Friends Boys School
Teacher: Mohammad Saleem
Past Papers
Math. Methods: Algebra
Paper 2
Answers
1.
(a)
(b)
Plan A: 1000, 1080, 1160...
Plan B: 1000, 1000(1.06), 1000(1.06)2…
2nd month: $1060, 3rd month: $1123.60
(A1)(A1)
For Plan A,
For Plan B,
(c)
(i)
(ii)
For Plan A,
For Plan B,
T12 = a + 11d
= 1000 + 11(80)
= $1880
(M1)
(A1)
T12 = 1000(1.06)11
= $1898 (to the nearest dollar)
(M1)
(A1)
12
[2000 + 11(80)]
2
= 6(2880)
= $17280 (to the nearest dollar)
S12 =
1000 (1.06 12  1)
1.06  1
= $16870 (to the nearest dollar)
S12 =
2
4
(M1)
(A1)
(M1)
(A1)
4
[10]
2.
360 240 3

 = 1.5
240 160 2
(a)
r=
(b)
2002 is the 13th year.
u13 = 160(1.5)13–1
= 20759 (Accept 20760 or 20800.)
(c)
5000 = 160(1.5)n–1
5000
= (1.5)n–1
160
 5000 
log 
 = (n – 1)log1.5
 160 
 5000 
log 

 160  = 8.49
n–1=
log 1.5
 n = 9.49  10th year
 1999
(A1)
1
(M1)
(M1)
(A1)
3
(M1)
(M1)
(A1)
(A1)
OR
Using a gdc with u1 = 160, uk+1 =
3
uk, u9 = 4100, u10 = 6150
2
1999
(d)
(e)
(M2)
(G2)
1.513  1 
S13 = 160 

 1.5  1 
= 61958 (Accept 61960 or 62000.)
Nearly everyone would have bought a portable telephone so there
would be fewer people left wanting to buy one.
4
(M1)
(A1)
2
(R1)
OR
Sales would saturate.
(R1)
1
[11]
3.
(a)
Ashley
AP
12 + 14 + 16 + ... to 15 terms
(M1)
S15 =
15
[2(12) + 14(2)]
2
(M1)
= 15 × 26
= 390 hours
(b)
(A1)
3
Billie
GP
12, 12(1.1), 12(1.1)2…
(i)
In week 3, 12(1.1)2
= 14.52 hours
(ii)
S15 =
(M1)
(A1)
(AG)
12[1.1 – 1]
1.1 – 1
(M1)
= 381 hours (3 sf)
(A1)
15
4
(c)
12 (1.1)n–1 > 50
(M1)
50
(1.1)n–1 >
12
(n – 1) ln 1.1 > ln
(A1)
50
12
50
12
n–1>
ln 1.1
ln
(A1)
n – 1 > 14.97
n > 15.97
 Week 16
(A1)
OR
12(1.1)n–1 > 50
By trial and error
12(1.1)14 = 45.6, 12(1.1)15 = 50.1
 n – l = 15
 n = 16 (Week 16)
(M1)
(A1)
(A1)
(A1)
4
[11]
4.
(a)
(i)
PQ =
=
(b)
AP 2  AQ 2
(M1)
42 = 2 2 cm
22  22 =
(A1)(AG)
(ii)
Area of PQRS = (2 2 )(2 2 ) = 8 cm2
(i)
Side of third square =
 2   2
2
2
Area of third square = 4 cm
(A1)
1st 16 2nd 8


2nd 8 3rd 4
(M1)
 Geometric progression, r =
10
(c)
(i)
1
2
u11 = u1r10 = 16   =
16
1024
3
= 4 = 2 cm
2
(ii)
(A1)
8 4 1
 
16 8 2
(A1)
(M1)
3
=
(ii)
S =
1
( = 0.015625 = 0.0156, 3 sf)
64
u1
16
=
1– r 1– 1
2
(A1)
(M1)
= 32
(A1)
4
[10]
5.
(a)
(b)
(i)
Area B =
1
,
16
(ii)
1
16  1
1 4
4
1
64  1 (Ratio is the same.)
1
4
16
(iii)
Common ratio =
1
4
(A1)
(i)
Total area (S2) =
1 1
5
= (= 0.3125) (0.313, 3 sf)
 
4 16 16
(A1)
(ii)
(c)
area C =
Required area = S8 =
1
64
1   1 
1  
4   4 
(A1)(A1)
8




1
1
4
= 0.333328 2(471...)
= 0.333328 (6 sf)
Note: Accept result of adding together eight areas
correctly.
1
4
Sum to infinity =
1
=
1
3
1
4
(M1)(R1)
5
(M1)
(A1)
(A1)
4
(A1)
(A1)
2
[11]
6.
(a)
(i)
$11400, $11800
(ii)
Total salary 
(A1)
10
(2 11000  9  400)
2
(A1)
= $128000
(b)
(A1) (N2)
(i)
$10700, $11449
(ii)
10th year salary  10 000(1.07)9
2
(A1)(A1)
(A1)
= $18384.59 or $18400 or $18385
(c)
1
(A1) (N2)
4
EITHER
Scheme A
SA 
n
 2 11000  (n  1) 400
2
(A1)
Scheme B
SB 
10 000(1.07 n  1)
1.07  1
(A1)
Solving SB  SA (accept SB  SA , giving n  6.33 ) (may be implied) (M1)
Minimum value of n is 7 years.
(A1) (N2)
OR
Using trial and error
(M1)
Arturo
Bill
6 years
$72 000
$71532.91
7 years
$85 400
$86 540.21
(A1)(A1)
Note: Award (A1) for both values for 6 years, and
(A1) for both values for 7 years.
Therefore, minimum number of years is 7.
(A1) (N2)
[11]
7.
(a)
(i)
r = 2
(ii)
u15 = 3 (2)14
= 49152 (accept 49200)
(b)
(i)
2, 6, 18
A1
N1
(A1)
A1
N2
A1
N1
4
(ii)
r=3
A1
N1
(c)
Setting up equation (or a sketch)
x 1 2 x  8
(or correct sketch with relevant information)

x  3 x 1
x2 + 2x + 1 = 2x2 + 2x  24
M1
A1
(A1)
x2 = 25
x = 5 or x = 5
x = 5
Notes: If “trial and error” is used, work must be
documented with several trials shown.
Award full marks for a correct answer with
this
approach.
If the work is not documented, award N2 for
a
correct answer.
(d)
1
2
(i)
r=
(ii)
For attempting to use infinite sum formula for a GP
S=
A1
N2
A1
N1
(M1)
8
1
1
2
S = 16
Note:
A1
N2
Award M0A0 if candidates use a value of r
where r > 1, or r < 1.
[12]
8.
(a)
(i)
S4 = 20
(ii)
u1 = 2, d = 2
A1
N1
(A1)
Attempting to use formula for Sn
M1
S100 = 10100
A1
N2
(b)
(c)
(i)
 1 4

M2 = 
 0 1
A2
(ii)
  1 2  1 4
 
 
For writing M3 as M2  M or M  M2  or 
  0 1  0 1 
M1
 1 0
M3 = 
0 0
A2
4  2

0  1
 1 6

M3 = 
 0 1
AG
N0
(i)
 1 8

M4 = 
 0 1
A1
N1
(ii)
 1 2  1 4  1 6  1 8
  
  
  

T4 = 
 0 1  0 1  0 1  0 1
 4 20 

= 
 0 4
(d)
N2
(M1)
A1A1
 1 2  1 4
 1 200 
  
  ...  

T100 = 
1
 0 1  0 1
0
100 10100 

= 
100 
 0
N3
(M1)
A1A1
N3
[16]
9.
Note: Throughout this question, the first and last terms are
interchangeable.
(a)
For recognizing the arithmetic sequence
u1 = 1, n = 20, u20 = 20 (u1 = 1, n = 20, d = 1)
Evidence of using sum of an AP
S20 =
1 20 20
S20 = 210
2
(or S 
20
2 119 1)
2
(M1)
(A1)
M1
A1
AG
N0
(b)
Let there be n cans in bottom row
Evidence of using Sn = 3240
eg
(M1)
1 n n  3240 , n 2  n 1  3240 , n 2n  n 1 1 3240
2
2
2
n2 + n  6480 = 0
A1
n = 80 or n = 81
(A1)
n = 80
(c)
(i)
A1
Evidence of using S =
1  n  n
(M1)
2
2S = n2 + n
A1
n2 + n  2S = 0
(ii)
N2
AG
N0
METHOD 1
Substituting S = 2100
eg n2 + n  4200 = 0,
2100 =
1  n  n
2
A1
EITHER
n = 64.3, n = 65.3
A1
Any valid reason which includes reference to integer being needed, R1
and pointing out that integer not possible here.
R1
N1
eg n must be a (positive) integer, this equation does not have
integer solutions.
OR
Discriminant = 16 801
A1
Valid reason which includes reference to integer being needed,
R1
and pointing out that integer not possible here.
R1
N1
eg this discriminant is not a perfect square, therefore no
integer solution as needed.
METHOD 2
Trial and error
S64 = 2080, S65 = 2145
Any valid reason which includes reference to integer
being needed,
and pointing out that integer not possible here.
A1A1
R1
R1
N1
[14]
10.
(a)
evidence of dividing two terms
eg 
1800
1800
,
3000
1080
r =  0.6
(b)
(M1)
A1
evidence of substituting into the formula for the 10th term
N2
(M1)
eg u10 = 3000( 0.6)9
u10 = 30.2 (accept the exact value 30.233088)
(c)
evidence of substituting into the formula for the infinite sum
e.g. S 
A1
N2
(M1)
3000
1.6
S = 1875
A1
N2
[6]
11.
evidence of using binomial expansion
(M1)
8
8
eg selecting correct term, a 8 b 0    a 7 b    a 6 b 2  ...
1
 2
evidence of calculating the factors, in any order
eg 56,
23
,  35 ,
3
3
A1A1A1
8  2 
   x   35
 5  3 
3
4032x3 (accept = 4030x 3 to 3 sf)
A1
N2
[5]