Course Support. Past Paper Questions by Topic. For Free.
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
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 \(3 - 1 = 2\)
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
16 \(P = 4x + 3y\)
\(x = 5\)
\(y = -2\)
| Answer | Mark | Mark scheme |
|---|---|---|
| 14 | M1 | for \(4 \times 5\) and \(3 \times -2\), the substitution may be seen in two separate calculations, eg \(4 \times 5\ (= 20)\) and \(3 \times -2\ (= -6)\) |
| A1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(4e^2 + 8e\) | B2 | for \(4e^2 + 8e\) |
| (B1 | for \(4e^2\) or \(8e\)) |
Note: \(4e^2 + 8e = 12e^3\) for example gets B1 only
| Answer | Mark | Mark scheme |
|---|---|---|
| 11 | M1 | for a correct first step eg \(3 \times m - 3 \times 4 = 21\) oe or \(m - 4 = 21 \div 3\ (= 7)\) oe |
| A1 | cao |
Showing \(\div 3\) by each side of equation is sufficient
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 |
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 |
No questions match these filters.