## The Number and Algebra unit

Please find links to key chapters from this area of the course below. In addition, here is a growing collection of practice questions for you to use as a reviw exercise for the whole unit. Currently there are 5 exam style questions worth 38 marks and should be completed in about 40 minutes, before looking at the solutions.

START QUIZ!

#### N & A Review ** 1/1 **

*Shorter P1 style question - 6 marks *

The Jacksonville football team played 18 home games last season and the club calimed a mean average attendance of 15100 people per game.

a) Based on this claim, calculate the total attendance at home games for the whole season.

Ticket sales for the season suggest that total attendance was actually 269325

b) Calculate the percentage error in the claim made by the club

c) Write down your answer to part b) in th form \(a\times { 10 }^{ k },\quad where\quad 1\le a<10,\quad k\in Z\)

a) Total attendance

b) Percentage error %

c) a = , k =

a) 15100 x 18 = 271800

b) \(\frac { 271800-269325 }{ 269325 } \times 100=0.9189640769\\ =0.919 % (3sf)\) = 0.919% (3sf)

Using the formula for percentage error, The key is to identify the Exact amount as the figure given from ticket sales

c) \(0.919\quad =\quad 9.19\times { 10 }^{ -1 }\)

*Shorter P1 style question - 6 marks *

On a particular stretch of railway line, there are said to be approximately \(9.62\times { 10 }^{ 5 }\)railway sleepers, spaced at an average of 62.5 cm from center to center.

a) Based on these calculations, what is the length of this stretch of railway to the nearest km ?

b) Given that the actual stretch of railway is exactly 625 km, calculate the percentage error in your answer to part a)

c) Given the exact length of 625km and an approximate spacing of 62.5cm between sleepers, how many gaps would you expect to find on the stretch of railway line? Give you answer in the form \(a\times { 10 }^{ k },\quad where\quad 1\le a<10,\quad k\in Z\)

a) km

b) %

c) a = , k =

a) \(\frac { 9.62\times { 10 }^{ 5 }\times 0.625 }{ 1000 } =601.25\\ =601\quad (Nearest\quad km)\)

One might argue about whether or not the track starts and finishes with a sleeper, but I wouldn't make a difference to the answer when given to the nearest km. Follow through marks would be awarded if you used an incorrrect answer to part a)

b)\(\frac { 601-625 }{ 625 } \times 100=3.84\\ =3.84%\)

Identify 625 as the Exact answer

c) \(\frac { 625000 }{ 0.625 } =1000000=1\times { 10 }^{ 6 }\)

Convert both distances to m (or otherwise)

*Shorter P1 style question - 6 marks *

Consider the following numbers

\(\pi ,\quad 7.5,\quad \sqrt { 7 } ,\quad -6,\quad 94,\quad 3.\dot { 6 } \)

Now decide, for each of the following statements about these numbers whether they are true or false. (enter either T of F in the cell)

a) All of the numbers listed are Real

b) the number, \(3.\dot { 6 } \) is rational

c) 3 of the nubers belong to the set of integers

d) 2 of the numbers are ratinal but not integers

e) \(\pi \) is the only number that is real, but not rational

f) All natural numbers are integers

a) Yes, all of the numbers are real. There are no examples of complex numbers given

b) Recurring decimas ARE rational, because they can be expressed as fractions

c) Only, -6 and 94 are whole numbers

d) \(7.5\quad and\quad 3.\dot { 6 } \) both belong in this section

e) \(\sqrt { 7 } \) also belongs in this category

f) True - Integers are whole numbers, positive and negative, Natural numbers are whole and positive and so a subset of integers

*Shorter P1 style question - 8 marks *

Gordie invests 2000 dollars (USD) in an account that pays a nominal interest rate of 5.5% compounded monthly for a period of 4 years. Chris invests the same amount of money at a rate of 5.6% compounding quarterly for the same period.

a) Calculate the amount of money Gordie's investment is worth after 4 years? (give your answer to the nearest dollar

b) Does chris have more or less money? (Enter M or L)

After 4 years, Chris takes out the interest he has earned (to the nearest dollar) and converts ot to Euros for a trip to Paris. the conversion rate is 1USD = *x*Euros and he receives 448 euros for his dollars

c) What is the value of *x*?

a) Gordie's investment is worth

b) Chris has (M for more, L for less)

c) *x* =

a) \(2000\times { \left( 1+\frac { \frac { 5.5 }{ 12 } }{ 100 } \right) }^{ 4\times 12 }=2490.901141\\ =2491\quad (nearest\quad dollar)\)

b)\(2000\times { \left( 1+\frac { \frac { 5.6 }{ 4 } }{ 100 } \right) }^{ 4\times 12 }=2498.257936.901141\\ =2498\quad (nearest\quad dollar)\\ Chris\quad has\quad more\quad money\)

c)\(Chris\quad has\quad 498\quad USD\quad in\quad interest\\ 498x=448\\ \frac { 448 }{ 498 } =x=0.8995983936\\ x=0.90\quad (2dp\quad -\quad money)\)

*Longer P2 style question - 14 marks (part e) and particularly part f) are a focus for students looking for 6s and 7s)*

In this question, give all of your answers to the nearest dollar

Ace and Billy opened a new restaurant opened in Jacksonville and made a profit of 1200 USD in their first week. The profit increased by by 150 USD every week there after.

a) What was the profit made in the 18th week?

b) Calculate the total profit made by the restaurant in the first 20 weeks

Across town, Teddy and Vern opened their own restaurant at the same time. They also made 1200 USD in the first week, but their profits increased by 6% every week.

c) What was the profit in the 18th week of Tedy and Vern's place?

d) What was their total profit for the first 20 weeks?

e) What week was the first week that Teddy and Vern's place made more profit than Ace and Billy's?

f) After how many weeks was Teddy and Vern's total profit bigger than that of Ace and Billy?

a) 18th week profits are USD

b) Total of first 20 weeks profits are USD

c) 18th week profits are USD

d) Total of first 20 weeks profits are USD

e) Teddy and Vern make more profit for the fiorst time in week

f) Tedy and Vern's total profit is bigger for the first time afetr week

a) \({ U }_{ n }={ U }_{ 1 }+d(n-1)\\ { U }_{ 18 }=1200+17x150=3750\)

b) \({ S }_{ n }=\frac { n }{ 2 } ({ 2U }_{ 1 }+d(n-1))\\ { S }_{ 20 }=\frac { 20 }{ 2 } (2\times 1200+150(20-1))\\ { S }_{ 20 }=52500\)

c) \({ U }_{ n }={ U }_{ 1 }\times { r }^{ n-1 }\\ { U }_{ 18 }=1200\times { 1.06 }^{ 18-1 }\\ { U }_{ 18 }=3231.327343=3231\quad (nearest\quad dollar)\)

d) \({ S }_{ n }={ U }_{ 1 }\left( \frac { { r }^{ n }-1 }{ r-1 } \right) \\ { S }_{ 20 }=1200\left( \frac { { 1.06 }^{ 20 }-1 }{ 1.06-1 } \right) \\ { S }_{ 20 }=44142.70944=44143\quad (nearest\quad dollar)\)

e) \(Enter\quad in\quad to\quad GDC\quad table\quad function\\ { U }_{ n }=1200+150(n-1)\quad or\quad y=1200+150(x-1)\\ And\\ { U }_{ n }=1200\times { 1.06 }^{ n-1 }\quad or\quad y=1200\times { 1.06 }^{ x-1 }\\ x=25\quad is\quad the\quad first\quad time\quad the\quad second\quad is\quad bigger\quad than\quad the\quad first\)

f) \(Enter\quad in\quad to\quad GDC\quad table\quad function\\ { S }_{ n }=\frac { n }{ 2 } (2\times 1200+150(n-1))\quad or\quad y=\frac { x }{ 2 } (2\times 1200+150(x-1))\\ And\\ { S }_{ n }=1200\left( \frac { { 1.06 }^{ n }-1 }{ 1.06-1 } \right) \quad or\quad y=1200\left( \frac { { 1.06 }^{ x }-1 }{ 1.06-1 } \right) \\ x=34\quad is\quad the\quad first\quad time\quad the\quad second\quad is\quad bigger\quad than\quad the\quad first\)

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