Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1 Attempt to multiply the numerator and denominator by
(8 3)k . For example,6 3 4 8 3
8 3 8 3
M1 1.1b 3rd
Rationalise the
denominator of a
fraction with a
simple surd
denominator
Attempt to multiply out the numerator (at least 3 terms
correct).
48 3 18 32 4 3
M1 1.1a
Attempt to multiply out the denominator (for example, 3 terms
correct but must be rational or 64 – 3 seen or implied).
64 8 3 8 3 3
M1 1.1b
p and q stated or implied (condone if all over 61).
44 143
61 61 or
44 14,
61 61p q
A1 1.1b
(4 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
2a Statement that discriminant is b2 – 4ac, and/or implied by
writing 2
8 4 1 8 1k k
M1 1.1a 4th
Understand and
use the
discriminant;
conditions for
real, repeated and
no real roots
Attempt to simplify the expression by multiplying out the
brackets. Condone sign errors and one algebraic error (but not
missing k term from squaring brackets and must have k2, k and
constant terms).
2 8 8 64 32 4k k k k o.e.
M1 1.1b
2 16 60k k A1 1.1b
(3)
2b Knowledge that two equal roots occur when the discriminant
is zero. This can be shown by writing b2 – 4ac = 0, or by
writing 2 16 60 0k k
M1 1.1b 5th
Solve problems
involving the
discriminant in
context and
construct simple
proofs involving
the discriminant
10, 6k k A1 1.1b
(2)
2c Correct substitution for k = 8: 2f( ) 16 65x x x B1 2.2a 3rd
Solve quadratic
equations by use
of formula
Attempt to complete the square for their expression of f(x).
2
f( ) 8 1x x
M1 1.1b
Statement (which can be purely algebraic) that f(x) > 0,
because, for example, a squared term is always greater than or
equal to zero, so one more than a square term must be greater
than zero or an appeal to a reasonable sketch of y = f(x).
A1 2.3
(3)
(8 marks)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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Notes
2a
Not all steps have to be present to award full marks. For example, the second method mark can still be awarded
if the answer does not include that step.
2b
Award full marks for k = 6, k = 10 seen. Award full marks for valid and complete alternative method (e.g.
expanding (x – a)2 comparing coefficients and solving for k).
2c
An alternative method is acceptable. For example, students could differentiate to find that the turning point of
the graph of y = f(x) is at (8, 1), and then show that it is a minimum. The minimum can be shown by using
properties of quadratic curves or by finding the second differential. Students must explain (a sketch will suffice)
that this means that the graph lies above the x-axis and reach the appropriate conclusion.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
3a 115 (m) is the height of the cliff (as this is the height of the
ball when t = 0). Accept answer that states 115 (m) is the
height of the cliff plus the height of the person who is ready to
throw the stone or similar sensible comment.
B1 3.2a 4th
Understand the
concepts of
domain and range
(1)
3b Attempt to factorise the – 4.9 out of the first two (or all)
terms.
2( ) 4.9 2.5 115h t t t or 2 5( ) 4.9 115
2h t t t
M1 3.1a 4th
Solve simple
quadratic
equations by
completing the
square
2 2( ) 4.9 1.25 ( 4.9) 1.25 115h t t
or
2 25 5
( ) 4.9 ( 4.9) 1154 4
h t t
M1 3.1a
2
( ) 122.65625 4.9 1.25h t t o.e.
(N.B. 122.65625 =3925
32)
Accept the first term written to 1, 2, 3 or 4 d.p. or the full
answer as shown.
A1 3.1a
(3)
3ci Statement that the stone will reach ground level when
h(t) = 0, or 24.9 12.25 115 0t t is seen.
M1 3.1a 4th
Form and solve
quadratic
equations in
context
Valid attempt to solve quadratic equation (could be using
completed square form from part b, calculator or formula). M1 3.1a
Clearly states that t = 6.25 s (accept t = 6.3 s) is the answer, or
circles that answer and crosses out the other answer, or
explains that t must be positive as you cannot have a negative
value for time.
A1 3.5a
(3)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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3cii hmax = awrt 123
ft A from part b.
B1ft 3.1a 4th
Form and solve
quadratic
equations in
context
t =5
4or t = 1.25
ft C from part b.
B1ft 3.2a
(2)
(9 marks)
Notes
3c
Award 4 marks for correct final answer, with some working missing. If not correct B1 for each of A, B and C
correct.
If the student answered part b by completing the square, award full marks for part c, providing their answer to
their part b was fully correct.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
4a Attempt to solve q(x) = 0 by completing the square or by using
the formula.
2
2
10 20 0
5 45 0
x x
x
or
10 100 4(1)( 20)
2(1)x
M1 1.1b 3rd
Solve quadratic
equations by use
of formula
5 3 5x and/or statement that says a = 5 and b = 5 A1 1.1b
(2)
4b Figure 1
q(0) = −20, so y = q(x)
intersects y-axis at (0, −20)
and x-intercepts labelled
(accept incorrect values from
part a).
B1ft 1.1b 3rd
Sketch graphs of
quadratic
functions
y = p(x) intersects y-axis
at (0, 3).
B1 1.1b
y = p(x) intersects x-axis
at (6, 0).
B1 1.1b
Graphs drawn as shown with
all axes intercepts labelled.
The two graphs should clearly
intersect at two points, one at
a negative value of x and one
at a positive value of x. These
points of intersection do not
need to be labelled.
B1 1.1b
(4)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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4c Statement indicating that this is the point where p(x) = q(x)
or 2 110 20 3
2x x x seen.
M1 2.2a 4th
Solve more
complicated
simultaneous
equations where
one is linear and
one is quadratic
Their equation factorised, or attempt to solve their equation by
completing the square.
2x2 −19x – 46 = 0
(2x – 23)(x + 2) = 0
M1 1.1b
23 11,
2 4
A1 1.1b
2,4 A1 1.1b
(4)
4d x < – 2 or
23
2x o.e.
B1 2.2a 4th
Represent
solutions to
quadratic
inequalities using
set notation
{ : , 2} { : , 11.5}x x x x x x ¡ ¡
NB: Must see “or” or (if missing SC1 for just the correct
inequalities).
B1 2.2a
(2)
(12 marks)
Notes
4a
Equation can be solved by completing the square or by using the quadratic formula. Either method is acceptable.
4b
Answers with incorrect coordinates lose accuracy marks as appropriate. However, the graph accuracy marks can
be awarded for correctly labelling their coordinates, even if their coordinates are incorrect.
4c
If the student incorrectly writes the initial equation, award 1 method mark for an attempt to solve the incorrect
equation. Solving the correct equation by either factorising or completing the square is acceptable.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
5 Figure 2
Asymptote drawn at x = 6
B1 1.1b 5th
Understand and
use properties of
asymptotes for
graphs of the
form y = a/x and y
= a/x2
Asymptote drawn at y = 5 B1 1.1b
Point13
0,3
labelled.
Condone 13
3clearly on y axis.
B1 1.1b
Point26
,05
labelled.
Condone26
5clearly on x axis.
B1 1.1b
Correctly shaped graph drawn
in the correct quadrants
formed by the asymptotes.
B1 1.1b
(5)
(5 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
6a Figure 3
Evidence of attempt to show
stretch of sf 1
2in x direction
(e.g. one correct set of
coordinates – not (0, –2)).
M1 1.1b 3rd
Transform graphs
using stretches
Fully complete graph with all
points labelled. A1 1.1b
(2)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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6b Figure 4
Evidence of attempt to show
reflection in y axis (e.g. one
correct set of coordinates – not
(0, –2)).
M1 1.1b 3rd
Transform graphs
using translations
Fully complete graph with all
points labelled. A1 1.1b
(2)
(4 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
7a Figure 5
Graph of y = 2x + 5 drawn.
B1
1.1b 4th
Represent linear
and quadratic
inequalities on
graphs
Graph of 2y + x = 6 drawn. B1 1.1b
Graph of y = 2 drawn onto the
coordinate grid and the
triangle correctly shaded.
B1 2.2a
(3)
7b Attempt to solve y = 2x + 5 and 2y + x = 6 simultaneously
for y. M1 2.2a 5th
Solve problems
involving linear
and quadratic
inequalities in
context
y = 3.4 A1 1.1b
Base of triangle = 3.5 B1 2.2a
Area of triangle = 1
2 (“3.4” – 2) 3.5
M1 2.2a
Area of triangle is 2.45 (units2). A1 1.1b
(5)
(8 marks)
Notes
7b
It is possible to find the area of triangle by realising that the two diagonal lines are perpendicular and therefore
finding the length of each line using Pythagoras’ theorem. Award full marks for a correct final answer using this
method.
In this case award the second and third accuracy marks for finding the lengths 2.45 and 9.8
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1a Use of the gradient formula to begin attempt to find k.
1 ( 2) 3
1 (3 4) 2
k
k
or
2 ( 1) 3
3 4 1 2
k
k
(i.e. correct
substitution into gradient formula and equating to 3
2 ).
M1 2.2a 1st
Assumed
knowledge.
2k + 6 = −15 + 9k
21 = 7k
k = 3* (must show sufficient, convincing and correct working).
A1* 1.1b
(2)
1b Student identifies the coordinates of either A or B. Can be seen
or implied, for example, in the subsequent step when student
attempts to find the equation of the line.
A(5, −2) or B(1, 4).
B1 1.1b 2nd
Find the equation
of a straight line
given the gradient
and a point on the
line. Correct substitution of their coordinates into y = mx + b or
y − y1 = m(x − x1) o.e. to find the equation of the line.
For example,
3
2 52
b
or 3
2 52
y x
or
3
4 12
b
or 3
4 12
y x
M1 1.1b
3 11
2 2y x or 3 2 11 0x y
A1 1.1b
(3)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
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1c Midpoint of AB is (3, 1) seen or implied. B1 2.2a 3rd
Find the equation
of a perpendicular
bisector.
Slope of line perpendicular to AB is2
3, seen or implied. B1 2.2a
Attempt to find the equation of the line (i.e. substituting their
midpoint and gradient into a correct equation). For example,
2
1 33
b
or 2
1 33
y x
M1 1.1b
2 3 3 0x y or 3 2 3 0y x . Also accept any multiple of
2 3 3 0x y providing a, b and c are still integers.
A1 1.1b
(4)
(9 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
2a 11 ( 7) 18 9
6 4 10 5m
B1 1.1b 2nd
Find the equation
of a straight line
given two points. Correct substitution of (4, −7) or (−6, 11) and their gradient
into y = mx + b or y − y1 = m(x − x1) o.e. to find the equation
of the line. For example,
9
7 45
b
or 9
7 45
y x or
9
11 65
b
or 9
11 65
y x .
M1 1.1b
5y + 9x − 1 = 0 or −5y − 9x + 1 = 0 only A1 1.1b
(3)
2b 10,
9y x so
1,0
9A
. Award mark for 1
9x seen.
B1 1.1b 3rd
Solve problems
involving length
and area in the
context of straight
line graphs.
10,
5x y so
10,
5B
. Award mark for 1
5y seen.
B1 1.1b
Area = 1 1 1 1
2 5 9 90
B1 1.1b
(3)
(6 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
3 y = mx − 2 seen or implied. M1 1.1b 4th
Use the
discriminant to
determine
conditions for the
intersection of
circles and
straight lines.
Substitutes their y = mx − 2 into 2 26 8 4x x y y
22 6 2 8( 2) 4x x mx mx o.e.
M1 3.1a
Rearranges to a 3 term quadratic in x
(condone one arithmetic error).
2 21 (6 12 ) 16 0m x m x
M1 1.1b
Uses 2 4 0b ac , 2 26 12 4 1 16 0m m
M1 3.1a
Rearranges to 220 36 7 0m m or any multiple of this. A1 1.1b
Attempts solution using valid method. For example,
236 36 4 20 7
2 20m
M1 2.2a
9 29
10 5m or
9 2 29
10m
o.e. (NB decimals A0).
A1 1.1b
(7)
(7 marks)
Notes
Elimination of x follows the same scheme.2y
xm
leading to
2
22 26 8 4
y yy y
m m
This leads to 2 2 2 2(1 ) (4 6 8 ) 4 12 4 0m y m m y m m
Use of 2 4 0b ac gives 2
2 2 24 6 8 4 1 4 12 4 0m m m m m which reduces to
2 24 20 36 7 0.m m m m cannot equal 0, so this must be discarded as a solution for the final A mark.
2 4 0b ac could be used implicitly within the quadratic equation formula.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
4a Student attempts to complete the square twice for the first
equation (condone sign errors).
2 2
2 2
5 25 6 36 3
5 6 64
x y
x y
M1 2.2a 4th
Find the centre
and radius of a
circle, given the
equation, by
completing the
square. Centre (−5, 6) A1 3.2a
Radius = 8 A1 3.2a
Student attempts to complete the square twice for the second
equation (condone sign errors).
2 2 2
2 2 2
3 9 9
3 18
x y q q
x y q q
M1 2.2a
Centre (3, q) A1 3.2a
Radius = 218 q A1 3.2a
(6)
4b Uses distance formula for their centres and 80 . For
example,
22 2
5 3 6 80q
M1 2.2a 5th
Solve coordinate
geometry
problems
involving circles
in context. Student simplifies to 3 term quadratic. For example, 2 12 20 0q q
M1 1.1b
Concludes that the possible values of q are 2 and 10 A1 1.1b
(3)
(9 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
5a Student completes the square twice. Condone sign errors.
2 2
2 2
4 16 5 25 1 0
4 5 40
x y
x y
M1 1.1b 4th
Find the centre
and radius of a
circle, given the
equation, by
completing the
square.
So centre is (4, −5) A1 1.1b
and radius is 40 A1 1.1b
(3)
5b Substitutes x = 10 into equation (in either form).
2 210 8 10 10 1 0y y or 2 2
10 4 5 40y
M1 2.2a 5th
Solve coordinate
geometry
problems
involving circles
in context.
Rearranges to 3 term quadratic in y 2 10 21 0y y
(could be in completed square form 2
5 4y )
M1 1.1b
Obtains solutions y = −3, y = −7 (must give both). A1 1.1b
Rejects y = −7 giving suitable reason (e.g. −7 < −5) or ‘it
would be below the centre’ or ‘AQ must slope upwards’ o.e.
B1 2.3
(4)
5c 3 ( 5) 1=
10 4 3AQm
B1 1.1b 5th
Find the equation
of the tangent to a
given circle at a
specified point. 2
3lm (i.e. −1 over their AQm ) B1ft 2.2a
Substitutes their Q into a correct equation of a line. For
example,
3 3 10 b or 3 3 10y x
M1 1.1b
y = −3x + 27 A1 1.1b
(4)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
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5d 6
2AQ
uuur o.e. (could just be in coordinate form).
M1 3.1a 5th
Solve coordinate
geometry
problems
involving circles
in context.
2
6AP
uuuro.e. so student concludes that point P has
coordinates (2, 1).
M1 3.1a
Substitutes their P and their gradient 1
3 ( AQm from 5c) into a
correct equation of a line. For example,
1
1 23
b
or 1
1 23
y x
M1 2.2a
1 1
3 3y x
A1 1.1b
(4)
5e 40PA B1 3.1a 5th
Solve coordinate
geometry
problems
involving circles
in context.
Uses Pythagoras’ theorem to find 40
9EP .
B1 2.2a
Area of EPA = 1 40
402 9 (could be in two parts).
M1 1.1b
Area = 20
3
A1 1.1b
(4)
(19 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1 Correctly shows that either
f(3) = 0, f(−2) = 0 or 1
f2
= 0
M1 3.1a 4th
Divide
polynomials by
linear expressions
with no
remainder Draws the conclusion that (x – 3), (x + 2) or (2x + 1) must
therefore be a factor. M1 2.2a
Either makes an attempt at long division by setting up the
long division, or makes an attempt to find the remaining
factors by matching coefficients. For example, stating
2 3 23 2 13 6x ax bx c x x x
or
2 3 22 2 13 6x rx px q x x x
or
2 3 22 1 2 13 6x ux vx w x x x
M1 1.1b
For the long division, correctly finds the the first two
coefficients.
For the matching coefficients method, correctly deduces that
a = 2 and c = 2 or correctly deduces that r = 2 and q = −3 or
correctly deduces that u = 1 and w = –6
A1 2.2a
For the long division, correctly completes all steps in the
division.
For the matching coefficients method, correctly deduces that
b = 5 or correctly deduces that p = −5 or correctly deduces
that v = –1
A1 1.1b
States a fully correct, fully factorised final answer:
(x – 3)(2x + 1)(x + 2)
A1 1.1b
(6 marks)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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Notes
Other algebraic methods can be used to factorise h(x). For example, if (x – 3) is known to be a factor then
3 2 22 13 6 2 ( 3) 5 ( 3) 2( 3)x x x x x x x x by balancing (M1)
2(2 5 2)( 3)x x x by factorising (M1)
(2 1)( 2)( 3)x x x by factorising (A1)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
2a States or implies the expansion of a binomial expression to
the 8th power, up to and including the x3 term.
8 8 8 8 7 8 6 2 8 5 30 1 2 3( ) ...a b C a C a b C a b C a b
or
8 8 7 6 2 5 3( ) 8 28 56 ...a b a a b a b a b
M1 1.1a 5th
Understand and
use the general
binomial
expansion for
positive integer n
Correctly substitutes 1 and 3x into the formula:
2 38 8 7 6 5(1 3 ) 1 8 1 3 28 1 3 56 1 3 ...x x x x
M1 1.1b
Makes an attempt to simplify the expression (2 correct
coefficients (other than 1) or both 9x2 and 27x3).
8 8 2 3(1 3 ) 1 24 28 9 56 27 ...x x x x
M1
dep
1.1b
States a fully correct answer:
8 2 3(1 3 ) 1 24 252 1512 ...x x x x
A1 1.1b
(4)
2b States x = 0.01 or implies this by attempting the substitution:
2 3
1 24 0.01 252 0.01 1512 0.01 ...
M1 2.2a 5th
Find
approximations
using the
binomial
expansion for
positive integer n
Attempts to simplify this expression (2 calculated terms
correct):
1 + 0.24 + 0.0252 + 0.001512
M1 1.1b
1.266712 = 1.2667 (5 s.f.) A1 1.1b
(3)
(7 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
3a States or implies the expansion of a binomial expression to
the 9th power, up to and including the x3 term.
9 9 9 9 8 9 7 2 9 6 30 1 2 3( ) ...a b C a C a b C a b C a b
or 9 9 8 7 2 6 3( ) 9 36 84 ...a b a a b a b a b
M1 1.1a 5th
Use the binomial
expansion to find
arbitrary terms
for positive
integer n
Correctly substitutes 2 and px into the formula.
9(2 )px
2 39 8 7 62 9 2 36 2 84 2 ...px px px
M1 1.1b
Makes an attempt to simplify the expression (at least one
power of 2 calculated and one bracket expanded correctly).
9 2 2 3 3(2 ) 512 9 256 36 128 84 64 ...px px p x p x
M1dep 1.1b
States a fully correct answer:
9 2 2 3 3(2 ) 512 2304 4608 5376 ...px px p x p x
A1 1.1b
(4)
3bi States that 35376 84p M1ft 2.2a 5th
Understand and
use the general
binomial
expansion for
positive integer n
Correctly solves for p:
3 1 1
64 4p p
A1ft 1.1b
3bii Correctly find the coefficient of the x term:
12304 576
4
B1ft 1.1b 5th
Understand and
use the general
binomial
expansion for
positive integer n
Correctly find the coefficient of the x2 term: 2
14608 288
4
B1ft 1.1b
(4)
(8 marks)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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Notes
ft marks – pursues a correct method and obtains a correct answer or answers from their 5376 from part a.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
4a Attempt is made at expanding 5.p q Accept seeing the
coefficients 1, 5, 10, 10, 5, 1 or seeing
5 5 5 5 4 5 3 2
0 1 2
5 2 3 5 4 5 53 4 5 o.e.
p q C p C p q C p q
C p q C pq C q
M1 1.1a 5th
Understand and
use the general
binomial
expansion for
positive integer n
Fully correct answer is stated:
5 5 4 3 2 2 3 4 55 10 10 5p q p p q p q p q pq q
A1 1.1b
(2)
4b States that p, or the probability of rolling a 4, is
1
4
B1 3.3 5th
Use the binomial
expansion to find
arbitrary terms
for positive
integer n
States that q, or the probability of not rolling a 4, is3
4
B1 3.3
States or implies that the sum of the first 3 terms (or 1 − the
sum of the last 3 terms) is the required probability.
For example,
5 4 3 25 10p p q p q or 1 − 2 3 4 5(10 5 )p q pq q
M1 2.2a
5 4 3 21 1 3 1 3
5 104 4 4 4 4
or 1 15 90
1024 1024 1024
or
2 3 4 51 3 1 3 3
1 10 54 4 4 4 4
or 270 405 243
11024 1024 1024
M1 1.1b
Either53
512o.e. or awrt 0.104
A1 1.1b
(5)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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(7 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 8 8
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
5a Makes an attempt to interpret the meaning of f(5) = 0.
For example, writing 125 + 25 + 5p + q = 0 M1 2.2a 5th
Solve non-linear
simultaneous
equations in
context
5p + q = −150 A1 1.1b
Makes an attempt to interpret the meaning of f(−3) = 8.
For example writing −27 + 9 – 3p + q = 8 M1 2.2a
−3p + q = 26 A1 1.1b
Makes an attempt to solve the simultaneous equations. M1ft 1.1a
Solves the simultaneous equations to find that p = −22 A1ft 1.1b
Substitutes their value for p to find that q = −40 A1ft 1.1b
(7)
5b Draws the conclusion that (x – 5) must be a factor. M1 2.2a 5th
Divide
polynomials by
linear expressions
with a remainder
Either makes an attempt at long division by setting up the
long division, or makes an attempt to find the remaining
factors by matching coefficients. For example, stating:
2 3 25 22 40x ax bx c x x x
(ft their −22 or −40)
M1ft 1.1b
For the long division, correctly finds the the first two
coefficients.
For the matching coefficients method, correctly deduces that
a = 1 and c = 8
A1 2.2a
For the long division, correctly completes all steps in the
division.
For the matching coefficients method, correctly deduces that
b = 6
A1 1.1b
States a fully correct, fully factorised final answer:
(x – 5)(x + 4)(x + 2)
A1 1.1b
(5)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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(12 marks)
Notes
Award ft through marks for correct attempt/answers to solving their simultaneous equations.
In part b other algebraic methods can be used to factorise:
x – 5 is a factor (M1)
3 2 222 40 ( 5) 6 ( 5) 8( 5)x x x x x x x x by balancing (M1)
2( 6 8)( 5)x x x by factorising (M1)
( 4)( 2)( 5)x x x by factorising (A1 A1) (i.e. A1 for each factor other than (x – 5))
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
6 Considers the expression 2 13
162
x x either on its own or as
part of an inequality/equation with 0 on the other side.
M1 3.1a 6th
Complete
algebraic proofs
in unfamiliar
contexts using
direct or
exhaustive
methods
Makes an attempt to complete the square.
For example, stating:
213 169 256
4 16 16x
(ignore any (in)equation)
M1 1.1b
States a fully correct answer:
213 87
4 16x
(ignore any (in)equation)
A1 1.1b
Interprets this solution as proving the inequality for all values
of x. Could, for example, state that
213
04
x
as a number
squared is always positive or zero, therefore 2
13 870
4 16x
. Must be logically connected with the
statement to be proved; this could be in the form of an
additional statement. So 2 16 18 2
2x x x (for all x) or by
a string of connectives which must be equivalent to “if and
only if”s.
A1 2.1
(4)
(4 marks)
Notes
Any correct and complete method (e.g. finding the discriminant and single value, finding the minimum point by
differentiation or completing the square and showing that it is both positive and a minimum, sketching the graph
supported with appropriate methodology etc) is acceptable for demonstrating that 2 1316 0
2x x for all x.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
7a Makes an attempt to expand the binomial expression
3
1 x (must be terms in x0, x1, x2, x3 and at least 2 correct).
M1 1.1a 6th
Solve problems
using the
binomial
expansion (for
positive integer n)
in unfamiliar
contexts
(including the link
to binomial
probabilities)
2 3 2 31 3 1 3 3x x x x x A1 1.1b
0 < 3x A1 1.1b
x > 0* as required. A1* 2.2a
(4)
7b Picks a number less than or equal to zero, e.g. x = −1, and
attempts a substitution into both sides. For example,
2 3 2 3
1 3 1 1 1 3 1 3 1 1
M1 1.1a 5th
Use the binomial
expansion to find
arbitrary terms for
positive integer n Correctly deduces for their choice of x that the inequaltity does
not hold. For example, 3 ≮ 0
A1 2.2a
(2)
(6 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
1a 45A o seen or implied in later working. B1 1.1b 5th
Solve problems
involving surds in
context and
complete simple
proofs involving
surds
Makes an attempt to use the sine rule, for example, writing
sin120 sin 45
8 3 4 1x x
o o
M1 1.1b
States or implies that 3
sin1202
o and 2
sin 452
o A1 1.2
Makes an attempt to solve the equation for x.
Possible steps could include:
3 2
16 6 8 2x x
or
6 1
16 6 4 1x x
or
3 6
16 6 8 2x x
8 3 2 3 16 2 6 2x x or 4 6 6 16 6x x or
24 6 16 6 6 6x x
6 2 2 3 16 2 8 3x or 4 6 6 16 6x x or
12 3 8 6 3 6x x
M1ft 1.1b
6 2 2 3
16 2 8 3x
or
6 6
16 4 6x
or
3 6 3
8 6 12x
o.e.
A1ft 1.1b
Makes an attempt to rationalise the denominator by
multiplying top and bottom by the conjugate.
Possible steps could include:
3 2 3 8 2 4 3
8 2 4 3 8 2 4 3x
48 12 6 8 6 12
128 48x
36 4 6
80x
M1ft 1.1b
States the fully correct simplifed version for x. A1* 2.1
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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9 6
20x
*
(7)
1b States or implies that the formula for the area of a triangle is
1sin
2ab C or
1sin
2ac B or
1sin
2bc A
M1 1.1a 3rd
Understand and
use the general
formula for the
area of a triangle.
1 9 6 9 64 1 8 3 sin15 or 0.259
2 20 20awrt
or 1
1.29 1.58 sin15 or 0.2592
awrt awrt awrt .
M1 3.1a
Finds the correct answer to 2 decimal places. 0.26 A1 1.1b
(3)
(10 marks)
Notes
1a
Award ft marks for correct work following incorrect values for sin 120° and sin 45°
1b
Exact value of area is 124 11 6 6 2 .
200 If 0.26 not given, award M1M1A0 if exact value seen.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
2a States or implies that the angle at P is 74° B1 2.2a 4th
Solve triangle
problems in a
range of contexts
States or implies the use of the cosine rule. For example,
2 2 2 2 cosp q r qr P
M1 1.1a
Makes substitution into the cosine rule.
2 2 27 15 2 7 15cos74p o
M1ft 1.1b
Makes attempt to simplify, for example, stating 2 216.11...p M1ft 1.1b
States the correct final answer. QR = 14.7 km. A1 1.1b
(5)
2b States or implies use of the sine rule, for example, writing
sin sinQ P
q p
M1 3.1a 4th
Solve triangle
problems in a
range of contexts
Makes an attempt to substitute into the sine rule.
sin sin 74
15 14.7
Q
o
M1ft 1.1b
Solves to find Q = 78.77…° A1ft 1.1b
Makes an attempt to find the bearing, for example, writing
bearing = 180° – 78.77…° – 33°
M1ft 1.1b
States the correct 3 figure bearing as 068° A1ft 3.2a
(5)
(10 marks)
Notes
2a
Award ft marks for correct use of cosine rule using an incorrect initial angle.
2b
Award ft marks for a correct solution using their answer to part (a).
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
3 States 2 2sin cos 1x x or implies this by making a
substitution.
28 7cos 6 1 cosx x
M1 2.1 5th
Solve more
complicated
trigonometric
equations in a
given interval
such as ones
requiring use the
tan identity
(degrees)
Simplifies the equation to form a quadratic in cos x.
26cos 7cos 2 0x x
M1 1.1b
Correctly factorises this equation.
3cos 2 2cos 1 0x x or uses equivalent method for
solving quadratic (can be implied by correct solutions).
M1 1.1b
Correct solution. cos x 2
3 or
1
2
A1 1.1b
Finds one correct solution for x. (48.2°,60°, 311.8° or 300°). A1 1.1b
Finds all other solutions to the equation. A1 1.1b
(6)
(6 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
4a 2 3 or awrt −3.46 B1 1.1b 4th
Determine exact
values for
trigonometric
functions in all
four quadrants
(1)
4b Figure 1
Sine curve with
max 2 and min
−2
B1 2.2a 4th
Transform the
graphs of
trigonometric
functions using
stretches and
translations
Sine curve
translated 60° to
the right.
B1 2.2a
Sin curve cuts
x-axis at (−120°,
0) and
(60°, 0) and the
y-axis (0, − 3 ).
B1 2.2a
Asymptotes for
tan curve at x =
90° and
x = −90°
B1 1.1b
Tangent curve is
‘flipped’. B1 2.2a
Uses the value
of
−2 tan (−120°)
to deduce no
intersection in
3rd quadrant
(can be
implied).
B1 2.2a
Tangent curve
cuts x-axis at
(−180°, 0),
(0, 0) and
B1 1.1b
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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(180°, 0).
(7)
4c States that solutions to the equation 2sin( 60 ) tan 0x x o
will occur where the two curves intersect.
B1ft 3.1a 4th
Use intersection
points of graphs
to solve equations
4d States that there are two solutions in the given interval. A1 2.2a 4th
Use intersection
points of graphs
to solve equations
(2)
(10 marks)
Notes
4b
Ignore any portion of curve(s) outside −180° ⩽ x ⩽ 180°
4c
Award both marks for correctly stating that there are two solutions even if explanation is missing.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
5 Makes an attempt to begin solving the equation. For example,
states that
sin 3 20 4
4 3cos 3 20
o
o
M1 2.1 5th
Solve more
complicated
trigonometric
equations in a
given interval
such as ones
requiring use the
tan identity
(degrees)
Uses the identity sin
tancos
to write,
4 1
tan 3 204 3 3
o
M1 2.1
States or implies use of the inverse tangent. For example,
1 13 20 tan
3
o or 3 20 30 o o
M1 1.1b
Shows understanding that there will be further solutions in the
given range, by adding 180° to 30° at least once.
3 20 30 , 210 , 390 ,... o o o o (ignore any out of range
values).
M1 1.1b
Subtracts 20 and divides each answer by 3.
10 190 370, , ,...
3 3 3
o o o
(ignore any out of range values).
M1 1.1b
States the correct final answers to 1 decimal place.
3.3°, 63.3°, 123.3°cao
A1 1.1b
(6)
(6 marks)
Notes
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
6a Any reasonable explanation.
For example, the student did not correctly find all values of 2x
which satisfy 3
cos22
x . Student should have subtracted
150° from 360° first, and then divided by 2.
N.B. If insufficient detail is given but location of error is
correct then mark can be awarded from working in part (b).
B1 2.3 4th
Solve simple
trigonometric
equations in a
given interval
(degrees)
(1)
6b x = 75° B1 2.2a 4th
Solve simple
trigonometric
equations in a
given interval
(degrees)
x = 105° B1 2.2a
(2)
(3 marks)
Notes
6a
Award the mark for a different explanation that is mathematically correct, provided that the explanation is clear
and not ambiguous.
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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Q Scheme Marks AOs Pearson
Progression Step
and Progress
descriptor
7a Figure 2
Correct
shape of sine
curve
through
(0, 0).
B1 3.1a 4th
Transform the
graphs of
trigonometric
functions using
stretches and
translations Sine curve
has max
value of 1
2and min
value of 1
2
B1 3.1a
Sine curve
has a period
of 2 (can be
implied by 5
complete
cycles) and
passes
through
(1,0),
(2,0),...,
(10,0).
B1 3.1a
(3)
7b Student states that the buoy will be 0.4 m above the still water
level 10 times. B1 3.2a 7th
Use functions in
modelling
(including
critiquing)
(1)
Mark scheme Pure Mathematics Year 1 (AS) Unit Test 4: Trigonometry
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7c Sensible and correct reason. For example:
A buoy would not move up and down at exactly the same rate
during each oscillation.
The period of oscillation is likely to change each oscillation.
The maximum (or minimum) height is likely to change with
time.
Waves in the sea are not uniform.
B1 3.2b 7th
Use functions in
modelling
(including
critiquing)
(1)
(5 marks)
Notes
7c
Award the mark for a different explanation that is mathematically correct. For example, stating that the buoy
would not move exactly vertically each time.
Mark scheme Statistics Year 1 (AS) Unit Test 1: Statistical Sampling
©Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 1
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1a Observation or measurement of every member of a population. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
1b Two from:
takes a long time/costly
difficult to ensure whole population surveyed
cannot be used if the measurement process destroys the
item
can be hard to manage and analyse all the data.
B1
B1
1.2
1.2
3rd
Comment on the
advantages and
disadvantages of
samples and
censuses.
(2)
1c The list of unique serial numbers. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
1d A circuit board. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
(5 marks)
Notes
Mark scheme Statistics Year 1 (AS) Unit Test 1: Statistical Sampling
©Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 2
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
2a A complete collection of relevant individual people or items. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
2b Opportunity (convenience). B1 1.2 3rd
Understand quota
and opportunity
sampling.
(1)
2c Systematic. B1 1.2 3rd
Understand and
carry out
systematic
sampling.
(1)
2d Two from:
not random
electoral register may have errors
there may not be enough (500) households on the
register.
B1
B1
2.4
2.4
5th
Select and
critique a
sampling
technique in a
given context.
(2)
2e Either: random sampling – it avoids bias.
Or: quota sampling – no sampling frame required, continue until
all quotas filled.
B1 2.4 5th
Select and
critique a
sampling
technique in a
given context.
Either: Random sampling from people buying kitchen cleaners
in a large store, as this would reduce potential bias.
Or: Quota sampling from people based on a chosen set of ages
and genders who use kitchen cleaners, continuing until all quotas
are filled, as this would avoid the need for a sampling frame and
allow for a more clearly representative sample.
B1 2.4
(2)
(7 marks)
Mark scheme Statistics Year 1 (AS) Unit Test 1: Statistical Sampling
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Notes
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
3a One of:
to obtain a representative sample
large number of students compared to staff so would be
unfair to take same numbers of both.
B1 2.4 5th
Select and
critique a
sampling
technique in a
given context.
(1)
3b A list of the names of staff and students. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
3c A member of staff or a student. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
3d Find proportions for different strata out of 60 (either explained
or some sensible calculation seen).
M1 3.1b 3rd
Understand and
carry out stratified
sampling.
250
280´ 60 » 54 students,
30
280´ 60 » 6 staff.
A1 1.1b
Select at random using a random number generator. B1 1.1b
(3)
Mark scheme Statistics Year 1 (AS) Unit Test 1: Statistical Sampling
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3e One of:
absence on the day of the survey
sampling frame may contain errors.
B1 2.2b 5th
Select and
critique a
sampling
technique in a
given context.
(1)
(7 marks)
Notes
3d
Must be whole numbers for A1.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
4a All readers of the online newspaper. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
4b A list of readers who subscribe to the extra content. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
4c The subscribers. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
Mark scheme Statistics Year 1 (AS) Unit Test 1: Statistical Sampling
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4d Advantage: accuracy of the data, unbiased. B1 1.2 3rd
Comment on the
advantages and
disadvantages of
samples and
censuses.
Disadvantage: difficult to get a 100% response to a survey. B1 1.2
(2)
4e Natural variation in a small sample. B1 1.2 3rd
Comment on the
advantages and
disadvantages of
samples and
censuses.
Bias. B1 1.2
(2)
(7 marks)
Notes
Mark scheme Statistics Year 1 (AS) Unit Test 1: Statistical Sampling
©Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 6
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
5a Quota. B1 1.2 3rd
Understand quota
and opportunity
sampling.
(1)
5b Advantages – two from:
easy to get sample size
inexpensive
fast
can be stratified if required.
B1
B1
2.4
2.4
5th
Select and
critique a
sampling
technique in a
given context.
Disadvantages – one from:
not random
could be biased.
B1 2.4
(3)
5c Allocate each of the males a number from 1 to 300 B1 3.1b 3rd
Understand and
carry out simple
random sampling.
Use calculator or number generator to generate 50 different
random numbers from 1 to 300 inclusive. B1 1.1b
Select males corresponding to those numbers. B1 1.1b
(3)
5d 300 ÷ 50 = 6 B1 3.1b 3rd
Understand and
carry out simple
random sampling.
Use a random number generator to select the first name (or one
of the first 6 names on the list) as a starting point and then select
every 6th name thereafter to get 50 names.
B1 1.1b
(2)
(9 marks)
Notes
Mark scheme Statistics Year 1 (AS) Unit Test 1: Statistical Sampling
©Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 7
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
6a There are a very large number of bags. B1 2.4 3rd
Comment on the
advantages and
disadvantages of
samples and
censuses.
Bags are tested to destruction – there would be no bags left. B1 2.4
(2)
6b One value is less than 12 kg B1 2.4 3rd
Comment on the
advantages and
disadvantages of
samples and
censuses.
therefore claim is not reliable. B1 2.3
(2)
6c Different samples can lead to different conclusions due to
natural variations. B1 2.3 3rd
Comment on the
advantages and
disadvantages of
samples and
censuses.
Only a small sample taken so unreliable. B1 2.3
(2)
6d Larger sample. B1 2.4 3rd
Comment on the
advantages and
disadvantages of
samples and
censuses.
(1)
(7 marks)
Notes
Mark scheme Statistics Year 1 (AS) Unit Test 1: Statistical Sampling
©Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 8
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
7a (Quantitative) continuous. B1 1.2 1st
Understand the
difference
between
qualitative and
quantitative data.
(1)
7b A list of the first two digits of the date. B1 1.2 2nd
Understand the
vocabulary of
sampling.
(1)
7c Simple random sample B1 3.1b 5th
Select and
critique a
sampling
technique in a
given context.
using a random number generator to select five dates. B1 1.1b
(2)
7d Number ordered list of data. B1 3.1b 3rd
Understand and
carry out
systematic
sampling.
Use random number generator is choose first selected piece of
data. B1 3.1b
Then take every 6th value
187
30
æ
èçö
ø÷
B1 1.1b
(3)
7e Some data may be missing or erroneous. B1 3.2b 5th
Select and
critique a
sampling
technique in a
given context.
(1)
(8 marks)
Notes
Mark scheme Statistics Year 1 (AS) Unit Test 2: Data presentation and interpretation
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1a All points correctly plotted.
B2 1.1b 2nd
Draw and
interpret scatter
diagrams for
bivariate data.
(2)
1b The points lie reasonably close to a straight line (o.e.). B1 2.4 2nd
Draw and
interpret scatter
diagrams for
bivariate data.
(1)
1c f B1 1.2 2nd
Know and
understand the
language of
correlation and
regression.
(1)
1d Line of best fit plotted for at least 2.2 ⩽ x ⩽ 8 with D and F
above and B and C below.
M1 1.1a 4th
Make predictions
using the
regression line
within the range
of the data.
26 to 31 inclusive (must be correctly read from x = 7 from the
line of best fit). A1 1.1b
(2)
Mark scheme Statistics Year 1 (AS) Unit Test 2: Data presentation and interpretation
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1e It is reliable because it is interpolation (700 km is within the
range of values collected). B1 2.4 4th
Understand the
concepts of
interpolation and
extrapolation.
(1)
1f No, it is not sensible since this would be extrapolation (as
180 km is outside the range of distances collected). B1 2.4 4th
Understand the
concepts of
interpolation and
extrapolation.
(1)
(8 marks)
Notes
1a
First B1 for at least 4 points correct, second B1 for all points correct.
1b
Do not accept ‘The points lie reasonably close to a line’. Linear or straight need to be noted.
1e
Also allow ‘It is reliable because the points lie reasonably close to a straight line’.
1f
Allow the answer ‘It is sensible since even though it is extrapolation it is not by much’ provided that the
answer contains both ideas (i.e. it IS extrapolation but by a small amount compared to the given range of
data).
Mark scheme Statistics Year 1 (AS) Unit Test 2: Data presentation and interpretation
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
2a 19.5 +
60 2910
43
= 26.7093… (Accept awrt 26.7 miles)
M1
A1
1.1b
1.1b
3rd
Estimate median
values, quartiles
and percentiles
using linear
interpolation.
(2)
2b 3552.5
120x = 29.6041… o.e. (Accept awrt 29.6 miles)
B1 1.1b 4th
Calculate
variance and
standard deviation
from grouped data
and summary
statistics.
2138 043.13
120x
or
2 2138 043.13
120x
or
2120
119s
M1 1.1a
σ = 16.5515… (Accept awrt 16.6 miles)
(or s = 16.6208… = 16.6 miles)A1 1.1b
(3)
2c Any sensible reason linked to the shape of the distribution.
For example:
The distribution is (positively) skewed.
A few large distances (values) distort the mean.
B1 2.4 4th
Calculate means,
medians, quartiles
and standard
deviation.
(1)
Mark scheme Statistics Year 1 (AS) Unit Test 2: Data presentation and interpretation
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
2d Comparison of the two means.
For example, the mean distance for London is smaller than for
Devon.
Sensible interpretation comparing a county to a city.
For example, distance to work into one city may not be as far
as travelling to different cities in a county.
For example, commuters need to travel further to the cities in
Devon for work.
Comparison of the two standard deviations:
For example, the standard deviation for London is larger than
for Devon.
Sensible interpretation relating to variability/consistency
For example, there is more variability (less consistency) in the
commute distances from the Greater London station than from
the Devon station.
B1
B1
B1
B1
1.1b
2.2b
1.1b
2.2b
4th
Compare data sets
using a range of
familiar
calculations and
diagrams.
(4)
(10 marks)
Notes
2a
Allow consistent use of n + 1 (i.e. for median 60.5th rather than 60th), median = 26.8
2c
Candidates must compare both the means and standard deviations with interpretations for full marks.
Mark scheme Statistics Year 1 (AS) Unit Test 2: Data presentation and interpretation
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
3ai 37 (minutes). B1 1.1b 2nd
Draw and
interpret box
plots.
(1)
3aii Upper quartile or Q3 or third quartile or 75th
percentile or P75 B1 1.2 2nd
Understand
quartiles and
percentiles.
(1)
3b Outliers.
Sensible interpretation:
For example:
Observation that are very different from the other
observations (and need to be treated with caution).
Possible errors.
These two children probably walked/took a lot longer.
B1
B1
1.2
2.4
3rd
Recognise
possible outliers
in data sets.
(2)
3c 50 + 1.5 × 20 = 80 or 30 − 1.5 × 20 =0
Maximum value =55 < 80 minimum value = 25 > 0
No outliers.
M1
A1
B1
1.1b
1.1b
1.1b
4th
Calculate outliers
in data sets and
clean data.
(3)
3d The scale must be the same as for school A.
Figure 1
B1 1.1b 2nd
Draw and
interpret box
plots.
Box & whiskers 30, 37, 50
B1 1.1b
25, 55 B1 1.1b
(3)
Mark scheme Statistics Year 1 (AS) Unit Test 2: Data presentation and interpretation
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3e Three comparisons in context.
Comment on comparing averages.
For example, children from school A took less time on
average.
B3 2.2b 4th
Compare data sets
using a range of
familiar
calculations and
diagrams. Comment comparing consistency of times.
For example, there is less variation in the times for school A
than school B.
Comment on comparing symmetry:
For example,both positive skew (or neither symmetrical or
median closer to LQ (o.e.) for both). (Most children took a
short time with a few taking longer.)
Comment on comparing outliers.
For example, school A has two children whose times are
outliers (or errors) where as school B has no outliers.
(3)
(13 marks)
Notes
3c
Allow horizontal line through box.
Mark scheme Statistics Year 1 (AS) Unit Test 2: Data presentation and interpretation
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
4 467
200y
= −2.335 (seen or implied)
= 2.5 + 755.0
= 749.1625 (Accept awrt 749)
σy =
= 6.3594…
σx = 2.5 × 6.3594…
= 15.8986… (Accept awrt 15.9)
B1
M1
M1
A1
M1
A1
A1
M1
A1
1.1b
3.1a
1.1b
1.1b
1.1b
1.1b
3.1a
1.1b
1.1b
5th
Calculate the
mean and
standard deviation
of coded data.
(9)
(9 marks)
Notes
0.7555.2 yx
200
467
2
200
467
200
9179
Mark scheme Statistics Year 1 (AS) Unit Test 2: Data presentation and interpretation
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
5a Order the data.
125, 160, 169, 171, 175, 186, 210, 243, 250, 258, 390, 420
M1 1.1b 2nd
Understand
quartiles and
percentiles. Q3 = (250 + 258) = 254
A1 1.1b
(2)
5b Q3 +1.5(Q3 – Q1) = 254 + 1.5(254 – 170) M1 1.1b 4th
Calculate outliers
in data sets and
clean data.
= 380 A1 1.1b
Patients F (420) and B (390) are outliers (so may be suspected
by the doctor as smoking more than one packet of cigarettes
per day).
B1 3.2a
(3)
(5 marks)
Notes
2
1
Mark scheme Statistics Year 1 (AS) Unit Test 2: Data presentation and interpretation
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
6 Three comparisons in context:
For example:
Very much warmer in Beijing than Perth.
Both consistent in the temperatures.
Less rainfall in Beijing.
Less likely to have high rainfall in Beijing.
Rainfall in Beijing is consistently less than in Perth.
Evidence of use of a statistic from the boxplots:
For example:
Medians
Measure of a difference in medians
Mention of a particular outlier
B3
B1
2.4
2.4
4th
Compare data sets
using a range of
familiar
calculations and
diagrams.
For accurately reading data from boxplots. B1 2.4
(5)
(5 marks)
Notes
Mark scheme Statistics Year 1 (AS) Unit Test 3: Probability
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1a
2 + 3
total number of students=
5
30=
1
6or awrt 0.167
M1
A1
1.1b
1.1b
1st
Calculate
probabilities for
single events.
(2)
1b
4 + 2 + 5+ 3
total
M1 1.1b 3rd
Understand and
use Venn
diagrams for
multiple events. =
14
30 or
7
15 or awrt 0.467
A1 1.1b
(2)
1c 0 B1 1.1b 3rd
Understand and
use the definition
of mutually
exclusive in
probability
calculations.
No student reads both magazine A and magazine C. B1 1.1b
(2)
1d
P(C reads at least one magazine) =
6 + 3
20=
9
20
B1 1.1b 3rd
Understand and
use Venn
diagrams for
multiple events.
(1)
1e
P(B) =
10
30=
1
3, P(C) =
9
30=
3
10
B1 2.1 4th
Understand and
use the definition
of independence in
probability
calculations. P(B and C) =
3
30=
1
10, and
M1 2.2a
P(B) ´ P(C) =
1
3´
3
10=
1
10= P(B and C)
So yes, they are independent.
A1 2.4
(3)
(10 marks)
Mark scheme Statistics Year 1 (AS) Unit Test 3: Probability
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Notes
1e
Allow alternative using formal conditional probability: P(B) = 13
(B1). Finding P(B|C) = 3 1(3 6) 3
and
comparing with P(B) (M1). Correct conclusion (A1).
Or P(C) = 310
(B1). Finding P(C|B) = 3 3(2 3 5) 10
and comparing with P(C) (M1). Correct conclusion (A1).
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
2a
For shape and labels: 3 branches followed by 3,2,2
with some R, B and G seen.
M1
3.1a
3rd
Draw and use tree
diagrams with
three branches
and/or three
levels.
First set of branches correct. A1 1.1b
Second set of branches correct. A1 1.1b
(3)
2b P(Blue bead and a green bead) =
1 1 1 1 2 1
4 3 4 3 12 6
.
M1
A1
3.4
1.1b
3rd
Draw and use tree
diagrams with
three branches
and/or three
levels.
(2)
(5 marks)
Mark scheme Statistics Year 1 (AS) Unit Test 3: Probability
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Notes
2a
Allow 3 branches followed by 3, 3, 3 if 0 properly placed on redundant branches.
R B G labels can be implied on second set but only if order is consistent and probabilities correct.
Further sets of branches max M1 A1 A0 (2/3).
2b
M1 for or …+
1
4´
1
3
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
3a
Three closed curves and four in centre.
Evidence of subtraction (any one of 31, 36, 24, 41, 17 or 11).
Any three of 31, 36, 24, 41, 17 or 11 correct.
All correct.
Labels on sets, 16 and closed curve or box outside.
M1
M1
A1
A1
B1
3.1a
3.3
1.1b
1.1b
1.1b
3rd
Understand and
use Venn
diagrams for
multiple events.
(5)
3bi P(None of the 3 options)
=
16
180=
4
45or awrt 0.0889
B1 3.4 3rd
Understand and
use Venn
diagrams for
multiple events.
(1)
3bii P(Networking only)
=
17
180or awrt 0.0944
B1 3.4 3rd
Understand and
use Venn
diagrams for
multiple events.
(1)
Mark scheme Statistics Year 1 (AS) Unit Test 3: Probability
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3c
P(Takes all three options takes S and N) =
4
40=
1
10or 0.1
M1
A1
3.4
1.1b
3rd
Understand and
use Venn
diagrams for
multiple events.
(2)
(9 marks)
Notes
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
4a
Correct tree structure.
All labels correct.
All probabilities correct.
B1
B1
B1
3.1a
1.1b
1.1b
3rd
Draw and use tree
diagrams with
three branches
and/or three
levels.
(3)
4bi
1
3´
1
10=
1
30or equivalent.
M1
A1
3.4
1.1b
3rd
Draw and use tree
diagrams with
three branches
and/or three
levels.
(2)
Mark scheme Statistics Year 1 (AS) Unit Test 3: Probability
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
4bii Car NL + Bike NL + Foot NL
=1
2´
4
5
æ
èçö
ø÷+
1
6´
3
5
æ
èçö
ø÷+
1
3´
9
10
æ
èçö
ø÷
M1 3.4 3rd
Draw and use tree
diagrams with
three branches
and/or three
levels. =
4
5or equivalent.
A1 1.1b
(2)
(7 marks)
Notes
4bii
ft from their tree diagram. Allow one error for M1.
Can also be found from
1-1
2´
1
5
æ
èçö
ø÷+
1
6´
2
5
æ
èçö
ø÷+
1
3´
1
10
æ
èçö
ø÷æ
èç
ö
ø÷
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
5a
Tree (both sections) and labels
0.85, 0.15
0.03, 0.97, 0.06, 0.94
B1
B1
B1
3.1a
1.1b
1.1b
2nd
Draw and use
simple tree
diagrams with
two branches and
two levels.
(3)
5b P(Not faulty) = (0.85 × 0.97) + (0.15 × 0.94)
= 0.9655
M1
M1dep
A1
3.4
1.1b
1.1b
2nd
Draw and use
simple tree
diagrams with
two branches and
two levels.
(3)
Mark scheme Statistics Year 1 (AS) Unit Test 3: Probability
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(6 marks)
Notes
5b
M1 for either 0.85 × 0.97 or 0.15 × 0.94 (ft from their tree diagram) and M1 (dep) for adding two such probabilities
(allow one error).
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
6a Find total frequency = åwidth ´ frequency density
= (5 2) + (4 4) + (4 6) + (7 5) + (15 1) = 100
P(Takes longer than 18 mins) =
35+15
“100”=
50
100=
1
2 or
equivalent.
M1
A1
M1
A1
3.1a
1.1b
3.1a
1.1b
2nd
Calculate
probabilities from
relative frequency
tables and real
data.
(4)
6b
1
3´15 = 5
P(Takes less than 30 mins) =
10 +16 + 24 + 35+ 5
100=
90
100=
9
10
or equivalent.
M1
M1
A1
2.2b
1.1b
1.1b
2nd
Calculate
probabilities from
relative frequency
tables and real
data.
(3)
(7 marks)
Notes
6a
M1 for attempt to find total frequency by adding at least three “width frequency density” terms (which may
contain errors).Alternative: M1 for 2
15 103
. M1 for 1 −10
100
" ".
" " A1 for
9
10 o.e.
Mark scheme Statistics Year 1 (AS) Unit Test 3: Probability
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
7a Total frequency = 120
P(Less than 17 cm) 52 5 57
120 120
or equivalent or 0.475
B1
M1
A1
3.1a
1.1b
1.1b
2nd
Calculate
probabilities from
relative frequency
tables and real
data.
(3)
7b P(Between 12 cm and 18 cm)
52 15 67
120 120
or awrt 0.558
Assumption: foot lengths between 17 and 19 are uniformly
distributed.
M1
A1
B1
2.2b
1.1b
3.5b
2nd
Calculate
probabilities from
relative frequency
tables and real
data.
(3)
(6 marks)
Notes
Mark scheme Statistics Year 1 (AS) Unit Test 4: Statistical Distributions
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1a k(16 – 9) + k(25 – 9) + k(36 – 9) (or 7k + 16k + 27k). M1 2.1 4th
Model simple
discrete random
variables as
probability
distributions.
= 1 M1 1.1b
Þ k =
1
50(answer given).
A1* 1.1b
(3)
1b
x 4 5 6
P(X = x)
7
50
16
50
27
50
Note: decimal values are 0.14, 0.32, 0.54 respectively.
B1
B1
2.5
1.1b
4th
Calculate
probabilities from
discrete
distributions.
(2)
(5 marks)
Notes
1b
Ignore any extra columns with 0 probability. Otherwise –1 for each. If 4, 5 or 6 missing B0B0.
Mark scheme Statistics Year 1 (AS) Unit Test 4: Statistical Distributions
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
2a 0.15 + 0.15 + + + 0.1 + 0.1 = 2 + 0.5 = 1 M1 1.1b 4th
Calculate
probabilities from
discrete
distributions.
= 0.25 A1 1.1b
(2)
2b P(–1 ⩽X < 2) = P(–1) + P(0) + P(1) = 0.6 B1 1.1b 4th
Calculate
probabilities from
discrete
distributions.
(1)
2c P(X > −2.3) = P(−2) + P(−1) + P(0) + P(1) + P(2) = 0.85 B1 1.1b 4th
Calculate
probabilities from
discrete
distributions.
(1)
(4 marks)
Notes
Mark scheme Statistics Year 1 (AS) Unit Test 4: Statistical Distributions
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
3a 2k + k + 0 + k = 1 M1 2.1 4th
Calculate
probabilities from
discrete
distributions.
Þ4k = 1, so k = 0.25 (answer given). A1* 1.1b
(2)
3b P(X1 + X2 = 5) = P(X1 = 3 and X2 = 2) + P(X1 = 2 and X2 = 3)
= 0 + 0 = 0 (answer given).
B1* 2.4 4th
Calculate
probabilities from
discrete
distributions.
(1)
3c
x1 + x2 0 1 2 3 4 5 6
P(X1 + X2) 0.25 0.25 0.0625 0.25 0.125 0 0.0625
M1
A1
A1
2.5
1.1b
1.1b
4th
Calculate
probabilities from
discrete
distributions.
(3)
3d P(1.3 ⩽ X1 + X2 ⩽ 3.2) = P(X1 + X2 = 2) + P(X1 + X2 = 3) M1 3.4 4th
Calculate
probabilities from
discrete
distributions. = 0.0625 + 0.25 = 0.3125 or
5
16
A1ft 1.1b
(2)
(8 marks)
Notes
3b
Must show that 5 can only be obtained from 2 and 3 or 3 and 2, and so must use P(X = 2) = 0 but condone
explanation in words.
3c
M1 for correct set of values for X1 + X2. Condone omission of 5 column.
A1 for correct probabilities for 0, 2 and 6. A1 for others. Equivalent fractions are 1 1 1 1 1 1, , , , ,4 4 16 4 8 16
Mark scheme Statistics Year 1 (AS) Unit Test 4: Statistical Distributions
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
4a Let X be the random variable the number of games Amir loses.
X ~ B(9, 0.2)
P(X = 3) = 0.17616… = 0.176 to 3 sf from calculator
B1
B1
3.3
1.1b
5th
Calculate
binomial
probabilities.
(2)
4b P( 4) X „ M1 3.4 6th
Use statistical
tables and
calculators to find
cumulative
binomial
probabilities.
= awrt 0.980 from calculator A1 1.1b
(2)
(4 marks)
Notes
4a
P( 3) P ) ( 2X X„ „ = 0.9144 – 0.7382
or
9!
3!6!(0.2)3(0.8)6
or 9 C
3´ 0.23 ´ 0.86
or 84 ´ 0.23 ´ 0.86
4b
0.98 is M1A0
Mark scheme Statistics Year 1 (AS) Unit Test 4: Statistical Distributions
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
5a X ~ B(20, 0.05)
B1 for binomial
B1 for 20 and 0.05
B1
B1
3.1b
3.1b
5th
Understand the
binomial
distribution (and
its notation) and
its use as a model.
(2)
5b P(X = 0) = 0.358 (awrt) B1
A1
3.4
1.1b
5th
Calculate
binomial
probabilities.
(2)
5c P(X > 4) = 1 – P( 4) X „
= 1 – 0.9974
M1 3.4 6th
Use statistical
tables and
calculators to find
cumulative
binomial
probabilities.
= 0.0026 (2 s.f.) (answer given) A1* 1.1b
(2)
(6 marks)
Notes
5b
P(X = 0) = 0.9520
Mark scheme Statistics Year 1 (AS) Unit Test 4: Statistical Distributions
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
6a X ~ B(15, 0.5)
B1 for binomial
B1 for 15 and 0.5
B1
B1
3.1b
3.1b
5th
Understand the
binomial
distribution (and
its notation) and
its use as a model.
(2)
6bi from calculator P(X = 8) = 0.19638… M1
A1
3.4
1.1b
5th
Calculate
binomial
probabilities.
(2)
6bii P(X …4) = 1 – P(X „ 3)
= 1 – 0.0176
M1 3.4 6th
Use statistical
tables and
calculators to find
cumulative
binomial
probabilities.
= awrt 0.982 or
503
512
A1 1.1b
(2)
(6 marks)
Notes
6bi
P(X = 8) = P(X … 8) – P(X „ 7) = 0.6964 – 0.5
or
15!
8!7!0.58(1- 0.5)7
or 15 C
8´ 0.58 ´ 0.57
or 6435´ 0.515
= awrt 0.196 or
6435
32768
Mark scheme Statistics Year 1 (AS) Unit Test 4: Statistical Distributions
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
7a Binomial (distribution). B1
1.2
5th
Understand the
binomial
distribution (and
its notation) and
its use as a model. Each plate is either blue or not blue, independently of each
other, with constant probability and there is a fixed number of
them.
B1 2.4
(2)
7b X ~ B(10, 0.06) (could be seen in part a) B1 2.5 5th
Calculate
binomial
probabilities.
P(X > 2) = 1 – P(X „
= 1 – 0.981162163… from calculator
M1 3.4
= awrt0.0188378… A1 1.1b
(3)
(5 marks)
Notes
7a
Ignore any parameter values given for the first B1. For second B1 all four points must be made with some context.
Mark scheme Statistics Year 1 (AS) Unit Test 4: Statistical Distributions
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
8a There is a fixed number of trials. B1 1.2 5th
Understand the
binomial
distribution (and
its notation) and
its use as a model.
Each trial results in 1 of 2 outcomes, ‘success’ and ‘failure’. B1 1.2
Probability of success on each trial is the same. B1 1.2
The trials are independent. B1 1.2
(4)
8bi P(X = 5) =
2
7; P(X 5) =
5
7
(either may be implied in either part)
B1
B1
3.3
1.1b
5th
Calculate
binomial
probabilities. Idea of five failures followed by a success (or 0 successes out of
five and then a success) seen or implied. M1 3.4
P(5 on sixth throw) =
55 2
7 7
M1 1.1b
= awrt 0.0531 A1 1.1b
(5)
8bii
B 8,2
7
æ
èçö
ø÷ – or
B 8,5
7
æ
èçö
ø÷– seen or implied.
M1 3.3 5th
Calculate
binomial
probabilities.
P(X = 3) = 0.24285… from calculator M1
A1
1.1b
1.1b
(3)
(12 marks)
Notes
8bii
P(exactly 3 fives in first eight throws) =
8!
5!3!
2
7
æ
èçö
ø÷
3
5
7
æ
èçö
ø÷
5
o.e.
Mark scheme Statistics Year 1 (AS) Unit Test 5: Statistical Hypothesis Testing
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1a Two from:
Each bolt is either faulty or not faulty.
The probability of a bolt being faulty (or not) may be
assumed constant.
Whether one bolt is faulty (or not) may be assumed
to be independent (or does not affect the probability
of) whether another bolt is faulty (or not).
There is a fixed number (50) of bolts.
A random sample.
B2
1.2
1.2
5th
Understand the
binomial
distribution (and
its notation) and
its use as a model.
(2)
1b Let X represent the number of faulty bolts.
X~B(50, 0.25)
P(X ⩽ 6) = 0.0194
P(X ⩽ 7) = 0.0453
P(X ⩾ 19) = 0.0287
P(X ⩾ 20) = 0.0139
M1
M1dep
3.4
1.1b
5th
Find critical
values and critical
regions for a
binomial
distribution.
Critical Region is X ⩽ 6 X ⩾ 20 A2 1.1b
1.1b
(4)
(6 marks)
Notes
1a
Each comment must be in context for its mark.
Mark scheme Statistics Year 1 (AS) Unit Test 5: Statistical Hypothesis Testing
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
2a The set of values of the test statistic for which the null
hypothesis is rejected in a hypothesis test.
B2
1.2
1.2
5th
Understand the
language of
hypothesis
testing.
(2)
2b P(X ⩾ 15) = 1− 0.9831 = 0.0169
P (X ⩾ 16) = 1 – 09936 = 0.0064
M1 1.1b 5th
Find critical
values and critical
regions for a
binomial
distribution.
Critical region is 16 ⩽ X (⩽ 30) A1 1.1b
Probability of rejection is 0.0064 A1 1.1b
(3)
2c Not in critical region therefore insufficient evidence to
reject H0. B1 2.2b 6th
Interpret the
results of a
binomial
distribution test in
context.
There is insufficient evidence at the 1% level to suggest
that the value of p is bigger than 0.3. B1 3.2a
(2)
(7 marks)
Notes
2c
Conclusion must be in context (i.e. use p), mention the significance level and be non-assertive.
Mark scheme Statistics Year 1 (AS) Unit Test 5: Statistical Hypothesis Testing
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
3a P(X ⩽ 1) = 0.0076 and P (X ⩽ 2) = 0.0355 M1 1.1b 5th
Find critical values
and critical regions
for a binomial
distribution.
P(X ⩾ 10) = 1 – 0.9520 = 0.0480 and
P(X ⩾ 11) = 1 – 0.9829 = 0.0171
A1 1.1b
Critical region is X ⩽ 1 11 ⩽ X (⩽ 20) A1 1.1b
(3)
3b Significance level = 0.0076 + 0.0171
= 0.0247 or 2.47%
B1 1.1b 6th
Calculate actual
significance levels
for a binomial
distribution test.
(1)
3c Not in critical region therefore insufficient evidence to reject
H0. B1 2.2b 6th
Interpret the results
of a binomial
distribution test in
context.
There is insufficient evidence at the 5% level to suggest that the
value of p is not 0.3. B1 3.2a
(2)
(6 marks)
Notes
3c
Conclusion must contain context and non-assertive for first B1.
Mark scheme Statistics Year 1 (AS) Unit Test 5: Statistical Hypothesis Testing
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
4a X~B(28, 0.37) M1 3.4 5th
Find critical values
and critical regions
for a binomial
distribution
P(X ⩾ 15) = 1 – 0.9454 = 0.0546 and
P(X ⩾ 16) = 1 – 0.9762 = 0.0238
M1dep 1.1b
Critical region is X ⩾ 16 A1 1.1b
(3)
4b In critical region therefore sufficient evidence to reject H0 B1 2.2b 6th
Interpret the results
of a binomial
distribution test in
context.
There is sufficient evidence at the 5% level to suggest that the
value of p is bigger than 0.37. B1 3.2a
(2)
(5 marks)
Notes
4a
First M1 for correct distribution seen or implied. Second M1 (dependent on first) for evidence that correct
probabilities for either critical value examined.
4b
Conclusion must contain context and non-assertive for first B1.
Mark scheme Statistics Year 1 (AS) Unit Test 5: Statistical Hypothesis Testing
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
5a H0: p = 0.2
H1: p > 0.2
B1 2.5 5th
Carry out 1-tail
tests for the
binomial
distribution. Let X represent the number of times the taxi is late.
X~B(5, 0.2) seen or implied.
M1 3.3
Either
P(X ⩾ 3) = 1 – P(X ⩽ 2) = 1 – 0.9421
= 0.0579
0.0579 > 0.05
There is insufficient evidence at the 5% significance level that
there is an increase in the number of times the taxi/driver is
late.
Or
P(X ⩾ 3) = 1 – P(X ⩽ 2) = 0.0579
P(X ⩾ 4) = 1 – P(X ⩽ 3) = 0.0067
So critical region is X ⩾ 4
3 < 4 or 3 is not in the critical region
So there is insufficient evidence at the 5% significance level
that there is an increase in the number of times the taxi/driver is
late.
M1
A1
B1
B1
M1
A1
B1
B1
1.1b
1.1b
1.1b
3.2a
1.1b
1.1b
1.1b
3.2a
(6)
5b Two sensible reasons. For example,
Different time of the day Linda travels to work.
More traffic on different days (e.g. Monday morning,
Friday afternoon).
Weather conditions.
Road works.
B2
2.2b
2.2b
5th
Understand the
binomial
distribution (and
its notation) and
its use as a model.
(2)
(8 marks)
Notes
Conclusion must be non-assertive.
Mark scheme Statistics Year 1 (AS) Unit Test 5: Statistical Hypothesis Testing
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
6a Let X represent the number of bowls with minor defects
(seen or implied).
XB(25, 0.2) may be implied
P(X ⩽ l) = 0.0274
P(X = 0) = 0.0038
P(X ⩽ 8) = 0.9532 P(X ⩾ 9) = 0.0468
P(X ⩽ 9) = 0.9827 P(X ⩾ 10) = 0.0173
Critical region is X = 0 X ⩾ 10
M1
M1dep
A1
M1
A2
3.4
1.1b
1.1b
1.1b
1.1b
1.1b
5th
Find critical
values and critical
regions for a
binomial
distribution.
(6)
6b Significance level = 0.0038 + 0.0173
= 0.0211 or 2.11%
B1 1.2 6th
Calculate actual
significance
levels for a
binomial
distribution test.
(1)
6c H0: p = 0.2; H1: p < 0.2 B1 2.5 5th
Carry out 1-tail
tests for the
binomial
distribution.
Let Y represent number of bowls with minor defects
(Under H0) Y~B(20, 0.2) (may be implied)
M1 3.4
Either
P(Y ⩽ 2) = 0.2061
0.2061 > 0.1 (or 10%)
Insufficient evidence at the 10% level to suggest that the
proportion of defective bowls has decreased.
Or
P(Y ⩽ 2) = 0.2061
P(Y ⩽ 1) = 0.0692 so critical region is Y ⩽ 1
Insufficient evidence at the 10% level to suggest that the
proportion of defective bowls has decreased.
0.2061 > 0.10 or 0.7939 < 0.9
B1
M1
A1
B1
M1
A1
1.1b
1.1b
3.2b
1.1b
1.1b
3.2a
(5)
(12 marks)
Mark scheme Statistics Year 1 (AS) Unit Test 5: Statistical Hypothesis Testing
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Notes
6a
M1 for examining probabilities for on both sides for either critical value, A1 for each correct pair.
6c
Conclusion must be non-assertive.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
7 H0: p = 0.25, H1: p> 0.25 B1 2.5 5th
Carry out 1-tail
tests for the
binomial
distribution.
Let X represent the number of seeds that germinate.
(Under H0) X~B(25, 0.25)
M1 3.4
P(X ⩾ 10) = 1 – P(X ⩽ 9) = 0.0713 M1 1.1b
> 0.05 A1 1.1b
10 is not in critical region therefore insufficient evidence to
reject H0. B1 2.2b
There is insufficient evidence at the 5% level to suggest that the
book has underestimated the probability. (o.e.) B1 3.2a
(6 marks)
Notes
Mark scheme Mechanics Year 1 (AS) Unit Test 6: Quantities and Units in Mechanics
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 1
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1a States correct answer: 5.3 (m s−1) B1 2.2a 4th
Understand the
difference
between a scalar
and a vector.
(1)
1b States correct answer: −4.8 (m s−1) B1 2.2a 4th
Understand the
difference
between a scalar
and a vector.
(1)
1c States correct answer: −30 (m) B1 2.2a 4th
Understand the
difference
between a scalar
and a vector.
(1)
(3 marks)
Notes
Mark scheme Mechanics Year 1 (AS) Unit Test 6: Quantities and Units in Mechanics
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
2ai States that x = 0 needs to be substituted or implies it by writing
2
1.7 0.18 0 0.01 0h
M1 3.1b 3rd
Understand how
mechanics
problems can be
modelled
mathematically.
Correctly substitutes x = 0 to get h = 1.7 (m) A1 1.1b
(2)
2aii States that x = 7 needs to be substituted or implies it by writing
h = 1.7 + 0.18(7) – 0.01(7)2 M1 3.1b 3rd
Understand how
mechanics
problems can be
modelled
mathematically.
Correctly substitutes x = 7 to get h = 2.47 (m)
Accept awrt 2.5 (m)
A1 1.1b
(2)
2b Understands that the ball will hit the ground when h = 0 or
writes 2
1.7 0.18 0.01 0x x
M1 3.1b 3rd
Understand how
mechanics
problems can be
modelled
mathematically.
Realises that the quadratic formula is needed to solve the
quadratic. For example a = 0.01, b = 0.18, c = 1.7 seen, or
makes attempt to use the formula:
20.18 0.18 4 0.01 1.7
2 0.01x
M1 1.1b
Simplifies the 2 4b ac part to get 0.1004 or shows
0.18 0.1004
0.02x
M1 1.1b
Calculates x = 24.84… (m)
Accept awrt 24.8 (m)
Does not need to show that 6.84...x (m)
A1 1.1b
States that the ball will be called ‘in’, or says, for example, yes
as 24.84… < 25. B1 3.2a
(5)
Mark scheme Mechanics Year 1 (AS) Unit Test 6: Quantities and Units in Mechanics
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2c 2 km 1000 m 1min
1min 1km 60sec
Award 1 method mark for multiplication by 1000 and 1
method mark for division by 60.
M2 1.1b 3rd
Understand how
mechanics
problems can be
modelled
mathematically.
33.3 (m s−1) or 33.3
(m s−1) A1 1.1b
(3)
(12 marks)
Notes
2ai
Award both marks for a correct final answer.
2aii
Award both marks for a correct final answer.
2b
0.01, 0.18, 1.7a b c is also acceptable.
2b
Award the third method mark even if this step is not seen, providing the final answer is correct.
Mark scheme Mechanics Year 1 (AS) Unit Test 6: Quantities and Units in Mechanics
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
3a Understands that the pole vaulter will land when h = 0 or
writes 21125 12 0
60x x
M1 3.1b 3rd
Understand how
mechanics
problems can be
modelled
mathematically.
Correctly factorises to get 125 12 0x x o.e. M1 1.1b
Solves to get 125
10.41...12
x (m)
Accept awrt 10.4 (m)
A1 1.1b
(3)
3b States that the greatest height will occur when x = 5.20…(m) M1 3.1b 3rd
Understand how
mechanics
problems can be
modelled
mathematically.
Makes an attempt to substitute x = 5.20…into the equation for
h. For example, 21125 5.20... 12 5.20...
60h seen.
M1 1.1b
h = 5.42…(m)
Accept awrt 5.4 (m)
A1 ft 1.1b
(3)
3c States h = 4.9 or states that 21
125 12 4.960
x x M1 3.1b 3rd
Understand how
mechanics
problems can be
modelled
mathematically.
Simplifies this to reach 2
12 125 294 0x x o.e. M1 1.1b
Realises that the quadratic formula is needed to solve the
quadratic. For example 12, 125, 294a b c seen, or
makes attempt to use the formula:
2125 125 4 12 294
2 12x
M1 1.1b
Simplifies the 2
4b ac part to get 1513 or shows
125 1513
24x
M1 1.1b
x = 6.82…(m)
Accept awrt 6.8 (m)
A1 1.1b
x = 3.58… (m)
Accept awrt 3.6 (m)
A1 1.1b
The pole vaulter can leave the ground between 3.6 m and
6.8 m from the bar. B1 3.2a
(7)
Mark scheme Mechanics Year 1 (AS) Unit Test 6: Quantities and Units in Mechanics
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3di Allows the person to be treated as a single mass and allows the
effects of rotational forces to be ignored. B1 3.4 3rd
Understand
assumptions
common in
mathematical
modelling.
(1)
3dii The effects of air resistance can be ignored. B1 3.4 3rd
Understand
assumptions
common in
mathematical
modelling.
(1)
(15 marks)
Notes
3b
For the first method mark, accept their answer to part a divided by 2. Continue to award marks for a correct answer
using their initial incorrect value.
3c
Accept 3.6 ⩽ x ⩽ 6.8
Mark scheme Mechanics Year 1 (AS) Unit Test 6: Quantities and Units in Mechanics
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 6
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
4a Makes an attempt to find the distance from A to B. For
example, 2 2
28 80 is seen.
M1 3.1b 4th
Find the
magnitude and
direction of a
vector quantity. Makes an attempt to find the distance from B to C. For
example, 2 2
130 15 is seen.
M1 3.1b
Demonstrates an understanding that these two values need to
be added. For example, 84.75… + 130.86… is seen. M1 1.1b
215.62… (m)
Accept anything which rounds to 216 (m)
A1 1.1b
(4)
4b States that 102 95AC i juuur
(m)
Award one point for each value.
B2 3.1b 4th
Find the
magnitude and
direction of a
vector quantity. States or implies that 95
tan102
M1 1.1b
Finds 42.96...
Accept awrt 43.0°
A1 1.1b
(4)
(8 marks)
Notes
Mark scheme Mechanics Year 1 (AS) Unit Test 6: Quantities and Units in Mechanics
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 7
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
5a Makes an attempt to find the absolute value. For example,
2 2
14 22 is seen.
M1 3.1b 4th
Find the
magnitude and
direction of a
vector quantity. Simplifies to 680 M1 1.1b
Finds speed = 26.07… (ms−1)
Accept awrt 26.1 (ms−1)
A1 1.1b
(3)
5b States that
22tan
14
M1 1.1b 4th
Find the
magnitude and
direction of a
vector quantity.
Finds the value of θ, θ = 57.52… A1 1.1b
Demonstrates that the angle with the unit j vector is
90 – 57.52… M1 1.1b
Finds 32.47… (°)
Accept awrt 32.5(°)
A1 1.1b
(4)
5c Ignore the value of friction between the hockey puck and the
ice. B1 3.4 3rd
Understand
assumptions
common in
mathematical
modelling.
(1)
5d
3
1.4 g 1kg 100cm 100cm 100cm1000g 1m 1m 1m1cm
Award 1 method mark for division by 1000 and 1 method
mark for multiplication by 100 only once and the final method
mark for multiplication by 100 three times.
M3 1.1b 4th
Know derived
quantities and SI
units.
1400 kg m−3 A1 1.1b
(4)
(12 marks)
Notes
5b
Award all 4 marks for a correct final answer. Award 2 marks for a student stating 14
tan22
, and then either
making a mistake with the inverse or subtracting that answer from 90.
Mark scheme Mechanics Year 1 (AS) Unit Test 7: Kinematics 1 (constant acceleration)
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Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
1a Figure 1
General shape of the
graph is correct. i.e.
horizontal line,
followed by negative
gradient, followed by
a positive gradient.
M1 3.3 4th
Use and interpret
graphs of velocity
against time.
Vertical axis labelled
correctly. A1 1.1b
Horizontal axis
labelled correctly. A1 1.1b
(3)
1b Makes an attempt to find the area of trapezoidal section where
the car is decelerating. For example, 15 104
T is seen.
M1 1.1b 4th
Calculate and
interpret areas
under velocity–
time graphs. Makes an attempt to find the area of the trapezoidal section
where the car is accelerating. For example, 3
10 204
T is
seen.
M1 1.1b
States that 25 90
15 1312.54 4
T TT
M1 1.1b
Solves to find the value of T: T = 30 (s). A1 1.1b
(4)
(7 marks)
Notes
1a
Accept the horizontal axis labelled with the correct intervals.
1b
Award full marks for correct final answer, even if some work is missing.
Mark scheme Mechanics Year 1 (AS) Unit Test 7: Kinematics 1 (constant acceleration)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 2
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
2a Velocity = acceleration × time seen or implied. M1 3.1b 4th
Use and interpret
graphs of velocity
against time.
Velocity = 11 × 8 = 88 m s−1 A1 1.1b
Figure 2
General shape of the
graph is correct. i.e.
positive gradient,
followed by horizontal
line, followed by
negative gradient not
returning to zero.
M1 3.3
Vertical axis labelled
correctly. A1 1.1b
Horizontal axis labelled
correctly. A1 1.1b
(5)
2b Makes an attempt to find the area of the trapezoidal section.
For example, 1
2 88 402
is seen.
M1 1.1b 4th
Calculate and
interpret areas
under velocity–
time graphs. Demonstrates an understanding that the three areas must total
1404. For example, 1 1
8 88 88 2 88 40 14042 2
T or
352 88 128 1404T is seen.
M1 2.1
Correctly solves to find T = 10.5 (s). A1 1.1b
(3)
(8 marks)
Notes
2a
Accept the horizontal axis labelled with the correct intervals.
2b
Award full marks for correct final answer, even if some work is missing.
Mark scheme Mechanics Year 1 (AS) Unit Test 7: Kinematics 1 (constant acceleration)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 3
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
3a v ua
t
seen or implied.
M1 3.1b 5th
Use equations of
motion to solve
problems in
familiar contexts. Finds the value of a:
20 6 140.4
35 35a
m s−2
A1 1.1b
(2)
3b Use the fact that 1
2
4
3
t
t to write 3t1 = 4t2 or 3t1 − 4t2 = 0 or
equivalent.
M1 1.1b 5th
Use equations of
motion to solve
problems in
familiar contexts. States or implies that t1 + t2 = 35 M1 3.1b
Solves to find t1 = 20 or t2 = 15. Could use substitution or
simultaneous equations. Does not need to find both values for
mark to be awarded as either value can be used going forward.
A1 1.1b
Use v = u + at to write either x = 6 + 0.4(20)
or 20 = x + 0.4(15) M1 2.2a
Finds x = 14 (m s−1). A1ft 1.1b
(5)
3c States or implies that
2
u vs t
M1 2.2a 5th
Use equations of
motion to solve
problems in
familiar contexts. Finds the value of s: 6 20
35 4552
s
(m). A1 1.1b
(2)
(9 marks)
Notes
3b
Award ft marks for a correct answer using their value from part a.
Mark scheme Mechanics Year 1 (AS) Unit Test 7: Kinematics 1 (constant acceleration)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 4
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
4a Demonstrates an understanding of the need to use
21
2s ut at
This can implied by using the equation in the next step(s).
M1 3.1b 5th
Use equations of
motion to solve
problems in
familiar contexts. Demonstrates the need to use (t – 3) when finding the
displacement of Q from A (or use (t + 3) when finding the
displacement of P from A). Can be implied in either of the
following steps.
M1 3.1b
Displacement of P: 22.8 0.06s t t A1 1.1b
Displacement of Q: 2
2.4 3 0.1 3s t t A1 1.1b
(4)
4b Writes 222.8 0.06 2.4 3 0.1 3t t t t M1 3.1b 5th
Use equations of
motion to solve
problems in
familiar contexts.
Makes an attempt to simplify this equation. For example,
2 22.8 0.06 2.4 7.2 0.1 6 9t t t t t
2 22.8 0.06 2.4 7.2 0.1 0.6 0.9t t t t t
20.04 6.3 0t t
M1 1.1b
Simplifies this expression to 22 50 315 0t t A1 1.1b
(3)
4c Makes an attempt to use the quadratic formula:
250 50 4 2 315
2 2t
M1 2.2a 5th
Use equations of
motion to solve
problems in
familiar contexts. Solves to find t = 30.21... (s).
Could also show that 5.21...t (s).
A1 1.1b
States or implies 21
2s ut at
M1 3.1b
Makes a substitution using their 30.21… into the formula:
21
2.8 30.2... 0.12 30.2...2
s
M1 1.1b
Finds s = 139.36... (m). Accept awrt 139 (m). A1 ft 1.1b
(5)
(12 marks)
Mark scheme Mechanics Year 1 (AS) Unit Test 7: Kinematics 1 (constant acceleration)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 5
Notes
4a
Award both accuracy marks if the following is seen:
Displacement of P from A: 2
2.8 3 0.06 3s t t
Displacement of Q from A: 22.4 0.1s t t
4c
Award ft marks for a correct answer using their ‘30.2’. They will have previously lost the first accuracy mark.
Q Scheme Marks AOs
Pearson
Progression Step
and Progress
descriptor
5a States or implies that s = −80 B1 3.1b 5th
Use equations of
motion to solve
problems
involving vertical
motion.
States or implies that a = −9.8 B1 3.1b
Writes 2 2 2v u as or makes a substitution
22 16 2 9.8 80v
M1 3.1b
Finds v = 42.70... (m s−1). Accept awrt 42.7 (m s−1). A1 1.1b
(4)
5b States or implies that s = 5 m. B1 3.1b 5th
Use equations of
motion to solve
problems
involving vertical
motion.
Simplifies 25 16 4.9t t to obtain 24.9 16 5 0t t M1 1.1b
Makes an attempt to use the quadratic formula:
216 16 4 4.9 5
2 4.9t
M1 1.1b
Solves to find t = 0.35… (s). Accept awrt 0.35 (s). A1 1.1b
Solves to find t = 2.91… (s). Accept awrt 2.92 (s). A1 1.1b
States that the ball is above 85 m for 2.56… (s). Accept awrt
2.6 (s). B1 3.2a
(6)
Mark scheme Mechanics Year 1 (AS) Unit Test 7: Kinematics 1 (constant acceleration)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 6
5c States or implies that at the greatest height v = 0 B1 3.1b 5th
Use equations of
motion to solve
problems
involving vertical
motion.
Finds the value of u: 1
42.7... 8.54...5
u (m s−1). Accept
awrt 8.5 (m s−1).
M1 3.1b
Writes 2 2 2v u as or makes a substitution
2 2
0 8.54... 2 9.8 s
M1 3.1b
Finds s = 3.72...(m). Accept awrt 3.7 (m). A1 ft 1.1b
(4)
(14 marks)
Notes
5c
Award ft marks for a correct answer using their answer from part a.