Download CHAPTER 2 REVIEW SOLUTIONS 1. (a) x + y = 10000 (A1) (C1) (b

CHAPTER 2 REVIEW SOLUTIONS
1.
(a) x + y = 10000
(A1) (C1)
(b) 2 × 12 + 3 × 5
(M1)
39 (39.0, 39.00) (AUD)
(A1) (C2)
(c) 12x + 5y = 108800
(A1) (C1)
(d) x = 8400, y = 1600
(A1)(ft)(A1)(ft) (C2)
Notes: Follow through from their equations.
If x and y are both incorrect then award (M1) for attempting to solve
simultaneous equations.
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2.
(a)
(b)
(c)
p + q = 47
(A1)
4p + q = 53
(A1) (C2)
Reasonable attempt to solve their equations
(M1)
p = 2, q = 45
(A1) (C2)
Note: Accept only the answers p = 2, q = 45.
C = 2 × 20.5(10) + 45
(M1)
C = 109
(A1)(ft) (C2)
Note: Award (M1) for substitution of 10 into the formula with their
values of p and q.
[6]
3.
(a)
(b)
(c)
(d)
50b + 20c = 260
(A1)
12b + 6c = 66
(A1)
Solve to get b = 4
(M1)(A1)(ft)(G2)
Note: (M1) for attempting to solve the equations
simultaneously
(i)
(ii)
(A1)(A1)(A1)
Notes: Award (A1) for labels and some idea of scale,
(A1)(ft)(A1)(ft) for each line
The axis can be reversed
(4, 3) or (3, 4)
(A1)(ft)
Note: Accept b = 4, c = 3
[8]
IB Questionbank Mathematical Studies 3rd edition
1
4.
(a)
(b)
Thursday’s sales, 6b + 9m = 23.40
(A1)
2b + 3m = 7.80
(A1) (C2)
m = 1.40
(accept 1.4)
(A1)(ft)
b = 1.80
(accept 1.8)
(A1)(ft) (C2)
Note: Award (A1)(d) for a reasonable attempt to solve by hand and
answer incorrect.
(c)
(A1)(A1)(ft) (C2)
Notes: (A1) each for two reasonable straight lines. The intersection point
must be approximately correct to earn both marks, otherwise penalize at
least one line.
The follow through mark is for candidate’s line from (a).
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5.
(a)
(b)
2925 = 12r + s
(M2)
4525 = 20r + s
(M2)
1600 = 8r
200 = r
(A2) (C6)
2925 = 12(200) + s
525 = s
(A2) (C2)
Note: Award (C2)(C2) if the candidate correctly solves an incorrect
system of equations.
[8]
6.
UP
Unit penalty applies in part (c)
(a) (x + 1)2–1 or x2 + 2x
(A1) (C1)
2
(b) (x + 1) – 1 = 109.25
(M1)
2
x + 2x – 109.25 = 0
(M1)
Notes: Award (M1) for writing an equation consistent with their
expression in (a) (accept equivalent forms), (M1) for correctly removing
the brackets.
OR
(x + 1)2 – 1 = 109.25
(M1)
x + 1 = 110.25
(M1)
Note: Award (M1) for writing an equation consistent with their
expression in (a) (accept equivalent forms), (M1) for taking the square
root of both sides.
OR
(x + 1)2 – 10.52 = 0
(M1)
(x – 9.5)(x + 11.5) = 0
(M1)
Note: Award (M1) for writing an equation consistent with their
expression in (a) (accept equivalent forms), (M1) for factorised left side
of the equation.
x = 9.5
(A1)(ft) (C3)
Note: Follow through from their expression in part (a). The last mark is
lost if x is non-positive.
If the follow through equation is not quadratic award at most
(M1)(M0)(A1)(ft).
(c) 4 × (9.5 + 1) = 42 m
(M1)(A1)(ft) (C2)
Notes: Award (M1) for correct method for finding the length of the fence.
Accept equivalent methods.
IB Questionbank Mathematical Studies 3rd edition
2
[6]
7.
(a)
(b)
(x – 2)(x – 4)
x = 2, x = 4
(c)
x = 0.807, x = 6.19
(A1)(A1) (C2)
Note: Award maximum of (A0)(A1) if coordinate pairs given.
OR
(M1) for an attempt to solve x2 – 7x + 5 = 0 via formula with
correct values substituted.
(M1)
7  29
x=
(A1) (C2)
2
(A1)(A1) (C2)
(A1)(ft)(A1)(ft) (C2)
[6]
8.
(a)
(b)
(c)
(d)
x(x – k)
(A1) (C1)
x = 0 or x = k
(A1) (C1)
Note: Both correct answers only
k=3
(A1) (C1)
 (3)
Vertex at x =
(M1)
2(1)
Note: (M1) for correct substitution in formula
x = 1.5
(A1)(ft)
y = –2.25
(A1)(ft)
OR
f′(x) = 2x – 3
(M1)
Note: (M1) for correct differentiation
x = 1.5
(A1)(ft)
y = –2.25
(A1)(ft)
OR
for finding the midpoint of their 0 and 3
(M1)
x = 1.5
(A1)(ft)
y = –2.25
(A1)(ft) (C3)
Note: If final answer is given as (1.5, –2.25) award a maximum of
(M1)(A1)(A0)
[6]
9.
(a)
(b)
(c)
6x + 3 – 6 + 2x = 13
8x = 16
(M1)
x=2
(A1) (C2)
(x + 3) (x – 1)
(A1)(A1) (C2)
x = 1.64575..
x = 1.65
(A2) (C2)
Note: If formula is used award (M1)(A1) for correct solution. If graph is
sketched award (M1)(A1) for correct solution.
[6]
10.
(a)
(b)
(c)
(d)
A = x2 + x or any equivalent unsimplified expression
(A1)(A1) (C2)
Note: Award (A1) for each term.
x2 + x = 30 or x2 + x – 30 = 0
(C1)
Note: The answer must be an equation.
(x – 5)(x + 6) = 0 or reasonable attempt to use formula.
(M1)(M1)
Note: Award (M1) for both signs wrong or one error in quadratic
formula (if used).
x = 5 or x = –6
(A1)(A1) (C4)
Note: Award (A2) d for x = 5 seen with no other working.
x = 5 because length must be positive (must have reason for the mark.)
(C1)
[8]
IB Questionbank Mathematical Studies 3rd edition
3
11.
(a)
(x – 5)(x  5)
(M1)(A1)(A1) (C3)
(b)
(x – 4)(x  1)
(M1)(A1)(A1) (C3)
(c)
x=4
x=–1
(A1)
(A1) (C2)
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12.
(a)
(b)
(x – 8)(x + 3) = 0
x = 8, x = –3
METHOD 1
(x – 5)(x + 2) = 0
x2 – 3x – 10 = 0
3x2– 9x – 30 = 0
a =3
METHOD 2
a(5)2 – 9(5) – 30 = 0
25a – 75 = 0
a =3
IB Questionbank Mathematical Studies 3rd edition
(M1)(M1)
(A1)(A1)(C2)(C2)
(M1)
(A1)
(A1)
(A1) (C4)
(M1)
(A2)
(A1) (C4)
4
METHOD 3
a(–2)2 – 9(–2) – 30 = 0
4a – 12 = 0
a =3
(M1)
(A2)
(A1) (C4)
[8]
IB Questionbank Mathematical Studies 3rd edition
5