IAS 1.1 Numeric Reasoning

IAS 1.1 – Numeric Reasoning
49
Compounding Rates Compounding Rates
Simple Interest
When money is borrowed from a bank the person borrowing the money must pay the money back to the bank plus an extra amount called interest.
If a person purchases an item on hire purchase they end up paying more for the item than the original price. For the convenience of being able to pay
the item off over a period of ‘x’ months they end up paying extra. This extra is interest.
The most basic form of interest is simple interest. Simple interest is interest paid on the principal only (the amount borrowed).
Consider the following. A person purchases a car and borrows $10 000 at 8% simple interest per annum for three years.
The simple interest paid can be calculated by using the formula I = PRT , where P is the principal
($10 000), R the interest rate (as a percentage, 8%) and T the time in years (3).
So using the formula I = (10 000 x 8 x 3) ÷ 100 we get $2400 as the interest paid on the loan over the three years.
The formula calculates the interest paid per annum on the amount borrowed (10 000 x 0.08) which is then multiplied by the period of the loan in years.
Compound Interest
Compound interest is calculated not just on the initial principal borrowed but also on any accumulated interest from prior compounding periods.
Consider the following. A person invests $10 000 with the bank at 8% compounded annually for three years.
The interest received can be calculated year by year. End of year 1: Balance = 10 000 + (10 000 x 0.08)
= $10 800
End of year 2: Balance = 10 800 + (10 800 x 0.08)
= $11 664
End of year 3: Balance = 11 664 + (11 664 x 0.08)
= $12 597.12
The person would have earned $2597.12
(12 597.12 – 10 000) in interest over the three years.
Interest is not always compounded annually. On loans such as mortgages it can be compounded monthly.
To calculate simple interest we use the formulaI = PRT
100
where I is the interest received P is the principal
R the interest rate and
T is the time in years
For simple interest, T must be in years if the interest rate is quoted as per annum.
Example
Find the amount of simple interest paid if Neil invests $35 000 for three years at 5% interest per annum.
Substituting P = 35 000, R = 5 and T = 3 into the
100
I = $5250
100
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
formula I = we get
PRT
100
I= 35000x5x3


50
IAS 1.1 – Numeric Reasoning
Compounding Interest cont...
If we look at another example, but this time set it out in a table it is possible to identify a formula for compound interest.
Doris invests $2500 at 6% per year for n years compounded annually. What is the total value of Doris’ investment after n years?
Beginning of year
Balance at start of year
$2500
End of year
1
$2500 x 1.061
2
$2650 x 1.061
$2650 x 1.062
3
$2650 x 1.062
$2650 x 1.063
4
$2650 x 1.063
$2650 x 1.064
.
.
.
n
$2650 x 1.06n–1
$2650 x 1.06n
Each year the balance has been multiplied by 1.06 which is equivalent to (1 + 6%) or (1 + 6 ).
Therefore to calculate the balance of an account receiving compound interest we use the formula
P x (1 + r%)n, where P is the initial amount invested (called the Principal), r is the interest rate and n is the number of periods the amount is compounded for.
If in the example above Doris invested her money for 4 years we would have P = $2500, r = 6 and
n = 4. Using a calculator we would enter:
On the TI-84 Plus:
which gives $3156 On the Casio 9750GII:
which gives $3156
In some compounding interest questions the amount is compounded every 3 months rather than annually, but the interest is quoted as ‘per annum’ or ‘yearly’. In this situation we need to adjust the interest rate accordingly for the number of times per year it is compounded.
Consider $2500 invested at 6% per annum, for five years, compounded every 3 months.
For this situation P = $2500, r = 1.5% (6% ÷ 4) and n = 20 (5 x 4).
100
2
5
0
0
x
(
1
+
6
÷
1
0
0
)
^
4
ENTER
2
5
1
0
0
+
0
0
6
x
(
÷
1
)
^
4
EXE
To calculate the balance of an account receiving compound interest we use the formula
A = P(1 + r%)n
where A is the account balance
P is the principal
r the interest rate and
n the number of compounding periods.
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
If $2500 is invested at 6% per annum, for five years, compounded every
3 months we make adjustments to
r and n to take into account that it is compounded quarterly, rather than annually.
So n = 5 x 4 = 20, since it is to be compounded 4 times a year for 5 years.
r = 6% ÷ 4 = 1.5%, since it is to be compounded 4 times a year.
So the calculation would be
2500(1 + 1.5%)20 = $3368 to the nearest dollar.


IAS 1.1 – Numeric Reasoning
51
Example
Carol invests $350 at 5% interest per annum, compounded annually. How much will she have at the end of the 10 years?
Substituting P = 350, r = 5 and n = 10 into the formula A = P(1 + r%)n
Example
David invests $5000 at 6% interest per annum, compounded six monthly. How much will he have at the end of the 15 years?
Substituting P = 5000, r = (6% ÷ 2) = 3% and
n = (15 x 2) = 30 into the formula A = P(1 + r%)n
we get
A = 350(1 + 5%)10
= 350(1 + 0.05)10
= 350(1.05)10
= $570 to the nearest dollar
we get
A = 5000(1 + 3%)30
= 5000(1 + 0.03)30
= 5000(1.03)30
= $12 136 to the nearest dollar
467.
469.
Find the amount of simple interest paid if 468. Ann was to invest $750 for one year at 4.5%
per annum.
Tabitha borrows $12 000 (simple interest 470. loan) to purchase a car and pays a rate of
16.5% per annum over 36 months. What is the simple interest on the loan?
Find the amount of simple interest paid if Paul invests $25 000 for 3 years at 4.5% per annum.
Tyler invests $50 000 for 5 years at 5.95% per annum. What is the simple interest earned on the investment?
Jacob borrows $7500 over 36 months at a simple interest rate of 14.25% per annum? How much are his monthly repayments?
Using the Casio 9750GII we would enter: 350x(1 5000x(
+5÷100 1+3÷10
) ^ 1 0 ENTER 0 ) ^ 3 0 EXE
which gives $570 which gives $12 316 Achievement – Answer the following simple interest questions.
Using the TI-84 Plus we would enter:
471. Matthew invested $25 000 for 5 years 472. and after 6 months received $1625 in
interest. What simple interest rate was he
earning on the money invested?
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


52 IAS 1.1 – Numeric Reasoning
Achievement – Answer the following compounding interest questions.
473. Jack invests $1200 at 8% per annum, compounded yearly. How much would he have after 10 years?
475. Courtney invests $50 000 at 4.5% per annum, compounded yearly. How much would she have after 20 years?
477. Tabitha invests $15 000 at 5% per annum, compounded every 6 months. How much would she have after 10 years?
479. Delia borrows $2500 to purchase a scooter. The interest rate is 9.6% per annum and it is compounded every 3 months for the period of the loan which is 2 years. How much in total does the scooter cost Delia?
481. Paul invests an inheritance of $24 000 at 4.2% per annum compounded every 4 months for 10 years. How much interest will he earn over this time?
474. Jill invests $8000 at 6.5% per annum, compounded yearly. How much would she have after 15 years?
476. Abe invests $125 000 at 3.5% per annum, compounded yearly. How much would he have after 25 years?
478. Hannah invests $80 000 at 4.4% per annum, compounded every 3 months. How much would she have after 10 years?
480. Jennifer borrows $1500 for a stereo at 14.5% per annum compounded six monthly for
two years. How much extra is Jennifer paying for the stereo?
482. Which is the better option? Invest $5000 at 4.5% per annum simple interest for 5 years or $4000 at 9% per annum compounded yearly for 5 years?
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


IAS 1.1 – Numeric Reasoning
53
Everyday Compounding Rates
Compounding rates can also apply to things
other than finance, hire purchase and loans. We can use compound rates to model population growth, exports and imports of a country, inflation, depreciation, radioactive decay and bacterial growth.
Inflation
Inflation is when the price of goods and services rise over time. We use the same formula as for compound interest to calculate the increased price of an article or service as a result of inflation, but use the variables I, O, r and n instead of A, P, r and n.
So I = O(1 + r%)n
whereIrepresentstheincreasedprice,Otheold price,rtherateofinflationandnthenumberof years.
Depreciation
Depreciation is when the value of an article decreases over time. We use the same basic formula as for compound interest, but because we are dealing with depreciation i.e. losing value we subtract rather than add. We use the variables D, O, r and n instead of A, P, r and n.
So D = O(1 – r%)n
where D represents the decreased or depreciated value, O the old price, r the rate of depreciation and n the number of years.
Example
A painting is initially purchased for $15 000 and appreciates at the rate
of 8.5% per annum.
What is the value of the painting in
10 years time?
We use the inflation formula with O = 15 000, r = 8.5
Example
The price of a litre of milk 20 years ago was 50¢. Today its price is $3.75. What has been the yearly rate of increase over that time?
We use the inflation formula with O = 50, I = 375 and n = 20.
I =Ox(1+r%)n 375 = 50x (1 + r%)20
7.5 = (1 + r%)20 (Dividing both sides by 50)
We now take the 20th root of both sides using the
and n = 10.
I =Ox(1+r%)n
I =15000x(1+0.085)10
I =15000x1.08510
I =$33915(tothenearestdollar)
The increased value of the painting is $33 915.
A car is initially purchased for $45 000 and depreciates at the rate of 25.2% per annum. What is the value of the car in 5 years time?
We use the depreciation formula with O = 45 000, r = 25.2 and n = 5.
D = O x (1 – r%)n
D = 45 000 x (1 – 25.2%)5
D = 45 000 x (1 – 0.252)5
D =45000x0.7485
D = $10 537 (to the nearest dollar)
The depreciated value of the car is $10 537.
Example
A computer was purchased for $6000. Four years later its value is estimated at $450. What has been the yearly percentage loss in value?
We use the depreciation formula with O=6000,D=450 andn=4.
D =Ox(1–r%)n 450 = 6000x (1 – r%)4
0.075 = (1 – r%)4
We now take the 4th root of both sides using the
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
TI-84 Plus calculator x√y
2
0 MATH 5 7 .
1.10560 =(1+r%) r% =0.10560
r =10.6%(1dp)
Example
5 ENTER
Casio 9750GII
^ 0 .
4 EXE
SHIFT
0 7 5
0.523 32 = (1 – r%)
r% =0.47668sor=47.7%(1dp)


54
IAS 1.1 – Numeric Reasoning
Example
The population of an endangered species
is declining at a rate of 2.5% per annum. It
is estimated that there are currently 850 of the species left in the wild. If the decline was to remain unchecked how long before there are only 100 of the species left?
We use the depreciation formula with O = 850, D = 100, r = 2.5.
D =Ox(1–r%)n 100 =850x(1–2.5%)n
0.1176 = (1 – 2.5%)n 0.1176 = 0.975n
n = 84.5 years
Note: Exponential equations such as 0.975n = 0.1176 can be solved for n using the solver on your graphics calculator (see below).
Using your Casio 9750GII
press MENU followed by 8 . Delete any existing equations then enter
It will list the variables and you enter all the known variables and place the cursor next to the unknown
(in this case N) and select SOLV F6 .
488. The population of a town is declining at a rate of 1.5% per annum. Currently the town has a population of 45 000. If the decline continues how long before the population of the town reaches a 35 000?
489. The premium on David’s life insurance policy when he took it out was $25.80 per month. It has increased yearly by an average of 4.8% per annum and is now $65.89 per month. How many years has David had the policy for?
Excellence – Answer the following questions.
483. A farmer pays $80 000 for a tractor which he can depreciate at the rate of 12% per annum. What is the value of the tractor in 10 years time?
484. Tori purchased her apartment 15 years ago for $195 000. Today it has a valuation of
$450 000. What has been the yearly percentage increase in value?
485. Michelle purchased a guitar for $250 and used it for 4 years before selling it on Trade Me for $85. What was the yearly percentage loss in value?
486. Stan’s starting salary when he began work after leaving university was $10 650. He estimates he has had a yearly increase of 7.5% per annum. What is his salary now, 25 years later?
487. The value of a company’s shares when it floated on the stock exchange were $7.50 each. Three years later they had a value of $4.95. What was the yearly percentage loss in value?
490. A company’s revenue has decreased yearly by 8.5%. Its initial value was $6.5 million and its current value is $2.7 million. How many years has it taken for it to drop to this level.
ALPHA
D
SHIFT
=
ALPHA
O
(
1
–
ALPHA
R
÷
1
0
0
)
^
ALPHA
N
EXE
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


IAS 1.1 – Numeric Reasoning 55
Practice Internal Assessment Task 1
Numeric Reasoning 1.1
Talia wants to put $50 000 on term deposit for a period of six months.
She looks up the investment options available from her bank and has to decide on one of the following options or combination of options to get the best return.
ACE Bank
3.95%
pa for five months
ACE Bank
Interest is paid at the end of each investment term and when Talia reinvests, she reinvests the total amount, principal and interest received to date.
What is the best investment option or combination of options for Talia to maximise her return over the six months and what is the value of her investment after this time?
Present your answer logically explaining each step and using appropriate mathematical calculations.
ACE Bank
3.75%
pa for one month
3.90%
pa for three months
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


56 IAS 1.1 – Numeric Reasoning
Practice Internal Assessment Task 2
Numeric Reasoning 1.1
Greg has $500 cash which he plans to put towards a Panasonic digital camera and an SD card.
He has identified the following three purchase options.
• Buy the camera for $818 plus the SD card for $39.98. Pay $500 cash and put the balance owing on his credit card which will cost him 2.5% per month interest on the outstanding balance. He
can afford to pay $100 a month towards repayments and interest.
• Buy the camera for $849 plus the SD card for $19.98. Pay $500 cash, receive three months’ interest free and pay interest of 2% per month after the interest free term. He can afford to
pay $100 a month towards repayments and interest.
• Buy the camera and SD card for $995, but get the GST of 15% refunded after the purchase. Pay $500 cash and a one off interest payment of 10.5% on the outstanding balance after 3 months. He can afford to pay $100 a month towards repayments and interest.
Identify which is the best option for Greg to go for.
Present you answer logically, investigating each option and using appropriate mathematical calculations.
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


IAS 1.1 – Numeric Reasoning 57
Practice Internal Assessment Task 3
Numeric Reasoning 1.1
Rewa does the fortnightly payroll for the firm she works for and is responsible for the wages calculation of a number of employees. The wage details of two employees are given below.
Employee One works 40 hours per week at an hourly rate of $14.75 per hour. The PAYE (tax) he pays can be calculated from the individual income tax rates in the table below and is deducted from his gross pay. In addition Employee One contributes 2% of his gross pay to KiwiSaver which is also deducted.
Employee Two works 45 hours per week at an hourly rate of $24.50 per hour. She pays PAYE (tax) as well as a student loan deduction of 8 cents per dollar earned. She also contributes 4% of her gross pay to KiwSaver which is also deducted.
What percentage difference is there in the fortnightly take home pay of Employee Two in comparison with Employee One?
Present you answer logically, calculating each employee’s fortnightly take home pay using appropriate mathematical calculations.
Salary (earnings per year)
PAYE tax rate
Up to $14 000
12.5 cents per dollar
over $14 000 to $48 000
over $48 000 to $70 000
21 cents per dollar
33 cents per dollar
over $70 000
38 cents per dollar
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


58
IAS 1.1 – Numeric Reasoning
Answers Page 5
Page 6 cont... 7.
Page 9
30. 6.0 (1 dp)
31. 9(1sf)
32. 3.22 (2 dp)
33. 39 (2 sf)
34. 2.62 (3 sf)
47
29
101
113
59
5
17
89
71
1. a) b) c)
2. a) b)
c)
3. a)
b)
c)
4. a)
b) c)
5. a) b)
c)
1,2,4,7,14,28
1,2,3,6,7,14,21,42
1,19
17, 34, 51, 68, ...
21,42,63,84,...
62, 124, 186, 248, ...
1,3,7,21 and1,2,4,7,8,14,28,56
HCF=7
1,3,5,9,15,45 and1,3,9,13,39,117
HCF=9 9. 1,5,19,95
and1,2,3,6,19,38,57,114
HCF=19 10.
12, 24, 36, 48, 60, 72, ...
20, 40, 60, ... 11.
LCM=60
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, ...
11, 22, 33, 44, 55, 66, .. LCM=66 13.
35. 12 (2 sf) 7=2+5,8=5+3, 36. 110 (2 sf)
8.
4=2+2,5=2+3,6=3+3,
9=2+7,10=5+5, 37. 0.13 (2 sf) 12=5+7, 13=11+2,
14=11+3,15=2+13 38. 22.8 m (1 dp) 16=13+3,18=7+11 39. 47 cm3 (2 sf) 19=2+17allcan. 40. 5.4m(1dp)
14 numbers less than 20. 41. 5.6L(2sf)
HCF of 448 and 616 is as only the 45 litres is
56 so greatest possible length measured (the 8 parts are is 56 cm. counted).
LCM of 28 and 24 which is 42. 43 cm2 (2 sf) 168. So 168 seconds 43. $353.50 (2 sf)
Page 6 6. 8,
14. 789.88 15. 0.0024 16. 50.0 17. 67 500 18. 0.0098 19. 2000 20. 655.0 21. 0.0480 22. 27 000 23. 44
24. 0.9 25. 5.23 26. 480 27. 8.4 28. 60 29. 7220
Page 10
45. 120 (accept 102) 46. 70 (accept 77) 47. 500
48. 9
49. 200
50. 60
51. 100
52. 250
53. $150
54. $10 000
55. $100
56. $3
57. $12 000
58. 1500 cm3
59. $300
60. 1250 km
9, 18, 27, 36, 45, 54, ... 15, 30, 45, ... LCM=45
prime number
138
2 x 69
3 x 23
2x3x23 140
10 x 14 2x52x7
2x5x2x7
16, 24, 32, 40, 48, 56, ... 14, 28, 42, 56, ...
56 seconds
so 16 pieces of chicken. 3 sf i.e 52.8 m3. Page 8 g) 18 hours.
12.
LCMof40,48and60=240, so 240 minutes (4 hours)
d) $170.95
e) $3897
48=2x2x2x2x3 44. a) 21 m2 108=2x2x3x3x3 b) 4 HCF=2x2x3=12 c) 55 cm3 (2 sf)
HCF of 96, 144 and 224, f) No should be rounded to
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


IAS 1.1 – Numeric Reasoning
59
Page 11
61. 8000 minutes
62. $180
63. $2400
64. Car $20 000. Deposit $4000.
Balance $16 000
+ interest = $18 000.
Cost per month = $1000. Other costs per month = $50. Fuel per month = $160. Sevices per month = $40. Total $1250 per month.
Page 12
65. 4.15 x 104 66. 5.91 x 102 67. 1.275 x 101 68. 4.5 x 10–2 69. 5.92 x 10–1 70. 7x100
71. 1.27 x 107 72. 9.56 x 10–6
Page 13
73. 9.32 x 103
74. 1.58 x 103
75. 4.53 x 10–5
76. 1.2 x 103
77. 4.44 x 101
78. 7.13 x 10–1
79. 5.0 x 102
80. 1.774 x 100
81. 5.025 x 106
82. 3.819 x 107
83. 5.0 x 10–3
84. 3.65 x 10–2
85. 0.000 485
86. 78 900 000
87. 0.325
88. 1.98
89. 987.6
90. 0.1423
91. 18.04
92. 0.0132
93. 499 seconds (3 sf)
94. 3.1 x 1010 years (2 sf)
Page 13 cont...
95. 9.47 x 1012 km (3 sf) 96. 14.0 x 1020 kg (1 dp)
=1.40x1021kg(2dp) 97. 1.9 x 1017 joules (2 sf) 98. 5.3 x 104 years
= 53 000 years
Page 14
99. 12.096 mm
=12mm(2sf) 100. $5263.15
$5300 per person (2 sf)
Page 16 cont... 122. 44
123. –27 124. 52 (0.4)
125. 2 5 (2.357 (3 dp)) 14
126. 3 (0.15) 20
127. 7 (0.778 (3 dp)) 9
128. 2 (0.667 (3dp)) 3
129. − 1 (–0.03125) 32
101. a) b) c) d) e)
f) g) h)
Page 16 102. 25 103. 21 104. 4 105. 64 106. 17 107. 60 108. 168 109. 168 110. 98 111. 540 112. 288 113. 24 114. 35 115. 177
116. 11 117. 124
130. 71–6x8+7x(4–4)–3 71–48+7x0–3
71–48+0–3 =20
131. 9+3–16+5x4–2 9+3–16+20–2
12–16+20–2 –4+20–2 16–2
=14
Page 18 132. 29 133. –129 134. –65 135. 183 136. –156 137. 240 138. –16 139. 27 140. 10 141. 648 142. 8 143. 3 144. –64 145. –39 146. –31 147. –5 148. 9 149. –1 150. –54 151. 0 152. –27 153. –114 154. 34
118. 17 25
120. 2 121. 73
73 800 000 000 000 7.38 x 1013
4.54 x 1023
8.88 x 10–10
5.35 x 1014 3.942 x 107 3.35 x 109 2.24 x 108
2
119. 27 70
120
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


60
IAS 1.1 – Numeric Reasoning
Page 18 cont... 155. –54
156. –34
157. 529
158. –89
159. 90
160. –14
161. 5074
162. –64
163. 167
164. –295
165. –1542
166. 60
167. –3
Page 19
168. $4750 Dr or –4750
169. $323 108 Dr or –$323 108 170. $32.5 million
171. 1290 m
172. 35 m below sea level or –35 m 173. 81 years
174. 1950 metres
175. 42 ˚C
Page 22 cont... 189. 13
Page 25 212. 19
24
50
190. 15 16
213. 111 20
191. 9 16
214. 7 40
192. 113 15
215. 3 4
193. 2 3
216. 3 50
194. 61
90 82
217. 1 200
195. 46 125 196. 464
218. 17 250
Page 23
219. 1 8
197. 198.
199.
200.
201.
202.
203.
204. 205.
206.
207.
208. 209.
210. 211.
7 16
15 minutes
3
20 7
20 4
220. 72% 221. 43.5% 222. 80% 223. 125% 224. 180% 225. 2.5% 226. 37.5% 227. 5% 228. $18 229. 135 kg 230. $323.75 231. 7.65 L 232. $22.50 233. 55
234. 4.74
235. 276
Page 26
236. $26.40
237. 12.9 kg
238. 756 girls 239. $105
240. 70.4% (1 dp) 241. 5%
242. $37.50
243. $17
244. 75.7% (1 dp) 245. 600 boys 246. a) $914.16
b) $68.56 c) $86.85 d) $83
176. a) b) c)
Page 22
177. 19
2370 years 1392 years 5414 years
15 11 15
4 15
178. 13
15 pieces 1890 products
6 32
29 284 $125
104 km
27 90
a) 95 vehicles b) 25 vehicles c) 20 cars
d) $900
e) 25 vehicles
f) 19 = 57 vehicles
179. 58
56
180. 27
6 21
181. 4 14
182. 40
183. 38
184. 31 15
9
185. 123 70
40
186. 1 35
187. 314 855
188. 311 15
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


IAS 1.1 – Numeric Reasoning
61
Page 33 cont...
298. $12, $20
299. 30L,10L
300. $49.78, $78.22
301. 4.75L,0.25L
302. 1.2m,2.4m,1.6m
303. $8.95, $5.37, $14.32, $7.16 304. $6.60, $11.00
305. $66, $55, $22 306. Alysia = $144
Barbara = $96 307. Clare = $26.67
Dennis = $20
Elliot = $33.33 308. $3400 and $5100
309. $480 and $200 Page 34
Page 28
247. $258.30
248. 126.1
249. $41.56
250. $47.25
251. $110.98
252. $3499.13
253. 506 pupils
254. $649 900
255. 1544 or 1545
256. $19 125
Page 29
257. 31.4 % (1 dp)
258. 23.1 % (1 dp)
259. 10.7% (1 dp)
260. 20.8% (1 dp)
261. 38.0 % (1 dp)
262. 46.6% (1 dp)
263. $182.61
264. $430.43
265. $45
266. $2086.96
267. $168.18
268. $133.90
269. $476.86
270. 12.0% (1 dp)
Page 30 Q272 cont...
b)
c) $402.50
d) 4.4%(1dp)
Page 31
Page 30
271. a) b) c)
d) e)
272. a)
2 16 3 3 :1
Page 35
315. 24 + 48 + 60 = 132
316. 10 men
317. 161.5 units of copper and 28.5 units of iron
318. 85 grams Page 36
319. 30 scientific and 10 algebraic calculators.
320. 24 males
321. Construction cost = $262 500
Total cost = $375 000 322. x=6
323. x=5
$314.50
$254.75
11.3% (1 dp)
Receives $266.00 now so better off by $11.25
77.6% (1 dp)
4.5% of $100 000
273. 274. 275. 276.
277.
$7024.64
$23 534.07
14.4% more (1 dp)
Commission – Company B
Total (including GST of 15%)
a) b) c) a)
b) c)
$30.52
$15.52
103.5%
Option 1 = $7693.12 Option 2 = $8525 Option 2 better
by $831.88
10.8% 11.3%
310. 311. 312. 313. 314.
$15
1.56 litres
12 cows
56 kg
a) 1333 g (1.333 kg) b) 67500L
c) $247.50
d) $224.24
e) 162.5 ml
f) 7.5 km
g) 3 gold medals.
Page 33
278. 1:
279. 1:
280. 1:
281. 2:
282. 16
283. 2:5
284. 3:4:10 285. 4:9:3 286. 3:5
287. 1:50
288. 1:6
289. 2:3:5 290. 10:6:2:1 291. 1:30
292. 1:100000 293. 3:10
294. w=35
295. x=144
296. y=2.33(2dp) 297. z=12
Commission – Company A
$210 000
Flat fee of $400
$400
$4500
2.5% of $110 000
$2750
Total (excluding GST)
$7650
Total (including GST of 15%)
$8797.50
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
$210 000
Flat fee
$250
4% of $80 000
$3200
3.5% of $130 000
$4550
Total (excluding GST)
$8000
$9200


62
Page 36 cont...
324. Printing = $36 000
Paper = $18 000 Covers = $9000 Total cost = $63 000
325. x=20
326. By 2400 books.
Page 38
327. 4.8 kg
328. 42 kg
329. 10 days
330. 20 days
331. 8 days
332. Uses an additional 300 L (1800 L in total).
Page 39
333. 4.2 ohms
334. 16 km/h
335. 22 500 baht 336. $500
337. 104.2 cm (1 dp) 338. $51.46
339. 2700 L
340. 1250
341. a) 33.3 g (1 dp)
b)1kg
c) 375 ml d) 187.5 g
Page 40
342. 1.6 hours
343. An extra 12 tradesmen
(27 in total)
344. For 33 days (33.3) 345. 11.7 minutes
346. 33.75 m
347. 7.5 minutes
348. $778 000 (3 sf) 349. 63 cm
IAS 1.1 – Numeric Reasoning
Page 42
Page 45 365. 1
Page 45 cont...
393. Between 3 and 4
394. Between 5 and 6 395. Between 6 and 7 396. Between 10 and 11 397. Between 12 and 13 398. Between 17 and 18 399. Between 30 and 31 400. Between 31 and 32 401. 6.083 (4 sf)
402. 14.07 (4 sf) 403. 19.49 (4 sf) 404. 24.90 (4 sf) 405. 12.25 (4 sf) 406. 17.32 (4 sf) 407. 30.82 (4 sf) 408. 31.62 (4 sf) 409. 39.75 (4 sf) 410. 50.12 (4 sf) 411. 98.89 (4 sf) 412. 113.5 (4 sf)
350.
351.
352.
353.
354. 355. 356.
357.
a) 7.5 seams/min.
b) 337 or 338 seams (337.5)
c) 133.3 mins. (1 dp) a) $14.75 per hour b) $3156.50
c) 340 hours
a) 427.5 km
b) 26.39 m/s
c) 3.8 seconds (1 dp) a) 18.7 mins.
b) 38.9 hours
c) 900 L/h
a) 25 kg/h
b) 23.3 kg/h
a) 0.354 cm/h
b) 23 days 13 hours a) 12 days
b) 1.2 days
c) 12 days x
a) 7.5 hours b) 4 hours
8
366. 1 25
367. 49 8
368. 125 369. 64
343
370. 625 371. 2187 372. 256 373. 1
374. 15 1
375. 16 1
375. 27
1 377. 6561
378. 4 3
Page 47
c) 60 hours Page 43 x
358. a) 12.5 days b) 22.1 days
359. a) 2 days b) 3.6 days
c) 15 people
360. a) 50 days b) 65 days
c) 45 people
361. a) 0.75 m track/ worker/day
b) 26.7 days
c) 100 workers
362. a) 200 seconds b) 54 km/h
c) 63 km/h 363. 4.8 hours 364. 4.5 hours
379. 2
64 413.
1 64
3
81
14
3 4
4
3 3
2
4 2
12
2
380. 27 414. 381. 20 415.
383. 85 416.
384. 96 417.
1 125
382. 28
1 100 216 1
385. 2 418.
1 419.
386. 4 420.
387. 2 421.
3 422. 388. 5
7 423. 389. 7
10 424. 390. 1 12 425. 2 426.
391. 13 427. 392. 13
5
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


IAS 1.1 – Numeric Reasoning
63
Page 48
1 428. 4
Page 48 cont... 463. 53 z4 464. 2x
465. 8 4 y3
466. 10 3 z2
Page 51
467. $33.75 468. $3375 469. $5940 470. $14 875 471. 13.0% 472. $297.40
Page 52
Page 55
Practice Assessment Task 1
Numeric Reasoning
Option 1
50 000 + 50 000 x 0.0395 x 5
= $50 822.92 12 1
50 822.92 + 50 822.92 x 0.0375 x 12 = $50 981.74
Option 2
Invest monthly at 3.75% for six months. 1
50 000 + 50 000 x 0.0375 x 12
= $50 156.25 1
50 156.25 + 50 156.25 x 0.0375 x 12 = $50 312.99 1
50 312.99 + 50 312.99 x 0.0375 x 12 = $50 470.22 1
50 470.22 + 50 470.22 x 0.0375 x 12 = $50 627.94 1
50 627.94 + 50 627.94 x 0.0375 x 12 = $50 786.15 1
50 786.15 + 50 786.15 x 0.0375 x 12 = $50 944.86
Option 3
429. 64 1
Invest for 5 months at 3.95% pa and then for 1 month at 3.75% pa.
430. 10
431. 8
1
432. 16
1
433. 25
1
434. 6
435. 64
436. 9
64
437. 27
438. 3
439. 16
27
440. 8
16
441. 9
8
442. 5
443. 15.625
444. 0.000 32
445. 10
446. 0.04
447. 16
448. 3.375
449. –5
450. 100
451. 0.16
452. –2
453. 0.027
454. 0.000 01
455. 3x
473. 474. 475. 476. 477. 478. 479. 480. 481. 482.
$2590.71
$20 574.73
$120 585.70
$295 405.62
$24 579.25
$123 918.52
$3022.31
$484.63
$12 420.83
$5000 at 4.5% simple = $6125.
$4000 at 9% compound = $6154.50.
So $4000 at 9% compounded by $29.50.
Invest for 3 months at 3.9% and then three single months at 3.75%.
50 000 + 50 000 x 0.039 x 3
= $50 487.50 12 1
Page 54
50 487.50 + 50 487.50 x 0.0375 x 12 = $50 645.27 1
50 645.27 + 50 645.27 x 0.0375 x 12 = $50 803.54 1
50 803.54 + 50 803.54 x 0.0375 x 12 = $50 962.30
Option 4
456. 3 y2
457. 4 z3
50 000 + 50 000 x 0.039 x 3
= $50 487.50 12 3
483. 484. 485. 486. 487. 488. 489. 490.
$22 280.08
5.7% (1 dp) 23.6% (1 dp)
$64 947.32 12.9% (1 dp) 16.6 years (1 dp) 20.0 years (1 dp) 9.9 years (1 dp)
Invest for three months at 3.9% and then three months at 3.9%.
458. 1x
459. 1
50 487.50 + 50 487.50 x 0.039 x 12 = $50 979.75
Best option is 5 months at 3.95% pa and 1 month at 3.75%pa. Total value of investment after this time is $50 981.74.
y3
460. 1 3 z2
461. 43 x
462. 3 y
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


64
IAS 1.1 – Numeric Reasoning
Page 56
Practice Assessment Task 2 Numeric Reasoning
Option 1
Cost after deposit of $500 = $357.98 End of month 1 =
357.98 x 1.025 – 100 = $266.93
End of month 2 =
266.93 x 1.025 – 100 = $173.60
End of month 3 =
173.60 x 1.025 – 100 = $77.94
End of month 4 =
77.94 x 1.025 – $79.89 = 0
Total paid $500 + $300 + 79.89
= $879.89
Option 2
Cost after deposit of $500 = $368.98 3 payments (interest free) = $300.00 End of month 4 =
68.98 x 1.02 – $70.36 = 0
Total paid $500 + $300 + 70.36
= $870.36
Option 3
$995 less GST = $865.22
After deposit of $500 = $365.22
3 payments (interest free) = $65.22 10.5% on outstanding balance = 65.22 x 1.105 – 72.07 = 0
Total paid $500 + $300 + $72.07 = $872.07
Best option is option is Option 2 for $870.36.
Page 57
Practice Assessment Task 3 Numeric Reasoning
Employee 1
Salary = 40 x 14.75 x 52 = $30 680 pa
PAYE = 0.125 x 14 000 = $1750.00 + 0.21 x 16 680 = $3502.80 = $5252.80
KiwiSaver 30 680 x 0.02 = $613.60 Total tax + KiwiSaver pa = $5866.40
Employee 2
Salary = 45 x 24.50 x 52 = $57 330 PAYE = 0.125 x 14 000 = $1750.00
Communicating solutions
At all levels there is a requirement relating to the communication of the solutions.
At Achieved level, the result of
a numerical calculation only is insufficient, working is expected and students need to indicate what the calculated answer represents. At Merit level, students need to clearly indicate what they are calculating, and their solutions need to be linked to the context.
At Excellence level, the response needs to be clearly communicated with correct mathematical statements, and students need to explain any decisions they make in the solution of the problem.
+ 0.21 x 34 000
+ 0.33 x 9 330 Loan = 57 330 x 0.08
= $7140.00 = $3078.90 = $4586.40
KiwiSaver 57 330 x 0.04 = $2293.20 Tax + KiwiSaver + loan = $18 848.50 Employee 1
Take home pay per fortnight =
($30 680 – $5866.40) ÷ 26 = $954.37
Employee 2
Take home pay per fortnight =
($57 330 –18 848.50) ÷ 26 = $1480.06
% difference = 55.1% (1 dp)
Assessment Clarifications:
Source: http://www.nzqa.govt.nz/quali- fications-standards/qualifications/ncea/ subjects/mathematics/clarifications/ level-1/as91026/
Expected evidence for Achieved
For the award of Achieved, the requirements include selecting and using a range of methods. The evidence for this aspect cannot come from a situation where students are told what method to use.
To be used as evidence, ‘methods’ must be relevant to the solution of the problem.
The ‘methods’ also need to be at the appropriate curriculum level for the standard, for example working with everyday fractions like one half, one quarter and one fifth is not at the appropriate curriculum level. For rounding with decimal places and significant figures, it is likely that there will need to be a holistic judgement on the evidence based on an understanding of rounding across the entire task
Extended abstract thinking
For the award of Excellence, there needs to be evidence that students are thinking beyond the problem. This could involve considering other identified factors and the effect of them on the solution of the problem. Alternatively, students could consider a change in one of the aspects involved in the solution and explore the consequences of that change on their solution.
IAS 1.1 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent


Other uLake titles
eas and ias series fOr Year 11, Year 12 and Year 13
Each EAS and IAS series comprises at least one External Achievement Standard (EAS) Workbook and
a number of Internal Achievement Standard (IAS) booklets. All are sold separately.
WOrkbOOks and hOmeWOrk bOOks fOr Year 8, Year 9 and Year 10
This (IAS) Internal Achievement Standard Booklet assists
Year 11 Mathematics students studying Numeric Reasoning 1.1.
It is designed to:
✰ Act as a guide by breaking the standard into bite-sized lumps.
✰ Provide reinforcement by giving examples and problems for each concept.
✰ Support students as they work to understand each concept by providing part solutions as well as full solutions for most problems.
✰ Assist students who miss class through sport or illness by providing detailed notes and examples.
✰ Be an excellent revision guide for the Internal Achievement Standard task.
Year 11 Mathematics
Year 11 Mathematics
EAS
IAS 1.1
Workbook
Numeric reasoNiNg NCEA
NCEA
1
1
NUL.IAS 1.1
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