AS 91027
Previous
Exams
1.2 Apply algebraic procedures in solving problems
4 credits
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1
Level 1 Mathematics and Statistics
91027 (1.2): Apply algebraic procedures in solving
problems
Credits: Four
You should answer ALL parts of ALL questions in this booklet.
You should show ALL working.
Electronic technology is not permitted in this examination.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
Check that this booklet has pages 2–9 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO YOUR TEACHER AT THE END OF THE ALLOTTED TIME.
For Assessor’s
use only
Achievement
Apply algebraic procedures in
solving problems.
Achievement Criteria
Achievement
with Merit
Apply algebraic procedures
involving relational thinking, in
solving problems.
Achievement
with Excellence
Apply algebraic procedures
involving extended abstract
thinking, in solving problems.
Overall Level of Performance
© New Zealand Qualifications Authority, 2010
All rights reserved. No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
2
You are advised to spend 60 minutes answering the questions in this booklet.
QUESTION ONE
(a)
Solve these equations:
(i)
5x  7  x  2
(ii)
x 4  81
(b)
Solve 5x 2  7 x  6  0
(c)
(i)
Factorise x 2  7 x  10
(ii)
Simplify:
x 2  7 x  10
x2  2x
Mathematics and Statistics 91027 (1.2)
Assessor’s
use only
3
(d)
Show that the square of any prime number cannot be a prime. Give the factors.
Mathematics and Statistics 91027 (1.2)
Assessor’s
use only
4
QUESTION TWO
(a)
Assessor’s
use only
Sara found the following equation:
A
w
g
(i)
If w  25 and g  16 , find A in terms of  .
(ii)
Sara now wants to use the formula to find w in a different situation.
Rewrite the formula with w as the subject.
(b)
(i)
Expand: 2 x3  x 
(ii)
Solve: 2 x3  x  8
Mathematics and Statistics 91027 (1.2)
5
(c)
Joey needs to make a path from the front of his house to the back as shown in the diagram
below.
The width of the path is x metres.
Jim only has enough to make a path with a total area of 9 m2.
Form equations and use these to find the width of the concrete path around his house.
Mathematics and Statistics 91027 (1.2)
Assessor’s
use only
6
QUESTION THREE
Assessor’s
use only
(a)
Solve: 3xx  4  0
(b)
Simplify:
(c)
Ari spent $45 buying some CDs in a sale. He bought R Rock CDs and B Blues CDs.
(i)
9x 5
12x 3

Ari writes an equation for the amount he spent as: 2R  B  45
Explain the terms of the equation.
(ii)
Ari bought four times as many Rock CDs as Blues CDs.
How many blues CDs did he buy all together?
Mathematics and Statistics 91027 (1.2)
7
(d)
Scenic School is using two vans to take a group of students on a field trip.
If two students move from van A to van B, then the two vans would have the same number of
students in each, but if two students moved from van B to van A, then van B would have half
the number of students that were then in van A.
Use the information above to find the total number of students on the field trip.
In your answer, you must give at least ONE equation that you would use to solve the problem.
Mathematics and Statistics 91027 (1.2)
Assessor’s
use only
8
Extra paper for continuing your answers, if required.
Clearly number the question.
Question
number
Mathematics and Statistics 91027 (1.2)
Assessor’s
use only
9
Extra paper for continuing your answers, if required.
Clearly number the question.
Question
number
Mathematics and Statistics 91027 (1.2)
Assessor’s
use only
SPYDER
The Spyder calculator is another grand design from Mahobe
Resources (NZ) Ltd. It is recommended by The New Zealand
Centre of Mathematics. Purchase it direct from the Mahobe
website and support more projects like this publication.
www.mahobe.co.nz.
NCEA Level 1 Mathematics and Statistics 91027 (1.2) — page 1 of 3
SAMPLE ASSESSMENT SCHEDULE
Mathematics and Statistics 91027 (1.2): Apply algebraic methods in solving problems
Achievement
Merit
Excellence
Apply algebraic methods in solving
problems.
Apply algebraic methods, using relational thinking, in solving problems.
Apply algebraic methods, using
extended abstract thinking, in solving problems.
Evidence Statement
One
(a)
(i)
Expected coverage
4 x  9
9
x
4
x  2.25
Achievement
Merit
Excellence
Apply algebraic methods
in solving problems by:
Apply algebraic
methods, using
relational thinking,
in solving problems by:
Apply algebraic
methods, using
extended abstract
thinking, in solving
problems by:
TWO of:
TWO of:
ONE of:
 solving the problem
x  3
 partially solving the
problem by finding x =
3
 finding both solutions of the
problem
(b)
(5x + 3)(x – 2) = 0
x = -3/5 = -0.6
or x = 2
 factorising the expression.
 finding all solutions of the
problem
(c)
(i)
x  2x  5
 factorising the expression
 simplifying fully
(ii)
x5
x
 beginning to make a
proof by using a particular value or values
to demonstrate the
principle.
 using an algebraic approach
well but not
drawing a conclusion, or drawing a valid general conclusion
but not using algebra.
(ii)
(d)
Let n be any prime number.
Any prime number squared =
2
n
2
2
n has factors of 1, n and n
2
hence n cannot be prime.
 developing a
chain of logical
reasoning to
show a complete proof.
NCEA Level 1 Mathematics and Statistics 91027 (1.2) — page 2 of 3
Two
Expected coverage
Achievement
Merit
Excellence
Apply algebraic methods in solving problems
by:
Apply algebraic
methods, using
relational thinking,
in solving problems
by:
Apply algebraic
methods, using
extended abstract
thinking, in solving
problems by:
TWO of:
TWO of:
ONE of:
5
 or 1.25 
4
 simplifying an expression
2
 beginning to make w
the subject by correctly dividing by π or
by correctly squaring
the equation
 rearranging the
formula successfully
 expanding correctly
 devising a strategy to completely
solve the problem
2 x  6 x  2 x 2  3x 2  9
 assembling a correct
algebraic expression
for one area in the
shape (ie one of the
terms in the top line).
x 2  8x  9  0
 CAO will be awarded
N
 connecting different concepts in a
partial algebraic
solution, e.g. stating only the positive solution
(a)
(i)
A
(ii)
 A
w  g 
 
Or equivalent.
(b)
(i)
6x  2x2
(ii)
2x 2  6x  8  0


2 x 2  3x  4  0
x  4x  1  0
x  1
(c)
x4
2 x  6  2 x x  3 x 2  9
or : 2 x  (6  x) x  2 x 2  9
( x  9)( x  1)
x = -9 or 1
x cannot be negative so the
width of the path is 1m.
 developing a
chain of logical
reasoning that is
used to comprehensively
solve the problem
NCEA Level 1 Mathematics and Statistics 91027 (1.2) — page 3 of 3
Three
Expected coverage
Achievement
Apply algebraic
methods, using relational thinking, in
solving problems by:
Apply algebraic
methods, using
extended abstract
thinking, in solving
problems by:
TWO of:
TWO of:
ONE of:
 describing one of
the terms in the
equation
 describing two of
the terms in the
equation
 describing all
three of the
terms in the
equation
 assembling an algebraic expression
for the second relationship, or solving
the problem with a
numerical method
 solving the problem
with an extended
algebraic method
 assembling an algebraic expression
for one relationship
only, or solving the
problem with a numerical method with
no equations given.
 stating an algebraic
expression for one
relationship and
solving the problem, or stating both
equations but not
solving the problem.
x  0 or -4
 solving
(b)
3x 2
4
2 R is the amount spent on
 simplifying
rock CDs or $2.00 per Rock
CD
Excellence
Apply algebraic
methods in solving
problems by:
(a)
(c)
(i)
Merit
1B is the amount spent on
blues CDs or $1.00 per Blues
CD
$45 is the total amount
spent.
(ii)
R = 4B
So 2(4B) + B = 45
9B = 45
B =5
(d)
A – 2 = B + 2, and
2(B – 2) = A + 2
Therefore,
2B – 6 = B + 4
B = 10
A = 14
 stating an algebraic expression
for both relationships and
solving the
problem.
Total number of students is
24.
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
Minimum of:
2A
Minimum of:
2M
Minimum of:
2E
RND#
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Purchase it direct from the Mahobe website:
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0.791
1
To be completed by Candidate and School:
Name:
NSN No:
School Code:
DAY 1
TUESDAY
SUPERVISOR’S USE ONLY
Level 1 Mathematics and Statistics CAT, 2012
91027 Apply algebraic procedures in solving problems
Tuesday 18 September 2012
Credits: Four
You should attempt ALL the questions in this booklet.
Calculators may NOT be used.
Show ALL working.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
You are required to show algebraic working in this paper. Guess and check methods do not
demonstrate relational thinking. Guess and check methods will limit grades to Achievement.
Check that this booklet has pages 2 – 8 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.
ASSESSOR’S USE ONLY
Achievement
Apply algebraic procedures in solving
problems.
Achievement Criteria
Achievement with Merit
Apply algebraic procedures, using
relational thinking, in solving problems.
Achievement with Excellence
Apply algebraic procedures, using
extended abstract thinking, in solving
problems.
Overall level of performance
© New Zealand Qualifications Authority, 2012. All rights reserved.
No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
2
You are advised to spend 60 minutes answering the questions in this booklet.
QUESTION ONE
(a)
Solve 3(2x + 9) = 15
(b)(i)
Factorise x2 – 3x – 28
(ii)
Solve x2 – 3x – 28 = 0
x 2 − 3x − 28
(iii) Simplify
x+4
(iv) Show that x = 12 is the only real solution to
x 2 − 3x − 28
=5
x+4
Mathematics and Statistics CAT 91027 (Day 1), 2012
ASSESSOR’S
USE ONLY
3
(v)
x2 – ax + 6 = 30, where a is a positive number.
The difference between the solutions to the equation is 10.
Find the value of a.
(c)
A square room and a hallway are to have carpet laid on the floor.
123 m2 of carpet is required to cover both the hallway and the room.
The width of the hallway is 6 m less than the length of the room.
The hallway is 5 m longer than the length of the room.
Write an equation showing this relationship and solve this equation to find the length of the
room.
Mathematics and Statistics CAT 91027 (Day 1), 2012
ASSESSOR’S
USE ONLY
4
QUESTION TWO
ASSESSOR’S
USE ONLY
Simplify fully 10x2y + 8xy2 – 5x2y
(a)
(i)
(ii) Factorise fully the following expression and write it in its simplest form.
10x2y + 8xy2 – 5x2y
(b) Expand and simplify (2x – 4)(3x – 5)
(c)
(i)
Simplify
(ii) Solve
x 2x − 1
−
5
2
x 2 x − 1 −3x
−
≥
5
2
5
Mathematics and Statistics CAT 91027 (Day 1), 2012
5
(d) The formula for the volume of a cylinder is
ASSESSOR’S
USE ONLY
V = π r 2 h
where r is the radius and h is the height of the cylinder.
Write the formula for the radius, r, of the cylinder in terms of V, h and π.
(i)
(ii) The length L of a straight straw that will just fit in a cylindrical can with
a height of 8 cm is given by:
L
2
2
L = 8 + (2r )
where r is the radius of the can.
A straight straw that is 3 times as long (3L) just fits in a larger can that
has the same height.
Write an expression for the radius of the larger can R in terms of r.
Mathematics and Statistics CAT 91027 (Day 1), 2012
r
8
6
QUESTION THREE
(a)
ASSESSOR’S
USE ONLY
Simplify fully (4x3)2
(b) Solve 3x + 6 = 7 – 2x
(c)
Solve 2x2 – 5x – 6 = 6
(d) Sarah borrows her friend’s car for a holiday.
She agrees to pay $7 a day and $1 per kilometre that she travels.
(i)
(ii) Sarah travelled 185 km and should pay her friend $213.
Write an equation for the amount P Sarah agrees to pay.
Use your equation to find the number of days Sarah borrowed the car for.
Mathematics and Statistics CAT 91027 (Day 1), 2012
7
(e)
Emma is 3 times as old as Tara.
In another 7 years Emma will be twice as old as Tara will be.
Write at least one equation and use algebra to find Emma’s age now.
(f)
If (x3)4 = (y2)3
Express x in terms of y.
Mathematics and Statistics CAT 91027 (Day 1), 2012
ASSESSOR’S
USE ONLY
8
91027
QUESTION
NUMBER
Extra paper if required.
Write the question number(s) if applicable.
Mathematics and Statistics CAT 91027 (Day 1), 2012
ASSESSOR’S
USE ONLY
The DS-742ET
Did you know that Mahobe added equation solving to
make this an even more powerful calculator?
eTOOL
-100
-50
0
www.mahobe.co.nz.
50
100
MAHOBE
1
To be completed by Candidate and School:
Name:
NSN No:
School Code:
DAY 2
WEDNESDAY
SUPERVISOR’S USE ONLY
Level 1 Mathematics and Statistics CAT, 2012
91027 Apply algebraic procedures in solving problems
Wednesday 19 September 2012
Credits: Four
You should attempt ALL the questions in this booklet.
Calculators may NOT be used.
Show ALL working.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
You are required to show algebraic working in this paper. Guess and check methods do not
demonstrate relational thinking. Guess and check methods will limit grades to Achievement.
Check that this booklet has pages 2 – 8 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.
ASSESSOR’S USE ONLY
Achievement
Apply algebraic procedures in solving
problems.
Achievement Criteria
Achievement with Merit
Apply algebraic procedures, using
relational thinking, in solving problems.
Achievement with Excellence
Apply algebraic procedures, using
extended abstract thinking, in solving
problems.
Overall level of performance
© New Zealand Qualifications Authority, 2012. All rights reserved.
No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
2
You are advised to spend 60 minutes answering the questions in this booklet.
QUESTION ONE
Simplify fully 12a2b + 6ab2 – 7a2b
(a)
(i)
(ii) Factorise fully the following expression and write it in its simplest form.
12a2b + 6ab2 – 7a2b
(b) Expand and simplify (3a – 2)(4a – 5)
(c)
(i)
Simplify
(ii) Solve
a 3a − 6
−
5
4
a 3a − 6 −2 a
−
≥
5
4
5
Mathematics and Statistics CAT 91027 (Day 2), 2012
ASSESSOR’S
USE ONLY
3
(d) The formula for the volume of a cylinder is
ASSESSOR’S
USE ONLY
V = π r 2 h
where r is the radius and h is the height of the cylinder.
Write the formula for the radius, r, of the cylinder in terms of V, h and π.
(i)
(ii) The length L of a straight straw that will just fit in a cylindrical can with
a height of 8 cm is given by:
L
2
2
L = 8 + (2r )
where r is the radius of the can.
A straight straw that is 3 times as long (3L) just fits in a larger can that
has the same height.
Write an expression for the radius of the larger can R in terms of r.
Mathematics and Statistics CAT 91027 (Day 2), 2012
r
8
4
QUESTION TWO
(a)
ASSESSOR’S
USE ONLY
Solve 3(2x + 7) = 9
(b)(i)
Factorise x2 – 4x – 32
(ii)
Solve x2 – 4x – 32 = 0
x 2 − 4 x − 32
x+4
(iii) Simplify
x 2 − 4 x − 32
(iv) Show that x = 13 is the only real solution to
=5
x+4
Mathematics and Statistics CAT 91027 (Day 2), 2012
5
(v)
x2 – ax + 11 = 23, where a is a positive number.
The difference between the solutions is 8.
Find the value of a.
(c)
In front of a garage there is a square concrete pad with a concrete path leading to it.
The total area of the concrete is 151 m2.
The width of the concrete path is 4 m less than the length of the concrete pad.
The concrete path is 5 m longer than the length of the concrete pad.
Write an equation showing this relationship, and solve this equation to find the length of the
square concrete pad.
Mathematics and Statistics CAT 91027 (Day 2), 2012
ASSESSOR’S
USE ONLY
6
QUESTION THREE
(a)
ASSESSOR’S
USE ONLY
Simplify fully (5x3)2
(b) Solve 5x + 8 = 9 – 4x
(c)
Solve 2x2 – 5x – 8 = 4
(d) Sarah borrows her friend’s car for a holiday.
She agrees to pay $6 a day and $1 per kilometre that she travels.
(i)
(ii) Sarah travelled 176 km and should pay her friend $218.
Write an equation for the amount P Sarah agrees to pay.
Use your equation to find the number of days Sarah borrowed the car for.
Mathematics and Statistics CAT 91027 (Day 2), 2012
7
(e)
George is 4 times as old as Leo.
In another 5 years George will be 3 times as old as Leo will be.
Write at least one equation and use algebra to find George’s age now.
(f)
If (x2)8 = (y4)2
Express x in terms of y.
Mathematics and Statistics CAT 91027 (Day 2), 2012
ASSESSOR’S
USE ONLY
8
91027
QUESTION
NUMBER
Extra paper if required.
Write the question number(s) if applicable.
Mathematics and Statistics CAT 91027 (Day 2), 2012
ASSESSOR’S
USE ONLY
Is there a piece missing in your Mathematics?
SPYDER
The Spyder calculator is another grand design
from Mahobe Resources (NZ) Ltd. Purchase it
direct from the Mahobe website and support
more projects like this publication.
www.mahobe.co.nz.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2012 — page 1 of 9
Assessment Schedule – 2012
Mathematics and Statistics CAT: Apply algebraic procedures in solving problems (91027, Day 1)
Evidence Statement
Question
Evidence
Achievement
u
Achievement with
Merit
r
Achievement
with Excellence
t
Comments
ONE
(a)
x = –2
Equation solved.
(b)(i)
(x – 7)(x + 4)
Expression
factorised.
Watch signs
(b)(ii)
x = 7 or x = –4
Equation solved
consistently giving
both solutions.
Watch signs
(b)(iii)
x 2 − 3x − 28 (x − 7)(x + 4)
=
x+4
x+4
= x−7
(b)(iv)
(b)(v)
(c)
x = 12
Fraction simplified.
Candidate does not
use their simplified
answer to (b)(iii) and
instead uses the full
equation and
multiplies.
AND
Tries substitution of
12 only rather than
solving the quadratic
OR
finds TWO solutions:
x = 12 and x = –4
.
Equation solved
using (b)(iii) giving
ONE solution only.
OR
Rearranging the full
equation generating
a quadratic –
solving giving 2
solutions and then
eliminating the
invalid solution.
x2 – ax – 24 = 0
(x – 6)(x + 4) = 0
x = 6 or –4
6 – (–4) = 10
a=2
Rearranged and = 0.
TWO values of x
identified.
a found.
Accept –2.
Common error –12
and +2 and then
giving an answer of
10. This gains r,
x2 +(x – 6)(x + 5) = 123
2x2 – x – 30 = 123
2x2 – x – 153 = 0
(2x + 17)(x – 9) = 0
x = 9 or – 8.5
Length 9
Any correct equations
demonstrating the full
relationship between
the two shapes. This
may involve the
lengths or widths and
areas including in
terms of W and L.
Equation simplified
to 2x2 – x – 153=0
OR
negative answer
given for length.
Problem solved
with at least one
equation being
given followed by
guess and check
where the
numbers used are
>7.
Some candidates
solve using x as the
width of the hall.
A=
2A =
M=
2M =
E = 1 of t
2E = 2 of t
2 of u
>2 of u
1 of r
> 1 of r
Straight substitution
of 12 into the
original equation
scores n
Used solution to
(b)(iii) and then
substituted 12,
ie 12 – 7 = 5 is
insufficient
evidence to show
that 12 is the only
solution hence
gains u.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2012 — page 2 of 9
5x2y + 8xy2
or xy(5x + 8y)
Simplifying an
expression.
(ii)
xy(5x + 8y)
OR
xy(10x + 8y – 5x)
Factorised
expression.
(b)
6x2 – 22x + 20
Expression expanded
and simplified.
2x − 10x + 5 −8x + 5
=
10
10
4x 1
or − +
5 2
Writing expression.
2x − 10x + 5 ≥ −6x
A correctly solved
simplified problem
from (c)(i), where the
fractions have the
same denominator
and where the
inequality does not
need reversing gains
u.
A correctly solved
equation where the
expression is simplified
so there are no
fractions involved but
the inequality needs
reversing.
OR
Consistent solution to
an equation that has
fractions with different
denominators but does
not require the change
of the inequality sign.
OR
Solves without the
inequality.
ONE variable not
correct in
rearrangement.
OR
not taken square root
V
r2 =
!h
Correct formula.
TWO
(a)(i)
(c)(i)
(ii)
−8x + 6x + 5 ≥ 0
−2x + 5 ≥ 0
2x ≤ 5
x ≤ 2.5
(d)(i)
r=
V
!h
Accepting
factorised without
simplifying
Common error –
failure to change to
+5. Fractions must
be combined – not
just changed to be
over a common
denominator.
Inequation
consistently
solved.
OR r = V ÷ ! × h
OR
+/– in front of sqrt.
(ii)
(3L)2 − 82 = 4R 2
9(82 + 4r 2 ) − 82 = 4R 2
9 × 64 + 36r 2 − 64 = 4R 2
2 × 64 + 9r 2 = R 2
Assembling a correct
algebraic expression
independent of L.
OR an equation
relating the two cans.
Developing a
chain of
logical
reasoning that
is used to
solve the
problem.
M=
2M =
E = 1 of t
2E = 2 of t
R = 9r 2 + 128
or equivalent
A=
2A =
2 of u
>2 of u
1 of r
> 1 of r
Take care with
change in inequality
with the division by
a negative.
Most consistent
equation will have
–5 rather than + 5.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2012 — page 3 of 9
THREE
(a)
(b)
(c)
16x6
x=
Simplified.
Solved.
1
5
(2x + 3)(x – 4) = 0
x = –3/2 (or –1.5)
or x = 4
Factorised
(d)(i)
P = 7d + k
Writing full equation
with equals sign.
(d)(ii)
213 = 7d +185
7d = 28
d=4
CAO from correct
equation in (d)(i).
OR
No equation in (d)(i)
or (ii) but clearly
demonstrated
working to find the
correct solution.
Solved equation
showing
candidate’s
working.
E = 3T
E + 7 = 2 (T + 7)
E = 2 T +7
3T=2T+7
T=7
Emma is 21
Setting up ONE
equation.
Set up both
equations or ONE
combined equation.
x12 = y 6
x12 = y6
Accept identifying of
x12 and y6 without
stating that they are
equal.
x2 = y
x = 12 y 6
A=
2A =
M=
2M =
(e)
(f)
x =y
2
x=
y
Solved with both
solutions given.
•
Solved with clear
logical chain of
reasoning.
The second
equation is very
rarely correct,
usually because the
brackets are
missing. Random
use of 7 does not
warrant r, ie if it is
used in guess and
check it must be
supported by clear
valid justification.
x = y Accept
with or without
+/– sign and
accept further
reasoning
supporting a valid
solution.
This question is
assessing level 6
manipulation of
indices
OR guess and
check from ONE
equation with clear
valid evidence.
2 of u
>2 of u
1 of r
>1 of r
E = 1 of t
2E = 2 of t
Overall sufficiency
Grade Boundaries
E
2E
Or higher
M
3M
M+E
Or higher
A
3A
A+M
Or higher
2M is a higher level of achievement than A + M, hence question grades of 2M or M + M gain an A overall for the paper.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2012 — page 4 of 9
Notes
1(c) where x as the width of the hall
x(x + 11) + (x + 6)2 = 123
x2 + 11x + x2 + 12x + 36 = 123
2x2 +23x – 87 = 0
(2x + 29)(x – 3) = 0
x = 3 or –14.5 (reject as x can’t be negative)
Length of room = 3 + 6 = 9 m
Gains t
2(a)(ii) Accept factorised answer with simplifying.
2(b) The student who writes the expansion correctly and then incorrectly writes +22x instead of –22x gains n
Or the student writes the expansion correctly and then incorrectly writes –20 instead of +20. This is considered as a transfer
error. Accept and write TE next to the error.
3(b) Written the correct answer x = 1/5, but then gone on to write x = 5. Gains n
3(d) If 3(d)(i) incorrect
and 3(d)(ii) deduct 185 from 213,
and divide by 7 and give 4 as the answer.
Gains u
3(e) E = 3T,
21 = 3 × 7
21 + 7 = 28
7 + 7 = 14
21 = 2 × 14
Gains u for first line.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2012 — page 5 of 9
Assessment Schedule – 2012
Mathematics and Statistics CAT: Apply algebraic procedures in solving problems (91027, Day 2)
Evidence Statement
Question
ONE
(a)(i)
Evidence
•
•
5a2b + 6ab2
or ab(5a + 6b)
Simplifying an
expression.
(ii)
•
•
•
ab(5a + 6b)
or
ab(12a +6b – 7a)
Factorised
expression.
(b)
•
12a2 – 23a + 10
Expression
expanded and
simplified.
•
+ 30 expression.
4a − 15a + 30 −11aWriting
=
20
20
−11a 3
OR
=
+
20
2
(c)(i)
(ii)
•
–11a + 30 ≥ −8a
−3a + 30 ≥ 0
3a ≤ 30
a ≤ 10
Achievement
with Merit
r
Achievement
u
A correctly solved
simplified problem
from (c)(i), where
the fractions have
the same
denominator and
where the
inequality does not
need reversing
gains u..
•
Achievement with
Excellence
t
Comments
•
•
•
•
Accept
factorised without
simplifying
Common error –
failure to change to
+30. Fractions must
be combined – not
just changed to be
over a common
denominator.
A correctly
solved equation
where the
expression is
simplified so
there are no
fractions
involved but the
inequality needs
reversing.
OR
Consistent
solution to an
equation that
has fractions
with different
denominators
but does not
require the
change of the
inequality sign.
OR
Solves without
the inequality.
Inequation
consistently solved.
Take care with
change in inequality
with the division by a
negative.
Most consistent
equation will have 30 rather than + 30
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2012 — page 6 of 9
(d)(i)
•
ONE variable not
correct in
rearrangement.
OR
not taken square
root
V
r2 =
!h
V
!h
r=
Correct formula.
OR r = V ÷ ! × h
OR
+/– in front of sqrt.
(3L)2 − 82 = 4R 2
(ii)
9(8 + 4r ) − 8 = 4R
2
•
2
2
Assembling a
correct algebraic
expression
independent of
L.
OR
An equation
relating to the
two cans.
2
9 × 64 + 36r 2 − 64 = 4R 2
2 × 64 + 9r 2 = R 2
R = 9r 2 + 128
•
or equivalent
A=
2A =
2 of u
>2 of u
M=
1 of r
•
2M =
> 1 of r
Developing a chain
of logical reasoning
that is used to solve
the problem.
•
•
E = 1 of t
2E = 2 of t
•
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2012 — page 7 of 9
•
•
Watch sign
Expression
factorised
•
•
Watch sign
Equation solved
consistently giving
both solutions
•
•
•
•
•
TWO
(a)
x = –2
Equation solved
(b)(i)
(x – 8)(x + 4)
(ii)
x = 8 or x = –4
•
x 2 − 4x − 32 (x − 8)(x + 4)
=
x+4
x+4
= x −8
•
x = 13
(iii)
(iv)
•
Fraction
simplified.
Candidate does not
use their simplified
answer to (b)(iii)
and instead uses the
full equation and
multiplies.
AND
Tries substitution of
13 only rather than
solving the
quadratic OR
finds TWO
solutions:
x = 13 and x = –4
Equation solved
using (b)(iii)
giving ONE
solution only.
OR
Rearranging the
full equation
generating a
quadratic –
solving giving 2
solutions and
then eliminating
the invalid
solution.
Straight substitution
of 13 into the original
equation scores n
Rearranged and = 0.
TWO values of
x identified.
a found.
Accept –4
Used solution to
(b)(iii) and then
substituted 13,
ie 13 – 8 = 5 is
insufficient evidence
to show that 13 is the
only solution, hence
gains u.
(v)
•
•
•
•
(c)
•
x2 +(x + 5)(x – 4)
= 151
•
2x2 + x – 20 = 151
•
2x2 + x – 171 = 0
•
(2x + 19)(x – 9) =
0
•
x = 9 or – 9.5
•
Length of
concrete = 9 m
Any correct
equations
demonstrating the
full relationship
between the two
shapes. This may
involve the lengths
or widths and areas
including in terms
of W and L.
Equation
simplified to
2x2 + x – 171 =
0
OR
negative
answer given
for length.
Problem solved with
at least one equation
being given followed
by guess and check
where the numbers
used are > 4.
Some candidates
solve the problem
using x as the width
of the path.
•
A=
2A =
M=
1 of r
•
2M =
> 1 of r
•
•
•
x2 – ax – 12 = 0
(x – 6)(x + 2) = 0
x = 6 or x = –2
a=4
2 of u
>2 of u
E = 1 of t
2E = 2 of t
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2012 — page 8 of 9
Simplified.
•
•
•
Solved.
•
•
•
(2x + 3)(x – 4) = 0
x = –3/2 (or –1.5)
or x = 4
Factorised
Solved with
both solutions
given.
(d)(i)
•
P = 6d + k
Writing full
equation with
equals sign.
(d)(ii)
•
•
•
218 = 6d +176
6d = 42
d=7
CAO from correct
equation in (d)(i)
OR
No equation in
(d)(i) or (ii), but
clearly
demonstrated
working to find the
correct solution.
Solved equation
showing
candidate’s
working.
Set up ONE
equation.
Set up both
equations or
ONE combined
equation.
THREE
(a)
(b)
(c)
(e)
•
•
25x6
x=
1
9
G = 4L
G + 5 = 3 (L + 5)
4 L + 5 = 3 L + 15
L = 10
George is 40.
•
•
Solved with clear
logical chain of
reasoning.
The second equation
is very rarely correct,
usually because of
missing brackets are
missing. Random use
of 5 does not warrant
r, ie if it is used in
guess and check, it
must be supported by
clear valid
justification.
This question is
assessing level 6
manipulation of
indices.
OR guess and
check from
ONE equation
with clear valid
evidence.
x16 = y 8
(f)
•
x2 = y
x=
y
•
x16 = y8
Accept identifying
of x16 and y8
without stating they
are equal.
x2 = y
x = 16 y 8
x = y Accept with
or without +/– sign
and accept further
reasoning supporting
a valid solution.
A=
2A =
M=
1 of r
•
2M =
> 1 of r
•
•
2 of u
>2 of u
E = 1 of t
2E = 2 of t
Overall sufficiency
Grade Boundaries
Or higher
E
2E
M
3M
M+E
Or higher
A
3A
A+M
Or higher
2M is a higher level of achievement than A + M, hence question grades of 2M or M + M gain an A overall for the paper.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2012 — page 9 of 9
Notes
2(b) The student who writes the expansion correctly and then incorrectly writes +23a instead of –23a gains n
Or the student writes the expansion correctly and then incorrectly writes –10 instead of +10. This is considered as a transfer
error. Accept and write TE next to the error.
3(b) written the correct answer x = 1/9, but then gone on the write x = 9. Gains n
3(d) If 3(d)(i) incorrect
and 3(d)(ii) deduct 176 from 218 ,
and divide by 6 and give 7 as the answer.
Gains u
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MAHOBE
1
To be completed by Candidate and School:
Name:
NSN No:
School Code:
DAY 1
TUESDAY
SUPERVISOR’S USE ONLY
Level 1 Mathematics and Statistics CAT, 2013
91027 Apply algebraic procedures in solving problems
Tuesday 17 September 2013
Credits: Four
You should attempt ALL the questions in this booklet.
Calculators may NOT be used.
Show ALL working.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
You are required to show algebraic working in this paper. Guess and check methods do not
demonstrate relational thinking. Guess and check methods will limit grades to Achievement.
Check that this booklet has pages 2 – 8 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.
ASSESSOR’S USE ONLY
Achievement
Apply algebraic procedures in solving
problems.
Achievement Criteria
Achievement with Merit
Apply algebraic procedures, using
relational thinking, in solving problems.
Achievement with Excellence
Apply algebraic procedures, using
extended abstract thinking, in solving
problems.
Overall level of performance
© New Zealand Qualifications Authority, 2013. All rights reserved.
No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
2
You are advised to spend 60 minutes answering the questions in this booklet.
QUESTION ONE
(a)
Solve 7 – 3x = 1
x=
(b) Solve 5(2m – 3) = 6(m – 4)
(c)
m=
Solve x2 + 4x – 12 = 0
x=
(d) Solve 3x – 5 > 7x + 15
Mathematics and Statistics CAT 91027 (Day 1), 2013
ASSESSOR’S
USE ONLY
3
(e)
(i)
Pierre does not want to tell Reece his weight.
He says he weighs at least 14 kg plus ¾ of Reece’s weight.
Write an expression for Pierre’s weight, P, in terms of Reece’s weight, R.
(ii) Reece weighs 56 kg.
Find Pierre’s weight and state how it compares with Reece’s weight.
Express the relationship between Pierre’s weight and Reece’s weight in words.
You must justify your statement using algebra and show that you have used your
expression from part (i).
(f)
h = n2 – 6n + 8
For what values of n will h be negative?
Mathematics and Statistics CAT 91027 (Day 1), 2013
ASSESSOR’S
USE ONLY
4
QUESTION TWO
(a)
ASSESSOR’S
USE ONLY
Expand (x + 4)(x – 2)
(b) Factorise x2 – 7x – 60
(c)
Factorise the following expression and write it in its simplest form.
6a2b – 4ab2 + 4a2b
(d) Solve (n – 4)(n + 3) = 0
n=
(e)
Write as a single fraction 2 x −
x−3
4
Mathematics and Statistics CAT 91027 (Day 1), 2013
5
(f)
Nick and Marnie are saving for a school holiday trip.
Marnie is paid $15 an hour and Nick is paid $13 an hour.
Altogether they earned a total of $166.
(i)
If Marnie worked for m hours and Nick worked for n hours, write an equation showing
the above information.
(ii) Altogether they worked a total of 12 hours.
Use algebra to show how many more hours Nick worked than Marnie.
(g) A cylinder and a cone have the same height.
The volume of the cone is half the volume of the cylinder.
The volume of the cylinder is π r 2 h and the volume of the cone is 13 π R 2 h.
Describe the relationship between the radius of the cone and the radius of the cylinder.
You must show algebraic working and then describe the relationship in words.
Mathematics and Statistics CAT 91027 (Day 1), 2013
ASSESSOR’S
USE ONLY
6
QUESTION THREE
(a)
ASSESSOR’S
USE ONLY
Simplify fully (3x4)2
(b) Solve:
(i)
p3 = 64
p=
r
(ii)2 = 16
r=
(c)
1 1 1
= +
f u v
Write an expression for f in terms of u and v.
(i)
(ii) Write an expression for v in terms of u and f.
Mathematics and Statistics CAT 91027 (Day 1), 2013
7
(d) Charlie has hired a bike for 3 hours.
If he is late returning the bike, he is fined $5 for the first hour late. He is fined 2 times as much
if he is 2 hours late, 4 times the original fine if he is 3 hours late, and so on.
The formula used to calculate the total fine, T, that Charlie has to pay is
T = 5 × 2(h – 1), where h is the number of hours late that the bike is returned.
How many hours late did he return the bike if the fine was $160?
You must show use of the formula.
(e)
(i)
Tama and Rani go out for dinner.
Tama’s meal costs 1.5 times as much as Rani’s.
Write an equation for the total cost of the dinner, in terms of R (the cost of Rani’s meal).
(ii) Sharee joins her two friends, Tama and Rani, at the same dinner.
Her meal costs $5 more than Rani’s meal.
The total cost of the dinner for the three of them is $75.
Use algebra to find the cost of Sharee’s meal.
Mathematics and Statistics CAT 91027 (Day 1), 2013
ASSESSOR’S
USE ONLY
8
91027
QUESTION
NUMBER
Extra paper if required.
Write the question number(s) if applicable.
Mathematics and Statistics CAT 91027 (Day 1), 2013
ASSESSOR’S
USE ONLY
Is there a piece missing in your Mathematics?
SPYDER
The Spyder calculator is another grand design
from Mahobe Resources (NZ) Ltd. Purchase it
direct from the Mahobe website and support
more projects like this publication.
www.mahobe.co.nz.
RND#
SPYDER
When it comes to buying a reliable calculator don’t rely on
chance. Only the Mahobe SPYDER calculator is
recommended by The New Zealand Centre of Mathematics.
Purchase it direct from the Mahobe website:
www.mahobe.co.nz.
0.791
1
To be completed by Candidate and School:
Name:
NSN No:
School Code:
DAY 2
WEDNESDAY
SUPERVISOR’S USE ONLY
Level 1 Mathematics and Statistics CAT, 2013
91027 Apply algebraic procedures in solving problems
Wednesday 18 September 2013
Credits: Four
You should attempt ALL the questions in this booklet.
Calculators may NOT be used.
Show ALL working.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
You are required to show algebraic working in this paper. Guess and check methods do not
demonstrate relational thinking. Guess and check methods will limit grades to Achievement.
Check that this booklet has pages 2 – 8 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.
ASSESSOR’S USE ONLY
Achievement
Apply algebraic procedures in solving
problems.
Achievement Criteria
Achievement with Merit
Apply algebraic procedures, using
relational thinking, in solving problems.
Achievement with Excellence
Apply algebraic procedures, using
extended abstract thinking, in solving
problems.
Overall level of performance
© New Zealand Qualifications Authority, 2013. All rights reserved.
No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
2
You are advised to spend 60 minutes answering the questions in this booklet.
QUESTION ONE
(a)
Expand (x – 5)(x + 3)
(b) Factorise x2 – 2x – 63
(c)
Solve (m + 6)(m – 2) = 0
m=
(d) Factorise the expression below and write it in its simplest form.
8ab2 – 3a2b + 4a2b
(e)
Write as a single fraction x −
x−2
5
Mathematics and Statistics CAT 91027 (Day 2), 2013
ASSESSOR’S
USE ONLY
3
(f)
Sophia and Rewa are being paid to tidy the grounds of their neighbour’s house.
Sophia, being the elder of the two, is paid $17 an hour, and Rewa is paid $13 an hour.
Altogether they earned a total of $176.
(i)
(ii) Altogether they worked a total of 12 hours.
If Sophia worked for s hours and Rewa worked for r hours, write an equation showing
the above information.
Use algebra to show how many more hours Rewa worked than Sophia.
(g) A cylinder and a cone have the same height.
The volume of the cylinder is half the volume of the cone.
The volume of the cylinder is π r 2 h and the volume of the cone is 13 π R 2 h.
Describe the relationship between the radius of the cone and the radius of the cylinder.
You must show algebraic working and then describe the relationship in words.
Mathematics and Statistics CAT 91027 (Day 2), 2013
ASSESSOR’S
USE ONLY
4
QUESTION TWO
(a)
ASSESSOR’S
USE ONLY
Solve 9 – 4x = 1
x=
(b) Solve 4(2n – 3) = 3(n – 11)
(c)
n=
Solve x2 + 5x – 14 = 0
x=
(d) Solve 6x – 3 > 8x + 9
Mathematics and Statistics CAT 91027 (Day 2), 2013
5
(e)
(i)
Sharee wants to know how much money Tama has saved for the holidays.
Tama says he has saved at least $18, plus ¾ of the amount Sharee has saved.
Write an equation for the amount, T, Tama has saved in terms of the amount, S, Sharee
has saved.
(ii) Sharee has saved $72.
Find the amount Tama has saved and show how it relates to Sharee’s amount.
You must justify your statement using algebra and show that you have used your
expression from part (i).
(f)
h = n2 – 7n + 10
For what values of n will h be negative?
Mathematics and Statistics CAT 91027 (Day 2), 2013
ASSESSOR’S
USE ONLY
6
QUESTION THREE
(a)
ASSESSOR’S
USE ONLY
Simplify fully (2a4)3
(b) Solve:
(i)
m3 = 64
m=
n
(ii)3 = 81
n=
(c)
1 1 1
= +
f u v
Write an expression for f in terms of u and v.
(i)
(ii) Write an expression for v in terms of u and f.
Mathematics and Statistics CAT 91027 (Day 2), 2013
7
(d) Sam has gone on holiday and forgotten to return a reference book to the library.
He is fined $3 for the first week he is late returning the book.
He is fined 2 times as much if he is 2 weeks late, 4 times the original fine if he is 3 weeks late,
and so on.
The formula used to calculate the total fine, T, that Sam has to pay is T = 3 × 2(w–1),
where w is the number of weeks he is late in returning the book.
How many weeks late did he return the book if the fine was $192?
You must show use of the formula.
(e)(i)
Jamie and Pippa go to a fun park for a day out.
They go on different rides.
Jamie’s day out costs 1.5 times as much as Pippa’s.
Write an equation for the total cost of the day out in terms of P (the cost of Pippa’s day
out).
(ii) Zack joins his two friends for the day out.
His day out costs $5 more than Pippa’s.
The total cost of the day out for the three of them is $75.
Use algebra to find the cost of Zack’s day out.
Mathematics and Statistics CAT 91027 (Day 2), 2013
ASSESSOR’S
USE ONLY
8
91027
QUESTION
NUMBER
Extra paper if required.
Write the question number(s) if applicable.
Mathematics and Statistics CAT 91027 (Day 2), 2013
ASSESSOR’S
USE ONLY
RND#
SPYDER
When it comes to buying a reliable calculator don’t rely on
chance. Only the Mahobe SPYDER calculator is
recommended by The New Zealand Centre of Mathematics.
Purchase it direct from the Mahobe website:
www.mahobe.co.nz.
0.791
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2013 — page 1 of 5
Assessment Schedule – 2013
Mathematics and Statistics CAT: Apply algebraic procedures in solving problems (91027, Day 1)
Evidence Statement
Evidence
Achievement with Merit
(r)
Achievement with
Excellence (t)
comments
One
(a)
x=2
Equation solved.
Fraction does not need to be simplified
(b)
10m – 15 = 6m – 24
4m = –9
m = –9/4 or –2.25
Equation solved. If fraction correct
and decimal incorrect accept answer.
Incorrect calculation of -24 + 15 giving +9
is not accepted a numerical error - gains n
(c)
(x – 2)(x + 6) = 0
x = 2 or x = –6
Equation factorised.
Or incorrectly factorised equation
consistently solved. Must have both
solutions
Equation correctly factorised
and solved.
Must have both solutions
(d)
–20 > 4x
x < –5
Generating solutions with incorrect
signs of x > –5 or x = –5
Inequality solved.
Accept –5 > x or –20 / 4 > x
¾ R + 14
Expression given.
Or
Consistent statement from wrong
expression
Statement that they both
weigh 56 kg.
Or Pierre weighs 56 kg
Expression factorised.
Or Guess and check finding the
values of 2 and 4
Expression factorised and
solved for h = 0.
Accept expression factorised
and roots found without
equating to zero
(e) i)
ii)
(f)
Achievement
(u)
P ≥ ¾ × 56 + 14
≥ 42 + 14
Pierre is at least as heavy (as
Reece. )
Or Pierre weighs more than 56kg
(n – 4)(n – 2) = 0
n = 2 or n = 4
h is negative between n = 2 and n =
4
The discarding of the negative solution is
not acceptable – gains u.
Be careful of incorrect factorising and
solving giving correct solution.
Correct statement
relating to the inequality
consistent with the
question.
Incorrect calculation of ¾ of 56 may be
counted as a numerical error for all levels
Solution found with
correct interval statement
in symbols or words. Do
not penalise use of less
than or equal to.
If the solution gives the correct roots and a
statement that then gives a value in the
range this must be supported by a further
statement e.g. the graph is a positive
parabola or a graph showing the feasible
region in order to gain t.
Working may be with an equation or
involve the incorrect inequality symbol.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2013 — page 2 of 5
Achievement
(u)
Evidence
x2 + 2x – 8
Correct expansion and correctly
simplified.
(b)
(x – 12)(x + 5)
Correct factorisation.
(c)
10a2b – 4ab2
2ab(5a – 2b)
Correctly simplified OR factorised.
Eg 2ab(3a – 2b+2a)
ab(10a – 4b)
(d)
n = 4 or –3
Correctly solved.
Must have both solutions
(e)
8x − x + 3 7x + 3
=
4
4
Omitted changing the sign of the 3 in
simplifying – (x – 3) ie answer
7x - 3
4
Correct expression, which does
not need to be simplified.
8x – x + 3
4
8x –( x – 3)
4
14x + 6
8
TWO
(a)
(f)i)
ii)
Achievement with Merit
(r)
15m + 13n = 166
n + m = 12
15(12 – n) + 13n= 166
180 – 15n + 13n = 166
2n = 14
n=7m=5
Nick works 2 hours more than
Marnie.
Achievement with
Excellence (t)
ab(6a – 4b +4a) scores n
Correctly simplified AND
factorised.
Having given the two solutions the
discarding of the negative solution
is to be ignored – gains u.
Must deal with negative times
negative.
Omitting denominator gains n
Forming an equation.
Or solved by guess and check.
Substitution of values that add to 12
into 15m + 13 n = 166 to find m
and n scores r.
Two equations given and m =5 or
n = 7 found.
Or one equation given and two
solutions found
Correct algebra leading to
m = 5 and n = 7
Or a statement that Nick
worked two hours longer
A numerical error in the
multiplication or subtraction is not
penalised.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2013 — page 3 of 5
(g)
2
r
2
= ⎛⎜ 2 ⎞⎟ R Or
⎝ 3⎠
r =
r =
Equation set up.
⎛ 1 ⎞ πr2 h = ⎛ 1 ⎞ πR2 h
⎜ ⎟
⎜ ⎟
⎝ 2⎠
⎝ 3⎠
2
R
2
= ⎛⎜ 3 ⎞⎟ r
⎝ 2⎠
⎛ 2 ⎞ R2 Or R =
⎜ ⎟
⎝ 3⎠
2 R Or R =
3
⎛ 3 ⎞ r2
⎜ ⎟
⎝ 2⎠
3 r
2
Relationship involving r2 or R2
given.
Or consistent solution with the
multiplier of ½ on the wrong side
of the equation
Relationship involving r =
or R = given.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2013 — page 4 of 5
Achievement
(u)
Evidence
THREE
(a)
(b)(i)
9x8
Achievement with Merit
(r)
Achievement with
Excellence
(t)
Simplified.
Solved.
Answer including -4 gains n
Accept 4 x 4 x 4
r=4
Solved.
Answer including -4 gains n
Accept 2 x 2 x 2 x 2
1 = u + v Or 1 = f ⎛ 1 + 1 ⎞
⎜
⎟
uv
v⎠
f
⎝u
Either fractions added or expression
multiplied by one or more of f, u and
v to eliminate a denominator.
p = 3 64
OR p = 4
(ii)
(c)(i)
f=
1
uv
Or f = u + v
u +v
uv
Correct expression for f.
(c)(ii)
Watch for f and u the wrong way
round in the denominator
Correct expression for v.
f(u + v) = uv
Use of incorrect solution for f that
still involves a fraction to find a
consistent “v =” scores r..
uf + v f = uv
v=
(d)
Comments
1
uf
Or v =
1–1
u – f
f u
T = 5. 2(h – 1)
160 = 5 . 2(h – 1)
32 = 2(h – 1)
32 = 25
Removal of factor of 5 in equation ie
32 = 2h-1.
Solution found including CAO or
repeated doubling or guess and
check
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 1) 2013 — page 5 of 5
h=6
(e)(i)
(ii)
Total cost of the two meals = 2.5R
S=R+5
75 = 3.5R + 5
3.5R = 70
R = $20
Sharee’s meal is $25.
Equation for the total cost given for
any two of the meals in terms of one
variable. (accept unsimplified.)
Equation for the total cost of the
three meals given in one variable
and simplified form.
Do not penalise numerical errors in
dividing 70 by 3.5.
Accept solution of S = 70/3.5 + 5
for t.
Cost of Sharee’s meal
found.
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 2) 2013 — page 1 of 5
Assessment Schedule – 2013
Mathematics and Statistics CAT: Apply algebraic procedures in solving problems (91027, Day 2)
Evidence Statement
Evidence
Achievement with
Merit
r
ONE
(a)
x2 – 2x – 15
Correct expansion and
correctly simplified.
(b)
(x – 9)(x + 7)
Correct factorisation.
(c)
m = 2 or –6
Correctly solved.
Must have both solutions
(d)
8ab2 + a2b
Correctly simplified OR
factorised.
Or
ab(8b – 3a + 4a)
Correctly simplified
and factorised.
Omitted changing the sign
of the 2 in simplifying – (x
– 2) ie answer
4x - 2
5
Correct expression
which does not need
to be simplified.
5x – x + 2
5
5x – (x -2)
5
8x + 4
10
ab(8b + a)
(e)
(f)i)
ii)
Achievement
u
5x – x + 2 = 4x + 2
5
5
17s + 13r = 176
s + r = 12
17s + 13(12 – s) = 176
Comments
Having given the two solutions then discarding the
negative solution gains u.
Forming equation.
Or solved by guess and
Achievement with
Excellence
t
Two equations
given and
Must deal with negative times negative.
Omitting denominator gains n
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 2) 2013 — page 2 of 5
4s = 176 – 156
4s = 20
s = 5, so r = 7
Rewa works 2 hours more than
Sophia
(g)
check
found s = 5 or r = 7
Or one equation
given and two
solutions found.
A numerical error in the multiplication or subtraction is
not penalised.
Correct algebraic
solution giving s = 5
and r = 7.
Or statement that
Rewa worked 2
hours longer.
2
2
2π.r h = ⎛⎜ 1 ⎞⎟ πR h Or e q u i v a l e n t
⎝ 3⎠
Equation set up.
2
r
2
= ⎛⎜ 1 ⎞⎟ R Or
⎝ 6⎠
r =
r = R
⎛ 1 ⎞ R2
⎜ ⎟
⎝ 6⎠
2
R
= (6)r
Or R =
1 Or R = r
6
6
2
(6)r
2
Relationship
involving r2 or R2
given.
Or consistent
solution with 2 on
the wrong side
Relationship
involving r =
or R = given.
Substitution of values that add to 12 into
17s + 13r = 176
to find s and r scores r
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 2) 2013 — page 3 of 5
Evidence
TWO
(a)
Achievement
U
Achievement with
Merit
r
Achievement with
Excellence
t
Comments
x=2
Equation solved.
Fraction does not need to be simplified
(b)
8n – 12 = 3n – 33
5n = –21
n = –21 / 5 or –4.2
Equation solved. If
fraction correct and
decimal incorrect accept
answer
Incorrect calculation of -33 + 12 giving + 21 is not
accepted as a numerical error gains n.
(c)
(x + 7)(x – 2) = 0
x = –7 or x = 2
Equation factorised or
Or incorrectly factorised
equation consistently
solved. Must have both
solutions
Equation correctly
factorised and
correctly solved.
Must have both
solutions
(d)
–12 ≥ 2x
x ≤ –6
Generating solutions with
incorrect signs x ≥ -6
Or x = -6
Inequality solved.
Accept – 6 ≥ x
Or -12/2 > x
(e) i)
ii)
¾ S + 18
Expression given
T > ¾ × 72 + 18
> 54 + 18
Tama has saved at least as much
as Sharee
Consistent solution from
incorrect expression
Statement that both
saved $72
Or Tama has saved
$72
The discarding of the negative solution gains u
Be careful of incorrect factorising and solving giving
correct solution.
Correct statement
relating to the
inequality consistent
with the question.
Incorrect calculation of ¾ of 72 may be counted as a
numerical error for all levels.
Working may be with an equation or involve the incorrect
inequality symbol.
Tama has saved at least $72
(f)
(n – 5)(n – 2) = 0
n = 5 or n = 2
h is negative if 2 < n < 5
Expression factorised.
Or
Guess and check finding
the values of 2 and 5
Expression factorised
and solved for h = 0.
Accept expression
factorised and roots
found without
equating to zero
Solution found with
correct interval
statement in words or
symbols. Do not
penalise us of less
than or equal to sign.
If the solution gives the correct roots and a statement that
then gives a value in the range this must be supported by a
further statement e.g. the graph is a positive parabola or by
a graph showing the feasible region in order to gain t
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 2) 2013 — page 4 of 5
Evidence
THREE
(a)
(b)(i)
(ii)
(c)(i)
Achievement
u
8a12
Simplified.
m = 3 64
OR m = 4
Solved.
Inclusion of m = -4
gains n
n=4
Solved.
1 = u + v Or 1 = f ⎛ 1 + 1 ⎞
⎜
⎟
uv
v⎠
f
⎝u
Either fractions added
or expression multiplied
by one or more of f, u
and v to eliminate a
denominator.
f=
1
uv
Or f = u + v
u +v
uv
Achievement with
Merit
r
Achievement with
Excellence
t
Accept 4 x 4 x 4
Answer including -4 gains n
Accept 3 x 3 x 3 x 3
Correct expression for f.
(c)(ii)
Watch for f and u the wrong way round in the denominator
Correct expression
for v.
f(u + v) = uv
uf + v f = uv
v=
(d)
1
uf
Or v =
1–1
u – f
f u
T =3. 2(w – 1)
192 = 3 . 2(w – 1)
64= 2(w – 1)
64 = 26
Removal of factor of 3
in equation
ie 64 = 2w-1.
Solution found
including CAO or
repeated doubling or
guess and check
Use of incorrect solution for f that still involves a fraction to
find a consistent “v =” scores r..
NCEA Level 1 Mathematics and Statistics CAT (91027, Day 2) 2013 — page 5 of 5
w=7
(e)(i)
(ii)
Cost = 2.5 P
Z=P+5
75 = 3.5 P + 5
3.5 P = 70
P = $20
Zack’s day out cost $25
Equation for the total
cost for any two of the
days out given in one
variable. (accept
unsimplified.)
Do not penalise numerical errors in dividing 70 by 3.5.
Accept solution of Z = 70/3.5 + 5 for t.
Equation for the total
cost of the three days
out in terms of one
variable and simplified
form.
Cost of Zack’s day
out found.
1
To be completed by Candidate and School:
Name:
NSN No:
School Code:
DAY 1
TUESDAY
SUPERVISOR’S USE ONLY
Level 1 Mathematics and Statistics CAT, 2014
91027 Apply algebraic procedures in solving problems
Tuesday 16 September 2014
Credits: Four
You should attempt ALL the questions in this booklet.
Calculators may NOT be used.
Show ALL working.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
You are required to show algebraic working in this paper. Guess and check methods do not
demonstrate relational thinking. Guess and check methods will limit grades to Achievement.
Check that this booklet has pages 2 – 8 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.
ASSESSOR’S USE ONLY
Achievement
Apply algebraic procedures in solving
problems.
Achievement Criteria
Achievement with Merit
Apply algebraic procedures, using
relational thinking, in solving problems.
Achievement with Excellence
Apply algebraic procedures, using
extended abstract thinking, in solving
problems.
Overall level of performance
© New Zealand Qualifications Authority, 2014. All rights reserved.
No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
2
QUESTION ONE
(a)
ASSESSOR’S
USE ONLY
Simplify 5m2n × m3n2
(b) Solve 4a3 = 32
a=
(c)
Solve 8 =
5x − 4
2
x=
(d) Factorise 3a2b + a3b2 – 5a2b, giving your answer in the simplest form.
Mathematics and Statistics CAT 91027 (Day 1), 2014
3
(e)
Mark had worked twice as many hours as James.
If James had worked another 48 hours, he would have worked twice as long as Mark.
Write an equation, and use this to find how many more hours Mark worked than James.
(f)Solve
4x − 6
> 2x + 1
3
(g) Simplify
r2 −1
r2 + r
Mathematics and Statistics CAT 91027 (Day 1), 2014
ASSESSOR’S
USE ONLY
4
QUESTION TWO
(a)
ASSESSOR’S
USE ONLY
Factorise x2 – 3x – 40
(b) Sam is paid to work at a chemist shop after school.
He receives an extra $2 for each delivery he makes.
One day he makes 5 deliveries and is paid a total of $25.
If d = the number of deliveries:
Give the formula for the wages, P, that he receives each day.
(i)
(ii)Make d the subject of the formula you wrote in part (i).
(c)
Emma says that her height is at least as much as her younger sister’s plus a quarter as much
again.
(i)
(ii) Emma’s sister’s height is 96 cm.
Write an inequation to express Emma’s height E, in terms of the height of her sister, S.
Find Emma’s height.
Mathematics and Statistics CAT 91027 (Day 1), 2014
5
(iii) Use your answer from (c)(ii) to describe, in words, how Emma’s height compares with
her sister’s height.
(d)An n-sided polygon has D diagonals, where D =
n
( n − 3) .
2
Use the formula to find how many sides the polygon has, if there are 20 diagonals.
Mathematics and Statistics CAT 91027 (Day 1), 2014
ASSESSOR’S
USE ONLY
6
QUESTION THREE
(a)
Simplify
ASSESSOR’S
USE ONLY
3x 2 x
+
7
5
(b) Simplify (2 x 2 ) 3
(c)
Solve 3 × 2a – 1 = 96
a=
(d) Sam is investigating sequences of numbers.
One of the sequences is listed below:
Number, n
Sequence, T
Prime Number?
1
23
yes
2
25
no = 5 × 5
3
29
yes
4
35
no = 5 × 7
5
43
yes
The formula for the nth term of this sequence is T = n2 – n + 23.
(i)
What is the value of T for the 12th term in the sequence?
Mathematics and Statistics CAT 91027 (Day 1), 2014
7
Some of the numbers in the sequence are prime numbers.
(A prime number is one that can only be divided by 1 and itself. 1 is not a prime number.)
(ii) For the sequence of numbers where T = n 2 – n + a,
show that for any value of n, if n = a, then T will never be a prime number.
Assume n > 1.
(iii) If T = n2 – n + 5 and R = (5n – 4)(n + 1) – 2n(2n + 3) + 4(n + 1) – 3,
write an equation for R in terms of T.
(iv) Using the formula T = n2 – n + 1, find the value of n when T = 91.
(v) Explain why T = n2 – n – 6 will never give a prime number value for T.
Mathematics and Statistics CAT 91027 (Day 1), 2014
ASSESSOR’S
USE ONLY
8
91027A
QUESTION
NUMBER
Extra paper if required.
Write the question number(s) if applicable.
Mathematics and Statistics CAT 91027 (Day 1), 2014
ASSESSOR’S
USE ONLY
1
To be completed by Candidate and School:
Name:
NSN No:
School Code:
DAY 2
THURSDAY
SUPERVISOR’S USE ONLY
Level 1 Mathematics and Statistics CAT, 2014
91027 Apply algebraic procedures in solving problems
Thursday 18 September 2014
Credits: Four
You should attempt ALL the questions in this booklet.
Calculators may NOT be used.
Show ALL working.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
You are required to show algebraic working in this paper. Guess and check methods do not
demonstrate relational thinking. Guess and check methods will limit grades to Achievement.
Check that this booklet has pages 2 – 9 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.
ASSESSOR’S USE ONLY
Achievement
Apply algebraic procedures in solving
problems.
Achievement Criteria
Achievement with Merit
Apply algebraic procedures, using
relational thinking, in solving problems.
Achievement with Excellence
Apply algebraic procedures, using
extended abstract thinking, in solving
problems.
Overall level of performance
© New Zealand Qualifications Authority, 2014. All rights reserved.
No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
2
QUESTION ONE
(a)
ASSESSOR’S
USE ONLY
Factorise x2 – x – 30
(b) Solve 2x3 = 250
x=
(c)
Marie is paid $35 for baby-sitting for up to 3 hours, plus $15 for each extra hour.
(i)
If Marie works for more than 3 hours, give the formula that she could use to work out
her wages, W, where h is the number of hours she works.
Simplify your answer.
(ii)Make h the subject of the formula you wrote in part (i).
(d) Simplify
m2 − m
m2 − 1
Mathematics and Statistics CAT 91027 (Day 2), 2014
3
(e)
Solve 2 × 3y – 2 = 162
ASSESSOR’S
USE ONLY
y=
n
( n + 1)
2
(f)
The sum of the numbers 1, 2, 3, 4, … n is given by T =
Use algebra to find how many numbers of the sequence, starting from 1, need to be added
together to give a sum of 21.
Mathematics and Statistics CAT 91027 (Day 2), 2014
4
QUESTION TWO
(a)
ASSESSOR’S
USE ONLY
Simplify 4 a2b × a4b3
(b) Solve 9 =
7x + 1
4
x=
(c)
Jasper is investigating sequences of numbers.
Some of the numbers in the sequence are composite numbers.
(A composite number is a number which is the product of two numbers, but not including one
and itself.)
One of the sequences is: 9, 11, 15, 21, 29, … and is listed below:
n
T
Composite number?
1
9
YES as 3 × 3 = 9
2
11
NO
3
15
YES as 3 × 5 = 15
4
21
YES as 7 × 3 = 21
5
29
NO
The formula for the n th term of this sequence is T = n2 – n + 9.
(i)
What is the value of T for the 11th term in the sequence?
Mathematics and Statistics CAT 91027 (Day 2), 2014
5
(ii) For a sequence of numbers where
T = n2 – n + m
and
•
the value of m = n
•
n > 1,
(iii)If T = n2 – n + 3 and R = (3n + 2)(n – 2) – 2n(n+1) + 5(n + 3) – 12,
show that for any value of n, T will always be a composite number.
write an equation for R in terms of T.
(iv) Using the formula T = n2 – n + 1, use algebra to find the value of n when T = 57.
Mathematics and Statistics CAT 91027 (Day 2), 2014
ASSESSOR’S
USE ONLY
6
(v)
If T is a positive number and T = n2 – n – 12,
explain why T is a composite number for all values of n > 5.
(d) Solve
x−3
> 3x + 4
2
Mathematics and Statistics CAT 91027 (Day 2), 2014
ASSESSOR’S
USE ONLY
7
QUESTION THREE
(a)
ASSESSOR’S
USE ONLY
Simplify:
3x 2 x
(i) +
4
3
(ii)(2x3)2
(b) Factorise 5x2 y + xy2 – 7x2y, giving your answer in the simplest form.
(c)
Michelle has twice as much money as Nicola.
If Nicola is given $60, she will now have twice as much money as Michelle.
After Nicola has been given the $60, how much money would Michelle need to be given so
they have the same amount?
Remember to show algebraic working, including at least one equation.
Mathematics and Statistics CAT 91027 (Day 2), 2014
8
(d) Jon wants to know how much his mother and father earn each week.
His mother says the amount she earns changes each week depending on her hours.
She says she earns at least as much as his father plus a quarter as much again.
(i)
(ii) Jon’s father earns $408 each week.
Write an inequation to express how much Jon’s mother earns, M, in terms of how much
his father earns, F.
Find how much Jon’s mother earns each week.
You must use your inequation from part (i), and algebra, to find your answer.
(iii) Use your answer from (d)(ii) to compare, in words, the amount his mother earns with
the amount his father earns.
Mathematics and Statistics CAT 91027 (Day 2), 2014
ASSESSOR’S
USE ONLY
9
QUESTION
NUMBER
Extra paper if required.
Write the question number(s) if applicable.
Mathematics and Statistics CAT 91027 (Day 2), 2014
ASSESSOR’S
USE ONLY
NCEA Level 1 Mathematics and Statistics CAT (91027A) 2014 — page 1 of 8
DAY 1
Assessment Schedule – 2014
Mathematics and Statistics: Apply algebraic procedures in solving problems (91027A)
Assessment Criteria
Achievement
Merit
Excellence
Apply algebraic procedures in solving
problems.
Apply algebraic procedures, using
relational thinking, in solving problems.
Apply algebraic procedures, using
extended abstract thinking, in solving
problems.
Evidence Statement
Question
One
Evidence
Achievement
(a)
5m5n3 accept 5m5 × n3
Simplified.
(b)
a3 = 8
a=2
accept a = 3√8
Equation solved.
16 = 5x – 4
x=4
Correct solution.
Correct Answer Only accepted.
(c)
OR
Merit
Including a = –2 gains n.
20
5
(d)
a3b2 – 2a2b
= a2b(ab – 2)
Fully factorised
a2b(3 + ab – 5)
or fully simplified
a3b2 – 2a2b or equivalent.
Fully simplified and
factorised.
(e)
M = 2J
J + 48 = 2M
= 4J
3J = 48
J = 16
M = 32
Mark worked 16 hours
more than James.
OR
M – J = 16
A correct equation relating
M and J
u-p
OR
Guess and check for the
difference in the number of
hours worked from no equation.
Equations solved
finding either J or M.
OR
Correct conclusion
from guess and check
from 2 correct
equations.
OR
Consistent conclusion
where 1 equation is
incorrect.
4x – 6 > 6x + 3
–9 > 2x
x < – 4.5
OR
Expression rearranged
OR
x = –9/2 or equivalent.
OR
Consistent solution to an
incorrect inequality that has x
on both sides.
Inequation solved.
Numerator or denominator
factorised.
OR
An incorrectly factorised
expression correctly simplified.
u-p
Numerator and
denominator correctly
factorised.
(f)
−9
9
or
2
−2
−9
OR
>x
2
x<
(g)
Excellence
(r + 1)(r − 1) r − 1
=
r(r + 1)
r
Accept
r −1
r −0
Conclusion correct
for the difference in
the number of hours
worked.
Expression
simplified.
Accept correct
answer only
provided there is no
incorrect working.
NCEA Level 1 Mathematics and Statistics CAT (91027A) 2014 — page 2 of 8
Sufficiency
Less than 2u scores N
2 u gives A
3 u gives 2A
1r gives M
2r gives 2M
1t gives E
2t gives 2E
NCEA Level 1 Mathematics and Statistics CAT (91027A) 2014 — page 3 of 8
Question
Two
(a)
(b)(i)
(ii)
Achievement
(x – 8)(x + 5)
Correct factorsation.
Do not penalise if the candidate
continues and solves as if it was
an equation resulting in x = 8 or
–5
P = 25 + 2(d – 5)
= 15 + 2d
Accept with $ in
equation.
Unsimplified equation.
Do not accept P = 2d + 25
d=
P − 15
2
E ≥ S + 0.25S
≥ 1.25 S
Or equivalent.
(ii)
96
E ≥ 96 +
4
≥ 96 + 24
≥ 120
(iii)
Emma is at least 24
cm taller than her
sister.
OR Emma is at least
120 cm.
(d)
40 = n2 – 3n
n2 – 3n – 40 = 0
(n + 5)(n – 8) = 0
n = –5 or 8 (do not
need evidence of –5 as
a solution from a
correctly factorised
quadratic.
Merit
Excellence
Correct expression.
Consistent rearrangement from
b(i).
Rearrangement of equation must
involve at least 2 steps.
Correct inequality.
Accept > instead of ≥
Accept unsimplified
u-p
OR
Calculation giving E = 120
from E = S + 0.25S
or E = S + 0.25
OR
Correct statement of equality
from E = S + 0.25S
or E = S + 0.25
OR
Arriving at E ≥ 120 from
E≥S+¼
u-p
Equation rearranged.
OR
Incorrectly factorised as
(n – 5)(n + 8) = 0
Or guess and check.
Calculation of E ≥ 120 OR
Statement giving
equality
Emma is 120cn tall
Or Emma is 24 cm
taller (than her sister)
OR
Correct statement for
ii) calculated from
E ≥ S + 0.25 ie
involving 120 or 24
cm.
Correct
statement from
correct
inequality
including
where an
equation is
used in the
working.
Equation factorised and
=0
Solution given as 5
from incorrectly
factorised expression.
Solution with only
positive value in
answer.
Number of sides must
be +ve; therefore 8
sides.
Sufficiency
Less than 2u scores N
2 u gives A
3 u gives 2A
One grade
(c)(i)
Evidence
1r gives M
2r gives 2M
1t gives E
2t gives 2E
NCEA Level 1 Mathematics and Statistics CAT (91027A) 2014 — page 4 of 8
Question
Three
Evidence
Achievement
(a)
15x + 14x 29x
=
35
35
Simplified.
Without the x gains n.
(b)
8x6
Simplified.
(c)
2a – 1 = 96/3
= 32
2a - 1 = 25
a=6
Division by 3.
OR
Correct answer only.
T = 155
Solution found
(ii)
T = n2 – n + n
= n2
n2 is always the product of a
number multiplied by itself, so n
cannot be a prime number.
OR other valid reasoning.
Accept with a substituted for n
or n substituted for a.
Correctly simplified.
(iii)
R = 5n2 + 5n – 4n– 4
– 4n2 – 6n + 4n + 4 – 3
Correct expansion of one
multiplication of factors
giving
5n2 + 5n – 4n – 4 or
equivalent
OR
–4n2 – 6n
u-p
Fully simplified.
OR
Consistent relationship
for incorrect expansion
or simplification.
91 = n2 – n + 1
n2 – n – 90 = 0
(n + 9)(n – 10) = 0
n = –9 or 10
n = 10
Correct rearrangement
=0
OR
Guess and check giving
answer of 10
Forming and solving
giving solution of -9
and 10 or just 10
T = (n – 3)(n + 2)
Hence T is the product of 2
numbers.
Or any other valid explanation
of generalisation
Factorised.
u-p
OR
Two substitutions for
guess and check and a
comment or validation.
Sufficiency
Less than 2u scores N
2 u gives A
3 u gives 2A
(d)(i)
= n2 – n – 3
R=T–8
Accept T = R + 8
(iv)
(v)
Sufficiency across the paper
Achievement – a total of 3 A or higher.
This may be gained from an A in each of 3 questions
Or an A in one question and 2A in another.
Or A and M from different questions
Any lower scores Not achieved, as does a single E or M
Merit – 3 M or higher Or 1E and 1M
Excellence – 3 E Or 2E and 2 M
Merit
Excellence
Solved
Accept 6 – 1 = 5
Clearly explained with
full justification.
Correctly equated.
Correct generalisation
not fully justified
Conclusion stated
with justification.
1r gives M
2r gives 2M
1t gives E
2t gives 2E
NCEA Level 1 Mathematics and Statistics CAT (91027A) 2014 — page 5 of 8
Guidelines for marking the MCAT 2014 Day 1
Since the introduction of NCEA there have been changes in the way in which external
assessments are graded.
Grading in general
1. In grading a candidate’s work the focus is on evidence required within the achievement
standard.
2. Where there is evidence of correct algebraic working and the answer is then destroyed by a
numerical error, the candidate should not be penalised except 3di, e.g. converting of a
fraction to a decimal. If it cannot be determined if it is a numerical or algebraic error, the
grade should not be awarded, e.g. factorising of a quadratic.
3. Units are not required anywhere in the paper.
Grading parts of questions
4. The grade for evidence towards the awarding of achievement is coded as u or u-p,
for merit the demonstrating of relational thinking is coded as r,
and for excellence the demonstrating of abstract thinking is coded as t.
5. This standard requires demonstration of the ability to use algebra in the solving of
problems. This year in preparation for 2015 a new code “u-p” has been introduced into the
marking. This grade is awarded when a candidate shows evidence of the use of algebra in
the solving of a problem. Demonstrating the ability to use an algebraic skill listed in the
explanatory notes of the standard gains u. While not required this year, in future years a
candidate will be required to demonstrate that they can use algebra in solving a problem ie
require u-p grades in order to demonstrate sufficiency. Further information about this will
be provided in the 2015 assessment specifications.
Note that there will need to be changes in the language for some of the questions to ensure
the criteria for solving a problem are met, e.g. the writing of equations in the 2014 paper will
not qualify for a u-p grade because candidates are instructed to write the equation. If they
had been told to use algebra to … then this would have met the requirements for u-p.
In this context “in solving problems” means correctly
1. Selecting and applying an appropriate algebraic procedure(s) from those listed in EN 4
2. Translating a word problem into mathematical terms
3. Translating mathematical terms into word statements
in progressing towards a solution of a problem.
6. When the highest level of performance for a part of a question is demonstrated in the
candidate’s work a code is recorded against that evidence.
Only the highest grade is recorded for each part of a question. Questions with linked
sections only have one grade is awarded across those sections. This occurs in question 2c.
7. There is no requirement to satisfy evidence for achievement before evidence for merit is
counted, or for sufficient evidence for merit to be shown before evidence for excellence is
able to be recognised.
8. Sufficiency for each question.
Less than 2u scores N
1r gives M
2u gives A
2r gives 2M
3u gives 2A
Sufficiency across the paper
Achievement – a total of 3A or higher.
1t gives E
2t gives 2E
NCEA Level 1 Mathematics and Statistics CAT (91027A) 2014 — page 6 of 8
This may be gained from an A in each of 3 questions OR an A in one question and 2A in
another.
Or A and M from different questions
Any lower scores Not achieved, as does a single E or M
Merit – 3 M or higher OR 1E and 1M
Excellence – 3 E OR 2E and 2 M
9. Guess and check can only gain u, unless an equation is asked for in order to gain u which
they have done correctly. From there, they may use guess and check to solve the problem
and gain r. This applies in question 1e.
Results
10. When loading school data, ensure you follow the instructions given on the NZQA schools’
secure web site (In high security features, Provisional and Final Results Entry, L1 MCAT
Instructions – School’s PN has access to this).
11. Please ensure that all registered candidates have a grade recorded on the website.
Verifying
12. Reminder that candidates’ work submitted for verification should not be scripts where
assessors have allocated final grades by professional judgement or on a holistic basis.
Sufficiency examples
Question One
A
n
u
u
n
u
n
n
B
n
u
u
u
u
n
n
C
n
u
u
n
n
n
n
D
n
u
r
n
n
n
n
e u-p
t
up
r
n
n
up
t
f
n
u
n
r
n
n
n
g u-p
up
n
up
up
n
n
T
Question Grade
E
2A
2M
M
A
N
2E
NCEA Level 1 Mathematics and Statistics CAT (91027A) 2014 — page 7 of 8
Question Two
A
n
u
u
n
u
n
n
bi
r
u
r
n
u
n
n
ii
n
u
u
n
n
n
n
ci u-p
Ii
up
ONE
GRADE
u
t
r
r
n
t
iii u-p
d u-p(some
responses
n
n
n
up
n
n
t
Question Grade
E
2A
2M
M
A
N
2E
A
n
u
u
n
u
n
n
B
n
u
u
u
u
n
n
C
n
u
u
n
n
n
n
Di
n
u
r
n
n
n
n
dii
r
u
r
r
n
u
r
diii u-p
t
n
u
u
n
n
t
Iv
r
n
n
u
n
n
r
v u-p
t
n
u
n
n
n
u
Question Grade
2E
2A
2M
M
A
N
E
Question Three
Note:
i)
ii)
On the grids above the shaded grades are not taken into account in awarding the
question grade.
Only one grade is awarded in some parts of questions – in this case there is no line
dividing the parts of the question in the schedule, and the numbers are shaded in
the assessment schedule. Question 2c as highlighted above.
13. Holistic decision. If a candidate’s work provides significant evidence towards the award of a
higher grade and the assessor believes it would be appropriate to award such a grade, the
assessor should review the entire script and determine if it is a minor error or omission that
is preventing the award of the higher grade. The question then needs to be asked “Is this
error preventing demonstration of the requirements of the standard?” The final grade
should then be determined in the basis of the response to this question.
NCEA Level 1 Mathematics and Statistics CAT (91027B) 2014 — page 1 of 9
Assessment Schedule – 2014
DAY 2
Mathematics and Statistics: Apply algebraic procedures in solving problems (91027B)
Assessment Criteria
Achievement
Merit
Excellence
Apply algebraic procedures in solving
problems.
Apply algebraic procedures, using
relational thinking, in solving problems.
Apply algebraic procedures, using
extended abstract thinking, in solving
problems.
Evidence Statement
Question
One
Evidence
Achievement
(a)
(x – 6)(x + 5)
Correct factorising.
Do not penalise if the
candidate continues and
solves as if it was an
equation resulting in x = 6
or x = –5.
(b)
x3 = 125
x=5
Equation solved
accept x = 3 125
Including x = –5 gains n
W = 35 + 15(h – 3)
= 15h – 10
Unsimplified equation.
(c)(i)
Accept if interpreted as $35 an
hour ie
W = 35 × 3 + 15(h – 3)
= 15h + 60
h = (W + 10) / 15
Consistently rearranged
from c(i).
Rearrangement of
equation must involve at
least 2 steps.
(d)
m(m − 1)
m
=
(m + 1)(m − 1) m + 1
Numerator or denominator
factorised.
OR
an incorrectly factorised
expression correctly
simplified
u-p
(e)
3y – 2 =
m+0
m +1
162
2
= 81
3 y – 2 = 34
y=6
OR 2 × 34 = 162
y=6
Excellence
Correct simplified
expression.
Do not accept
W = 15h + 35
(ii)
Accept
Merit
Numerator and
denominator correctly
factorised.
Accept correct
answer only as long
as there is no
incorrect working.
Divided by 2.
OR correct answer only.
Expression
simplified.
m ≠ -1
not required
Solved.
Accept 6 – 2 = 4
NCEA Level 1 Mathematics and Statistics CAT (91027B) 2014 — page 2 of 9
(f)
n
21 = (n + 1)
2
42 = n2 + n
2
n + n − 42 = 0
(n + 7)(n − 6) = 0
n is a positive integer,
therefore n = 6
Equation rearranged.
OR
Incorrectly factorised as
(n – 7)(n + 6) = 0 equation
consistently solved
u-p
OR used guess and check.
Sufficiency
Less than 2u scores N
2 u gives A
3 u gives 2A
Factorised and = 0.
Solution with only
positive value in
answer.
Solution given as 7 from
incorrectly factorised
expression.
1r gives M
2r gives 2M
1t gives E
2t gives 2E
NCEA Level 1 Mathematics and Statistics CAT (91027B) 2014 — page 3 of 9
Question
Two
Evidence
Achievement
(a)
4a6b4 accept 4a6 × b4
Simplified
(b)
36 = 7x + 1
x=5
35
or
7
Correct solution.
Correct Answer Only
accepted.
T = 119
Solution found
(ii)
T = n2 – m + m OR n2 – n + n
= n2
n2 is always the product of a
number multiplied by itself.
Or other valid reasoning.
Correctly simplified.
Clearly explained
with full justification.
(iii)
R = 3n2 + 2n – 6n – 4
– 2n2 – 2n + 5n + 15 – 12
= n2 – n – 1
Correct expansion of one
multiplication of factors
giving
3n2 + 2n – 6n – 4 or
equivalent
OR
–2n2 – 2n
u-p
Fully simplified.
OR
Consistent
relationship between
R and T for incorrect
expansion or
simplification.
Forming and solving
giving solution of –7
and 8 or just 8.
(c)(i)
R=T–4
Accept T = R + 4
(iv)
57 = n2 – n + 1
n2 – n – 56 = 0
(n + 7)(n – 8) = 0
n = –7 or 8
n=8
Correct rearrangement = 0
OR
Guess and check giving
answer of 8.
(v)
T = (n – 4)(n + 3)
Hence T is a product of 2
numbers.
Factorised.
u-p
OR
Two substitutions for
guess and check and
comment or validation.
x – 3 > 6x + 8
–11 > 5x
x < – 11 / 5
Expression rearranged.
OR
(d)
OR
−11
>x
5
11
OR x <
−5
OR x < –2.2
Sufficiency
x=
Merit
Excellence
Correctly equated.
Correct generalisation
not fully justified.
Conclusion stated
with justification.
Inequality solved.
−11
5
OR
Consistent solution to an
incorrect inequality with x
on both sides.
Less than 2u scores N
2 u gives A
3 u gives 2A
1r gives M
2r gives 2M
1t gives E
2t gives 2E
NCEA Level 1 Mathematics and Statistics CAT (91027B) 2014 — page 4 of 9
Evidence
Achievement
(a)(i)
17x
12
Simplified.
Without x gains n.
(a)(ii)
4x6
OR 22x6
Simplified.
(b)
xy2 – 2x2y
= xy(y – 2x)
Fully factorised
xy(5x + y – 7x)
OR fully simplified
xy2 – 2x2y or
equivalent
Factorised and simplified.
(c)
M = 2N
2M = N + 60
4N = N + 60
N = 20
Nicola = 20 + 60 = 80
Michelle 40
Therefore need to give
Michelle $40.
A correct equation
relating M and N. u-p
OR
Guess and check for
the correct amount that
needs to be given from
no equation.
Equations solved finding
either M or N.
OR
Correct conclusion from
guess and check from 2
equations.
OR
Consistent conclusion
where 1 equation is
incorrect.
M≥F+F/4
≥ 5F / 4
Or equivalent
Correct inequality.
Accept > instead of ≥
Accept unsimplified
u-p
OR
Calculation giving
M = 510
from M = F + 0.25F
OR M = F + 0.25
OR
Correct statement of
equality
M = F + 0.25 F
OR M = F + 0.25
OR
Arriving at M ≥ 510
from equation
M≥F+¼
u-p
Examples
at the end
of the
schedule.
(d)(i)
(ii)
(iii)
F = 408
M ≥ 408 + 102
M ≥ 510
Mother earns at least $102
more than father.
OR
Mother earns at least $510.
Merit
Calculation of M ≥ 510
OR statement giving
equality.
Mother earns $510.
OR Mother earns $102
more (than father).
OR
Correct statement
involving 102 or 510
calculated from
M ≥ F + 0.25.
Excellence
Conclusion correct for
the amount Michelle
needs to be given.
Correct
statement from
correct
inequality
including
where an
equality was
used in the
working.
OR correct statement
with no mathematical
statement involving F
and ¼ (F).
Sufficiency
Less than 2u scores N
2 u gives A
3 u gives 2A
Sufficiency across the paper
Achievement – a total of 3 A or higher.
This may be gained from an A in each of 3 questions
Or an A in one question and 2A in another.
Or A and M from different questions
1r gives M
2r gives 2M
1t gives E
2t gives 2E
One grade
Question
Three
NCEA Level 1 Mathematics and Statistics CAT (91027B) 2014 — page 5 of 9
Any lower scores Not achieved, as does a single E or M
Merit – 3 M or higher Or 1E and 1M
Excellence – 3 E Or 2E and 2 M
From
U
R
M ≥ F + 0.25F
M ≥ 1.25 F
M > F + 0.25F
Inequation as given.
Calculated the 510 even if
an equality sign is used in
the calculation.
M = F + 0.25F
M ≤ F + 0.25F
M < F + 0.25F
Calculation giving
M = 510
OR
M ≤ 510
M < 510
Correct statement:
i) Mother earns at least
$510.
ii) Mother earns at least
$102 more than her
Father.
OR a consistent statement:
iii) Mother earns $510.
iv) Mother earns $102 more
than her Father.
v) Mother earns less than
$510.
vi) Mother earns at least
$102 less than her
Father.
M ≥ F + 0.25
Calculation of M ≥ 510
Statement:
i) Mother earns at least
$510.
ii) Mother earns at least
$102 more than her
Father.
M = F + 0.25
Calculation giving
M = 510
Statement giving:
i) Mother earns at least
$510.
ii) Mother earns at least
$102 more than her
Father.
iii) Mother earns $510.
iv) Mother earns $102 more
than her Father.
Any other expression for M
Treat as guess and check
gaining u for a calculation
≥ 510 and / or a correct
statement.
t
If the statement is correct ie
Mother is at least $510.
OR Mother earns at least
$102 more than her Father
This statement from the
correct inequality gains t
even if an = sign is used
throughout the calculation.
NCEA Level 1 Mathematics and Statistics CAT (91027B) 2014 — page 6 of 9
Guidelines for marking the MCAT 2014 Day 2
Since the introduction of NCEA there have been changes in the way in which external
assessments are graded.
Grading in general
1. In grading a candidate’s work the focus is on evidence required within the achievement
standard.
2. Where there is evidence of correct algebraic working and the answer is then destroyed by a
numerical error, the candidate should not be penalised except 2c(i), e.g. converting of a
fraction to a decimal is not penalised. If it cannot be determined if it is a numerical or
algebraic error, the grade should not be awarded, e.g. factorising of a quadratic.
3. Units are not required anywhere in the paper.
Grading parts of questions
4. The grade for evidence towards the awarding of achievement is coded as u or u-p,
for merit the demonstrating of relational thinking is coded as r,
and for excellence the demonstrating of abstract thinking is coded as t.
5. This standard requires demonstration of the ability to use algebra in the solving of
problems. This year in preparation for 2015 a new code “u-p” has been introduced into the
marking. This grade is awarded when a candidate shows evidence of the use of algebra in
the solving of a problem. Demonstrating the ability to use an algebraic skill listed in the
explanatory notes of the standard gains u. While not required this year, in future years a
candidate will be required to demonstrate that they can use algebra in solving a problem ie
require u-p grades in order to demonstrate sufficiency. Further information about this will
be provided in the 2015 assessment specifications.
Note that there will need to be changes in the language for some of the questions to ensure
the criteria for solving a problem are met, e.g. the writing of equations in the 2014 paper will
not qualify for a u-p grade because candidates are instructed to write the equation. If they
had been told to use algebra to … then this would have met the requirements for u-p.
In this context “in solving problems” means correctly
1. Selecting and applying an appropriate algebraic procedure(s) from those listed in EN 4
2. Translating a word problem into mathematical terms
3. Translating mathematical terms into word statements
in progressing towards a solution of a problem.
6. When the highest level of performance for a part of a question is demonstrated in the
candidate’s work a code is recorded against that evidence.
Only the highest grade is recorded for each part of a question. Questions with linked
sections only have one grade is awarded across those sections. This occurs in question 2c.
There is no requirement to satisfy evidence for achievement before evidence for merit is
counted, or for sufficient evidence for merit to be shown before evidence for excellence is
able to be recognised.
Coreect answer only gains u, except in question 1d.
7. Sufficiency for each question.
Less than 2u scores N
1r gives M
2u gives A
2r gives 2M
3u gives 2A
1t gives E
2t gives 2E
NCEA Level 1 Mathematics and Statistics CAT (91027B) 2014 — page 7 of 9
Sufficiency across the paper
Achievement – a total of 3A or higher.
This may be gained from an A in each of 3 questions OR an A in one question and 2A in
another.
Or A and M from different questions
Any lower scores Not achieved, as does a single E or M
Merit – 3 M or higher OR 1E and 1M
Excellence – 3 E OR 2E and 2 M
8. Guess and check can only gain u, unless an equation is asked for in order to gain u which
they have done correctly. From there, they may use guess and check to solve the problem
and gain r. This applies in question 3c.
Results
9. When loading school data, ensure you follow the instructions given on the NZQA schools’
secure web site (In high security features, Provisional and Final Results Entry, L1 MCAT
Instructions – School’s PN has access to this).
10. Please ensure that all registered candidates have a grade recorded on the website.
Verifying
11. Reminder that candidates’ work submitted for verification should not be scripts where
assessors have allocated final grades by professional judgement or on a holistic basis.
Sufficiency examples
Question One
a
n
u
u
n
u
n
n
b
n
u
u
u
u
n
n
c(i)
n
u
r
n
n
n
n
c(ii)
n
u
u
n
n
n
n
d u-p
t
up
r
n
n
up
t
e
n
u
n
r
n
n
n
f u-p
up
n
up
up
n
n
t
Question Grade
E
2A
2M
M
A
N
2E
NCEA Level 1 Mathematics and Statistics CAT (91027B) 2014 — page 8 of 9
Question Two
a
n
u
u
n
u
n
n
b
u
u
u
n
u
n
n
c(i)
n
u
u
n
n
n
n
c(ii) u-p
r
up
up
r
n
u
u
c(iii)
t
n
r
u
n
n
t
c(iv) u-p
r
u
r
u
n
n
r
c(v)
n
n
n
up
n
n
t
d u-p
r
n
n
u
n
u
r
Question Grade
E
2A
2M
M
A
N
2E
a(i)
n
u
u
n
u
n
n
a(ii)
n
u
u
u
u
n
n
b
n
u
u
n
n
n
r
c
t
up
r
r
n
u
t
Question Three
d(i) u-p
d(ii) u-p
ONE
GRADE
u
d(iii)
t
n
r
Question Grade
2E
2A
2M
M
n
n
r
A
N
E
Note:
i)
ii)
On the grids above the shaded grades are not taken into account in awarding the
question grade.
Only one grade is awarded in some parts of questions – in this case there is no line
dividing the parts of the question in the schedule, and the numbers are shaded in
the assessment schedule. Question 3d as highlighted above.
12. Holistic decision. If a candidate’s work provides significant evidence towards the award of a
higher grade and the assessor believes it would be appropriate to award such a grade, the
assessor should review the entire script and determine if it is a minor error or omission that
is preventing the award of the higher grade. The question then needs to be asked “Is this
error preventing demonstration of the requirements of the standard?” The final grade
should then be determined in the basis of the response to this question.
40
MAHOBE
www.mahobe.co.nz.
When they collide, the DS-742ET will be there calculating it for you.