 = 14 | B1 |
| No because –15 is positive in Fahrenheit | B0 |
16 \(\mathrm{f}(x) = x^2 - x^3\)
Circle the value of \(\;\mathrm{f}(-3)\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| 36 | B1 |
12 \(a - b = 5\)
| Answer | Mark | Comments |
|---|---|---|
| 10 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 35 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| –5 | B1 |
3 Circle the point that does not lie on the curve \(\quad y = x^3\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(\left(\dfrac{1}{3}, \dfrac{1}{9}\right)\) | B1 |
19 The value of \(x\) can be 2 or 5
The value of \(y\) can be 3 or 12
You must show your working. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 6, 15, 24, 60 in any order | B2 | B1 for 6, 15, 24, 60 with no more than one additional value or three correct values with no more than one incorrect value |
| Ignore repeated values for B2 and B1 | |
| 6, 10, 15, 24, 60 | B1 |
| 6, 10, 15, 24 | B1 |
| 6, 10, 15, 24, 36 | B0 |
| \(2 \times 3,\ 5 \times 3,\ 2 \times 12,\ 5 \times 12\) | B0 |
| \(6xy,\ 15xy,\ 24xy,\ 60xy\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{2 - 12}{2}\) or one correctly evaluated trial with correct substitutions for \(x = 2\) or 5 and \(y = 3\) or 12 or two correct values from \(-\dfrac{10}{2},\ -\dfrac{1}{2},\ -\dfrac{7}{5},\ \dfrac{2}{5}\) oe or two correct values from \(-5,\ -0.5,\ -1.4,\ 0.4\) oe | M1 | \(\dfrac{2 - 3}{2} = -\dfrac{1}{2}\) oe or \(\dfrac{5 - 12}{5} = -\dfrac{7}{5}\) oe or \(\dfrac{5 - 3}{5} = \dfrac{2}{5}\) oe |
| \(-\dfrac{10}{2}\) or \(-5\) | A1 |
| Two separate correct values can be in either fraction or decimal form | |
| \(2 - 12 \div 2 = -5\) (recovered) | M1A1 |
| \(2 - 12 \div 2\) | M0A0 |
| An example of an incorrect substitution with different values of \(x\) eg \(\dfrac{5 - 12}{2} = -\dfrac{7}{2}\) |
12 A football team has \(P\) points.
\[P = 3W + D\]\(W\) is the number of wins
\(D\) is the number of draws
How many points does the team have? [1 mark]
11 games were draws.
How many games has this team lost? [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| 20 | B1 | allow \(P = 20\) |
| Answer | Mark | Comments |
|---|---|---|
| \(53 - 11\) or 42 or \(33 \times 3\) or 99 or \(11 \times 2\) or \(33 - 11\) or 22 | M1 | |
| their 42 \(\div\) 3 or 14 or their 99 \(-\) 53 \(-\) their 22 or (their 22 \(\times\) 3) \(-\) their 42 or 24 | M1dep | oe eg build up - allow one error |
| \(33 - 11 -\) their 14 or their 24 \(\div\) 3 | M1dep | dep on M1M1 |
| 8 | A1 |
| \(3 \times 14 + 11 = 53\) | M2 |
7 Here is a number machine.
| Answer | Mark | Comments |
|---|---|---|
| 10 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| –14 | B1 |
23 A pot of tea is made using boiling water and left to cool.
The temperature, \(T\) °C, of the pot of tea after \(n\) minutes is given by the formula
\(T = ab^n\).
The graph shows some information about the temperature of the pot of tea.
Not to scale
Find the value of \(a\) and the value of \(b\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(a\) = ] 98 [\(b\) =] 0.949[9…] or 0.95[0] | 3 | B1 for [\(a\) = ] 98 AND B2 for [\(b\) =] 0.949[9…] or 0.95[0] or M1 for \(84 = 98b^3\) or better or \(84 = ab^3\) | SC2 if correct answers transposed unless correctly identified in body |
20 A container takes 8 minutes to be filled with water.
The graph shows the rate of flow, in litres per minute, of the water into the container during the 8 minutes.
Not to scale
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 18 × (7.5 – 0.5) or 18 × 7 | 1 | 126/18 = 7 only scores if 7 is related to the context 126/7 = 18 only scores if 18 is related to the context | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 16.87[5] or 16.88 or 16.9 oe | 4 | M3 for \(\frac{126 + 2 \times \frac{1}{2} \times 0.5 \times 18}{8}\) oe or M2 for \(126 + 2 \times \frac{1}{2} \times 0.5 \times 18\) oe may be implied by 135 or M1 for \([2\ \times]\ \frac{1}{2} \times 0.5 \times 18\) oe may be implied by 4.5 for one triangle or 9 for two triangles If 0 scored, SC1 for \(\frac{126 + 2 \times 9}{8}\) or \(\frac{126 + 18}{8}\) | e.g. \(\frac{\frac{1}{2} \times (7 + 8) \times 18}{8}\) e.g. \(\frac{1}{2} \times (7 + 8) \times 18\) or \(18 \times 8 - 2 \times \frac{1}{2} \times 0.5 \times 18\) |
12
\(v = u + at\)
to find the final velocity, when:
[2]
\(p = \dfrac{1}{2}qr\)
to find the value of \(q\) when \(p = 39\) and \(r = 6\). [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 99 | 2 | B1 for \(8 + 6.5 \times 14\) | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 13 | 2 | M1 for \(2 \times 39 = 6q\) or better or for \(39 \div 6 = \dfrac{1}{2}q\) or better or for \(39 \div 3\) or \(\dfrac{1}{2} \times 6 \times 13 = 39\) | Do not accept embedded answer for 2 marks |
12 You may use these kinematics formulae to answer this question.
\(v = u + at\)
\(s = ut + \frac{1}{2}at^2\)
where \(a\) is the constant acceleration, \(u\) is the initial velocity, \(v\) is the final velocity, \(s\) is the displacement and \(t\) is the time taken.
A particle has an initial velocity of 2 m/s.
After 6 seconds the particle has a velocity of 11 m/s.
The particle had constant acceleration.
Work out the distance the particle has travelled in the 6 seconds. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 39 | 4 | B2 for [\(a\) =] 1.5 oe or M1 for \(11 = 2 + 6a\) or better | |
| M1 for \(2 \times 6 + \frac{1}{2} \times\) their a \(\times 6^2\) or better | their a must come from use of \(v = u + at\) | ||
2 You may use these kinematics formulae to answer these questions.
\(v = u + at\)
\(v^2 = u^2 + 2as\)
A moving particle accelerates at 2 m/s2 for 8 seconds.
The particle’s final velocity after the 8 seconds is 21 m/s.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(v = u + at\) written or better e.g. \(21 = u + 2 \times 8\) | B1 | written means selected and better includes \(u = v - at\) | |
| [\(u\) =] 21 – 2 × 8 [= 5] or 21 – 16 [ = 5 ] | B1 | If 0 scored SC1 for 5 + 2 × 8 or 5 + 16 [ = 21 ] | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 104 | 3 | M2 for \(21^2 = 5^2 + 2 \times 2 \times s\) or better oe or M1 for \(v^2 = u^2 + 2as\) seen | Equivalents include e.g. 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 |
14 Use the formula
\(v^2 = u^2 + 2as\)
to find the final velocity when
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 nfww | 3 | M2 for \(6^2 + 2 \times 4 \times 8\) oe or M1 for [\(u^2 =\)] \(6^2\) or [\(2as =\)] \(2 \times 4 \times 8\) | Accept answer \(-10\) or \(\pm\) 10 M2 implied by [\(v^2 =\)] 100 or \(\sqrt{100}\) M1 implied by [\(u^2 =\)] 36 or [\(2as =\)] 64 |
8 Here is a formula.
\(a = b(c + d)\)
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 36 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1.5 | 3 | M1 for \(30 = 4(c + 6)\) M1FT for \(7.5 = c + 6\) or for \(30 = 4c + 24\) OR Without formal algebra: M1 for \(30 \div 4\) implied by 7.5 M1 for (their 7.5) – 6 | First correct step only apply FT to an equivalent M1 expression Accept any letter for \(c\) \(c\) must not be blank |
14
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(v - u = at\) seen as single statement | 1 | Accept e.g. \(-v + u = -at\) but not e.g. \(\dfrac{v - u}{t} = at \div t\) | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1.5 oe | 2 | M1 for correct substitution \(\dfrac{9 - 3}{4}\) oe | e.g. \(1\frac{1}{2}\) or \(\frac{6}{4}\) or \(\frac{3}{2}\) For M1 accept a two-step evaluation with an arithmetic slip eg \(9 - 3 = 5\) followed by \(5 \div 4\) Do not accept \(\dfrac{3 - 9}{4}\) for M1 |
10
Finley’s working is shown below.
\[\begin{aligned} 5x + 4 &= 19 \\ 5x &= 19 + 4 \\ 5x &= 23 \\ x &= 4.6 \end{aligned}\]Write down the error that Finley has made. [1]
to find the initial velocity, when
Charlie’s working is shown below.
\[\begin{aligned} v &= 29 + (5 \times 3) \\ v &= 29 + 15 \\ v &= 44 \end{aligned}\]Write down the error that Charlie has made. [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| He has added 4 | 1 | Do not allow contradictions See appendices | |
Exemplar responses for 10a
| Response | Mark | |
|---|---|---|
| Added 4 | 1 | |
| The error Finley has made is he has added 4 whereas he needed to do the opposite | 1 | |
| He moved the equation to the other side \(5x = 19 + 4\) he should have taken 4 off 19 | 1 | |
| \(5x = 19 + 4\) | 1 | |
| He has added 19+4 instead of taking 4 off 19 | 1 | |
| \(5x\) isn’t equal to 19+4, \(5x\) plus 4 is equal to 19 | first part is correct, second part is irrelevant not a contradiction | 1 |
| Finley hasn’t worked out the right way as he has added 4 onto the 19 | 1 | |
| It should be \(5x = 19 - 4\) | 1 | |
| On the 2nd part where it says \(5x=19+4\) is wrong he shouldn’t have added 4 to 19 | 1 | |
| He should have subtracted 4 from 19 | 1 bod | |
| Finley did not subtract 4 | 0 | |
| \(5x = 19 - 4 \quad 5x = 15 \quad x = 3\) with no further explanation | 0 | |
| He should have subtracted 4 | 0 | |
| You need to do the same to both sides | 0 | |
| Finley got the answer wrong because it is \(x = 3\) | need to see a reason | 0 |
| He’s added 4 on to 19 when it’s supposed to equal 19 | 0 | |
| He wasn’t supposed to add them | do they mean 5x+4 or 19+4 | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| He has used 29 as the initial velocity | 1 | Accept \(u\) and \(v\) if clear \(v = 29\) not \(u = 29\) | See appendices |
Exemplar responses for 10b
| Response | Mark |
|---|---|
| \(v = 29\) not \(u\) | 1 |
| He put 29 in the place of the u and not the v. Its meant to be 29 = u + (5x3) not v=29+(5x3) | 1 |
| Initial velocity not 29 | 1 |
| v not initial velocity | 1 |
| He didn’t substitute properly he put 29m/s for initial velocity | 1 |
| He has written the final velocity down in the equation in the wrong place which therefore makes his calculation wrong | 1 bod |
| He has used the 29m/s when that is the velocity, he was using the wrong numbers | 0 |
| He substituted the formula wrong not enough | 0 |
| It should equal 29 not v ‘it’ isn’t enough | 0 |
7 The diagram shows the graph of \(y = kx - x^2 + 2\), where \(k\) is an integer.
Give your answers to 1 decimal place. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Correct substitution of \((x, y)\) from integer point on curve into equation leading to \(k = 3\) e.g. (2, 4) \(4 = 2k - 2^2 + 2\) or \(4 = 2k - 4 + 2\) leading to \(k = 3\) with at least one correct intermediate step | 2 | M1 for correct substitution of \((x, y)\) from integer point on curve into \(y = kx - x^2 + 2\) or \(y = 3x - x^2 + 2\) OR M1 for e.g. \(x = 2\) correctly substituted in \(y = 3x - x^2 + 2\) and finding \(y = 4\) Max M1 if \(k = 3\) substituted | (-1, -2) : \(-2 = -[1]k - (-1)^2 + 2\) (1, 4) : \(4 = [1]k - 1^2 + 2\) (2, 4) : \(4 = 2k - 2^2 + 2\) (3, 2): \(2 = 3k - 3^2 + 2\) (4, -2): \(-2 = 4k - 4^2 + 2\) Use of (0, 2) scores 0 but may be replaced with another point (ie do not treat as a choice) Examples of intermediate steps \(4 = 2k - 2^2 + 2\) then \(4 = 2k - 4 + 2\) is a sufficient int step or \(4 = 2k - 2\) is a sufficient int step or \(6 = 2k\) is a sufficient int step \(3 = k\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 0.4 and 2.6 | 2 | B1 for 0.4 or 2.6 or M1 for line \(y = 3\) drawn or for (0.4, 3) and (2.6, 3) indicated | Line to cut curve twice Treat \(x = 3\) drawn or multiple horizontal lines as choice unless \(y = 3\) clearly chosen Condone good freehand line eg circled or lines drawn down to \(x\)-axis |
2
Using part (a), or otherwise, calculate the rocket’s initial velocity in m / s. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(u = [\pm]\sqrt{v^2 - 2as}\) as final answer | 2 | M1 for [\(u^2\) =] \(v^2 - 2as\) |  |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3924 to 3924.3 | 3 | M2 for \(\sqrt{8000^2 - 2 \times 90 \times 270000}\) or M1 for \(\sqrt{8000^2 - 2 \times 90 \times \text{figs } 27}\) or \(8000^2 - 2 \times 90 \times 270000\) or \(8000^2 = u^2 + 2 \times 90 \times 270000\) or for correct substitution seen of \(v = 8000\), \(a = 90\) and \(s\) = figs 27 in their final answer to part (a) If 0 scored SC1 for 7996 to 7997 as final answer | Condone 3920 as final answer for full marks, following M2 For SC1, condone 8000 following 7996 to 7997. |
16
to find the final velocity, when
.................... m/s [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(d = \dfrac{f - 4}{5}\) or \(d = \dfrac{f}{5} - \dfrac{4}{5}\) | 2 | M1 for \(f - 4 = 5d\) or \(\dfrac{f}{5} = d + \dfrac{4}{5}\) or \(\dfrac{f - 4}{5}\) or \(\dfrac{f}{5} - \dfrac{4}{5}\) as answer | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 50 | 2 | M1 for \(5 + 7.5 \times 6\) | |
14 The number of bees, \(P\), in a colony is given by the formula
\[P = ab^x\]where \(x\) is the number of months after the start of July.
At the start of July, there were 25 000 bees in the colony.
After one month, there were 23 500 bees in the colony.
Find the value of \(a\) and the value of \(b\).
Give the value of \(b\) as a decimal. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(a\) = ] 25 000 [\(b\) = ] [0].94 | 4 | B1 for 25 000 AND B3 for 0.94 oe or B2 for \(\frac{23500}{25000}\) or 94% oe or \(\frac{25000 - 23500}{25000}\) or 0.06 oe or M1 for 23500 = \(ab^{[1]}\) or 23500 = \(25000b^{[1]}\) | |
10 You may use these kinematics formulae to answer this question.
\[v = u + at\]\[s = ut + \frac{1}{2}at^2\]A particle has an initial velocity of 3 m/s.
After 20 seconds the particle has a velocity of 11 m/s.
Work out the distance the particle has travelled after 20 seconds, in m. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 140 | 4 | M1 for 11 = 3 + 20\(a\) oe implied by [0].4 A1 for [\(a\) = ] [0].4 M1 for 3 × 20 + \(\frac{1}{2}\) × their 0.4 × \(20^2\) Alternative method M3 for 3 × 20 + \(\frac{1}{2}\) × (11 – 3) × 20 oe OR M1 for 3 × 20 M1 for \(\frac{1}{2}\) × (11 – 3) × 20 M1 for their (3 × 20) + their (\(\frac{1}{2}\) × (11 – 3) × 20) | Condone one error in second M1. If their \(a\) is clearly stated do not count as an error. Equivalent includes \(\frac{1}{2}\)(11 + 3) × 20 may be implied by 60 may be implied by 80 |
7 Find the value of \(4x + 5y\) when \(x = 3\) and \(y = -2\). [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2 | 2 | M1 for 12 or -10 or \(4 \times 3 + 5 \times -2\) oe | Not \(12x\) or \(-10y\) |
14 Here is a rule to work out the time, in minutes, needed to cook a turkey.
Use the rule to work out the weight of Ling’s turkey. [2]
Use the rule to work out at what time James should put his turkey in the oven.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3 | 2 | M1 for \((150 - 90) \div 20\) | Condone \(150 - 90 \div 20\) for M1 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [0]9:45 [am] with correct working | 5 | M1 for \(6 \times 20 + 90\) oe A1 for 210 M1 for their 210 correctly converted into hrs and mins or decimal hours soi by 3 h 30 m or 3.5 [h] | “correct working” requires evidence of at least M1 M1 Provided their 210 is not a multiple of 60 Do not accept 3.3 hours as a correct conversion from 3 h 30 m, but condone 3.3 used in the next step |
| M1 for 1:15 – their time If 0, 1 or 2 scored, instead award SC3 for answer 9:45[am] with no working or insufficient working If 0 or 1 scored, instead award SC2 for 3 h 30 m or 3.5 [h] with no working or insufficient working If 0 scored, SC1 for 210 with no working or insufficient working | Their time may be in whole hours, minutes, hrs and mins or decimal hours | ||
7
Harper rounds each value to 1 significant figure.
Harper uses the rounded values and the formula
to estimate the distance travelled in the 10.1 seconds.
Harper’s answer is 430 metres.
Using Harper’s method, show that their answer is wrong. [4]
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(s = 230\) with 4, 3 and 10 or 100 seen | 4 | B2 for 4, 3 and 10 or 100 or B1 for two correct | For all marks condone e.g. 3.00, 4.0, 10.0 used These values may be written in the stem of the question |
| AND M1 for \((3 \times 10) + \tfrac{1}{2}(4 \times 10^2)\) or correct substitution of unrounded or incorrectly rounded values If 0 scored then SC1 for sight of 230 | For M1 e.g. allow a mixture \((2.93 \times 10.1) + \tfrac{1}{2}(4.1 \times 10^2)\) | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(t = [\pm]\sqrt{\dfrac{2s}{a}}\) oe final answer | 3 | Square root must dip below fraction line in final answer unless \(\left(\frac{2s}{a}\right)\) bracketed For 3 marks oe e.g. \(t = [\pm]\sqrt{\dfrac{2 \times s}{a}}\), \(t = \sqrt{\dfrac{s}{\frac{1}{2}a}}\) | |
| M2 for first two steps correctly completed e.g. \(\frac{2s}{a} = t^2\) or answer \([\pm]\sqrt{\dfrac{2s}{a}}\) (no \(t =\) ) | M2 for e.g. \(\dfrac{s}{\frac{a}{2}} = t^2\), \(\dfrac{s}{\frac{1}{2}a} = t^2\), \(\dfrac{s}{0.5a} = t^2\) | ||
| or M1 for first step correctly completed e.g. \(2s = at^2\) or \(\frac{s}{a} = \frac{1}{2}t^2\) | Allow M1 for \(\frac{s}{0.5} = at^2\) oe | ||
| If 0 scored, SC1 for final answer \(t = [\pm]\sqrt{\dfrac{\frac{1}{2}s}{a}}\) oe | oe for SC1 e.g. \(t = [\pm]\sqrt{\dfrac{s}{2a}}\) | ||
20
\[x^2 - 2y = 5 \quad \text{and} \quad 4y + z = 7.\]Write \(z\) in terms of \(x\).
Give your answer in its simplest form. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(z = 17 - 2x^2\) final answer | 4 | B3 for answer \(17 - 2x^2\) OR M3 for \(7 = 4\left(\dfrac{x^2 - 5}{2}\right) + z\) oe or \(2x^2 + z = 17\) oe or \(x^2 - 2\left(\dfrac{7 - z}{4}\right) = 5\) or better | Correct unsimplified formula in \(x\) and \(z\) |
| or M2 for \(y = \dfrac{2x^2 - 10}{4}\) oe or \(y = \dfrac{7 - z}{4}\) oe or \(4y = 2x^2 - 10\) or \((2x^2 - 4y) + (4y + z) = 10 + 7\) oe | M2 sets up for substitution with \(y\) explicit or for method for elimination by equating coefficients of \(y\) and correct method to eliminate \(y\) | ||
| or M1 for \(2x^2 - 4y = 10\) or \(-2y = 5 - x^2\) or \(4y = 7 - z\) or better | M1 equates coefficients of \(y\) or first step in rearrangement to eliminate | ||
16 The following kinematics formulas may be used in this question.
\[v = u + at \qquad s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as\]The initial velocity of a particle is 20 m/s.
The acceleration of the particle is −8 m/s².
After \(t\) seconds, the particle has travelled 25 m.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Subst into correct formula (may be implied) and partial simplification | 2 | B2 for \(25 = 20t - \frac{1}{2} \times 8 \times t^2\) oe or \(25 = 20t + (-4)t^2\) or B1 for subst eg \(25 = 20t + \frac{1}{2}(-8)t^2\) | Only accept \(25 = 20t - 4t^2\) if subst seen For B1 condone ambiguity caused by missing brackets |
| \(25 = 20t - 4t^2\) seen and correct completion to \(4t^2 - 20t + 25 = 0\) | 1dep | Dep on previous 2 marks | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.5 oe | 3 | M2 for \((2t - 5)(2t - 5)\) or M1 for any two factors that give two correct terms when expanded or for partial factorisation \(2t(2t - 5) - 5(2t - 5)\) OR M2 for [\(t =\)] \(\dfrac{20 \pm \sqrt{400 - 400}}{8}\) or better or M1 for [\(t =\)] \(\dfrac{-(-20) \pm \sqrt{(-20)^2 - 4 \times 4 \times 25}}{2 \times 4}\) with at most one error | eg a sign error, short fraction line, short root, but condone missing brackets |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Shows \(v = 0\) and concludes “stationary” | 3 | M1 for [\(v^2 =\)] \(20^2 + 2(-8)25\) or [\(v =\)] \(20 + (-8) \times\) their (b) A1 \(v = 0\) If 0 scored, instead award SC2 for \(v = 0\) and other values substituted into a relevant equation as a correct check or SC1 for \(v = 0\) | |
12
Use the formula \(v = u + at\) to calculate the velocity of the train after the 6 seconds. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 18 | 2 | M1 for 15 + 0.5 × 6 or better | Do not accept use of \(0.5^2\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(a = \dfrac{v - u}{t}\) oe final answer | 2 | M1 for correct first step or \(\dfrac{v - u}{t}\) | Accept \(\dfrac{v - u}{t} = a\) |
16 Martina has answered some questions on algebra.
In each question, she has made an error.
Describe her error and give the correct answer to each problem.
Question 1
Simplify. \(2a \times a \times a\)
Martina’s answer \(4a\)
Martina’s error is ..................
Correct answer = .................. [2]
Question 2
Simplify. \(\dfrac{x^{10}}{x^2}\)
Martina’s answer \(x^5\)
Martina’s error is ..................
Correct answer = .................. [2]
Question 3
\(s = ut + \tfrac{1}{2}at^2\)
Find \(s\) when \(u = 0\), \(t = 5\) and \(a = 6\).
Martina’s solution \(s = 0 \times 5 + \tfrac{1}{2} \times 6 \times 5^2\)
\(s = 0 + 15^2\)
\(s = 225\)
Martina’s error is ..................
Correct answer = .................. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| She added the terms oe \(2a^3\) | 1 1 | In all 3 parts any incorrect statement treat as choice Allow correct descriptions of what Martina should have done in each part See AG | |
Exemplar responses for Q16a
| Response | Mark | |
|---|---|---|
| 1 | She did not multiply [the terms] | 1 |
| 2 | She added up all the a | 1 |
| 3 | She added [to] the 2a’s | 1BOD |
| 4 | She hasn’t multiplied the a’s | 1 |
| 5 | it is × not + | 1 |
| 6 | she would be right if they were plus signs and not times | 1 |
| 7 | Not 2a + a + a | 1 |
| 8 | a would equal 2a + a + a. When you times you add them 2a x a x a = 2a3 | 1 |
| 9 | She added a to the multiplication instead of using index laws | 0 |
| 10 | She has added the 2a’s then times by the 2a | 0 |
| 11 | She added the a’s to the 2 instead of multiplying them | 0 |
| 12 | She added the 2a to the a x a | 0 |
| 13 | 2a is different to a times a = a2 | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| She divided the powers oe \(x^8\) | 1 1 | See AG | |
Exemplar responses for Q16b
| Response | Mark | |
|---|---|---|
| 1 | She divided the 10 and the 2/she divided the powers (must refer to ‘numbers’ or ‘indices’) | 1 |
| 2 | she done 10 ÷ 2 = 5 | 1 |
| 3 | She should have done 10 – 2 | 1 |
| 4 | She should take away the indices | 1 |
| 5 | She divided 10 by 2 instead of subtracting 2 | 1 |
| 6 | Laws of indices it should be taken away | 1 |
| 7 | Divided the numbers | 1 |
| 8 | She divided instead of taking away (‘indices’ implied by referring to division and subtraction) | 1 |
| 9 | She has used division | 0 |
| 10 | She divided it when it should be timesd | 0 |
| 11 | She didn’t use the laws of indices | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| She squared \((\tfrac{1}{2} \times 6 \times 5)\) oe 75 | 1 1 | See AG | |
Exemplar responses for Q16c
| Response | Mark | |
|---|---|---|
| 1 | didn’t do the power first | 1 |
| 2 | She did 5 x 6 and then x ½ but the 2 is near the 52 not all of it | 1 |
| 3 | She should have squared 5 first and then divide 6 by 2 and times them together | 1 |
| 4 | She didn’t do the 52 bit first | 1 |
| 5 | She didn’t square 5 and times 6 by 25 | 1 |
| 6 | She hasn’t squared the 5 | 1 |
| 7 | Doing 152 she needs to do 52 | 1 |
| 8 | She did 152 | 0 |
| 9 | She halved 30 when she meant to halve 6 earlier in the equation | 0 |
| 10 | She didn’t square the number 5 correctly nor did she times by 6 or halve it by half | 0 |
| 11 | She square after, when you square before | 0 |
| 12 | She halved 6 before multiplying by 52 | 0 |
| 13 | Didn’t use BODMAS | 0 |
| 14 | 6 x 25 = 150 | 0 |
8 Find the value of \(3g - h\) when \(g = 4\) and \(h = 5\). [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7 | 2 | M1 for \(3 \times 4 - 5\) | May be in steps Allow 12 – 5 for M1 |
12 Use the formula
\[v = u + at\]to find the final velocity, when
[2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 23 | 2 | M1 for 8 + 3 \(\times\) 5 or better | |
11 Gill uses the formula
\[h = 2fg\]| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 6 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1.5 or \(1\dfrac{1}{2}\) or \(\dfrac{3}{2}\) oe | 2 | M1 for \(18 = 2 \times 6[\times]g\) or better | May be (eg) \(\dfrac{18}{6} = 2g\) or \(\dfrac{18}{2 \times 6}\) etc. |
9 The graph of \(y = x^3 - 7x - 12\) is shown below.
The root of the equation \(x^3 - 7x - 12 = 0\) is \(p\).
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| −6 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(x = 4\),] \(y = 24\) Change of sign, so \(p\) lies between 3 and 4 oe | 2 | B1 for 24 seen If using \(3.27 \lt x \lt 4\) rather than 4: SC2 evaluate \(y\) correctly (see table in (c)), state change of sign oe and that because \(3 \lt p \lt\) their \(x\)-value, then so \(3 \lt p \lt 4\). 0 for just evaluating y. | After \(x = 4\), \(y = 24\) scored: Examples just sufficient for second mark include: change of sign \(-6 \lt 0 \lt 24\) \(x = 3\) gives an answer < 0 and \(x = 4\) gives an \(\gt 0\) Examples insufficient for second mark: so \(p\) lies between 3 and 4 |
| Answer | Marks | Part marks and guidance | |||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Examples: when \(x = 3.5\), \(y = 6.4\), so \(3 \lt p \lt 3.5\) when \(x = 3.1\), \(y = -3.9\), so \(3.1 \lt p \lt 4\) when \(x = 3.1\), \(y = -3.9\) and when \(x = 3.5\), \(y = 6.4\), so \(3.1 \lt p \lt 3.5\) | 3 | M2 for one further value of \(y\) evaluated correctly, possibly rot to 2 or more sf, for a value of \(x\) such that \(3 \lt x \lt 4\) OR M1 for working shown to calculate one further value of \(y\) for a value of \(x\) such that \(3 \lt x \lt 4\) Note after SC considered in (b): if SC2 was awarded then they must use a value of \(x\) that produces a smaller interval than \(3 \lt x \lt\) their x-value in (b); if SC2 was not awarded then \(3 \lt x \lt 4\) applies If 0 scored, award SC1 or SC2 if evidence for M1 or M2 has not yet been credited in (b) | Solution is approx. 3.2670 Common values:
eg M2 only for when \(x = 3.1\), \(y = -3.9\) so \(3.1 \lt p \lt 3.5\) (as 3.5 has not been correctly justified) Calculations in support of \(x = 3\) or \(x = 4\) need not be repeated from parts (a) or (b). | ||||||||||||||||||||||||||||
6 A leopard is running with a velocity of 3 m/s.
It then accelerates at 2 m/s\(^2\) for 4 seconds.
Use the formula
\[v = u + at\]to work out the final velocity of the leopard. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 11 | 2 | M1 for 3 + 2 \(\times\) 4 | |
6 You are given that \(5y = 4x\).
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 8 | 2 | M1 for \(5y = 4 \times 10\) oe | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{4}{5}x\) or \([0].8x\) final answer | 1 | Accept alternative fractions and forms such as \(4x \div 5\) | |
5 The diagram shows a straight line that passes through points A and B, and a curve that passes through points P and Q.
Not to scale
Find the value of \(k\). [3]
Triangle ABQ is isosceles.
Is Diann correct?
You must show all your working. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(y = 0.75x + 2\) oe | 3 | B2 for \(y = 0.75x\) [\(+ c\)] or answer \(0.75x + 2\) OR | ISW after a correct equation if attempting rearrangement Accept oe throughout eg B2 for \(4y = 3x\) |
| M1 for attempt at \(\dfrac{\text{change in } y}{\text{change in } x}\) soi by \(\dfrac{\pm(5 - 2)}{\pm(4 - 0)}\) or ±0.75 and B1 for \(y = kx + 2\) with \(k \ne 0\) | Examples: M1B1 for \(y = -0.75x + 2\) M1B0 for 0.75, \(0.75x\), −0.75, \(-0.75x\) If gradient inverted: M0B1 for \(y = 1.3x + 2\) M0B0 for \(1.3x + 2\), \(y = 1.3x\) Condone poorly written \(\frac{3}{4}x\) unless clearly 3 over \(4x\). | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3 nfww | 3 | M2 for \(12 = 16 - 4k + 8\) or better OR M1 for \(12 = -4^2 + -4 \times k + 8\) or sign errors in \(12 = 16 - 4k + 8\) or better or \(k = \dfrac{y - x^2 - 8}{x}\) | Condone −4 not in brackets but \(12 = -4^2 + k - 4 + 8\) with no times sign or dot between \(k\) and −4 scores 0 unless subsequently clarified. |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Using symmetry: | |||
| Q is (0, 8) | 1 | dep mark is always dependent on 3 marks being achieved | For first mark in all methods, condone [Q =] 8 or [QA =] 8-2 or 6, seen in working or on diagram. |
| Midpoint, M, of AQ is at (0, 5) | 1 | Accept implied symmetry | eg 8 – 5 = 3 and 5 – 2 = 3 so B is in the middle of A and Q May see “midpoint” or any other letter for M |
| MB is perpendicular to QA | 1 | ||
| So isosceles/Diann is correct | 1dep | ||
| OR | |||
| Using Pythagoras: | |||
| Q is (0, 8) | 1 | ||
| \(\text{AB}^2 = 4^2 + 3^2\) oe or AB = 5 nfww or \(\text{QB}^2 = 4^2 + (\textit{their } 8 - 5)^2\) or QB = 5 nfww | 1 | ||
| AB = 5 and QB = 5 or \(\text{AB}^2 = 25\) and \(\text{QB}^2 = 25\) | 1 | ||
| AB = QB or “two sides are equal” oe so isosceles/Diann is correct | 1dep | ||
| OR | |||
| Using trig: | |||
| Q is (0, 8) | 1 | ||
| tan BAQ = 4/3 [=53.1] | 1 | ||
| tan BQA = 4/3 [= 53.1] | 1 | ||
| BAQ = BQA or “two angles are equal” oe so isosceles/Diann is correct | 1dep | ||
1 for Q is (0, 8)
1 for gradients/vectors/descriptions of translations for both AB and QB (must be seen together in part (c)): eg
gradients: AB = 3/4 and QB = −3/4 (may be implied from the equations of the two lines)
descriptions: AB is 4 along (treat as in positive sense) and 3 up and QB is 4 along and 3 down oe
To score more than 2 marks, the approach needs to be developed to justify isosceles, such as by switching to the 3rd and 4th marks of the Pythagoras or trig methods.
Condone poor notation, such as missing vector brackets or fraction lines in vectors if intention is clear.
eg gradient AB = ¾ and gradient QB = −3/4 scores a max of 1 1 0 0
eg gradient AB = ¾ and gradient QB = −3/4, so triangle is isosceles also scores a max of 1 1 0 0
Warnings:
dimensions of triangle shown as (8 – 2), 4, 4 and isosceles stated is B1 only;
blank answer space but BQ drawn on diagram is 0 not NR.
4
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(4x - 3y\) final answer | 2 | B1 for \(4x\) or \(-3y\) in final answer | \(4x + -3y\) scores B1 only |
| (ii) \(w^6\) final answer | 1 | ||
| (iii) \(15c^3d\) final answer | 1 | Accept \(15dc^3\) Do not accept eg \(15 \times c^3 \times d\) | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 13 | 1 | ||
| (ii) 3 | 1 | ||
8 This is a rule to find the time, in minutes, needed to roast lamb.
Use the rule to work out the weight of this piece of lamb. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 140 isw | 2 | B1 for 120 seen | Accept 2 h[ours] 20 m[inutes] |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.5 oe | 2 | B1 for 75 seen or M1 for their 75 \(\div\) 30 correctly evaluated | To 2 significant figures or better |
1 Use the formula \(s = ut + \dfrac{1}{2}at^2\).
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 200 | 2 | B1 for 50 or 150 soi | Eg. answer 500 or 275 with (5 × 10) + … seen |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(a = \dfrac{2(s - ut)}{t^2}\) oe | 2 | M1 for \(s - ut = \frac{1}{2}at^2\) | |
5
to find the final velocity, when
[2]
[1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 62 cao | 2 | B1 for 48 or 14 or M1 for \(3 \times 16 + 2 \times 7\) | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 11 cao | 2 | M1 for \(2 + 6 \times 1.5\) If 0 scored SC1 for answer \(-7\) | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(d = \dfrac{c}{7}\) oe | 1 | Accept \(d = c \div 7\) and \(\dfrac{c}{7} = d\) | \(\dfrac{c}{7}\) or \(c \div 7\) with no subject scores 0 |
4 Find the value of \(s\) when \(u = 12\), \(a = 10\) and \(t = 4\).
\[s = ut + \tfrac{1}{2}at^2\][2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 128 | 2 | M1 for \(12(4) + \frac{1}{2}(10)(4)^2\) or B1 for 48 or 80 | |
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