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Solving Simple Equations
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More questions on this topic: IGCSE Algebra: Solving Simple Equations (256) Level 2 FM Algebra: Solving Basic Equations (6)
13
| Answer | Mark | Mark scheme |
|---|---|---|
| \(n^3\) | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 6 | M1 | for isolating \(k\) terms, eg \(5k = 26 + 4\) or \(5k = 30\) or \(k - \dfrac{4}{5} = \dfrac{26}{5}\) oe or \(26 + 4 \div 5\) |
| A1 | cao |
For M mark step must be carried out not just intention shown.
For example, if you see
\(5k - 4 = 26\)
\(+4 \qquad +4\)
award M1 for \(5k = r\) where \(r \gt 26\)
10 Solve \(\dfrac{14 - x}{3} = 3x\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.4 | M1 | for a correct first step eg \(\dfrac{14 - x}{3} \times 3 = 3x \times 3\) or \(\dfrac{14}{3} - \dfrac{1}{3}x + \dfrac{1}{3}x = 3x + \dfrac{1}{3}x\) oe |
| A1 | oe, eg \(\dfrac{14}{10}\) |
For M mark, step must be carried out not just intention shown.
For example if you see \[\begin{array}{ccc} \dfrac{14 - x}{3} &=& 3x \\ \times 3 && \times 3 \end{array}\] Only award M1 when you see \(14 - x = kx\) where \(k \gt 3\)
9 Greg is \(x\) years old.
Greg is 5 years older than Katy.
Carl is twice as old as Greg.
| Answer | Mark | Mark scheme |
|---|---|---|
| \(x - 5\) | B1 | cao |
Allow \(-5 + x\)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(2x\) | B1 |
Allow \(x + x\) or \(x2\) or \(x \times 2\) or \(2 \times x\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 3 | B1 | cao |
6 Four numbers add up to 76
One of the numbers is 13
The other three numbers are each the same number.
Work out the value of each of the other three numbers. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 21 | P1 | for a beginning process, eg \(76 - 13\ (= 63)\) or \(3x + 13 = 76\) oe |
| P1 | for \(\text{``}63\text{''} \div 3\) or \((76 - 13) \div 3\ (= x)\) | |
| A1 | cao |
May be implied by \(a + b + c = 63\) or \(a + b + c + 13 = 76\)
\(a\), \(b\) and \(c\) may be seen in a list
\(13 + 21 + 21 + 21 = 76\), 76, 63 or 13 on answer line scores P1P1A0
21, 21, 21 (13) on the answer line will score P1P1A1
15
| Answer | Mark | Mark scheme |
|---|---|---|
| \(3(2a + 5)\) | B1 |
| Answer | Mark | Mark scheme |
|---|---|---|
| 2 | M1 | for correct expansion of brackets, ie \(12y + 4\) or dividing throughout by 4 as a first step to solve equation, eg \(3y + 1 = 28 \div 4\) |
| M1 | for isolating terms in \(y\), eg \(12y = 28 - 4\) or \(3y = 7 - 1\) | |
| A1 | cao |
For M marks step must be carried out not just intention shown.
For example, if you see
\(\begin{array}{rcl} 4(3y + 1) &=& 28 \\ \div 4 && \div 4 \end{array}\)
Award M1 for:
\(3y + 1 = k\) with \(k \neq 28\) or 112
ft their equation of the form \(ay \pm b = c\)
For example, if you see
\(\begin{array}{rcl} 12y + 4 &=& 28 \\ -4 && -4 \end{array}\)
Award M1 for:
\(12y = k\) with \(k \neq 28\) or 32
6
| Answer | Mark | Mark scheme |
|---|---|---|
| 5 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 17 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 8 | B1 | cao |
17 Solve \(\quad 2(4x - 5) = 18\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 3.5 | M1 | for correct expansion of brackets, ie \(8x - 10\) or dividing throughout by 2 as a first step to solve equation, eg \(4x - 5 = 9\) |
| M1 | for isolating terms in \(x\), eg \(8x = 18 + 10\) or \(4x = 9 + 5\) | |
| A1 | for 3.5 or \(3\dfrac{1}{2}\) oe or \(\dfrac{7}{2}\) oe |
For M marks step must be carried out not just intention shown.
For example, if you see\[\begin{array}{c} 2(4x - 5) = 18 \\ \div 2 \qquad\qquad \div 2 \end{array}\]Award M1 for:\[4x - 5 = k \quad \text{with } k \ne 18, 36\]
ft their equation of the form \(ax \pm b = c\)
For example, if you see\[\begin{array}{c} 8x - 10 = 18 \\ +10 \qquad\qquad +10 \end{array}\]Award M1 for:\[8x \quad = k \quad \text{with } k \ne 8, 18\]
9 Here is a number machine.
| Answer | Mark | Mark scheme |
|---|---|---|
| 16 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 19 | M1 | starts method to find input using inverse operations eg \(28 + 10\ (=38)\) or sight of \(+10\) and \(\div 2\) |
| A1 | cao |
\(+10\) and \(\div 2\) could be seen in a flow diagram
Working may be next to number machine.
| Answer | Mark | Mark scheme |
|---|---|---|
| Shown | M1 | for carrying out at least one trial or for forming a suitable equation, eg \(2x - 10 = x\) or for identifying 10 |
| C1 | for showing that an input of 10 gives an output of 10 |
Trial can be for any value, must be correctly evaluated.
Accept correct inverse function trial, correctly evaluated.
If working seen on the number machine provided in the question allow for a trial other than input 13 or output 28.
Allow \(10 \times 2 = 20 - 10 = 10\) for M1C1
25 Solve \(\ 5x - 14 = 52 - x\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 11 | M1 | for one correct step to isolate \(x\) term or constant term on one side, eg adds \(x\) to both sides to get \(5x - 14 + x = 52 - x + x\) or adds 14 to both sides to get \(5x - 14 + 14 = 52 - x + 14\) oe |
| M1 | for both correct steps to isolate terms in \(x\) on one side and constant term on one side, eg \(\text{``}6x\text{''} - 14 + 14 = 52 + 14\), or \(5x + x = \text{``}66\text{''} + x - x\) | |
| A1 | cao |
May be seen in different equivalent forms but must be carried out, not just intention seen.
Can be implied by
eg \(4x = 66\) or \(6x = 38\)
24 Mano has three shelves of books.
There are \(x\) books on shelf A.
There are \((3x + 1)\) books on shelf B.
There are \((2x - 5)\) books on shelf C.
There is a total of 44 books on the three shelves.
All the books have the same mass.
The books on shelf B have a total mass of 7500 g.
Work out the total mass of the books on shelf A. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 2400 | P1 | for setting up an equation in \(x\) eg \(x + (3x + 1) + (2x - 5) = 44\) or \(6x - 4 = 44\) or \(x = 48 \div 6\ (= 8)\) |
| P1 | for substituting “8” into either \((3x + 1)\) or \((2x - 5)\) eg \(3 \times \text{``}8\text{''} + 1\ (= 25)\) or \(2 \times \text{``}8\text{''} - 5\ (= 11)\) | |
| P1 | for finding the mass of one book eg \(7500 \div \text{``}25\text{''}\ (= 300)\) | |
| P1 | for finding the mass of the books on shelf A eg \(\text{``}300\text{''} \times \text{``}8\text{''}\) | |
| A1 | cao |
16 Solve \(\ \dfrac{x}{7} + 9 = 4\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-35\) | M1 | for a correct first step, eg shows \(\dfrac{x}{7} + 9 - 9 = 4 - 9\) or \(\dfrac{x}{7} = 4 - 9\) or \(\dfrac{7x}{7} + 9 \times 7 = 4 \times 7\) or \(x + 63 = 28\) |
| A1 | cao |
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
7 Mano has three shelves of books.
There are \(x\) books on shelf A.
There are \((3x + 1)\) books on shelf B.
There are \((2x - 5)\) books on shelf C.
There is a total of 44 books on the three shelves.
All the books have the same mass.
The books on shelf B have a total mass of 7500 g.
Work out the total mass of the books on shelf A. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 2400 | P1 | for setting up an equation in \(x\), eg \(x + (3x + 1) + (2x - 5) = 44\) or \(6x - 4 = 44\) or \(x = 48 \div 6\ (= 8)\) |
| P1 | for substituting "8" into either \((3x + 1)\) or \((2x - 5)\) eg \(3 \times \text{``}8\text{''} + 1\ (= 25)\) or \(2 \times \text{``}8\text{''} - 5\ (= 11)\) | |
| P1 | for finding the mass of one book, eg \(7500 \div \text{``}25\text{''}\ (= 300)\) | |
| P1 | for finding the mass of the books on shelf A, eg \(\text{``}300\text{''} \times \text{``}8\text{''}\) | |
| A1 | cao |
4 Solve \(5x - 14 = 52 - x\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 11 | M1 | for one correct step to isolate \(x\) term or constant term on one side, eg adds \(x\) to both sides to get \(5x - 14 + x = 52 - x + x\) or adds 14 to both sides to get \(5x - 14 + 14 = 52 - x + 14\) oe |
| M1 | for both correct steps to isolate terms in \(x\) on one side and constant term on one side, eg \(\text{``}6x\text{''} - 14 + 14 = 52 + 14\), or \(5x + x = \text{``}66\text{''} + x - x\) | |
| A1 | cao |
May be seen in different equivalent forms but must be carried out, not just intention seen.
Can be implied by eg \(4x = 66\) or \(6x = 38\)
18 Solve \(\quad 4(2x - 3) = 20\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 4 | M1 | for a correct first step eg shows \(4 \times 2x - 4 \times 3\) or \(8x - 12\) or \(2x - 3 = \dfrac{20}{4}\) |
| M1 | for isolating terms in \(x\) eg \(2x = 5 + 3\) | |
| A1 | cao |
5 Solve \(\quad p - 2 = 3\) (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 5 | B1 | cao |
14
There are \(x\) sweets in a box.
There are \(y\) sweets in a packet.
| Answer | Mark | Mark scheme |
|---|---|---|
| \(7c + 6d\) | M1 | for \(7c\) or \(6d\) |
| A1 | for \(7c + 6d\) |
| Answer | Mark | Mark scheme |
|---|---|---|
| 7 | M1 | for correct method to expand, eg \(5 \times 2m - 5 \times 6\), or divides both sides by 5 as a first step. |
| M1 | for correct method to isolate terms in \(m\), eg \(10m - 30 + 30 = 40 + 30\) | |
| A1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(3x + 2y\) | M1 | for \(3x\) or \(2y\) |
| A1 | cao |
Condone use of \(b\) and \(p\)
8
| Answer | Mark | Mark scheme |
|---|---|---|
| 7 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 6 | M1 | for \(4n\ (= 24)\) or \(24 \div 4\) |
| A1 | cao |
24 Rick, Selma and Tony are playing a game with counters.
Rick has some counters.
Selma has twice as many counters as Rick.
Tony has 6 counters less than Selma.
In total they have 54 counters.
the number of counters Rick has : the number of counters Tony has \(= 1 : p\)
Work out the value of \(p\). (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.5 | P1 | for process to develop 3 algebraic expressions, eg (\(R =\)) \(n\), (\(S =\)) \(2n\), (\(T =\)) \(2n - 6\), oe, at least two must be correct. or for selecting 3 values satisfying the given criteria, eg (\(R =\)) 10, (\(S =\)) 20, (\(T =\)) 14 |
| P1 | for process to sum 3 algebraic expressions and equating to 54, eg \(n + \text{``}2n\text{''} + \text{``}2n - 6\text{''} = 54\) or for finding the correct sum of their values eg \(\text{``}10\text{''} + \text{``}20\text{''} + \text{``}14\text{''} = 44\) | |
| P1 | for start of process to solve the correct linear equation, eg \(5n = 54 + 6\ (n = 12)\) or for 12, 24, 18 | |
| P1 | for \(\text{``}12\text{''} : 2 \times \text{``}12\text{''} - 6\) oe eg \(12 : 18\) oe or \(18 : 12\) linked to \(T\), \(R\) | |
| A1 | for 1.5 or \(\dfrac{3}{2}\) or \(1\frac{1}{2}\) |
Accept \(1 : 1.5\) etc as answer
17
| Answer | Mark | Mark scheme |
|---|---|---|
| \(12 - 6x\) | B1 | for \(12 - 6x\) (accept \(-6x + 12\)) |
| Answer | Mark | Mark scheme |
|---|---|---|
| 16 | M1 | for a correct first step, eg \(3y = 12 \times 4\ (= 48)\) or \(\dfrac{y}{4} = \dfrac{12}{3}\) |
| A1 | cao |
Do not accept ambiguous algebraic expressions
| Answer | Mark | Mark scheme |
|---|---|---|
| \(2(2p + 3)\) | B1 | cao |
Do not accept equivalent expressions not fully factorised
12 Here is a number machine.
Here is a different number machine.
The number machine is not complete.
When the input is 8, the output is 154
| Answer | Mark | Mark scheme |
|---|---|---|
| 9 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 6 | M1 | starts to find input using inverse operations eg \(154 \div 11\ (= 14)\) or indicates \(\div 11\) and \(-\ 8\) or derivation of equation eg \((8 + n) \times 11 = 154\) or starting to solve for unknown eg \(154 - 8 \times 11\ (= 66)\) |
| A1 | cao |
\(\div 11\) and \(-8\) could be seen in a flow diagram
Evidence could be provided by algebraic statement, numerical statements or by diagram
3 Rick, Selma and Tony are playing a game with counters.
Rick has some counters.
Selma has twice as many counters as Rick.
Tony has 6 counters less than Selma.
In total they have 54 counters.
the number of counters Rick has : the number of counters Tony has \(= 1 : p\)
Work out the value of \(p\). (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.5 | P1 | for process to develop 3 algebraic expressions, eg (\(R =\)) \(n\), (\(S =\)) \(2n\), (\(T =\)) \(2n - 6\), oe, at least two must be correct. or for selecting 3 values satisfying the given criteria, eg (\(R =\)) 10, (\(S =\)) 20, (\(T =\)) 14 |
| P1 | for process to sum 3 algebraic expressions and equating to 54, eg \(n + \text{``}2n\text{''} + \text{``}2n - 6\text{''} = 54\) or for finding the correct sum of their values eg \(\text{``}10\text{''} + \text{``}20\text{''} + \text{``}14\text{''} = 44\) | |
| P1 | for start of process to solve the correct linear equation, eg \(5n = 54 + 6\ (n = 12)\) or for 12, 24, 18 | |
| P1 | for \(\text{``}12\text{''} : 2 \times \text{``}12\text{''} - 6\) oe eg \(12 : 18\) oe or \(18 : 12\) linked to T, R | |
| A1 | for 1.5 or \(\dfrac{3}{2}\) or \(1\frac{1}{2}\) |
Accept \(1 : 1.5\) etc as answer
17
| Answer | Mark | Mark scheme |
|---|---|---|
| \(y^2 + 5y\) | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(2(2a - 3)\) | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 2.9 | M1 | for a correct first stage eg expanding the brackets, \(2 \times 5x - 2 \times 4\ (= 10x - 8)\) or division of both sides by 2, eg \(\dfrac{2(5x - 4)}{2} = \dfrac{21}{2}\) |
| M1 | for isolating terms in \(x\) eg \(10x = 21 + 8\) | |
| A1 | oe |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(20e^3f^4\) | M1 | for any two of \(4 \times 5\ (= 20)\), \(e^{2+1}\ (= e^3)\), \(f^{1+3}\ (= f^4)\) in a product or written as individual terms |
| A1 | cao |
Do not award if there is contradiction
16 \(ABCD\) is a kite.
\(AB = (4x - 2)\) cm
Jasper says that \(x\) could be 0.5
\(AD = 3AB\)
The kite has a perimeter of 64 cm.
| Answer | Mark | Mark scheme |
|---|---|---|
| Explanation | C1 | for explanation, eg \(AB\) cannot be zero (cm) or shows \(AB\) to be zero, eg \(4 \times 0.5 - 2 = 0\) |
Accept say ‘\(AB\) would then be 0’
| Answer | Mark | Mark scheme |
|---|---|---|
| 2.5 | P1 | for a correct expression for \(AD\), eg \(3(4x - 2)\) or \(12x - 6\) OR \(2(3AB + AB) = 64\) oe or \(3AB + AB = 32\) oe or \(AB = 8\) OR for an equation with mixed variables, eg \(6AB + 2(4x - 2) = 64\) |
| P1 | for forming a correct equation in \(x\), eg \(4x - 2 + 4x - 2 + 3(4x - 2) + 3(4x - 2) = 64\) or \(4x - 2 = 8\) or \(4x - 2 + 3(4x - 2) = 32\) | |
| A1 | cao |
May be seen on diagram
15
| Answer | Mark | Mark scheme |
|---|---|---|
| \(2a + 2d\) | B1 | cao |
Accept \(2 \times a + 2 \times d\)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(y(6y - 5)\) | B1 | cao |
Accept \(y \times (6y - 5)\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 11 | M1 | for isolating \(x\) terms, eg \(4x = 37 + 7\) or \(4x = 44\) or for \(x - \frac{7}{4} = \frac{37}{4}\) or for \(37 + 7 = 44\) followed by \(\text{``}44\text{''} \div 4\ (= 11)\) |
| A1 | cao |
19 Solve \(4(x - 6) = 44\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 17 | M1 | for correctly expanding the bracket, as part of an equation to get \(4x - 24 = 44\) or for dividing both sides of the equation by 4 as a first step, eg \(\frac{4(x-6)}{4} = \frac{44}{4}\) oe |
| A1 | cao |
Award M1 for an embedded value of 17 if not identified as the answer
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 \(3 - 1 = 2\)
10
| Answer | Mark | Mark scheme |
|---|---|---|
| 8.5 | M1 | for multiplying both sides by 7 as a first step eg \(9 + x = 7(11 - x)\) or dividing each term on the left hand side by 7 eg \(\dfrac{9}{7} + \dfrac{x}{7} = 11 - x\) |
| M1 | (dep M1) for method to isolate the \(x\) terms on one side | |
| A1 | oe |
\(\times 7\) written near the equation is not enough for this mark
| Answer | Mark | Mark scheme |
|---|---|---|
| \(4(y + 3)\) | B1 | \(4(y + 3)\) or \(4y + 12\) |
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
10
| Answer | Mark | Mark scheme |
|---|---|---|
| 4 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 8 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 3 | M1 | for a correct first step eg subtracting 2 from both sides or dividing all terms by 6 |
| A1 | cao |
Division by 6 must be ALL terms
19
| Answer | Mark | Mark scheme |
|---|---|---|
| 8 | M1 | for a correct first step eg \(3x - 12 = 12\) or \(3(x - 4) \div 3 = 12 \div 3\) |
| A1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(3b(3 - b)\) | M1 | for \(3(3b - b^2)\) or \(b(9 - 3b)\) or \(3b\)(two term linear expression) |
| A1 | cao |
17 The diagram shows a pentagon.
The pentagon has one line of symmetry.
\(AE = 4x\)
\(AB = 2x + 1\)
\(BC = x + 2\)
All these measurements are given in centimetres.
The perimeter of the pentagon is 18 cm.
| Answer | Mark | Mark scheme |
|---|---|---|
| Full working seen | M1 | for an initial step with the expressions eg doubling \(2x + 1\) or \(x + 2\) or halving \(4x\) or for identifying \(CD\) as \(x + 2\) or for identifying \(DE\) as \(2x + 1\) |
| M1 | for an expression for the total perimeter, eg \(4x + 2 \times (2x + 1) + 2 \times (x + 2)\) | |
| C1 | for full simplification and equating to 18 |
May be seen in working or on diagram
| Answer | Mark | Mark scheme |
|---|---|---|
| 1.2 | M1 | for isolating terms in \(x\) can ft an equation stated in (a) provided in form \(ax + b = c\) |
| A1 | for 1.2 oe |
\(10x = 18 - 6\)
Accept \(\dfrac{12}{10}\) or \(\dfrac{6}{5}\)
10 Here is a number machine.
| Answer | Mark | Mark scheme |
|---|---|---|
| 38 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 6 | M1 | starts process to find input using inverse operations eg \(28 + 2\) or sight of \(+2\ \ \div 5\) or by forming an equation eg \(x \times 5 - 2 = 28\) |
| A1 | cao |
\(+2\ \ \div 5\) could be seen in a flow diagram
25 Solve \(\dfrac{5 - x}{2} = 2x - 7\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 3.8 | M1 | for a correct first step, eg \(5 - x = 2(2x - 7)\) or \(5 - x = 4x - 14\) or \(\dfrac{5}{2} - \dfrac{x}{2} = 2x - 7\) |
| M1 | (dep) for isolating terms in \(x\) eg \(4x + x = 14 + 5\) or \(-\dfrac{x}{2} - 2x = -7 - \dfrac{5}{2}\) | |
| A1 | oe |
Method must show LHS \(\times 2\) and both terms on RHS \(\times 2\) or \(5 - x\) and both terms on RHS \(\times 2\)
eg \(-4x\) both sides with \(-5\) both sides
or \(+x\) both sides with \(+14\) both sides
Accept \(\dfrac{19}{5}\), \(3\dfrac{4}{5}\) oe but not \(\dfrac{-19}{-5}\) oe
11
| Answer | Mark | Mark scheme |
|---|---|---|
| 17 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 12 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| 5.5 | B1 | Accept \(\dfrac{11}{2}\), \(5\dfrac{1}{2}\) oe |
7 Solve \(\dfrac{5 - x}{2} = 2x - 7\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 3.8 | M1 | for a correct first step, eg \(5 - x = 2(2x - 7)\) or \(5 - x = 4x - 14\) or \(\dfrac{5}{2} - \dfrac{x}{2} = 2x - 7\) |
| M1 | (dep) for isolating terms in \(x\) eg \(4x + x = 14 + 5\) or \(-\dfrac{x}{2} - 2x = -7 - \dfrac{5}{2}\) | |
| A1 | oe |
Method must show LHS \(\times 2\) and both terms on RHS \(\times 2\) or \(5 - x\) and both terms on RHS \(\times 2\)
eg \(-4x\) both sides with \(-5\) both sides
or \(+x\) both sides with \(+14\) both sides
Accept \(\dfrac{19}{5}\), \(3\dfrac{4}{5}\) oe but not \(\dfrac{-19}{-5}\) oe
16 Solve \(5x - 6 = 3(x - 1)\) (3)
| Answer | Mark | Notes |
|---|---|---|
| \(1\dfrac{1}{2}\) | M1 | for correct expansion of the bracket or dividing all terms by 3 as a first step eg \(3x - 3\) or \((5x - 6)/3 = 3(x - 1)/3\) |
| M1 | for isolating terms in \(x\) on one side of an equation eg \(5x - 6 - 3x = -3\) or both constants on one side of an equation, eg \(5x = 3x - 3 + 6\), ft \(5x - 6 = 3x - 1\) | |
| A1 | for \(1\dfrac{1}{2}\) oe |
3 Solve \(\dfrac{y}{4} = 10.5\) (1)
| Answer | Mark | Notes |
|---|---|---|
| 42 | B1 | cao |
1 Solve \(5x - 6 = 3(x - 1)\) (3)
| Answer | Mark | Notes |
|---|---|---|
| \(1\dfrac{1}{2}\) | M1 | for correct expansion of the bracket or dividing all terms by 3 as a first step eg \(3x - 3\) or \((5x - 6)/3 = 3(x - 1)/3\) |
| M1 | for isolating terms in \(x\) on one side of an equation eg \(5x - 6 - 3x = -3\) or both constants on one side of an equation, eg \(5x = 3x - 3 + 6\), ft \(5x - 6 = 3x - 1\) | |
| A1 | for \(1\dfrac{1}{2}\) oe |
19
\(-3 \lt t \leqslant 2\)
\(t\) is an integer.
| Answer | Mark | Notes |
|---|---|---|
| 9.5 | M1 | expands brackets or divides by 4 as a first step |
| A1 | oe |
| Answer | Mark | Notes |
|---|---|---|
| \(-2, -1, 0, 1, 2\) | B2 | cao |
| (B1) | (for the numbers \(-2, -1, 0, 1\) (accept with \(-3\) and/or 2 only), or 4 correct with no incorrect) |
11 Solve \(\dfrac{3x - 2}{4} - \dfrac{2x + 5}{3} = \dfrac{1 - x}{6}\) (4)
| Answer | Mark | Notes |
|---|---|---|
| \(9\dfrac{1}{3}\) | M1 | for writing at least 2 fractions with a common denominator eg. \(\dfrac{3(3x - 2)}{12}\), \(\dfrac{4(2x + 5)}{12}\), \(\dfrac{2(1 - x)}{12}\) with at least one correct numerator or for \(\dfrac{3x}{4} - \dfrac{2}{4} - \dfrac{2x}{3} - \dfrac{5}{3} = \dfrac{1}{6} - \dfrac{x}{6}\) (accept \(+\dfrac{5}{3}\) instead of \(-\dfrac{5}{3}\)) |
| M1 | (dep) for a method to eliminate all fractions in an equation, ignore errors in any expanded terms eg. \(3(3x - 2) - 4(2x + 5) = 2(1 - x)\) or \(6 \times [3(3x - 2) - 4(2x + 5)] = 12 \times [1 - x]\) or \(3 \times 3x - 3 \times 2 - 4 \times 2x - 4 \times 5 = 2 \times 1 - 2 \times x\) OR for the correct expansion of brackets leading to \(\dfrac{9x - 6 - 8x - 20}{12} = \dfrac{2 - 2x}{12}\) | |
| M1 | (dep on M2) for correctly isolating terms in \(x\) and number terms of their linear equation e.g. \(9x - 8x + 2x = 2 + 6 + 20\) | |
| A1 | for \(9\dfrac{1}{3}\) oe |
3
| Answer | Mark | Notes |
|---|---|---|
| \(56ef\) | B1 | cao |
| Answer | Mark | Notes |
|---|---|---|
| 12.5 | B1 | oe |
21 Solve \(4(x + 3) = 2x + 8\) (3)
| Answer | Mark | Notes |
|---|---|---|
| \(-2\) | 3 | M1 for expanding brackets e.g. \(4x + 12\ (= 2x + 8)\) or divide by 4 as a first step e.g. \(x + 3 = \dfrac{2x}{4} + \dfrac{8}{4}\) M1 ft their equations which have to be of the form \(ax + b = 2x + 8\) or \(x + 3 = ax + b\) for isolating terms in \(x\) and numbers e.g. \(4x - 2x = 8 - 12\) seen as part of their solution oe A1 cao |
17
| Answer | Mark | Notes |
|---|---|---|
| \(\dfrac{3}{x + 1}\) | 1 | B1 oe |
| Answer | Mark | Notes |
|---|---|---|
| \(-2.5\) | 3 | M1 for attempting to multiply both sides by 5 as a first step and cancelling the LHS (can be implied by equations of the form \(15 - x = 15x + ?\) or \(15 - x = ?x + 55\)) M1 for method to isolate terms in \(x\) and number terms from \(ax + b = cx + d\) A1 for \(-2.5\) oe OR M1 for splitting the LHS correctly into two parts \(\dfrac{15}{5} - \dfrac{x}{5} = 3x + 11\) M1 for method to isolate terms in \(x\) and number terms from \(ax + b = cx + d\) A1 for \(-2.5\) oe |
| Answer | Mark | Notes |
|---|---|---|
| \(m = \dfrac{2E}{v^2}\) | 3 | M1 for a method to remove square root or clear fraction as a first step M1 for a method to remove square root and clear fraction A1 for \(m = \dfrac{2E}{v^2}\) |
17 Solve \(\dfrac{x + 1}{3} + \dfrac{2x + 5}{4} = 2\) (4)
| Answer | Mark | Notes |
|---|---|---|
| 0.5 | 4 | M1 for identifying a common denominator in a correct expression or equation eg \(\dfrac{4(x + 1)}{12} + \dfrac{3(2x + 5)}{12}\) (= 2) M1 (dep on M1) for expanding both brackets in an equation eg \(\dfrac{4x + 4}{12} + \dfrac{6x + 15}{12} = 2\) eg \(4x + 4 + 6x + 15 = 12 \times 2\) M1 (dep on M1) for isolating “like terms” on each side of an equation eg \(4x + 6x = 24 - 15 - 4\) A1 for 0.5 oe |
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 |
12
Hassan thinks of a number.
He multiplies the number by 3
He then adds 12
His answer is 60
| Answer | Mark | Notes |
|---|---|---|
| 12 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 45 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 16 | 2 | M1 for correct order of inverse operations \(-12\) then \(\div 3\) or for forming the equation \(3x + 12 = 60\) and showing intention to subtract 12 from both sides or divide each term by 3 as a first step A1 cao |
4 Solve \(4(x + 3) = 2x + 8\) (3)
| Answer | Mark | Notes |
|---|---|---|
| \(-2\) | 3 | M1 for expanding brackets eg \(4x + 12\) (\(= 2x + 8\)) or dividing by 4 as a first step eg \(x + 3 = \dfrac{2x}{4} + \dfrac{8}{4}\) M1 ft their equation of the form \(ax + b = 2x + 8\) or \(x + 3 = ax + b\) for isolating terms in \(x\) and numbers e.g. \(4x - 2x = 8 - 12\) seen as part of their solution oe A1 cao |
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 |
9
| Answer | Mark | Notes |
|---|---|---|
| 6 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 8 | 1 | B1 cao |
17
| Answer | Mark | Notes |
|---|---|---|
| \(x \lt 7\) | 2 | M1 for isolating term in \(x\) eg \(3x \lt 16 + 5\) or \(3x \lt 21\) or for \((x =)\ 7\) or \(x \gt 7\) etc A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(\dfrac{7}{5}\) | 3 | M1 for multiplying by 4 or adding \(\dfrac{w}{4}\) or subtracting \(\dfrac{11}{4}\) or subtracting 1 [all applied to both sides and as a first step] M1 for isolating terms in \(w\) on one side and number terms on the other side of the equation A1 for \(\dfrac{7}{5}\) oe |
8
| Answer | Mark | Notes |
|---|---|---|
| 15 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 6 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 3 | 2 | M1 for intention to subtract 7 from both sides or divide all terms by 2 as a first step 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 |
5
| Answer | Mark | Notes |
|---|---|---|
| 5 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 2.8 | 1 | B1 oe |
| Answer | Mark | Notes |
|---|---|---|
| 1.5 | 2 | M1 for the intention to subtract 3 from both sides or divide each term by 4 as a first step A1 1.5 oe |
19
| Answer | Mark | Notes |
|---|---|---|
| \(-0.5\) | 2 | M1 for intention to subtract 19 from both sides or divide all terms by 8 as a first step A1 for \(-0.5\) oe |
| Answer | Mark | Notes |
|---|---|---|
| 3 | 2 | M1 for a correct operation to collect the \(c\) terms or the number terms on one side of the equation e.g. \(2c - c + 5 = 8\), \(2c + 5 - 5 = c + 8 - 5\) A1 cao |
13
| Answer | Mark | Notes |
|---|---|---|
| 9 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 50 | 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 \(\tfrac{10}{4}\)) |
24
| Answer | Mark | Notes |
|---|---|---|
| \(e(3e + 5)\) | 1 | B1 for \(e(3e + 5)\) |
| Answer | Mark | Notes |
|---|---|---|
| 4 | 3 | M1 for intention to expand brackets eg \(7k - 21\) or division of all terms on RHS by 7 as a first step M1 for correct method to isolate terms in \(k\) in an equation A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(a = 2f - 1\) | 2 | M1 for a correct first step eg intention to multiply both sides by 2 A1 cao |
12
| Answer | Mark | Notes |
|---|---|---|
| \(e(3e + 5)\) | 1 | B1 for \(e(3e + 5)\) |
| Answer | Mark | Notes |
|---|---|---|
| 4 | 3 | M1 for intention to expand brackets, eg \(7k - 21\) or division of all terms on RHS by 7 as first step, eg \(\tfrac{3}{7}k - \tfrac{5}{7}\) M1 for correct method to isolate terms in \(k\) in an equation A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(2x^2 - 13x - 24\) | 2 | M1 for 4 terms correct ignoring signs or 3 out of no more than 4 terms correct A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(-\dfrac{1}{3}\) | 3 | M1 for clear intention to multiply both sides by 4 or split into individual fractions on LHS M1 for correct method to isolate term in \(f\) in an equation, ft from equations of form \(a + bf = c\), where \(a, b, c \neq 0\) A1 cao |
12
| Answer | Mark | Notes |
|---|---|---|
| 22 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 18 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 3.4 | 2 | M1 for intention to subtract 7 from both sides or divide all terms by 5 as a first step. A1 for 3.4 oe |
26
\(-5 \lt y \leqslant 0\)
\(y\) is an integer.
| Answer | Mark | Notes |
|---|---|---|
| \(\dfrac{2}{3}\) | 2 | M1 for intention to subtract 4 from both sides or divide each term by 3 or \(3p = 2\) written in the workspace A1 for \(\frac{2}{3}\) accept answer to two decimal places rounded or truncated |
| Answer | Mark | Notes |
|---|---|---|
| \(-4\), \(-3\), \(-2\), \(-1\), 0 | 2 | B2 for all 5 correct values; ignore repeats, any order (B1 for 4 correct (and no incorrect) values or all five correct values and \(-5\)) |
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 |
23
| Answer | Mark | Notes |
|---|---|---|
| 9 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 5 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 17 | 2 | M1 for clear intention to expand bracket or divide both sides by 2 as the first step eg. \(2y - 2 \times 5 = 24\) or \(y - 5 = 24 \div 2\) A1 for 17 |
| Answer | Mark | Notes |
|---|---|---|
| \(5(3p + 8)\) | 1 | B1 cao |
20
| Answer | Mark | Notes |
|---|---|---|
| 4 | 3 | M1 for correct expansion to \(32x - 8\) or multiplying both sides by \(3x\) or dividing both sides by 4 M1 for a compete and correct method to isolate the \(x\) terms and the number terms (condone one arithmetic error in multiplying out the bracket) A1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\dfrac{2(y - 6) - (y + 3)}{(y + 3)(y - 6)}\) | \(\dfrac{y - 15}{(y + 3)(y - 6)}\) | 3 | M1 for common denominator of \((y + 3)(y - 6)\) M1 for \(\dfrac{2(y - 6)}{(y + 3)(y - 6)} - \dfrac{y + 3}{(y + 3)(y - 6)}\) oe or \(\dfrac{2(y - 6) - (y + 3)}{(y + 3)(y - 6)}\) oe A1 for \(\dfrac{y - 15}{(y + 3)(y - 6)}\) or \(\dfrac{y - 15}{y^2 - 3y - 18}\) |
19
| Answer | Mark | Notes |
|---|---|---|
| 14 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 5 | 2 | M1 for intention to subtract 4 from each side or divide each term by 3 as a first step or embedded answer. A1 cao |
16
| Answer | Mark | Notes |
|---|---|---|
| 6.25 | 3 | M1 for clear intention to expand bracket or divide both sides of the equation by 5 as first step M1 for correct method to isolate terms in \(f\) A1 for 6.25 oe |
| Answer | Mark | Notes |
|---|---|---|
| \(-0.75\) | 4 | M1 for correct method to clear fractions eg. multiply all terms by 6 M1 for expansion of brackets oe M1 (dep on M1) for isolating the terms in \(h\) and the constant terms A1 for \(-0.75\) oe |
26 Solve \(3(x - 2) = x + 7\) (3)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(3x - 6 = x + 7\) \(2x = 13\) | 6.5 | 3 | M1 for \(3 \times x - 3 \times 2\) (\(= 3x - 6\)) or \(\dfrac{x}{3} + \dfrac{7}{3}\) seen M1 for correct method to isolate the terms in \(x\) or the number terms on opposite sides of an equation A1 for 6.5 oe |
23
| Answer | Mark | Notes |
|---|---|---|
| 10 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 8.5 | 1 | B1 accept \(\dfrac{17}{2}\) or \(8\frac{1}{2}\) |
| Answer | Mark | Notes |
|---|---|---|
| 32 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(6 + 3t\) | 1 | B1 for \(6 + 3t\) |
12
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(3x - 6 = x + 7\) \(2x = 13\) | 6.5 | 3 | M1 for \(3 \times x - 3 \times 2\) (\(= 3x - 6\)) or \(\frac{x}{3} + \frac{7}{3}\) seen M1 for correct method to isolate the terms in \(x\) or the number terms on opposite sides of an equation A1 for 6.5 oe |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(2 - y = 1 \times 5\) | \(-3\) | 2 | M1 for intention to multiply both sides by 5 (to give \(2 - y = 1 \times 5\)) A1 cao |
17 Solve \(\dfrac{4x - 1}{5} + \dfrac{x + 4}{2} = 3\) (3)
| Answer | Mark | Notes |
|---|---|---|
| \(\dfrac{12}{13}\) | 3 | M1 for multiplying throughout by 10 oe or writing LHS as a single fraction e.g \(2(4x - 1) + 5(x + 4) = 3 \times 10\) or \(\dfrac{2(4x - 1) + 5(x + 4)}{10}\) or \(\dfrac{2(4x - 1)}{10} + \dfrac{5(x + 4)}{10}\) M1 (dep) for a complete correct method to obtain linear equation of the form \(ax = b\) (condone one arithmetic error in multiplying out the bracket) A1 for \(\dfrac{12}{13}\) oe (decimal equivalent is 0.923…) |
9
| Answer | Mark | Notes |
|---|---|---|
| \(4a\) | 1 | B1 for \(4a\) oe as a single term |
| Answer | Mark | Notes |
|---|---|---|
| \(3cd\) | 1 | B1 for \(3cd\) oe as a single term |
| Answer | Mark | Notes |
|---|---|---|
| \(7ef\) | 1 | B1 for \(7ef\) oe as a single term |
| Answer | Mark | Notes |
|---|---|---|
| 3 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 2 | 2 | M1 for intention to subtract 7 from each side or divide each term by 5 or embedded method A1 cao |
7
| Answer | Mark | Notes |
|---|---|---|
| \(6x - 3y\) | 2 | M1 for an attempt to combine terms in \(x\) or terms in \(y\) correctly eg \(5x + x\) (\(= 6x\)), \(4y - 7y\) (\(= -3y\)) A1 for \(6x - 3y\) oe |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(7x + 14 = 7\) or \(x + 2 = 1\) \(7x = -7\) | \(x = -1\) | 2 | M1 for correctly expanding the bracket or an attempt to divide both sides by 7 e.g. \(7x + 14\) or \(x + 2 = 7 \div 7\) oe A1 cao |
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 |
17
| Answer | Mark | Notes |
|---|---|---|
| 7 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 12 | 1 | B1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(5w = 10 + 6\) \(w = 16 \div 5\) or \(w - \frac{6}{5} = \frac{10}{5}\) oe | 16/5 oe | 2 | M1 for \(5w - 6 + 6 = 10 + 6\) oe or \(w - \frac{6}{5} = \frac{10}{5}\) oe A1 for \(\frac{16}{5}\), \(3\frac{1}{5}\), 3.2, oe |
19
| Answer | Mark | Notes |
|---|---|---|
| 19 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| 8 | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(2\frac{1}{4}\) | 2 | M1 for \(4m = 15 - 6\) or clear attempt to subtract 6 from both sides of the equation A1 for \(2\frac{1}{4}\) or 2.25 or \(\frac{9}{4}\) |
14
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(p^2 - 4p + 9p - 36\) | \(p^2 + 5p - 36\) | 2 | M1 for all 4 terms correct (condone incorrect signs) or 3 out of 4 terms correct with correct signs A1 cao |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(5w - 8 = 3(4w + 2)\) \(5w - 8 = 12w + 6\) \(-8 - 6 = 12w - 5w\) \(-14 = 7w\) | \(-2\) | 3 | M1 for attempting to multiply both sides by 3 as a first step (this can be implied by equations of the form \(5w - 8 = 12w + ?\) or \(5w - 8 = ?w + 6\) i.e. the LHS must be correct M1 for isolating terms in \(w\) and the number terms correctly from \(aw + b = cw + d\) A1 cao OR M1 for \(\dfrac{5w}{3} - \dfrac{8}{3} = 4w + 2\) M1 for isolating terms in \(w\) and the number terms correctly A1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \((x + 7)(x - 7)\) | 1 | B1 cao |
| Answer | Mark | Notes |
|---|---|---|
| \(3x^4y^{\frac{3}{2}}\) | 2 | B2 for \(3x^4y^{\frac{3}{2}}\) or \(3x^4y^{1.5}\) or \(3x^4y^{1\frac{1}{2}}\) (B1 for any two terms correct in a product eg. \(3x^4y^n\)) |
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 |
16 Here is an equilateral triangle.
All measurements are in centimetres.
Not drawn accurately
Work out the value of the perimeter of the triangle. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(6x - 8 = 2x + 12\) | M1 | oe equation in terms of \(x\) any letter |
| \(6x - 2x = 12 + 8\) or \(-8 - 12 = 2x - 6x\) or \(4x = 20\) or \((x =)\ 5\) | M1dep | oe collection of terms eg \(4x - 20 = 0\) |
| \(6 \times\) their \(5 - 8\) or \(2 \times\) their \(5 + 12\) or 22 | M1dep | a combination of sides may be used eg \(14 \times 5 - 4\) or \(10 \times 5 + 16\) |
| 66 | A1 |
| 5 or 22 may be shown on diagram | |
| Trial and improvement with 22 selected as length of side | M3 |
| Trial and improvement with 5 selected as value of \(x\) | M2 |
| \(8x + 4 = 180\) and \(x = 22\) | M0 |
4 Here is an equilateral triangle.
All measurements are in centimetres.
Not drawn accurately
Work out the value of the perimeter of the triangle. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(6x - 8 = 2x + 12\) | M1 | oe equation in terms of \(x\) any letter |
| \(6x - 2x = 12 + 8\) or \(-8 - 12 = 2x - 6x\) or \(4x = 20\) or \((x =)\ 5\) | M1dep | oe collection of terms eg \(4x - 20 = 0\) |
| \(6 \times\) their \(5 - 8\) or \(2 \times\) their \(5 + 12\) or 22 | M1dep | a combination of sides may be used eg \(14 \times 5 - 4\) or \(10 \times 5 + 16\) |
| 66 | A1 |
| 5 or 22 may be shown on diagram | |
| Trial and improvement with 22 selected as length of side | M3 |
| Trial and improvement with 5 selected as value of \(x\) | M2 |
| \(8x + 4 = 180\) and \(x = 22\) | M0 |
4 Here is a number machine.
| Answer | Mark | Comments |
|---|---|---|
| 7 | B1 |
If answer line blank check the output oval
| Answer | Mark | Comments |
|---|---|---|
| \(19 + 5\) or 24 or \(n \div 3\) or \(\dfrac{x + 5}{3}\) | M1 | oe may be seen on diagram where \(14 \leqslant n \leqslant 24\) \(x\) can be any letter |
| 8 | A1 |
| \(8 \times 3 - 5\) written in working or 8 adjacent to or in the input oval | M1 |
| \(\dfrac{14}{3}\) | M1 |
| Answer | Mark | Comments |
|---|---|---|
| \(\div 6\) or \(-15\) or \(\times \dfrac{1}{6}\) or \(+{-15}\) | B1 | oe fraction or decimal for \(\dfrac{1}{6}\) |
| Mark the rectangle on number machine | |
| Divide by 6 or subtract 15 or multiply by \(\dfrac{1}{6}\) or add negative 15 | B1 |
| More than one operation | B0 |
1
| Answer | Mark | Comments |
|---|---|---|
| 11 | B1 |
| Condone \(11 + 8 = 19\) on the answer line or in the working if the answer line is blank | B1 |
| \(11 + 8 = 19\) in the working and 19 on the answer line | B0 |
| Answer | Mark | Comments |
|---|---|---|
| 8 | B1 |
| Condone \(3 \times 8 = 24\) on the answer line or in the working if the answer line is blank | B1 |
| \(3 \times 8 = 24\) in the working and 24 on the answer line | B0 |
| Answer | Mark | Comments |
|---|---|---|
| 13 | B1 |
| Condone \(20 - 13 = 7\) on the answer line or in the working if the answer line is blank | B1 |
| \(20 - 13 = 7\) in the working and 7 on the answer line | B0 |
26 Here is a rectangle.
All measurements are in centimetres.
Not drawn accurately
\(AB : BC = 1 : 3\)
Work out the value of the area. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4x + 1 = 2x + 17\) | M1 | oe equation in terms of \(x\) any letter |
| \(4x - 2x = 17 - 1\) or \(1 - 17 = 2x - 4x\) or \((x =)\ 8\) | M1dep | oe collection of terms |
| Correctly substitutes their 8 into a correct expression for the length or width of the rectangle | M1 | their \(8 \gt 0\) and their \(8 \ne 1\) |
| Correct method for both the length and the width of the rectangle using their 8 | M1dep | their \(8 \gt 0\) and their \(8 \ne 1\) dep on 3rd M |
| 3267 | A1 | SC1 \(12x + 3\) or \(6x + 51\) |
| The first M1 or SC1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Trial and improvement to find \((x =)\ 8\) is M2 | |
| Using an incorrect value of \(x\) for 3rd and 4th marks eg when \(x = 10\) \(4 \times 10 + 1 = 41\) and \(41 \times 3 = 123\) or \(12 \times 10 + 3 = 123\) and \(123 \div 3 = 41\) or \(2 \times 10 + 17 = 37\) and \(6 \times 10 + 51 = 111\) | M0M0M1M1dep M0M0M1M1dep M0M0M1M1dep |
11 Here is a diagram.
Work out the value of \(b\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 21.5 in second circle and 29.5 in third circle | B1 | oe |
| Answer | Mark | Comments |
|---|---|---|
| 14 in circle | B1 | |
| \(+\,9\) in box | B1ft | ft their 14 |
| Accept other operators eg 14 in circle with \(\times\,\dfrac{14}{5}\) (oe) | B1B1 |
| 36 in circle \(+\,31\) in box | B0B1ft |
| 36 in circle \(+\,9\) in box | B0B1 |
| Answer | Mark | Comments |
|---|---|---|
| \(a + 3a = 28\) or \(4a = 28\) or \(28 + 2a = b\) or \(6a = b\) or \(28 \div 4\) | M1 | oe |
| 7 | A1 | may be seen on diagram |
| 42 | B1ft | ft \(28 + 2 \times\) their 7 or \(6 \times\) their 7 |
| Answer 42 | M1A1B1 |
| \(4a = 28\) followed by \(a = 24\) and \(b = 76\) or \(b = 144\) | M1A0B1ft |
| \(a = 24\) (no working) and \(b = 76\) or \(b = 144\) | M0A0B1ft |
| Do not accept \(a + 3a = 28 + 2a = b\) for \(a + 3a = 28\) or \(28 + 2a = b\) | M0 |
10
| Answer | Mark | Comments |
|---|---|---|
| 45 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: Expands bracket first | ||
| \(8d - 20\) | M1 | oe expression eg \(8 \times d - 20\) |
| their \(8d = 28 + \text{their } 20\) or \(8d = 48\) | M1 | their 20 cannot be zero |
| 6 | A1ft | ft M0M1 or M1M0 |
| Alternative method 2: Divides by 4 first | ||
| \(2d - 5 = \dfrac{28}{4}\) or \(2d - 5 = 7\) | M1 | |
| \(2d =\) their \(7 + 5\) or \(2d = 12\) | M1 | their 7 cannot be 28 |
| 6 | A1ft | ft M0M1 or M1M0 |
| Alt 1 \(6d - 5 = 28\) \(6d = 33\) \(d = \dfrac{33}{6}\) | M0 M1 A1ft |
| Alt 2 \(2d - 5 = 24\) \(2d = 29\) \(d = 14.5\) | M0 M1 A1ft |
| Answer 6 from trial | M1M1A1 |
| Embedded answer \(4(2 \times 6 - 5) = 28\) | M1M1A0 |
| ft answers correct to 1 dp or better |
3
| Answer | Mark | Comments |
|---|---|---|
| 6 | B1 |
| Embedded answer without 6 being selected eg \(5 \times 6 = 30\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| 12 | B1 |
| Embedded answer without 12 being selected eg \(-2 + 12 = 10\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| 5 | B2 | B1 correct partial simplification eg \(\dfrac{20}{4}\) or \(\dfrac{5}{1}\) or \(\dfrac{10w}{2w}\) or \(\dfrac{5w}{w}\) SC1 \(5w\) |
| Correct partial simplification followed by incorrect further work eg \(\dfrac{10w}{2w}\), Answer \(8w\) | B1 |
| Correct answer followed by further work | B1 |
24 Solve \(\quad 7x - 22 = 4x + 29\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(7x - 4x\) or \(3x\) or \(4x - 7x\) or \(-3x\) or \(-22 - 29\) or \(-51\) or \(22 + 29\) or 51 | M1 | |
| \(3x = 51\) or \(-3x = -51\) | A1 | \(\dfrac{51}{3}\) or \(\dfrac{-51}{-3}\) implies M1A1 implied by correct answer |
| 17 | A1ft | ft M1A0 from an equation of the form \(\pm 3x = a\) or \(bx = \pm 51\) |
| Trial and improvement scores 0 or 3 | |
| If a follow through answer does not simplify to an integer, accept it as a fraction, mixed number or decimal to at least 1dp. eg from \(3x = 7\) accept \(\dfrac{7}{3}\) or \(2\dfrac{1}{3}\) or 2.3 or better Ignore any attempt to convert a correct ft fraction | M1A0A1ft |
| Embedded answer | M1A1A0 |
8 Solve \(\quad 7x - 22 = 4x + 29\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(7x - 4x\) or \(3x\) or \(4x - 7x\) or \(-3x\) or \(-22 - 29\) or \(-51\) or \(22 + 29\) or 51 | M1 | |
| \(3x = 51\) or \(-3x = -51\) | A1 | \(\dfrac{51}{3}\) or \(\dfrac{-51}{-3}\) implies M1A1 implied by correct answer |
| 17 | A1ft | ft M1A0 from an equation of the form \(\pm 3x = a\) or \(bx = \pm 51\) |
| Trial and improvement scores 0 or 3 | |
| If a follow through answer does not simplify to an integer, accept it as a fraction, mixed number or decimal to at least 1dp. eg from \(3x = 7\) accept \(\dfrac{7}{3}\) or \(2\dfrac{1}{3}\) or 2.3 or better Ignore any attempt to convert a correct ft fraction | M1A0A1ft |
| Embedded answer | M1A1A0 |
7
Complete the number machine. [1 mark]
Complete the number machine. [1 mark]
When \(\quad x = 5 \quad y = 13\)
and
when \(\quad x = 10 \quad y = 28\)
Complete the number machine. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 40 in correct position in number machine | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \(+\,11\) in correct position in number machine | B1 | oe operation to reach 18 eg \(-\,-11\) or \(\times\,\dfrac{18}{7}\) |
| Answer | Mark | Comments |
|---|---|---|
| 3 and 2 in correct positions in number machine | B2 | B1 correct operations for input 5, output 13 or correct operations for input 10, output 28 |
| B1 may be awarded for correct work, with no or incorrect answer, even if this is seen amongst multiple attempts | |
| Examples of correct operations for input 5, output 13 include \(\times\,2.6\) and \(-\,0\) or \(\times\,4\) and \(-\,7\) or \(\times\,5\) and \(-\,12\) | B1 |
| Examples of correct operations for input 10, output 28 include \(\times\,2.8\) and \(-\,0\) or \(\times\,4\) and \(-\,12\) or \(\times\,5\) and \(-\,22\) | B1 |
15 Here are two number machines.
The outputs are the same.
Work out the value of \(a\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(30 \div 6 + 18\) or \(5 + 18\) | M1 | oe |
| 23 | A1 | 23 may be in either output oval |
| \(3 \times 30 - a =\) their 23 or \(90 - a =\) their 23 or \(3 \times 30 -\) their 23 \((= a)\) or \(90 -\) their 23 \((= a)\) | M1 | oe |
| 67 | A1ft | ft \(90 -\) their 23 |
| Answer 67 | M1A1M1A1 |
| \(90 - 67 = 23\) shown in working without 67 on answer line | M1A1M1A0 |
| For the ft marks allow decimal or fractional answers eg \(30 \div 6 + 18 = 1.25\) \(90 - 1.25 = 88.75\) | M1A0M1A1ft |
| \(\dfrac{23 + a}{3} = 30\) | M1A1M1 |
| \(23 + a \div 3 = 30\) | M1A1M0 |
| Allow their 23 to come from working or the first output oval eg 12 written in the first output oval, then answer 78 | M0A0M1A1ft |
13 Multiplying \(y\) by 6 gives the same result as adding 15 to \(y\)
| Answer | Mark | Comments |
|---|---|---|
| \(6y = y + 15\) | B2 | correct single equation with \(6y\) and \(y + 15\) eg \(15 + y = y \times 6\) B1 \(6y\) or \(y + 15\) or rearranged equation eg \(6y - 15 = y\) or \(5y = 15\) but not \(y = 3\) only |
| B1 may be awarded for a correct term even if this is seen amongst multiple attempts or embedded in an incorrect equation or incorrect term eg \(6y + 15\) or \(6y + 15y\) or \(6(y + 15)\) | |
| Allow any variable for B1 but must be consistent for B2 | |
| Allow unprocessed terms for B1 or B2 eg \(6 \times y\) or \(y6\) | |
| \(6y = y + 15\) seen, but then correctly simplified or solved | B2 |
| \(6y = y + 15\) seen, but then incorrectly simplified or solved | B1 |
| \(6y = 18\) or \(y + 15 = 18\) or both (unless combined to a single equation) | B1 |
| No work worth B2 or B1 and answer \(y = 3\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: substitutes \(y = 4\) into both sides | ||
| \((6y =)\ 24\) and \((y + 15 =)\ 19\) | B1ft | oe eg \(4 \times 6 = 24\) and \(4 + 15 = 19\) correct or ft their equation if their equation has a term in \(y\) on each side |
| Alternative method 2: solves equation | ||
| \((y =)\ 3\) | B1ft | oe eg \(3 \times 6 = 18\) and \(3 + 15 = 18\) correct or ft their equation if their equation has a term in \(y\) on each side |
| Allow any variable | |
| Only allow \((y =)\ 3\) seen in (a) if referenced in (b) and not contradicted | B1 |
| For Alt 1, accept substituting into one side and then equating and solving the other eg \(4 \times 6 = 24\) and \(24 - 15 = 9\) | B1 |
4
| Answer | Mark | Comments |
|---|---|---|
| 18 | B1 |
| Embedded answer with no or incorrect answer eg \(\dfrac{18}{3} = 6\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(2x = 27 - 3\) or \(2x = 24\) or \(\dfrac{27 - 3}{2}\) or \(\dfrac{24}{2}\) or \(x + 1.5 = 13.5\) | M1 | oe eg \(2x + 3 = 27\) then \(-3\) from both sides or \(-2x = 3 - 27\) or \(-2x = -24\) |
| 12 | A1 |
| Embedded answer with no or incorrect answer eg \(2 \times 12 + 3 = 27\) without 12 selected or with incorrect answer | M1A0 |
| Trial and improvement with answer 12 | M1A1 |
| Trial and improvement with no answer or answer other than 12 | M0A0 |
17 Solve \(\quad \dfrac{x + 8}{2} + \dfrac{9 - x}{5} = 4\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – multiplies through by 10 or common denominator of 10 | ||
| \(5(x + 8) + 2(9 - x)\) or \(5x + 40 + 18 - 2x\) | M1 | oe numerator on the left-hand side if written as a fraction allow one error or omission in the expansion if brackets not seen eg \(5x + 18 - 2x\) |
| \(3x + 58\) | A1 | may be implied by eg \(3x + 18 = 0\) or \(3x = -18\) |
| their \((3x + 58) = 4 \times\) (their 10) or their \((3x + 58) = 40\) or \(3x + 18 = 0\) or \(3x = -18\) | M1 | oe allow an unsimplified expression for their \((3x + 58)\) equation may be implied by answer |
| \(-6\) | A1ft | ft M1A0M1 |
| Alternative method 2 – collects terms with fractions | ||
| \(\dfrac{x}{2} + 4 + \dfrac{9}{5} - \dfrac{x}{5}\) | M1 | oe eg \(0.5x + 4 + 1.8 - 0.2x\) allow one error |
| \(\dfrac{3}{10}x + \dfrac{29}{5}\) | A1 | oe eg \(0.3x + 5.8\) |
| \(\dfrac{3}{10}x = \dfrac{20}{5} - \dfrac{29}{5}\) or \(\dfrac{3}{10}x = -\dfrac{9}{5}\) | M1 | oe eg \(0.3x = -1.8\) terms must be collected |
| \(-6\) | A1ft | ft M1A0M1 |
| Accept decimal answers for follow through correct to 1 dp or better | |
| Apply the principles of alt 1 for any use of other common denominators eg common denominator of 20 (or multiplication through by 20) | |
| \(10(x + 8) + 4(9 - x) = 6x + 116\) | M1A1 |
| \(6x + 116 = 80 \qquad x = -6\) | M1A1 |
| An incorrect simplification of \(5x + 40 + 18 - 2x\) may still gain the third and fourth marks | |
| eg \(5x + 40 + 18 - 2x = 3x + 68\) followed by \(3x + 68 = 40\) and \(x = -\dfrac{28}{3}\) | M1A0M1A1ft |
| eg \(5x + 40 + 18 - 2x = 2x + 68\) followed by \(2x + 68 = 40\) and \(x = -14\) | M1A0M1A1ft |
| An incorrect denominator may still gain the third and fourth marks \(\dfrac{5x + 40 + 18 - 2x}{7}\) followed by \(5x + 40 + 18 - 2x = 28\) and \(x = -10\) | M1A0M1A1ft |
| Denominator not processed \(3x + 58 = 4\) followed by \(3x = -54\) and \(x = -18\) | M1A1M0A0 |
| \((x + 8) + (9 - x) = 40\) | M0A0M1A0 |
| Two errors in the expansion but with brackets seen may go on to get the third and fourth marks \(5(x + 8) + 2(9 - x) = 5x + 8 + 18 - x\) | 1st M1A0 |
| Two errors in the expansion and no brackets seen, no follow through allowed \(5x + 8 + 18 - x\) followed by \(4x + 26 = 40\) and \(x = \dfrac{14}{4}\) | M0A0M1A0 |
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 \(W\) |
3 Solve \(\quad 5x + 11 = 3x + 19\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(5x - 3x\) or \(2x\) or \(3x - 5x\) or \(-2x\) or \(19 - 11\) or 8 or \(11 - 19\) or \(-8\) | M1 | |
| 4 | A1 |
| Answer 4 with no working or no incorrect working | M1A1 |
| Embedded answer eg \(5 \times 4 + 11 = 3 \times 4 + 19\) | M1A0 |
1
| Answer | Mark | Comments |
|---|---|---|
| 3 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 43 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 32 | B1 |
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 |
15
| Answer | Mark | Comments |
|---|---|---|
| \(11x - 6x\) or \(6x - 11x\) or \(\pm 5x\) or \((+)1 + 3\) or \(-3 - 1\) or \(\pm 4\) | M1 | oe terms in \(x\) or constant terms collected |
| \(5x = 4\) or \(-5x = -4\) | A1 | may be implied eg \(4 \div 5\) or \(-4 \div -5\) or \(\dfrac{-4}{-5}\) |
| \(\dfrac{4}{5}\) or 0.8 | A1ft | oe ft any equation of the form \(5x = a\) or \(-5x = a\) or \(bx = 4\) or \(bx = -4\) |
| Ignore attempt to convert or simplify after correct answer seen | |
| Trial and improvement scores 3 or 0 | |
| \(5x - 4\ (= 0)\) with no further work | M1A0A0 |
| \(\dfrac{4}{5}\) and \(5x = 4\) on answer line | M1A1A1 |
| Embedded answer eg \(11 \times 0.8 - 3 = 6 \times 0.8 + 1\) | M1A1A0 |
| ft answers must be exact or rounded to 2 dp or better eg \(17x = 4\), answer \(\dfrac{4}{17}\) eg \(17x = -4\), answer \(-0.24\) | M1A0A1ft M1A0A1ft |
| \(5x + 4\) or \(5x + 4 = 0\) or \(17x - 4\) or \(17x - 4 = 0\) etc with no further work | M1 |
| \(\pm 5x\) or \(\pm 4\) must not have come from incorrect working |
| Answer | Mark | Comments |
|---|---|---|
| \(2x = 14 \times 5\) or \(2x = 70\) or \(\dfrac{x}{5} = 14 \div 2\) or \(\dfrac{x}{5} = 7\) or \(14 \times 5 \div 2\) or \(70 \div 2\) | M1 | oe eg \(14 \div 0.4\) |
| 35 | A1 |
| Trial and improvement scores 2 or 0 | |
| Embedded answer eg \(\dfrac{2 \times 35}{5}\) | M1A0 |
| \(\dfrac{2x}{5} = \dfrac{14 \times 5}{5}\) | M1 |
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
7
| Answer | Mark | Comments |
|---|---|---|
| 12 | B1 |
| Answer \(12 - 12 = 0\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| 0 | B1 |
| \(\dfrac{0}{7}\) | B0 |
| Answer \(7 \times 0 = 0\) | B0 |
28 Solve \(5(2x - 1) = 6x + 9\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(10x - 5\) | M1 | may be seen in a grid |
| their \(10x - 6x = 9 +\) their 5 or \(4x = 14\) or \(14 \div 4\) or \(7 \div 2\) | M1 | oe eg their \(-5 - 9 = 6x -\) their \(10x\) or \(4x - 14 = 0\) collecting two terms in \(x\) and two constant terms correctly |
| \(\dfrac{14}{4}\) or \(3\dfrac{2}{4}\) or \(\dfrac{7}{2}\) or \(3\dfrac{1}{2}\) or 3.5 | A1ft | oe ft M1M0 or M0M1 with exactly one error |
| Alternative method 2 | ||
| \(\dfrac{6x}{5} + \dfrac{9}{5}\) | M1 | oe two terms eg \(1.2x + 1.8\) |
| \(2x -\) their \(\dfrac{6x}{5} =\) their \(\dfrac{9}{5} + 1\) or \(\dfrac{4x}{5} = \dfrac{14}{5}\) | M1 | oe eg \(-1 -\) their \(\dfrac{9}{5} =\) their \(\dfrac{6x}{5} - 2x\) or \(\dfrac{4x}{5} - \dfrac{14}{5} = 0\) collecting two terms in \(x\) and two constant terms correctly |
| \(\dfrac{14}{4}\) or \(3\dfrac{2}{4}\) or \(\dfrac{7}{2}\) or \(3\dfrac{1}{2}\) or 3.5 | A1ft | oe ft M1M0 or M0M1 with exactly one error |
| Ignore simplification or conversion if correct answer seen | |
| Correct answer from trial and improvement | M1M1A1 |
| Correct equation with terms collected or division with no or incorrect answer | M1M1A0 |
| Embedded 3.5 with no or incorrect answer | M1M1A0 |
| \(10x - 5 = 6x + 9\) \(10x - 6x = 9 - 5\) \(x = 1\) (exactly one error in line 2) | M1 M0 A1ft |
| \(7x - 5 = 6x + 9\) \(7x - 6x = 9 + 5\) \(x = 14\) (exactly one error in line 1) | M0 M1 A1ft |
| \(10x - 5 = 6x + 9\) \(10x + 6x = 9 - 5\) \(x = \dfrac{4}{16}\) (two errors in line 2) | M1 M0 A0ft |
| \(10x - 1 = 6x + 9\) \(10x - 6x = 9 + 1\) \(x = 3\) (exactly one error in line 1 but answer does not ft) | M0 M1 A0ft |
| \(7x - 6 = 6x + 9\) \(7x - 6x = 9 + 6\) \(x = 15\) (two errors in line 1) | M0 M1 A0ft |
| \(10x + 4 = 6x + 9\) \(10x - 6x = 9 + 4\) \(x = 3.25\) (neither M mark scored) | M0 M0 A0ft |
| \(10x - 5 = 30x + 45\) | M1M0A0ft |
| Any ft answer must be rounded or truncated to 1 dp or better | |
| The last two marks can be implied without the collection of terms seen eg \(10x - 1 = 6x + 9\) and \(x = 2.5\) | M0M1A1ft |
| Collecting terms before the bracket has been expanded | M0M0A0ft |
26 Solve \(\dfrac{2w}{15} = \dfrac{4}{5}\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2w = \dfrac{4}{5} \times 15\) or \(2w = \dfrac{60}{5}\) or \(2w = 12\) or \(\dfrac{2w}{15} = \dfrac{12}{15}\) or \(\dfrac{w}{3} = \dfrac{2}{1}\) or \(\dfrac{w}{2} = \dfrac{3}{1}\) or \(\dfrac{w}{15} = \dfrac{4}{5} \div 2\) or \(\dfrac{w}{15} = \dfrac{2}{5}\) or \(2w \times 5 = 4 \times 15\) or \(10w = 60\) or \(\dfrac{4}{5} \div \dfrac{2}{15}\) | M1 | oe in the form \(aw = n\) where \(a\) is an integer and \(n\) is an integer, fraction or decimal oe in the form \(\dfrac{bw}{x} = \dfrac{c}{x}\) where \(x\) is a common denominator oe calculation |
| 6 | A1 |
| Embedded answer 6 eg \(\dfrac{2 \times 6}{15} = \dfrac{4}{5}\) | M1A0 |
13 Solve \(\quad \dfrac{2w}{15} = \dfrac{4}{5}\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2w = \dfrac{4}{5} \times 15\) or \(2w = \dfrac{60}{5}\) or \(2w = 12\) or \(\dfrac{2w}{15} = \dfrac{12}{15}\) or \(\dfrac{w}{3} = \dfrac{2}{1}\) or \(\dfrac{w}{2} = \dfrac{3}{1}\) or \(\dfrac{w}{15} = \dfrac{4}{5} \div 2\) or \(\dfrac{w}{15} = \dfrac{2}{5}\) or \(2w \times 5 = 4 \times 15\) or \(10w = 60\) or \(\dfrac{4}{5} \div \dfrac{2}{15}\) | M1 | oe in the form \(aw = n\) where \(a\) is an integer and \(n\) is an integer, fraction or decimal oe in the form \(\dfrac{bw}{x} = \dfrac{c}{x}\) where \(x\) is a common denominator oe calculation |
| 6 | A1 |
| Embedded answer 6 eg \(\dfrac{2 \times 6}{15} = \dfrac{4}{5}\) | M1A0 |
5 Solve \(\quad 5(2x - 1) = 6x + 9\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(10x - 5\) | M1 | may be seen in a grid |
| their \(10x - 6x = 9 +\) their 5 or \(4x = 14\) or \(14 \div 4\) or \(7 \div 2\) | M1 | oe eg their \(-5 - 9 = 6x -\) their \(10x\) or \(4x - 14 = 0\) collecting two terms in \(x\) and two constant terms correctly |
| \(\dfrac{14}{4}\) or \(3\dfrac{2}{4}\) or \(\dfrac{7}{2}\) or \(3\dfrac{1}{2}\) or 3.5 | A1ft | oe ft M1M0 or M0M1 with exactly one error |
| Alternative method 2 | ||
| \(\dfrac{6x}{5} + \dfrac{9}{5}\) | M1 | oe two terms eg \(1.2x + 1.8\) |
| \(2x -\) their \(\dfrac{6x}{5} =\) their \(\dfrac{9}{5} + 1\) or \(\dfrac{4x}{5} = \dfrac{14}{5}\) | M1 | oe eg \(-1 -\) their \(\dfrac{9}{5} =\) their \(\dfrac{6x}{5} - 2x\) or \(\dfrac{4x}{5} - \dfrac{14}{5} = 0\) collecting two terms in \(x\) and two constant terms correctly |
| \(\dfrac{14}{4}\) or \(3\dfrac{2}{4}\) or \(\dfrac{7}{2}\) or \(3\dfrac{1}{2}\) or 3.5 | A1ft | oe ft M1M0 or M0M1 with exactly one error |
| Ignore simplification or conversion if correct answer seen | |
| Correct answer from trial and improvement | M1M1A1 |
| Correct equation with terms collected or division with no or incorrect answer | M1M1A0 |
| Embedded 3.5 with no or incorrect answer | M1M1A0 |
| \(10x - 5 = 6x + 9\) \(10x - 6x = 9 - 5\) \(x = 1\) (exactly one error in line 2) | M1 M0 A1ft |
| \(7x - 5 = 6x + 9\) \(7x - 6x = 9 + 5\) \(x = 14\) (exactly one error in line 1) | M0 M1 A1ft |
| \(10x - 5 = 6x + 9\) \(10x + 6x = 9 - 5\) \(x = \dfrac{4}{16}\) (two errors in line 2) | M1 M0 A0ft |
| \(10x - 1 = 6x + 9\) \(10x - 6x = 9 + 1\) \(x = 3\) (exactly one error in line 1 but answer does not ft) | M0 M1 A0ft |
| \(7x - 6 = 6x + 9\) \(7x - 6x = 9 + 6\) \(x = 15\) (two errors in line 1) | M0 M1 A0ft |
| \(10x + 4 = 6x + 9\) \(10x - 6x = 9 + 4\) \(x = 3.25\) (neither M mark scored) | M0 M0 A0ft |
| \(10x - 5 = 30x + 45\) | M1M0A0ft |
| Any ft answer must be rounded or truncated to 1 dp or better | |
| The last two marks can be implied without the collection of terms seen | |
| eg \(10x - 1 = 6x + 9\) and \(x = 2.5\) | M0M1A1ft |
| Collecting terms before the bracket has been expanded | M0M0A0ft |
18 Solve \(\quad 10x - 3 = 21\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(10x = 21 + 3\) or \(10x = 24\) or \((21 + 3) \div 10\) or \(24 \div 10\) | M1 | oe eg \(-10x = -3 - 21\) |
| 2.4 | A1 | oe eg \(\dfrac{24}{10}\) or \(\dfrac{12}{5}\) or \(2\dfrac{4}{10}\) or \(2\dfrac{2}{5}\) SC1 1.8 oe |
| \(10x - 3 + 3 = 21 + 3\) | M1 |
| \(10x - 3 = 21 + 3\) or \(10x - 3 + 3 = 21\) unless recovered | M0 |
| \(10x \div 10 - 3 \div 10 = 21 \div 10\) | M1 |
| \(10x \div 10 - 3 = 21 \div 10\) unless recovered | M0 |
| Embedded answer eg \(10 \times 2.4 - 3 = 21\) with no or incorrect answer | M1A0 |
5 Solve \(\qquad 10x = 62.4 - 3x\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(10x + 3x\) or \(13x\) or \(-3x - 10x\) or \(-13x\) | M1 | may be implied eg \(62.4 \div 13\) or \(-62.4 \div -13\) |
| 4.8 or \(4\dfrac{4}{5}\) or \(\dfrac{24}{5}\) | A1 | oe eg \(\dfrac{624}{130}\) |
| \(\dfrac{-24}{-5}\) | M1A0 |
| Correct answer embedded eg \(10 \times 4.8 = 62.4 - 3 \times 4.8\) | M1A0 |
| Ignore conversion attempt after correct answer seen |
1 Solve \(\quad 4 + x = 12\)
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(x = 8\) | B1 |
17 Here are two rectangles.
Not drawn accurately
The area of the shaded rectangle is \(\dfrac{1}{4}\) the area of the large rectangle.
Work out the value of \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2(12 - x)\) or \(24 - 2x\) or \(12(x + 2)\) or \(12x + 24\) or \(12x + 2x\) or \(14x\) or \(2x + x^2 + x(12 - x)\) or \(2x + x^2 + 12x - x^2\) | M1 | oe correct area of small rectangle or large rectangle or unshaded section may be seen on diagram |
| \(\dfrac{12(x + 2)}{4} = 2(12 - x)\) or \(12x + 2x = 6(12 - x)\) | M1dep | oe equation eg \(3(x + 2) = 2(12 - x)\) \(3x + 6 = 24 - 2x\) \(12(x + 2) = 8(12 - x)\) \(12x + 24 = 96 - 8x\) |
| \(3x + 2x = 24 - 6\) or \(14x + 6x = 72\) | M1dep | oe equation with brackets expanded and terms collected eg \(5x = 18\) \(12x + 8x = 96 - 24\) \(20x = 72\) |
| \(\dfrac{18}{5}\) or \(3\dfrac{3}{5}\) or 3.6 | A1 | oe |
| \(3x + 6\) | M1 |
| Trial and improvement with \(x = 3.6\) chosen | M1M1M1A1 |
| Trial and improvement without \(x = 3.6\) chosen | M0M0M0A0 |
14
| Answer | Mark | Comments |
|---|---|---|
| \(6x = 13 + 11\) or \(6x = 24\) or \(\dfrac{24}{6}\) | M1 | oe eg \(-6x = -13 - 11\) or \(-6x = -24\) or \(\dfrac{-24}{-6}\) |
| 4 | A1 |
| Embedded answer, eg \(6 \times 4 - 11 = 13\) | M1A0 |
| 24 with no other working | M0A0 |
| Flow chart method, if 4 not given as the answer. \(x \to \times 6 \to -11 \to 13\) and \(13 \to +11 \to \div 6 \to x\) | M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| \((2 \times 4a =)\ 8a\) | B1 | |
| \(\left(\dfrac{15a}{3} =\right) 5a\) | B1 | |
| \(13a + 2\) | B1ft | ft B1B0 or B0B1 for their \(8a +\) their \(5a + 9 - 7\) is in the form \(pa + q\) do not award with further incorrect work eg \(13a + 2 = 15a\) |
| \(13a + c\) could come from incorrect working eg \(8a + 4 + 9 + 5a - 7 = 13a + 16\) (their \(8a\) is \(8a + 4\)) eg \(8a + 4 + 9 + 5a - 7 = 13a + 6\) (their \(8a\) is \(8a + 4\)) eg \(8a + 9 + 5a - 7 = 13a + 16\) eg \(13a + 16\) (no other working) | B0B1B0ft B0B1B1ft B1B1B0ft B1B1B0ft |
| \(6a + 9 + 5a - 7 = 11a + 2\) | B0B1B1ft |
| \(8a + 9 + 12a - 7 = 20a + 2\) | B1B0B1ft |
| \(8a + 9 + 5 - 7 = 8a + 7\) | B1B0B1ft |
| \(8a + \dfrac{15a}{3} + 7\) | B1B0B0ft |
| \(6a + 9 + 12a - 7 = 18a + 2\) | B0B0B0ft |
| \(6a + 5a + 16 = 11a + 16\) | B0B1B0ft |
5
| Answer | Mark | Comments |
|---|---|---|
| 8 | B1 |
| \(56 \div 7 = 8\) | B1 |
| Answer of \(\;\times 8\) (unless recovered) | B0 |
| Answer of \(\;8x\) (unless recovered) | B0 |
| Award the mark for an embedded answer only if the answer is selected eg1 \(7 \times 8 = 56\) with no answer or with incorrect answer eg2 \(7 \times 8 = 56\) (with 8 circled) with no contradictory answer | B0 B1 |
| Answer | Mark | Comments |
|---|---|---|
| 7 | B1 |
| \(25 - 18 = 7\) | B1 |
| \(18 - 25 = 7\) (allow recovery) | B1 |
| Answer of \(\;-7\) (unless recovered) | B0 |
| Answer of \(\;7y\) (unless recovered) | B0 |
| Award the mark for an embedded answer only if the answer is selected eg1 \(25 - 7 = 18\) with no answer or with incorrect answer eg2 \(25 - 7 = 18\) (with 7 circled) with no contradictory answer | B0 B1 |
21 Solve \(\quad 8x + 7 = 2x + 10\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\pm 6x\) or \(\pm 3\) or \(8x - 2x = 10 - 7\) or \(7 - 10 = 2x - 8x\) | M1 | oe terms in \(x\) or constant terms collected |
| \(6x = 3\) or \(-6x = -3\) | A1 | oe implied by correct answer |
| 0.5 or \(\dfrac{1}{2}\) | A1ft | oe eg \(\dfrac{3}{6}\) ft any equation of form \(6x = a\) or \(-6x = a\) or \(bx = 3\) or \(bx = -3\) |
| \(\dfrac{-3}{-6}\) | M1A1A0 |
| Trial and Improvement scores 0 or 3 |
18 Solve \(\quad \dfrac{x + 15}{3} = 2(x + 10)\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(2x + 20\) | M1 | correct expansion |
| \(x + 15 = 6x + 60\) | M1dep | multiplication by 3 |
| \(15 - 60 = 6x - x\) or \(-45 = 5x\) or \(60 - 15 = x - 6x\) or \(45 = -5x\) | M1dep | collects terms |
| \(-9\) | A1 | SC2 \(-3\) from \(2x + 10\) or 1 from \(6x + 10\) |
| Alternative method 2 | ||
| \(2x + 20\) | M1 | correct expansion |
| \(\dfrac{x}{3} + 5 = 2x + 20\) and \(5 - 20 = 2x - \dfrac{x}{3}\) or \(-15 = \dfrac{5x}{3}\) or \(20 - 5 = \dfrac{x}{3} - 2x\) or \(15 = -\dfrac{5x}{3}\) | M1dep | splits the fraction and collects terms |
| \(15 - 60 = 6x - x\) or \(-45 = 5x\) or \(60 - 15 = x - 6x\) or \(45 = -5x\) | M1dep | multiplication by 3 |
| \(-9\) | A1 | SC2 \(-3\) from \(2x + 10\) or 1 from \(6x + 10\) |
| Alternative method 3 | ||
| \(6(x + 10)\) or \(6x + 60\) | M1 | multiplication of rhs by 3 |
| \(x + 15 = 6x + 60\) | M1dep | correct expansion |
| \(15 - 60 = 6x - x\) or \(-45 = 5x\) or \(60 - 15 = x - 6x\) or \(45 = -5x\) | M1dep | collects terms |
| \(-9\) | A1 | SC2 \(-3\) from \(2x + 10\) or 1 from \(6x + 10\) |
8 Here is an identity.
\(\quad a(3x - 10) \equiv 21x + 2b\)
Work out the values of \(a\) and \(b\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(a = 7\) | B2 | B1 \(3ax - 10a\) or \(3ax = 21x\) or \(3ax - 21x = 0\) or \(3a = 21\) or \(3a - 21 = 0\) or \(21 \div 3\) oe or \(-10a = 2b\) oe |
| \(b = -35\) | B1ft | ft \(-5 \times\) their \(a\) where \(a \ne 0\) |
| Ignore collection error if correct expansion seen eg \(3ax - 10a - 21x + 2b = 0\) (should be \(-2b\)) | B1 |
| Ignore incorrect simplification if correct expansion seen eg \(3ax - 10a = -7ax\) | B1 |
| Allow eg \(a \times 3x\) for \(3ax\) | |
| Allow eg \(a3x\) for \(3ax\) | |
| Embedded 7 with \(a = 7\) not stated eg \(7(3x - 10)\) or \(7 \times 3x = 21x\) or \(21 \div 7 = 3\) | B1 |
| Allow B1 even if not subsequently used |
7
| Answer | Mark | Comments |
|---|---|---|
| \(-5\) | B1 |
| \(-5 + 17 = 12\) or \(17 - 5 = 12\) but \(-5\) not selected as answer | B0 |
| Answer | Mark | Comments |
|---|---|---|
| 48 | B1 |
| 48 seen but 12 given as answer | B0 |
| Answer \(\dfrac{48}{4}\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{3}{4}\) or 0.75 | B2 | B1 partial simplification eg \(\dfrac{3m}{4m}\) or \(\dfrac{0.75m}{m}\) or \(\dfrac{9}{12}\) |
| eg \(\dfrac{3m}{4m}\) seen but answer given as \(0.75m\) | B1 |
2 Solve \(\quad 3x = 2x\)
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(x = 0\) | B1 |
2 Solve \(\quad 3x = 6\)
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(x = 2\) | B1 |
25 Towns \(P\), \(Q\) and \(R\) are connected by roads \(PQ\), \(PR\) and \(QR\).
\(PR\) is 10 km longer than \(PQ\).
\(QR\) is twice as long as \(PR\).
The total length of the three roads is 170 km
Not drawn accurately
Work out the length of \(PQ\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – PQ as the unknown | ||
| \(x + 10\) or \(2(x + 10)\) | M1 | any unknown |
| \(x + x + 10 + 2(x + 10) = 170\) | M1dep | oe any consistent unknown \(x\) + their two expressions (with at least one correct) = 170 |
| \(4x + 30 = 170\) | M1dep | oe \(4x = 140\) must be correct |
| 35 | A1 | |
| Alternative method 2 – PR as the unknown | ||
| \(x - 10\) or \(2x\) | M1 | any unknown |
| \(x + x - 10 + 2x = 170\) | M1dep | oe any consistent unknown \(x\) + their two expressions (with at least one correct) = 170 |
| \(4x - 10 = 170\) or \(x = 45\) | M1dep | oe \(4x = 180\) must be correct |
| 35 | A1 | |
| Alternative method 3 – QR as the unknown | ||
| \(\dfrac{x}{2}\) or \(\dfrac{x}{2} - 10\) | M1 | any unknown |
| \(x + \dfrac{x}{2} + \dfrac{x}{2} - 10 = 170\) | M1dep | oe any consistent unknown \(x\) + their two expressions (with at least one correct) = 170 |
| \(2x - 10 = 170\) or \(x = 90\) | M1dep | oe \(2x = 180\) must be correct |
| 35 | A1 | |
| Alternative method 4 – trial and improvement with addition of three lengths | ||
| A correctly evaluated trial with a difference of 10 (km) between the two shorter lengths and the longest length twice the length of the middle length | M1 | may be seen as a subtraction of three numbers from 170 |
| A different correctly evaluated trial with a difference of 10 (km) between the two shorter lengths and the longest length twice the length of the middle length | M1dep | may be seen as a subtraction of three numbers from 170 |
| 35, 45 and 90 | A1 | |
| 35 | A1 | |
| Alternative method 5 – trial and improvement with subtraction from 170 | ||
| A correctly evaluated trial of two lengths subtracted from 170 with a difference of 10 (km) between the two lengths or one length twice the length of the other | M1 | |
| A different correctly evaluated trial of two lengths subtracted from 170 with a difference of 10 (km) between the two lengths or one length twice the length of the other | M1dep | |
| 35, 45 and 90 | A1 | |
| 35 | A1 | |
| If the student attempts more than one method, mark each method and award the highest mark | |
| Alt 1 \(\quad PQ + PQ + 10 + 2(PQ + 10) = 170\) | M1M1 |
| Alt 1 \(\quad PQ + PQ + 10 + 2PR = 170\) | M1 |
| Alt 2 \(\quad x\), \(x + 10\) and \(2x\) seen on diagram, \(4x + 10 = 170\) | M1M1M0A0 |
| Alt 4 \(\quad 35 + 45 + 90\) with no choice made | M1M1A1A0 |
| Alt 4 \(\quad 170 - 30 - 40 - 80 = 20\) | M1 |
| Alt 4 \(\quad 170 - 30 - 40 - 60 = 40\) incorrect number is doubled | M0 |
| Alt 5 \(\quad 170 - 30 - 60 = 80\) | M1 |
17 In a bag there are 10p coins, 20p coins and 50p coins.
There are two fewer 20p coins than 10p coins.
There are five more 50p coins than 10p coins.
| Coin | Number of coins |
|---|---|
| 10p | \(n\) |
| 20p | \(n - 2\) |
| 50p |
Work out the total value of the 20p coins. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(n + 5\) or \(5 + n\) | B1 | oe eg \(N - 2 + 7\) |
| Letters other than \(n\) or \(N\) eg \(x + 5\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(n + n - 2 +\) their \((n + 5)\) or \(3n + 3\) | M1 | condone any letter ft their algebraic expression in (a) |
| \(3n + 3 = 60\) or \((n =)\ 19\) or \((n - 2 =)\ 17\) | M1dep | ft their algebraic expression in (a) correct ft equation with terms on LHS collected 19 10p coins or 17 20p coins or 19, 17, 24 chosen implies M2 |
| (their \(19 - 2) \times 0.2\) or their \(17 \times 0.2\) or 3.4 or (their \(19 - 2) \times 20\) or their \(17 \times 20\) or 340 | M1dep | ft their algebraic expression in (a) 3.4 or 340 implies M3 |
| 3.40 | A1 | condone 3.40p SC2 answer 17 |
| Allow a restart in this part ie answer £3.40 scores full marks | |
| Working may be seen by the table | |
| Answer 340p | M1M1M1A0 |
| £3.40 with answer eg £17.30 (total of all coins) | M1M1M1A0 |
| Only follow through their algebraic expression from (a) if an expression and / or equation for the total number of coins is used in this part | |
| Award the M mark(s) for a correct ft expression or equation even if not subsequently used | |
| The solution to an equation derived from an incorrect expression in (a) can score the first three marks eg answer in (a) \(n - 5\) then working in (b) \(\quad n + n - 2 + n - 5 = 60 \quad n = [22, 23]\) \(([22, 23] - 2) \times 0.2 = [4, 4.20]\) | M1M1 M1A0 |
9 Towns \(P\), \(Q\) and \(R\) are connected by roads \(PQ\), \(PR\) and \(QR\).
\(PR\) is 10 km longer than \(PQ\).
\(QR\) is twice as long as \(PR\).
The total length of the three roads is 170 km
Not drawn accurately
Work out the length of \(PQ\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – PQ as the unknown | ||
| \(x + 10\) or \(2(x + 10)\) | M1 | any unknown |
| \(x + x + 10 + 2(x + 10) = 170\) | M1dep | oe any consistent unknown \(x\) + their two expressions (with at least one correct) = 170 |
| \(4x + 30 = 170\) | M1dep | oe \(4x = 140\) must be correct |
| 35 | A1 | |
| Alternative method 2 – PR as the unknown | ||
| \(x - 10\) or \(2x\) | M1 | any unknown |
| \(x + x - 10 + 2x = 170\) | M1dep | oe any consistent unknown \(x\) + their two expressions (with at least one correct) = 170 |
| \(4x - 10 = 170\) or \(x = 45\) | M1dep | oe \(4x = 180\) must be correct |
| 35 | A1 | |
| Alternative method 3 – QR as the unknown | ||
| \(\dfrac{x}{2}\) or \(\dfrac{x}{2} - 10\) | M1 | any unknown |
| \(x + \dfrac{x}{2} + \dfrac{x}{2} - 10 = 170\) | M1dep | oe any consistent unknown \(x\) + their two expressions (with at least one correct) = 170 |
| \(2x - 10 = 170\) or \(x = 90\) | M1dep | oe \(2x = 180\) must be correct |
| 35 | A1 | |
| Alternative method 4 – trial and improvement with addition of three lengths | ||
| A correctly evaluated trial witha difference of 10 (km) between the two shorter lengthsandthe longest length twice the length of the middle length | M1 | may be seen as a subtraction of three numbers from 170 |
| A different correctly evaluated trial witha difference of 10 (km) between the two shorter lengthsandthe longest length twice the length of the middle length | M1dep | may be seen as a subtraction of three numbers from 170 |
| 35, 45 and 90 | A1 | |
| 35 | A1 | |
| Alternative method 5 – trial and improvement with subtraction from 170 | ||
| A correctly evaluated trial of two lengths subtracted from 170 witha difference of 10 (km) between the two lengthsorone length twice the length of the other | M1 | |
| A different correctly evaluated trial of two lengths subtracted from 170 witha difference of 10 (km) between the two lengthsorone length twice the length of the other | M1dep | |
| 35, 45 and 90 | A1 | |
| 35 | A1 | |
| If the student attempts more than one method, mark each method and award the highest mark | |
| Alt 1 \(\ PQ + PQ + 10 + 2(PQ + 10) = 170\) | M1M1 |
| Alt 1 \(\ PQ + PQ + 10 + 2PR = 170\) | M1 |
| Alt 2 \(\ x,\ x + 10\) and \(2x\) seen on diagram, \(4x + 10 = 170\) | M1M1M0A0 |
| Alt 4 35 + 45 + 90 with no choice made | M1M1A1A0 |
| Alt 4 170 – 30 – 40 – 80 = 20 | M1 |
| Alt 4 170 – 30 – 40 – 60 = 40 incorrect number is doubled | M0 |
| Alt 5 170 – 30 – 60 = 80 | M1 |
2 Solve \(\quad x - 8 = 5\)
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(x = 13\) | B1 |
2 Solve \(\quad 4x = 8\)
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(x = 2\) | B1 |
25 This triangle is equilateral.
Not drawn accurately
Is the perimeter of the triangle greater than one metre?
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4x + 5 = 6x - 10\) or \(4x + 5 = 10(x - 4)\) or \(6x - 10 = 10(x - 4)\) | M1 | oe eg \(4x + 5 + 6x - 10 = 2 \times 10(x - 4)\) condone \(10x - 4\) for \(10(x - 4)\) |
| \(4x - 6x = -10 - 5\) or \(-2x = -15\) or \(4x - 10x = -40 - 5\) or \(-6x = -45\) or \(6x - 10x = -40 + 10\) or \(-4x = -30\) | M1dep | oe collection of terms eg \(4x + 6x - 20x = -80 - 5 + 10\) or \(-10x = -75\) condone \(10x - 4\) for \(10(x - 4)\) eg \(4x - 10x = -4 - 5\) or \(6x - 10x = -4 + 10\) |
| \((x =)\) 7.5 | A1 | oe may be implied by (side length =) 35 or (perimeter =) 105 |
| \((6 \times \text{their } 7.5 - 10) \times 3\) or \((4 \times \text{their } 7.5 + 5) \times 3\) or \(10 \times (\text{their } 7.5 - 4) \times 3\) or \(35 \times 3\) or \(6 \times \text{their } 7.5 - 10 + 4 \times \text{their } 7.5 + 5 + 10 \times (\text{their } 7.5 - 4)\) or \(20 \times \text{their } 7.5 - 45\) or 105 | M1dep | oe dep on M1M1 condone \(10x - 4\) for \(10(x - 4)\) must show working if M1M1A0 |
| 105 and Yes | A1 | oe eg 1.05 and Yes |
| \(4x + 5 = 6x - 10 = 10(x - 4)\) | M1 |
| Condone \(10x - 4\) for \(10(x - 4)\) for up to M3 |
22 Solve \(\quad \dfrac{x}{x + 4} + \dfrac{7}{x - 2} = 1\)
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Any two of \(x(x - 2)\) and \(7(x + 4)\) and \((x - 2)(x + 4)\) | M1 | oe \(x(x - 2)\) and \(7(x + 4)\) cannot be denominators |
| correct equation including \(x(x - 2)\) and \(7(x + 4)\) and \((x - 2)(x + 4)\) | M1dep | |
| \(x^2 - 2x + 7x + 28 = x^2 + 4x - 2x - 8\) | M1dep | oe all brackets must be expanded |
| \(-12\) | A1 | |
| Alternative method 2 | ||
| \(\dfrac{x(x - 2)}{x + 4} + 7 = x - 2\) | M1 | |
| \(\dfrac{x(x - 2)}{x + 4} = x - 9\) or \(x(x - 2) = (x - 9)(x + 4)\) | M1dep | |
| \(x^2 - 2x = x^2 - 9x + 4x - 36\) | M1dep | oe all brackets must be expanded |
| \(-12\) | A1 | |
| Alternative method 3 | ||
| \(x + \dfrac{7(x + 4)}{x - 2} = x + 4\) | M1 | |
| \(\dfrac{7(x + 4)}{x - 2} = 4\) or \(7(x + 4) = 4(x - 2)\) | M1dep | |
| \(7x + 28 = 4x - 8\) | M1dep | oe all brackets must be expanded |
| \(-12\) | A1 | |
In Alt 1, do not allow \(x \times x - 2\) or \(7 \times x + 4\) unless recovered
20 Solve \(\quad 3x - 8 = 19\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(3x = 19 + 8\) or \(3x = 27\) or \((19 + 8) \div 3\) or \(\dfrac{27}{3}\) | M1 | accept in ‘flow chart’ eg \((x \rightarrow) \times 3 \rightarrow -8 \rightarrow 19\) and \(\leftarrow \div 3 \leftarrow +8 \leftarrow 19\) enough for M1 |
| 9 | A1 | |
| Alternative method 2 | ||
| \(x - \dfrac{8}{3} = \dfrac{19}{3}\) | M1 | |
| 9 | A1 | |
| \(3 \times 9 - 8\ (= 19)\) | M1A0 |
10 This triangle is equilateral.
Not drawn accurately
Is the perimeter of the triangle greater than one metre?
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4x + 5 = 6x - 10\) or \(4x + 5 = 10(x - 4)\) or \(6x - 10 = 10(x - 4)\) | M1 | oe eg \(4x + 5 + 6x - 10 = 2 \times 10(x - 4)\) condone \(10x - 4\) for \(10(x - 4)\) |
| \(4x - 6x = {-10} - 5\) or \(-2x = {-15}\) or \(4x - 10x = {-40} - 5\) or \(-6x = {-45}\) or \(6x - 10x = {-40} + 10\) or \(-4x = {-30}\) | M1dep | oe collection of terms eg \(4x + 6x - 20x = {-80} - 5 + 10\) or \(-10x = {-75}\) condone \(10x - 4\) for \(10(x - 4)\) eg \(4x - 10x = {-4} - 5\) or \(6x - 10x = {-4} + 10\) |
| (\(x =\)) 7.5 | A1 | oe may be implied by (side length =) 35 or (perimeter =) 105 |
| \((6 \times \text{their } 7.5 - 10) \times 3\) or \((4 \times \text{their } 7.5 + 5) \times 3\) or \(10 \times (\text{their } 7.5 - 4) \times 3\) or \(35 \times 3\) or \(6 \times \text{their } 7.5 - 10 + 4 \times \text{their } 7.5 + 5\) \(+\; 10 \times (\text{their } 7.5 - 4)\) or \(20 \times \text{their } 7.5 - 45\) or 105 | M1dep | oe dep on M1M1 condone \(10x - 4\) for \(10(x - 4)\) must show working if M1M1A0 |
| 105 and Yes | A1 | oe eg 1.05 and Yes |
| \(4x + 5 = 6x - 10 = 10(x - 4)\) | M1 |
| Condone \(10x - 4\) for \(10(x - 4)\) for up to M3 |
5
| Answer | Mark | Comments |
|---|---|---|
| 96 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 72 | B1 |
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 |
2 Solve \(\qquad 3x = 2\)
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(x = \dfrac{2}{3}\) | B1 |
27 Solve \(\qquad 4(3x - 2) = 2x - 5\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(12x - 8\) | M1 | May be seen in a grid |
| their \(12x - 2x = -5 +\) their 8 or \(10x = 3\) or their \(-8 + 5 = 2x -\) their \(12x\) or \(-3 = -10x\) | M1 | Collecting two terms in \(x\) and two constant terms correctly oe eg \(10x - 3 = 0\) |
| 0.3 or \(\dfrac{3}{10}\) | A1ft | ft M1M0 or M0M1 with exactly one error |
| Alternative method 2 | ||
| \(\dfrac{x}{2} - \dfrac{5}{4}\) | M1 | |
| \(3x -\) their \(\dfrac{x}{2}\) = their \(-\dfrac{5}{4} + 2\) or \(\dfrac{5}{2}x = \dfrac{3}{4}\) or \(-2 +\) their \(\dfrac{5}{4}\) = their \(\dfrac{x}{2} - 3x\) or \(-\dfrac{3}{4} = -\dfrac{5}{2}x\) | M1 | Collecting two terms in \(x\) and two constant terms correctly oe eg \(\dfrac{5}{2}x - \dfrac{3}{4} = 0\) |
| 0.3 or \(\dfrac{3}{10}\) | A1ft | ft M1M0 or M0M1 with exactly one error |
| \(12x - 2 = 2x - 5\) \(10x = -3\) \(x = -0.3\) | M0 M1 A1ft |
| \(12x - 8 = 2x - 5\) \(10x = -5\) \(x = \dfrac{-5}{10}\) | M1 M0 A1ft |
| \(12x - 8 = 2x - 5\) \(14x = 3\) \(x = \dfrac{3}{14}\) | M1 M0 A1ft |
| \(12x - 8 = 2x - 5\) \(14x = -13\) \(x = -\dfrac{13}{14}\) (two errors) | M1 M0 A0ft |
| \(12x - 8 = 8x - 20\) | M1M0A0 |
| Any ft answer must be exact or rounded or truncated to at least 2 dp | |
| The last two marks can be implied without the collection of terms seen | |
| eg \(12x - 6 = 2x - 5\) and answer 0.1 | M0M1A1ft |
| Collecting terms before the bracket has been expanded | Zero |
17 Circle the equation which has the solution \(\qquad x = 6\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(x - 3 = \dfrac{x}{2}\) | B1 |
5 Solve \(\quad 4(3x - 2) = 2x - 5\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(12x - 8\) | M1 | May be seen in a grid |
| their \(12x - 2x = -5 +\) their 8 or \(10x = 3\) or their \(-8 + 5 = 2x -\) their \(12x\) or \(-3 = -10x\) | M1 | Collecting two terms in \(x\) and two constant terms correctly oe eg \(10x - 3 = 0\) |
| 0.3 or \(\dfrac{3}{10}\) | A1ft | ft M1M0 or M0M1 with exactly one error |
| Alternative method 2 | ||
| \(\dfrac{x}{2} - \dfrac{5}{4}\) | M1 | |
| \(3x -\) their \(\dfrac{x}{2} =\) their \(-\dfrac{5}{4} + 2\) or \(\dfrac{5}{2}x = \dfrac{3}{4}\) or \(-2 +\) their \(\dfrac{5}{4} =\) their \(\dfrac{x}{2} - 3x\) or \(-\dfrac{3}{4} = -\dfrac{5}{2}x\) | M1 | Collecting two terms in \(x\) and two constant terms correctly oe eg \(\dfrac{5}{2}x - \dfrac{3}{4} = 0\) |
| 0.3 or \(\dfrac{3}{10}\) | A1ft | ft M1M0 or M0M1 with exactly one error |
| \(12x - 2 = 2x - 5\) \(10x = -3\) \(x = -0.3\) | M0 M1 A1ft |
| \(12x - 8 = 2x - 5\) \(10x = -5\) \(x = \dfrac{-5}{10}\) | M1 M0 A1ft |
| \(12x - 8 = 2x - 5\) \(14x = 3\) \(x = \dfrac{3}{14}\) | M1 M0 A1ft |
| \(12x - 8 = 2x - 5\) \(14x = -13\) \(x = -\dfrac{13}{14}\) (two errors) | M1 M0 A0ft |
| \(12x - 8 = 8x - 20\) | M1M0A0 |
| Any ft answer must be exact or rounded or truncated to at least 2 dp | |
| The last two marks can be implied without the collection of terms seen eg \(12x - 6 = 2x - 5\) and answer 0.1 | M0M1A1ft |
| Collecting terms before the bracket has been expanded | Zero |
5
| Answer | Mark | Comments |
|---|---|---|
| 17 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| 9 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| –2 | B1 |
11 Solve \(\qquad 4x - 3 = 14\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4x = 14 + 3\) or \(4x = 17\) or \((14 + 3) \div 4\) or \(17 \div 4\) or \(x - \dfrac{3}{4} = \dfrac{14}{4}\) | M1 | |
| 4.25 or \(\dfrac{17}{4}\) or \(4\dfrac{1}{4}\) | A1 |
| Embedded answer of 4.25 with 4.25 not selected on answer line eg \(4 \times 4.25 - 3 = 14\) with no answer given or answer of 14 or 17 | M1A0 |
| \(14 + 3\) and answer 4.25 | M1A1 |
| \(14 + 3\) only | M0A0 |
| Trial and improvement with answer 4.25 | M1A1 |
| Trial and improvement with no answer or answer other than 4.25 | M0A0 |
| 4.25 or \(\dfrac{17}{4}\) or \(4\dfrac{1}{4}\) seen and then answer 4 given | M1A1 |
| Answer of \(\times 4.25\) | M1A0 |
| \(17 \div 4\) (and no further) | M1A0 |
4 Solve \(\quad x - 3 = 0\)
Circle your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(x = 3\) | B1 |
26
\(8x - 11 = 6x + 3\) [3]
\(x^2 + 15x - 16 = 0\) [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7 | 3 | M2 for \(8x - 6x = 3 + 11\) oe OR M1 for \(2x - 11 = 3\) or for \(2x = k\) or for \(8x = 14 + 6x\) or for \(kx = 14\) M1 for \(x = \dfrac{b}{a}\) FT their \(ax = b\) seen | Embedded 7 scores M2 e.g. \(8 \times 7 - 11 = 6 \times 7 + 3\) [= 45] identified as final working or written on answer line e.g. \(2x = -8\) and answer line \(-4\) with no other working scores M1 and M1FT |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((x + 16)(x - 1)\) | M2 | Condone final bracket missing Accept \(x + 16 = 0\) and \(x - 1 = 0\) for M2 | |
| M1 for \((x + a)(x + b)\) where \(ab = -16\) or \(a + b = 15\) or \(x(x + 16) - [1](x + 16)\) or \(x(x - 1) + 16(x - 1)\) | Allow correctly completed grid for M1 | ||
| \(-16\) and 1 | B1FT | For correct solutions from their quadratic factors (strict FT) If 0 scored, instead award SC1 for answers \(\pm 16\) and \(\pm 1\) | \((x - 16)(x + 1)\) with answers \(-16\) and 1 scores M1B0 |
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 \(v = u + at\) | ||
10
\(a\) and \(b\) are both positive integers.
When \(a\) is 7 the value of \(b\) is 26.
Find another pair of values for \(a\) and \(b\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 7.5 or \(7\dfrac{1}{2}\) oe cao | 1 | Ignore trailing zeros | |
| (ii) 43 | 2 | B1 for \(24 \times 2\) soi 48 | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [a =] 6 [b =] 13 | 3 | M2 for \(26 \div 2\) or \(26 \div 13\) or M1 for \(\dfrac{26}{8 - a}\) oe where \(a = 1, 2, 3, 4, 5\) | M2 may be implied by seeing 2, 13 and 26 linked e.g \((8 - 6) \times 13 = 26\) or factor tree For M1 “8 – a number” must be seen oe may be e.g. \(8 - 4 = k\) then \(k \times 6.5\) or \(8 - 4 = k\) then \(k\) doesn’t go into 26 oe |
If 0 or 1 scored, instead award SC2 for
| |||
9
\(8x - 11 = 6x + 3\) [3]
\(x^2 + 15x - 16 = 0\) [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7 | 3 | M2 for \(8x - 6x = 3 + 11\) oe OR M1 for \(2x - 11 = 3\) or for \(2x = k\) or for \(8x = 14 + 6x\) or for \(kx = 14\) M1 for \(x = \frac{b}{a}\) FT their \(ax = b\) seen | Embedded 7 scores M2 e.g. 8 × 7 – 11 = 6 × 7 + 3 [= 45] identified as final working or written on answer line eg \(2x = -8\) and answer line –4 with no other working scores M1 and M1FT |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((x + 16)(x - 1)\) | M2 | Condone final bracket missing Accept \(x + 16 = 0\) and \(x - 1 = 0\) for M2 | |
| M1 for \((x + a)(x + b)\) where \(ab = -16\) or \(a + b = 15\) or \(x(x + 16) - [1](x + 16)\) or \(x(x - 1) + 16(x - 1)\) | Allow correctly completed grid for M1 | ||
| –16 and 1 | B1FT | For correct solutions from their quadratic factors If 0 scored, instead award SC1 for answers ±16 and ±1 | \((x - 16)(x + 1)\) with answers –16 and 1 scores M1B0 |
7
\(6k - 3 = 177\) [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(a^7\) | 1 | ||
| (ii) \(20b^3\) as final answer | 2 | B1 for \(20b^k\) or \(kb^3\) \((k \ne 0)\) or correct answer seen and spoilt | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 30 as final answer | 2 | M1 for \(6k = 177 + 3\) or better or \(k - \dfrac{3}{6} = \dfrac{177}{6}\) or better or \(6 \times 30 - 3 = 177\) oe | Accept flow diagram |
22 In this question, all lengths are in centimetres.
The diagram shows an isosceles triangle ABC.
AB = AC.
Not to scale
Find the perimeter of the triangle.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 28[cm] with correct working | 6 | Correct working requires evidence of at least M1M1A1M1 or M1M1A1M2 | |
| M1 for \(5x + 4 = 3x + 7\) M1 for \(5x - 3x = 7 - 4\) oe | FT their equation if wrong sides equated Accept only: \(4x - 1 = 5x + 4\) or \(4x - 1 = 3x + 7\) | ||
| A1 for \(x = 1.5\) oe | After M1, \(x = 1.5\) implies M1M1A1 Do not penalise if their value of \(x\) is not subsequently used in their work leading to an algebraic final answer e.g. \(12x + 10\). | ||
| M1 for \(5x + 4 + 3x + 7 + 4x - 1\) soi | implied by \(12x + 10\) | ||
| M1 for substitution of their \(x\) into \(5x + 4\) or \(3x + 7\) or \(4x - 1\) or their \(12x + 10\) | their \(x\) must be > 0 and clearly stated as \(x =\)…. Substitution of their \(x\) into \(5x + 4\), \(3x + 7\) and \(4x - 1\) and then adding implies M1 M1. | ||
| If 0 or 1 scored, instead award SC2 for answer 28 If 0 scored, instead award SC1 for \(x = 1.5\) | Alternative method using trials: In all trials \(x\) must be > 0 M1M1A1 for both \(5x + 4\) and \(3x + 7\) correctly evaluated with \(x = 1.5\) or M1M1 for three correctly evaluated trials of both \(5x + 4\) and \(3x + 7\) with consistent value of \(x\) or M1 for two correctly evaluated trials of both \(5x + 4\) and \(3x + 7\) with consistent value of \(x\) AND M2 dep on at least M1 for their \(x\) substituted into their \(12x + 10\) oe or M1 dep on at least M1 for their \(x\) substituted into \(4x - 1\) M1 dep on previous M1 adding their three lengths | ||
14 Five adults and two children go to a theme park.
The cost of an adult ticket is £6 more than the cost of a child ticket.
The total cost of the seven tickets is £142.
Work out the cost of an adult ticket.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 22 with correct working | 5 | Correct answer from trials scores 5 “Correct working” requires evidence of at least 3 method marks | |
| Finding adult cost (a) directly: M3 for \(5a + 2a - 12 = 142\) or M2 for \(5a + 2(a - 6)\) or M1 for \(a - 6\) M1 for \(7a = 154\) OR M2 for 142 + 12 or M1 for \(6 \times 2\) used, may be implied by use of 12 M1 for their\((142 + 12) \div 7\) | Finding child cost (c) first: M3 for \(5c + 2c + 30 = 142\) or M2 for \(5(c + 6) + 2c\) or M1 for \(c + 6\) M1 for \(7c = 112\) or \(c = 16\) OR M2 for 142 – 30 or M1 for \(6 \times 5\) used, may be implied by use 30 M1 for their \((142 - 30) \div 7\) M1 for their\((142 - 30) \div 7 + 6\) | ||
| Simultaneous Equations M1 for \(5a + 2c = 142\) M1 for \(a - c = 6\) M1 for correct method to equate coefficients of one variable, allow one arithmetic error M1 for correct method to eliminate one variable, allow one arithmetic error | Trials: M3 for correctly evaluated trial using [\(a =\)] 22 and [\(c =\)] 16 or M2 for two correctly evaluated trials where \(c = a - 6\) or M1 for one correctly evaluated trial where \(c = a - 6\) | ||
| If 0 or 1 scored, instead award SC2 for 22 with no working or insufficient working If 0 scored instead award SC1 for 16 with no working or insufficient working | |||
9 Solve.
\(3x + 12 = 9 - 7x\) [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| -0.3 oe | 3 | M1 for correct first step | Embedded answer scores M2 max If not shown, M1 implied by \(10x = b\) or \(ax = -3\) |
| M1 FT their first step to to \(ax = b\) | \(10x = -3\) or \(-10x = 3\) | ||
9 The next term in a Fibonacci sequence is found by adding together the previous two terms.
In a particular Fibonacci sequence:
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(3 + x\) | 1 | ||
| \(3 + x + 3 + 2x\) [\(= 6 + 3x\)] | 1 | Intent to add correct successive algebraic terms must be clearly conveyed eg  | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 13 nfww | 4 | M2 for \(9 + 5x = 74\) or M1 for \(3 + 2x + 6 + 3x\) or better M1 FT for [\(x =\)] \(\frac{74 - \textit{their } 9}{\textit{their } 5}\) | M2 may be subsequently implied by arithmetic working FT must be from \(ax + b = c\) May be seen in stages as two arithmetic steps or as formal algebra Correct answer from trials scores 4 |
8 For each statement, tick (✓) whether the value of \(x\) is true or false.
The first one is done for you.
| Statement | Value of \(x\) | True | False |
|---|---|---|---|
| \(x > {}^{-}1\) | 5 | ✓ | |
| \(x \leqslant {}^{-}1\) | \({}^{-}1\) | ||
| \(\dfrac{x}{10} = 0.7\) | 70 | ||
| \(x - 2 \ne 5\) | 3 | ||
| \({}^{-}1 < x < 0.7\) | 0 |
[3]
| Answer | Marks | Part marks and guidance | |||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| |||||||||||||||||||||||||||
| 3 | B2 for three correct or B1 for two correct | ||||||||||||||||||||||||||
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 |
23 The diagram shows an equilateral triangle and a rectangle.
Not to scale
The equilateral triangle has the same perimeter as the rectangle.
Find the value of \(a\). [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 8 | 4 | M2 for \(a + a + a + 5 + a + 5 = 42\) oe or M1 for \(3 \times 14\) oe or \(a + a + a + 5 + a + 5\) oe AND M1 for \(a = \frac{42 - 10}{4}\) oe or FT their equation of the form \(ka + c = d\) oe to \(a = \frac{d - c}{k}\) oe | Allow M2 if correct expression seen first and then incorrectly simplified before equation = 42 M1 implied by 42 or \(4a + 10\) oe FT only from written equation Where \(k\), \(c\) and \(d\) are positive integers and \(k \gt 1\) Allow M1 for one trial into \(2(a + 5) + 2a\) evaluated correctly |
21 The diagram shows a rectangle with length \((3x - 4)\) cm and width \((x + 2)\) cm.
Not to scale
The length of the rectangle is twice the width of the rectangle.
Calculate the area of the rectangle.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 200 with correct working | 6 | M1 for \(2(x + 2) = 3x - 4\) oe M1 for reaching \(ax = b\) FT their equation A1 for \(x = 8\) OR Trials M1 for one trial into \(2(x + 2)\) and \(3x - 4\) evaluated correctly M1 for two trials of \(2(x + 2)\) and \(3x - 4\) evaluated correctly A1 for [\(w =\)] 10 and [\(l =\)] 20 AND M2 for (their \(x\) \(+ 2) \times (3 \times\) their \(x\) \(- 4)\) or M1 for (their \(x\) \(+ 2)\) or \((3 \times\) their \(x\) \(- 4)\) If 0 or M1 scored, instead allow SC2 for answer 200 with no or insufficient working If 0 scored, instead allow SC1 for \(x = 8\) with no or insufficient working | “Correct working” requires evidence of at least M1M1A1 or alternate convincing approach FT from an equation with \(x\) on both sides only. their \(x\) must be a positive integer and stated their \(x\) or their length/width may be on the diagram |
15 Here are two functions.
When the input of each function is 6, the output of Function A is equal to the output of Function B.
Work out the value of \(p\) in Function B. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7 nfww | 4 | M3 for \(6p - 8\) and their \((6 \times 5 + 4)\) correctly used together or \(34 + 8 = 6p\) or \((34 + 8) \div 6\) or \(6 \times 7 - 8 = 34\) OR M1 for \(6 \times 5 + 4\) may be implied by 34 M1 for \(6p - 8\) or \(34 + 8\) or for one correctly evaluated trial | eg an equation with a first step to solve |
12 The costs to hire a motorhome are shown below.
| Motorhome hire | |
|---|---|
| First day | £130 |
| Each extra day | £90 |
| Each mile travelled | 60p |
Finley hired a motorhome and travelled 560 miles.
The total cost of hiring the motorhome was £916.
Calculate how many days Finley hired the motorhome for.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 6 with correct working | 5 | M1 for \(560 \times 0.6\) or \(560 \times 60\) M1 for \(916 - 130\) or \(916 -\) their \((560 \times 0.6)\) M1 for \(916 - (130 +\) their \((560 \times 0.6))\) M1 for \((916 - (130 +\) their \((560 \times 0.6))) \div 90\) Trials M4 for \(560 \times 0.6 + 5 \times 90 + 130 = 916\) oe or M3 for two trials of the form \(560 \times 0.6 + n \times 90 + 130\) oe seen and correctly evaluated where \(n \geqslant 1\), \(n \ne 5\) or M2 for one trial of the form \(560 \times 0.6 + n \times 90 + 130\) seen and correctly evaluated where \(n \geqslant 1\), \(n \ne 5\) or M1 for \(560 \times 0.6\) [\(+\ldots\)] If 0, 1 or 2 scored, instead award SC3 for answer 6 with no working or insufficient working If 0 or 1 scored, instead award SC2 for answer 5 with no working or insufficient working If 0 scored, SC1 for 450 with no working or insufficient working | “Correct working” requires evidence of at least M3 or M1 M1 M1 Their \(560 \times 0.6\) may be \(560 \times 60\) with an attempt to convert to pounds \(\div 90\) may be repeated subtraction or addition egs for Trial and +90s M4 for \(560 \times 0.6 + 3 \times 90 + 130 = 736\) and \(736 + 90 + 90 = 916\) M3 for \(560 \times 0.6 + 3 \times 90 + 130 = 736\) and \(736 + 90 = 826\) M2 for \(560 \times 0.6 + 3 \times 90 + 130 = 736\) M1 for \(560 \times 0.6\) [\(+\ldots\)] For trials and trials + 90s, accept correct rearrangements that have a target of 786, 580 or 450 rather than 916 |
9 Here are two pieces of work.
Each shows a question and an incorrect solution.
For each part, describe the error made and write out a correct solution.
| Question: Factorise. \(x^2 + x - 20\) Solution: \((x + 4)(x - 5)\) |
The error is ……
A correct solution is …… [2]
| Question: Solve. \(4x + 5 = x + 2\) Solution: \(4x + 5 = x + 2\) \(3x + 5 = 2\) \(3x = 5 - 2\) \(3x = 3\) \(x = 1\) |
The error is ……
A correct solution is …… [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Expanding the brackets gives \(-x\) oe | 1 | Accept any correct response, look for answers in the working space (see appendix) | |
| \((x - 4)(x + 5)\) | 1 | ||
| Response | Mark |
|---|---|
| Expanding brackets gives \(-x\) | 1 |
| The signs are the wrong way round (condone inverted) | 1 |
| The minus sign is on the wrong one | 1 |
| They are the wrong factors | 0 |
| The answer is \((x - 4)(x + 5)\) | 0 |
| The answer is 4 or – 5 | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| The third line should be 2 – 5 | 1 | Accept any correct response, look for answers in the working space (see appendix) | |
| \(3x = 2 - 5\) \(3x = -3\) \(x = -1\) | 1 | ||
| Response | Mark |
|---|---|
| [The third line] should be 2 – 5 | 1 |
| She did 5 – 2 it should be 2 – 5 | 1 |
| It should be – 5 not – 2 | 1 |
| 5 should be subtracted by 2 | 0 |
| The error is in \(3x = 5 - 2\) | 0 |
8
\(p - 5 = -4\) [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(7a\) final answer | 1 | Accept \(7 \times a\) and \(a7\) and \(a \times 7\) | |
| (ii) \(-x + 4 - 3y\) final answer | 2 | B1 for \(-x\) seen or [\(+\)]\(4 - 3y\) seen in final answer | Accept answer in any order |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 1 | 1 | ||
21 The length of a word is classed as short, medium or long depending on the number of characters in the word.
This table shows the percentage, in terms of \(x\), of each word length found in a book.
| Length of word | Percentage (%) |
|---|---|
| Short | \(2x + 5\) |
| Medium | \(3x - 1\) |
| Long | \(x\) |
Find the percentage of words in this book which are classed as short words.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 37 with correct working | 5 | M1 for \(2x + 5 + 3x - 1 + x = 100\) may be implied by a subsequent correct equation M1 for simplifying their equation to \(ax + b = c\) or \(6x + 4 = 100\) M1 for the first correct step in solving their \(ax + b = c\) and M1 for substituting their 16 into \(2x + 5\) OR M3 for one complete correct evaluation of \(2x + 5 + 3x - 1 + x = 100\) or M2 for one complete correct evaluation of \(2x + 5 + 3x - 1 + x =\) a total or M1 for one correct evaluation of each expression \(2x + 5\) [+] \(3x - 1\) [+] \(x\) If 0, 1 or 2 scored, instead award SC3 for answer 37 with no or insufficient working If 0 or 1 scored, instead award SC2 for \(x = 16\) with no or insufficient working | “Correct working” requires evidence of first method and at least M1 and \(x = 16\) or M3 if using trials or M2M1 for non-algebraic method Three separate expressions are not enough Ignore inclusion of % in working \(6x + 4 = 100\) scores M1M1 e.g. \(6x = 100 - 4\) \(6x = 96\) scores M1M1M1 For M marks with trials: allow each term or each expression from the question evaluated separately e.g. (Use 10) 20 + 5 + 30 – 1 + 10 = 64 or 25 (+) 29 (+) 10 = 64 Alternative method (Non algebraic) M2 for 100 – 5 + 1 oe or M1 for – 5 + 1 or – 4 AND M1 for their \((100 - 5 + 1) \div 6\) AND M1 for substituting their 16 into \(2x + 5\) |
12 Here are two functions.
The input to function C is \(x\).
The output from function C is \(2x + 1\).
Find the value of \(x\).
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(3(x + 5)\) oe final answer | 1 | Equivalent includes \(3x + 15\) | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(-7\) with correct working | 5 | accept any correct method B1 for [output = ] 2 × their{\(3(x + 5)\)} – 1 oe or output through inverse of B as \(x + 1\) B1dep for their {2× \(3(x + 5)\) – 1} = \(2x + 1\) oe or output through inverse of A as \(\frac{x + 1}{3} - 5\) or \(x + 1\) = their {\(3(x + 5)\)} M1 for e.g. their \(6x + 30 - 1 = 2x + 1\) or their \(3x = x + 1 - 15\) or inverse method \(\frac{x + 1}{3} - 5 = x\) M1 for their {\(4x = -28\) or \(-2x = 14\)} oe Alternative method using trials M2 for at least two complete correct evaluations of both \(6x + 29\) oe(or functions) and \(2x + 1\) or M1 at least one complete correct evaluation of both \(6x + 29\) oe and \(2x + 1\) If 0 or 1 scored SC3 for answer \(-7\) with no or insufficient working | “Correct working” requires evidence of at least B1B1 or B1M1 or M2 if using trials B1 implied by \(2 \times 3(x + 5) - 1\) or better e.g. \(6x + 30 - 1\) dep. on previous B1 Note: \(3x + 15 + 2x - 1 = 2x + 1\) scores B0B0 M1 B1 B1 implied by \(6x + 29 = 2x + 1\) The first M1 is for dealing with bracket(s) correctly in a linear equation and the second M1 is for correctly getting the linear equation, with \(x\)’s on both sides, into the form \(ax = b\). We can only follow through an equation with \(x\)’s on both sides for M1 or \(x\)’s on both sides and a bracket for M2. |
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 |
5 Solve.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 18 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(-7\) | 1 | ||
14 Solve.
\[6x - 9 = 27 - 4x\]\(x =\) .................... [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3.6 oe | 3 | M2 for \(6x + 4x = 27 + 9\) oe OR M1 for \(10x - 9 = 27\) or \(10x = k\) or \(6x = 36 - 4x\) or \(kx = 36\) AND M1 for \(x = \dfrac{b}{a}\) FT their \(ax = b\) seen | e.g. \(x = \dfrac{36}{10}\) oe |
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 |
5 Solve.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 32 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 | 1 | ||
20 This list represents four numbers.
\[127 \qquad x \qquad x + 1 \qquad 2x\]The mean of the four numbers is 180.
Work out the numbers.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [127] 148 149 296 in any order With correct working | 5 | M3 for \(4x\) and 592 or for \((180 \times 4 - 127 - 1) \div 4\) A1 for 148 | “correct working” requires at least M2 \(180 \times 4\) can be implied by 720 throughout |
| or M2 for a correct algebraic expression and correct constant or for \(180 \times 4 - 127\ [-\ 1]\) or for \(127 + 148 + 149 + 296 = 180 \times 4\) oe or for trials leading to the correct answer | eg \(127 + x + x + 1 + 2x\) and 720; \(4x + 1\) and 593 Do not accept \(x\), \(2x\) or \(x + 1\) as correct algebraic expressions for M2 Condone e.g. \(127 + x + x + 1 + 2x \div 4\) for M1 | ||
| or M1 for \(127 + x + x + 1 + 2x\) nfww or \(180 \times 4\) or 720 | Mark to candidates advantage | ||
| If 0 or 1 scored, instead award SC2 for 148 149 296 in any order with no or insufficient working If 0 scored, SC1 for [\(x =\)]148 | |||
15 In this question, all lengths are in centimetres.
The diagram shows an equilateral triangle and a square.
Not to scale
The perimeter of each shape is 36 cm.
Find the value of \(b\). [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 5 nfww | 4 | B2 for [\(a =\)] 4 or M1 for \(9a = 36\) or better and M1 for showing substitution/use of their \(a\) e.g. \(4 \times\) their \(4 + 4b = 36\) oe or better or [\(b =\)] \(9 -\) their 4 | If another value for \(a\) is used to find \(b\) do not award B2 T&I only scores if ending at 4 Or better may be e.g. \(a + b = 9\) or \(\dfrac{36 - 4 \times \textit{their } 4}{4}\) |
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 | ||
10 Alex, Blake and Charlie play a computer game.
Alex goes first and scores \(n\) points.
Work out how many points they each score.
You must show your working. [7]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 87 253 278 with correct working | 7 | B1 for \(3n - 8\) B1 for \(3n - 8 + 25\) or better M1 for writing a correct equation equal to 618 using their expressions e.g. \(n + 3n - 8 + 3n - 8 + 25 = 618\) or better | “Correct working” requires evidence of at least B1 B1 Expressions could start from B or C. |
| M1 for simplifying their equation e.g. \(7n + 9 = 618\) A1FT for correctly solving their equation e.g. \(n = 87\) M1 for substituting their 87 into both expressions e.g. 3×87 – 8 and 3×87 – 8 + 25 oe | |||
| Trials: B1 for one complete trial with \(n \geqslant 3\) B1 for second complete trial \(n \geqslant 3\) If 0 or 1 scored SC3 for 87, 253, 278 with B1 only or SC2 for 87, 253, 278 with no working | See appendix for a more complete set of trials | ||
| \(n\) | \(3n - 8\) | \(3n - 8 + 25\) | Total |
|---|---|---|---|
| 10 | 22 | 47 | 79 |
| 20 | 52 | 77 | 149 |
| 30 | 82 | 107 | 219 |
| 40 | 112 | 137 | 289 |
| 50 | 142 | 167 | 359 |
| 60 | 172 | 197 | 429 |
| 70 | 202 | 227 | 499 |
| 80 | 232 | 257 | 569 |
| 90 | 262 | 287 | 639 |
| 100 | 292 | 317 | 709 |
| increments | +1 | +3 | +3 | +7 |
|---|---|---|---|---|
| e.g. | add 4 | add 12 | add 12 | add 28 |
| 24 | 64 | 89 | 177 |
9 A student thinks of a number.
They square it and then add 6.
Their answer is 295.
What number is the student thinking of? [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 17 | 2 | B1 for 289 or \(17^2\) or M1 for \(\sqrt{295 - 6}\) | Accept \(-17\) |
8 Here is a function.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 35 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 6 | 2 | M1 for \(\div 7\) and \(-3\) soi or for \(63 \div 7 = 9\) soi | |
7 Solve.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 44 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3.5 final answer | 2 | M1 for \(6y = 28 - 7\) or better | Accept \(3\frac{1}{2}\), \(\frac{7}{2}\) or \(\frac{21}{6}\) for 2 marks |
15 Solve.
\(\dfrac{x}{x + 6} = 5\) [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| −7.5 or \(-7\frac{1}{2}\) or \(-\frac{15}{2}\) | 3 | M1 for \(x = 5(x + 6)\) M1 for \(x - 5x = 30\) oe FT their first step | Condone \(-\frac{30}{4}\) as final answer Embedded answer scores M2 maximum |
15 A rectangle is three times as long as it is wide.
It has a perimeter of 44 cm.
Find the length of the rectangle. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 16.5 | 4 | B3 for 5.5 [cm] nfww or M2 for \(3x + x + 3x + x = 44\) or better or \(44 \div 8\) oe or M1 for \(3x\) [as length] and \(x\) [as width] or \(4x\) [as length + width] or \(8x\) [as perimeter] OR Using trial length and width with length \(= 3 \times\) width M1 for a perimeter found M1 for a second perimeter closer to 44 If 0 scored SC1 for answer 33 | May be other letters or in words for 2 or 1 mark \(3x\) and \(x\) may be on diagram |
15
[2]
[2]
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 20 | 2 | M1 for \(\dfrac{x}{2} = 15 - 5\) or better or \(x + 10 = 30\) | For M1 must be an equation in \(x\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(5a(a - 2)\) final answer | 2 | M1 for \(5(a^2 - 2a)\) or \(a(5a - 10)\) as answer | Condone missing final bracket |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((x + 7)(x + 8)\) | M2 | M1 for \((x + a)\) and \((x + b)\) where \(ab = 56\) or \(a + b = 15\) | |
| \({}^{-}7\) and \({}^{-}8\) final answer | B1FT | for correct solutions from their quadratic factors If 0 scored SC1 for answers \(\pm 7\) and \(\pm 8\) | |
10 Nadia thinks of a number.
She finds the square root and then divides by 5.
Her answer is 20.
What number is she thinking of? [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 000 | 2 | M1 for 20 × 5 | M1 may be implied by 100 |
2
\(4x + 3 = 13\) [2]
\(5(2x + 3) + 2(x - 4)\) [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.5 oe | 2 | M1 for \(4x = 13 - 3\) or for \(x + \frac{3}{4} = \frac{13}{4}\) | Accept \(\frac{10}{4}\) or \(\frac{5}{2}\) isw Embedded answer scores M1 max |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(12x + 7\) final answer | 3 | M1 for \(10x + 15\) M1 for \(2x - 8\) | |
24 In this question, all lengths are in centimetres.
Here is a square.
Not to scale
Find the length of one side of the square when \(b = 4\). [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 18 nfww | 6 | B4 for \(2a = 5\) OR M3 for \(6a - 4a = 8 - 4 + 1\) or better or M2 for \(6a + 4 - 1 = 4a + 2 \times 4\) or better or \(6a - 4a = 2b - b + 1\) or better or M1 for \(6a + b - 1 = 4a + 2b\) AND M1dep for correct substitution of 4 and their 2.5 in \(4a + 2b\) or \(6a + b - 1\) | Isolating \(a\) Expect \(6a + 3 = 4a + 8\) Expect \(2a = b + 1\) Dependent on at least M1 or B4 May be implied by 10 + 8 or 15 + 3 |
5 Solve.
\(6x - 10 = 4x + 1\) [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 5.5 oe | 3 | M1 for correct first step e.g. \(6x - 4x - 10 = 1\) M1 for \(6x - 4x = 1 + 10\) FT their first step or FT their \(ax = b\) to \(x = \frac{b}{a}\) | isw \(\frac{11}{2}\) converted to decimal Embedded answer scores M2 max If not shown, M1 implied by \(\pm 2x = b\) or \(ax = \pm 11\) e.g. M1 for \(2x = -9\) leading to \(x = -\frac{9}{2}\) oe |
5 Dan thinks of a number.
He adds 3 and divides the result by 2.
His answer is 16.
What number is Dan thinking of? [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 29 | 2 | M1 for 16 × 2 soi 32 | May be 16 × 2 – 3 |
12 Solve.
\[4x + 5 = 35\][2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7.5 or \(7\dfrac{1}{2}\) or \(\dfrac{15}{2}\) final answer | 2 | M1 for first correct step \(4x = 35 - 5\) or better or \(x + \dfrac{5}{4} = \dfrac{35}{4}\) or better | Do not accept embedded answers Accept a fully correct flowchart or working for M1 eg \(x \to \times 4 \to +5 \to 35\) \(\phantom{eg }x \leftarrow \div 4 \leftarrow -5 \leftarrow 35\) or \((35 - 5) \div 4\) may be in stages |
12 Here are two functions.
Find the value \(x\) when \(z = 4x\). [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{x + 2}{3}\) or \(\dfrac{x}{3} + \dfrac{2}{3}\) final answer | 2 | M1 for \(y + 2 = 3x\) or \(x + 2 = 3y\) or for \(x = 3y - 2\) or for \([x =]\ \dfrac{y + 2}{3}\) If 0 scored, SC1 for answer \(y = \dfrac{x}{2} - 7\) | For 2 marks, condone answer \(y = \dfrac{x + 2}{3}\) Allow M1 for correct reverse flowchart with arrows reversed ← ÷ 3 ← + 2 ← |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| −5 | 5 | For method marks, condone inclusion of multiplication signs Method must be seen in working space for part (b) If brackets omitted then allow recovery for method | |
| M3 for \(2(3x - 2 + 7) = 4x\) oe or \(\dfrac{2x - 7 + 2}{3} = x\) oe or M2 for \(2(3x - 2 + 7)\) oe seen or \(\dfrac{2x - 7 + 2}{3}\) oe seen or M1 for \(3x - 2\) or \(3x + 5\) oe seen or \(2x - 7\) or \(2x - 5\) oe seen | For M3 eg \(2x - 7 = 3x - 2\) | ||
| M1dep for correct rearrangement of their eqn with at least 2 terms in \(x\) to \(ax + b = 0\) or better | M1dep on at least M1 earned previously e.g. \(2(3x - 2) + 7\) [\(= 4x\)] scores M1 and then can earn M1 dep if they then correctly rearrange to \(ax = b\) | ||
2
[3]
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(6a^{11}\) | 3 | B2 for answer \(ka^{11}\) or \(\dfrac{6a^{13}}{a^2}\) or \(6 \times a^6 \times a^5\) or \(6 \times a^8 \times a^3\) shown in working OR B1 for answer \(6a^k\) or \(ka^{13}\) or for \(3a^8 \times 2a^3\) or \(3a^6 \times 2a^5\) shown in working | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.5 or \(2\frac{1}{2}\) or \(\frac{5}{2}\) | 3 | M1 for correct first step e.g. \(6x - 10 = 5\) | Do not accept \(\frac{15}{6}\) as final answer but allow to imply M1M1 Embedded answer scores M2 max |
| M1 for \(6x = 5 + 10\) FT their first step or for FT their \(ax = b\) to \(x = \dfrac{b}{a}\) | where \(\frac{b}{a}\) is a simplified fraction (improper or mixed number) or an integer e.g. M1 for \(6x = 1\) leading to \(x = \frac{1}{6}\) | ||
25 Marcin buys 7 rulers and 15 crayons for £7.
A ruler costs 12p more than a crayon.
Find the cost of one crayon. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 28 or [£][0].28 | 5 | B1 for \(7r + 15c = 7[00]\) or [\(r =\)] \(c + [0.]12\) M1 for \(7(c + [0.]12) + 15c = 7[00]\) or better oe or \(r - c = [0.]12\) M1 for \(7c + 84 + 15c = 7[00]\) or better oe or \(7r - 7c = [0.]84\) M1 for \(15c + 7c = 7[00] - [0.]84\) or better | Allow any pair of letters, Trial-and-improvement will score 0 or 5 only allow work in pence or pounds i.e. removing brackets i.e. rearranging their equation |
6 Solve.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 10 final answer | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 4 final answer | 1 | ||
17 Here is a function.
Write an algebraic expression, in terms of \(x\), for the input of function A. [2]
Find the value of \(k\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{x}{5} - 14\) oe | 2 | M1 for \(\dfrac{x}{5}\) If 0 scored then SC1 for \(\dfrac{x - 14}{5}\) oe | Condone use of another letter for M1 max Must use \(x\) in SC1 0 for \(x - 14 \div 5\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| −17.5 or \(-\dfrac{35}{2}\) oe nfww | 3 | M1 for 5(‘\(k\)’ + 14) = ‘\(k\)’ or ‘\(k\)’ = \(\dfrac{k}{5} - 14\) M1FT for 4’\(k\)’ = −70 or better or re-arrangement of their comparable f(\(k\)) = g(\(k\)) equation into the form \(ak = b\). M1FT solving their \(ak = b\) | eg \(5k + 14 = k\) becomes \(4k = -14\) and then \(k = -3.5\) scores M0 M1FT M1FT \(k + 70 = k\) is not comparable Answers may be in decimal or fractional form but fractions equating to integers should be simplified |
| Alternative (FT as above): M1 for ‘\(k\)’ = \(\dfrac{k}{5} - 14\) M1FT for \(\dfrac{4k}{5}\) = −14 or better M1FT solving their \(ak = b\) Trials or no working: SC3 for −17.5 | |||
7
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 14 | 1 | ||
| (ii) 18 | 1 | ||
| (iii) 6.5 final answer | 2 | M1 for \(8x = 46 + 6\) or better or \(x = \dfrac{b}{a}\) from their \(ax = b\) \(a \ne 1\) | Accept \(6\frac{1}{2}\) or \(\dfrac{13}{2}\) must be an equation Accept a fully correct flow chart for M1 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(-6\) and \(-5\) final answer | 3 | B2 for \((x + 6)(x + 5)\) Or M1 for \((x \pm a)(x \pm b)\) where \((a + b) = 11\) or \((ab) = 30\) or pairs of factors giving two correct terms may be implied in a table And B1 for correct solutions FT their quadratic factors | |
14 Halina cycled from A to B at an average speed of 26 km per hour.
She then cycled from B to C at an average speed of 20 km per hour.
Not to scale
She left A at 10.00 am, did not stop at B and arrived at C at 3.00 pm.
Find the distance from A to B. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) Valid explanation | 1 | Such as ‘distance is time times speed’ | Need to see ‘multiply’ oe See Appendix |
| (ii) \(5 - x\) | 2 | M1 for time to travel from A to C = 5 [hours] soi | Must be seen in this part |
| (iii) \(20(5 - x) = 100 - 20x\) | 1 | ||
Exemplar responses for Q14ai
| Response | Mark |
|---|---|
| Because it’s the average speed x the number of hours it takes | 1 |
| Because the average speed to A to B was 26km per hour so it’s 26 x hours | 1 |
| Because you need to work at how many hours it took and times it by the speed | 1 |
| Because she cycled at 26 kmph and the x is how long it took so you multiply them (26kmph implies speed) | 1 bod |
| Since to find the average speed you do distance x time so 26 x x \(\to\) 26x | 0 |
| Because she did 26 km per hour and we don’t know how many hours yet so we put x to show the number of hours | 0 |
| Because it’s 26x the 1 hour | 0 |
| Because her average speed is 26 | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 78 | 4 | M1 for \(26x + 100 - 20x = 118\) M1 for their \(6x\) = their 18 M1 for \(x = \dfrac{\textit{their } 18}{\textit{their } 6}\) soi | Simplifying their equation to \(ax = b\) Simplifying their \(ax = b\) to \(x = \dfrac{b}{a}\) |
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 |
6 This rectangle has length \((4x - 5)\) cm and width \((x + 3)\) cm.
Not to scale
The perimeter of the rectangle is 46 cm.
Calculate the area of the rectangle.
........................................ cm2 [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 120 | 5 | B3 for \(x = 5\) OR M1 for \(x + 3 + x + 3 + 4x - 5 + 4x - 5\) [= 46] oe M1 for \(10x = 46 + 4\) FT their linear eqn M1 for 50 ÷10 FT their \(ax = b\) M1 for (4 × their \(x\) – 5) × (their \(x\) + 3) | \(10x - 4\) [= 46] oe eg \(x + 3 + 4x - 5 = 23\) |
10
[2]
[3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(x \gt 4\) | 2 | M1 for a correct first step e.g. \(3x \gt 10 + 2\), \(3x - 12 \gt 0\) | For M1, condone wrong inequality symbol or equals |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{3}{10}\) oe isw | 3 | M1 for correct first step e.g. \(6x + 4x + 2 = 5\) M1 for \(6x + 4x = 5 - 2\) or FT their \(ax = b\) to \(x = \frac{b}{a}\) | Embedded answer scores M2 max If not shown, M1 implied by \(\pm 10x = b\) or \(ax = \pm 3\) e.g. M1 for \(2x = 7\) leading to \(x = \frac{7}{2}\) oe |
5
[2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(3x^2 + 6xy\) final answer | 2 | B1 for \(3x^2\) or \(6xy\) seen | Condone \(6yx\) Do not accept eg \(6 \times x \times y\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 4 | 1 | ||
| (ii) 33 | 2 | M1 for \(\dfrac{x}{3} = 9 + 2\) or better or M1 for \(x = a \times b\) following \(\dfrac{x}{b} = a\) | Alternative method M1 for \(x - 6 = 27\) or M1 for \(x = a + b\) following \(x - a = b\) |
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