HSC 2016 General Maths Marathon (1 Viewer)

davidgoes4wce · Feb 27, 2016

$$ This question was obtained from the New Century Maths Yr 12 book by: Bootsma, Hamper, Willard and Yen. Exercise 9-01 Q8e.$

$Calculate the difference in longitude between each pair of cities, and state whether the second city is east or west of the first.$

$Manila, Philippines $ (14 \degree N, 121 \degree E) $ and Port Moresby, Papua New Guinea $ (9 \degree S, 147 \degree E)$

$My answer was $ 26 \degree. $ The books answer was $ 86 \degree . $ (Since we are looking at East-West values) Could someone confirm?$

davidgoes4wce · Mar 3, 2016

I said in my answer it should be $0 \degree latitude$ . The answer has 'Equator'

For Q2a from the New Century Textbook

davidgoes4wce · Mar 17, 2016

$$ The great circle distance between Sydney and Honolulu, Hawaii is 4170 km. What angle is made at the centre of the Earth by these two cities? Answer to the nearest degree.$

I got an answer of 38 degrees. The books answer, is 37 degrees.

davidgoes4wce · Mar 17, 2016

This was my diagram, I then used the cosine rule in order to solve for $\theta$

jathu123 · Mar 17, 2016

davidgoes4wce said:
This was my diagram, I then used the cosine rule in order to solve for $\theta$

$$ you can't really use the cosine rule here cause it's just a sector of a circle, not a triangle. Since you know the radius and the arc length, you can do this: $ \\ \\ \frac{\theta}{360}\cdot2\cdot \pi \cdot 6400=4170 \\ \\$ now if you solve for theta, you'll get $37^{\circ}$

davidgoes4wce · Mar 18, 2016

Why did I think of using the cosine rule? My excuse was I've been working & studying from 5.30am through to midnight the last few days. Sometimes burnout can lead to forgetfulness.

Green Yoda · Mar 19, 2016

davidgoes4wce said:
This question is from the New Century Maths Year 12 textbook by : Klaas Bootsma, Sarah Hamper, Margaret Willard and Robert Yen. I am not sure how to answer Q1, so I thought I would bring it to an open forum for discussion.

sounds more like my yr 8 geo project ahahahaha

davidgoes4wce · Mar 24, 2016

New Century Yr 12 textbook

$It is possible to fly from Perth $ (32^\circ} S, 115^\circ E) $ to Tierra del Fuego $ (54^\circ S, 65^\circ W) $ on the southern trip of South America by going directly over the South Pole along a line of longitude. Calculate the great circle distance between Perth and Tierra del Fuego, correct to the nearest kilometre$

Answer: 10,500 km

Also as an interest, has anyone ever been to Tierra del Fuego before?

davidgoes4wce · Mar 24, 2016

What I am thinking initially was the difference in latitude is 22 degrees, the difference in longitude is 180 degrees.

I guess the part that threw me off is they say they travel directly over the South Pole, (which from what I know is $90^\circ S$ )

davidgoes4wce · Mar 24, 2016

InteGrand · Mar 25, 2016

davidgoes4wce said:
New Century Yr 12 textbook

$It is possible to fly from Perth $ (32^\circ} S, 115^\circ E) $ to Tierra del Fuego $ (54^\circ S, 65^\circ W) $ on the southern trip of South America by going directly over the South Pole along a line of longitude. Calculate the great circle distance between Perth and Tierra del Fuego, correct to the nearest kilometre$

Answer: 10,500 km

Also as an interest, has anyone ever been to Tierra del Fuego before?

The hardest part is visualising the diagram correctly, unless you are accustomed to spherical geometry (which may well be the case for General Maths students). I'll try my best to describe my diagrams.

$\noindent Consider this diagram:$

https://upload.wikimedia.org/wikipedia/commons/8/83/Latitude_and_longitude_graticule_on_a_sphere.svg.

$\noindent Let $O$ be the sphere's centre.$

$\noindent The angle marked $\phi$ there refers to the \textit{latitude} angle for a point on the sphere (Earth). So for a point $A$ on the sphere (in the picture, it's the point on the sphere's surface labelled with a dot), if it has latitude $\phi ^\circ$ South say ($0\leq\phi \leq 90$), that means it is in the Southern Hemisphere, and the angle $AOA_{\text{eq.}}$ is $\phi^\circ$, where $A_{\text{eq.}}$ is the point on the equator with the same \emph{longitude} as $A$ (in other words, on that diagram, and in general, $A_{\text{eq.}}$ is the point where the great circle through $A$ and the North/South Poles hits the equator (well there are two such points, and $A_{\text{eq.}}$ is the one nearer $A$.). If we denote by $O^\prime$ the centre of the circle of latitude through $A$ (i.e. the line of latitude through point $A$), then using alternate angles in parallel lines, the angle $\phi^\circ$ is also $\angle AOO^\prime$.$

$\noindent (In that picture, the point $A$ is in the upper hemisphere, but we can just call that the Southern Hemisphere for our visualisation purposes.)$

$\noindent Now, the angle marked $\lambda$ in the diagram refers to longitude, which is essentially angle $A_{\text{eq.}} OP_{\text{ref.}}$, where $P_{\text{ref.}}$ is a reference point chosen on the equator, defined as having $0^\circ$ latitude (it's the `left' point of the sector in the diagram involving the angle $\lambda$.) In geography, we conventionally use latitude West and East, with the angle being between $0^\circ$ and $180^\circ$, but it is mathematically equivalent to just say this latitude angle is between $0^\circ$ and $360^\circ$. Thus, any point on the sphere's surface can be located using these two numbers (angles) $\phi$ and $\lambda$, which tell us how far `up/down' (with respect to the equator) and how far `East/West' we are (with respect to $P_{\text{ref.}}$), respectively.$

$\noindent Now, for our particular problem, let the initial destination (Perth) be point $A$, and the final destination (Tierra del Fuego) be point $B$. On a diagram, draw the great circle through $A$ and $B$ (in other words, just a circle for now). Label the centre of this circle $O$, the centre of the sphere/Earth (since a great circle shares its centre with the sphere). Draw a dotted vertical line through $O$ (where this hits the `bottom' of the circle is the South Pole). Draw the point $A$ somewhere on the lower half of the circle ($A$ lies on the circle, since Perth is on the great circle), and draw a horizontal dotted line through $A$. This meets the vertical dotted line at a point; this point is the point $O^\prime$, namely the centre of the circle of latitude through $A$ (to see this, you will essentially need to draw a diagram of a sphere).$

$\noindent Now, $B$ also lies on this great circle, but it would be on the other side of $A$ as the vertical line, due to the large difference in longitude given. (Horizontal movement along the great circle corresponds to difference in longitude, while movement down corresponds to latitude. Note that the point $A^\prime$ with same latitude as $A$ but with $180^\circ$ greater longitude corresponds to the point diametrically opposite $A$ on the circle of latitude through $A$; in terms of our diagram, $A^\prime$ is the reflection of the point $A$ about $O^\prime$.) Since $B$ is more South than $A$, draw $B$ also lower than $A$ (i.e. a bit further down the circle than the point $A^\prime$). The diagram should look roughly like this for where $A$ and $B$ are:$

http://www.zefdamen.nl/CropCircles/Constructions/threepointscircle/3pointscircle05.gif.

$\noindent In the above picture (ignore the point in the upper half), the left point can refer to $A$, and the right one to $B$. Our diagram would also have the point $O$ and the dotted lines etc.$

$\noindent Now, also draw in the point $O''$ and join $BO''$, where $O''$ is to $B$ what $O^\prime$ is to $A$, i.e. $O''$ lies on the vertical dotted line and is horizontal; in other words, it is the centre of the circle of latitude through $B$.$

$\noindent Now, the path our plane will take is the minor arc $AB$ (the arc $AB$ that goes through the South Pole, which is the bottom of the circle we have drawn). We know that the radius of this great circle is simply the radius of the Earth, $R_{\oplus}$, which is $6400$ km, according to the HSC General Maths 2 formula sheet:$ ( https://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths-general-2-formulae-14.pdf ).

$\noindent So to calculate the required arc length, we need to find $\angle AOB$ (this is the angle $\theta$ in your formula). Now, recall that we said earlier that for a point $A$ with latitude angle $\phi$, we have $\angle OAO^\prime = \phi$, where $O^\prime$ is the centre of the circle of latitude through $A$. Hence we have in our diagram that $\angle OAO^\prime = 32^\circ$ and $\angle OBO'' = 54^\circ$, due to the given data on the latitudes of $A$ and $B$. Since triangles $AOO^\prime$ and $BOO''$ are right-angled, it follows from angle sum of a triangle that $\angle AOO^\prime = 58^\circ$ and $\angle BOO'' = 36^\circ$ (note that this is the same as $\angle BOO^\prime$, since $O^\prime$ lies between $O$ and $O''$). Now, by adjacent angles, $\angle AOB = \angle AOO^\prime + \angle BOO^\prime = 58^\circ + 36^\circ = 94^\circ$. It follows that:$

$\begin{align*}\text{minor arc } AB &= \frac{94}{360}\times 2\pi R_{\oplus} \\ &=\frac{94}{360}\times 2\pi \times 6400 \text{ km} \\ &= 10499.90078\ldots \text{ km} \\ &\approx 10500 \text{ km} \quad (\text{nearest km}).\end{align*}$

$\noindent Hence the answer is approx. 10500 km.$

Drongoski · Mar 25, 2016

Jathu's is the way to go - but no need for Cosine Rule.

The required distance is equivalent to traversing an arc of circle(called the Great Circle - radius here assumed 6400 km) subtending an angle of (36 + 58) = 94 degrees at the centre of the Earth. You fly south along the longitude 115 deg E and upon arriving at the South Pole, fly up north along the longitude 65 deg W. The length of this arc is simply:

$= r\theta = 6400 * \frac {94}{180} * \pi = 10499.9 \dots km$

PS

I'm bemused how our Grandmaster finds the advanced questions so easy, and has to labour over a mere General Maths question, from first principles.No offence meant, but I do find it so ironical.

davidgoes4wce · Mar 26, 2016

Multiple Choice Question

A. Increased in Both Year 10 and Year 11
B. Decreased in Both Year 10 and Year 11
C. Increased In Year 10 and decreased in Year 11
D. Decreased in Year 10 and increased In Year 11

The Answer was C. I wasn't too sure about the process they got to that answer.

All I know with 'Stratified Sampling'- a random sample is taken from each stratum or layer of a population, for example, if a population contains 70% adults and 30% children, the stratified sample must contain 70% adults and 30% children.

InteGrand · Mar 26, 2016

davidgoes4wce said:
Multiple Choice Question

A. Increased in Both Year 10 and Year 11
B. Decreased in Both Year 10 and Year 11
C. Increased In Year 10 and decreased in Year 11
D. Decreased in Year 10 and increased In Year 11

The Answer was C. I wasn't too sure about the process they got to that answer.

All I know with 'Stratified Sampling'- a random sample is taken from each stratum or layer of a population, for example, if a population contains 70% adults and 30% children, the stratified sample must contain 70% adults and 30% children.

$\noindent There are 1200 students in total, so the teacher needs to survey $120$ students (since this is $10\%$). Since she did the same number of students in each year group, she surveyed $20$ students from each year group (as there are six year groups).$

$\noindent Under the stratified sampling method, she would have to survey the right proportion of each year group. In other words, she would sample $S_k=\frac{n_k}{1200}\times 120 = \frac{n_k}{10}$ students from Year $k$ $\big{(}$$k \in \left\{7,8,\ldots, 12 \right\}$$\big{)}$, where $k$ is the number of students total in Year $k$ (given in the table). So the students sampled from Year 10 would be $S_{10}=\frac{n_{10}}{10}=\frac{230}{10}=23$, which is \emph{more} than the 20 sampled under the previous method. Similarly, the number of Year 11's sampled would be $S_{11}=15$, which is \emph{fewer} than the 20 sampled before. Hence the answer is (C).$

BLIT2014 · Mar 26, 2016

Two methods, I can see to approach this problem.

1. Mathematical way

Total Population of School is 1200

Total surveyed population= Total population of school * 10%
= 1200 * 10%
=120

Number of people surveyed in each age group = 120/6
= 20

Stratified Proportion

Proportion of year group (10) in school=230/1200

Proportion of year group(11) in school =150/1200

Number of students surveyed if started sample is chosen(Year 10)=230/1200 *120
= 23 (number of students being surveyed would increase, 23>20)

Number of students surveyed if started sample is chosen(Year 11) =150/1200 *120
= 15 (number of students being surveyed would decrease, 15<20)

2. Common Sense way?

We know that an equal proportion of students would be sampled from each year group in this case (20 per year group)

We know that 200 (1200/6) students would be need to be in a year group if the number of students were proportion in each year group

In year 10, 230>200, so therefore the number of students surveyed would increase
In year 11, 150<200, so therefore the number of students surveyed would decrease

davidgoes4wce · Mar 26, 2016

OK that makes a bit more sense now, i wasn't aware that we were comparing the benchmark of 20. (or 200 if we looked at the population of 1200 )

davidgoes4wce · Apr 3, 2016

2015 HSC Question 3

Gayle's gross pay each week is $952.25.

The following deductions are taken from her gross pay each week:

* Tax $180.93
*Superannuation $85.70
* Union membership $21.40
* Health Fund $38.15

What is Gayle's net pay each week?

A) $326.18
B) $626.07
C) $771.32
D) $952.25

I think there is a trick to the question, than meets the eye.

davidgoes4wce · Apr 3, 2016

OK I had an attempt of this question and am quite prepared to be laughed at (again) if I get this wrong:

I think the answer is 952.25-180.93-21.40-38.15-85.70=$626.07

Final answer is B. Anyone in agreeance with me?

InteGrand · Apr 3, 2016

davidgoes4wce said:
OK I had an attempt of this question and am quite prepared to be laughed at (again) if I get this wrong:

I think the answer is 952.25-180.93-21.40-38.15=$626.07

Final answer is B. Anyone in agreeance with me?

That is correct. The answers are here: https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/guides/2015-hsc-mg-maths-general-2.pdf .

Nailgun · Apr 3, 2016

davidgoes4wce said:
2015 HSC Question 3

Gayle's gross pay each week is $952.25.

The following deductions are taken from her gross pay each week:

* Tax $180.93
*Superannuation $85.70
* Union membership $21.40
* Health Fund $38.15

What is Gayle's net pay each week?

A) $326.18
B) $626.07
C) $771.32
D) $952.25

I think there is a trick to the question, than meets the eye.

what a random question lol
the 'trick' is that super isnt really a deduction?

HSC 2016 General Maths Marathon (1 Viewer)

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