Author Topic: Standard Math Q+A Thread (Read 182414 times) Tweet Share

Neilab · « **Reply #240 on:** March 27, 2017, 02:23:27 pm »

What is the best way to manage time in exams?

jamonwindeyer · « **Reply #241 on:** March 27, 2017, 02:24:52 pm »

Quote from: emilybrooks99 on March 27, 2017, 01:32:08 pm

Another question I got stuck with in my maths exam is this one below. Can anyone explain what the answer is and how to get it? (Image removed from quote.) thanks!

Sure! So the measure of kWh is really just a measure of time spent using the device. For example, the price of 12.5 cents per kWh. If we use a 1W device for an hour, we'd use 1kWh and it would cost 12.5 cents.

Here, we use the kettle 10 minutes a day (4 times 2.5 minutes). That is $10\times365=3650$ minutes per year, which is $3650\div60=60.83$ hours per year. The device is rated for 1800W. So, the final answer is obtained by multiplying the W of the device, with the time it is active to get Wh:

$1800\times60.83=109494$

Dividing this by 1000 to get kWh:

$109494\div1000=109.494$

Then multiplying by a price of $0.125:

$109.494\times0.125\approx$13.70$

So the answer is C (I'm not sure why it said nearest 50 cents though, because that would be $13.50?)

jamonwindeyer · « **Reply #242 on:** March 27, 2017, 02:28:55 pm »

Quote from: Neilab on March 27, 2017, 02:23:27 pm

What is the best way to manage time in exams?

Hey! I think it depends a lot on your personal style - Some people will like to spend more time on the hard stuff first. Some people will want the easy stuff out of the way. You need to figure out what works for you - The only golden rule is to never get stuck on one question for more than a minute, you can always come back for it!

olr1999 · « **Reply #243 on:** April 18, 2017, 03:18:03 pm »

Hi! Just wondering how to do question 5c and 6i in the attachment. Thanks!

RuiAce · « **Reply #244 on:** April 19, 2017, 08:29:53 am »

Quote from: olr1999 on April 18, 2017, 03:18:03 pm

Hi! Just wondering how to do question 5c and 6i in the attachment. Thanks!

$\text{I will only look at class A}\\ \text{For your data set, the median is 72}\\ \text{because when you count, you collide at }71\text{ and }73$
$\text{Further counting shows that the lower quartile is }62.5?\\ \text{The values below 62.5 are }62, 61, 58, 52\\ \text{There are 4 such values}$
Note: When I submitted the values into R (programming language) it told me that the lower quartile is actually 62.75. I'm not too sure why as I haven't had to do stem-and-leaf plots by hand in a while. It doesn't affect the final answer, but it could mess up working out.
$\text{But if you count the total number of values, you'll see that you have 16 values}\\ \text{As you'd expect, }\frac{4}{16}=\frac{1}{4}=25\text{ percent}\\ \text{and 25 percent of our data is }\textit{supposed to be below the lower quartile}$
$\text{In a similar way, 75 percent of our data will be below the upper quartile}$

olr1999 · « **Reply #245 on:** April 23, 2017, 01:49:46 pm »

Quote from: RuiAce on April 19, 2017, 08:29:53 am

$\text{I will only look at class A}\\ \text{For your data set, the median is 72}\\ \text{because when you count, you collide at }71\text{ and }73$
$\text{Further counting shows that the lower quartile is }62.5?\\ \text{The values below 62.5 are }62, 61, 58, 52\\ \text{There are 4 such values}$
Note: When I submitted the values into R (programming language) it told me that the lower quartile is actually 62.75. I'm not too sure why as I haven't had to do stem-and-leaf plots by hand in a while. It doesn't affect the final answer, but it could mess up working out.
$\text{But if you count the total number of values, you'll see that you have 16 values}\\ \text{As you'd expect, }\frac{4}{16}=\frac{1}{4}=25\text{ percent}\\ \text{and 25 percent of our data is }\textit{supposed to be below the lower quartile}$
$\text{In a similar way, 75 percent of our data will be below the upper quartile}$

I would have thought so too! But in actual fact (and I should have mentioned this before), the answer for question 5c is 30% for both data sets. Not sure how this answer was reached as the answers do not provide working out ...

RuiAce · « **Reply #246 on:** April 23, 2017, 02:04:40 pm »

Quote from: olr1999 on April 23, 2017, 01:49:46 pm

I would have thought so too! But in actual fact (and I should have mentioned this before), the answer for question 5c is 30% for both data sets. Not sure how this answer was reached as the answers do not provide working out ...

There's 16 numbers and 16 is divisible by 4 so I don't see how that works. Just in case I'm missing something in general maths I'll ask for some confirmation/correction

jamonwindeyer · « **Reply #247 on:** April 23, 2017, 02:15:52 pm »

Quote from: olr1999 on April 23, 2017, 01:49:46 pm

I would have thought so too! But in actual fact (and I should have mentioned this before), the answer for question 5c is 30% for both data sets. Not sure how this answer was reached as the answers do not provide working out ...

Hey! I'm with Rui, I don't get that answer of 30% - By definition each quartile represents 25% of your data, so the upper and lower quartiles should encompass 50% in both cases!

it could be that the answer is incorrect?

olr1999 · « **Reply #248 on:** April 23, 2017, 04:27:26 pm »

Quote from: jamonwindeyer on April 23, 2017, 02:15:52 pm

Hey! I'm with Rui, I don't get that answer of 30% - By definition each quartile represents 25% of your data, so the upper and lower quartiles should encompass 50% in both cases! it could be that the answer is incorrect?

I agree with both of you, I'm not sure how it works either but I think it means something different to how it sounds. The answer for question 6i is also different to what is expected (The answers say 56.25% for boys and 64% for girls)... I'll have to take it up with my maths teacher! Thanks anyway

jamonwindeyer · « **Reply #249 on:** April 23, 2017, 05:15:56 pm »

Quote from: olr1999 on April 23, 2017, 04:27:26 pm

I agree with both of you, I'm not sure how it works either but I think it means something different to how it sounds. The answer for question 6i is also different to what is expected (The answers say 56.25% for boys and 64% for girls)... I'll have to take it up with my maths teacher! Thanks anyway

Yeah I'm thinking we are misinterpreting the question somehow, let us know how you go!

tahliamag · « **Reply #250 on:** May 15, 2017, 10:35:34 am »

hey guys!
This is more a general question but does anyone have any tips on how to revise for prelim and hsc content at the same time? Thanks in advance!

Hi guys,
Does anyone have any tips on how to approach studying for multipe choice questions for exams? It's something I struggle with, thankyou!

Moderator action: Posts merged. At times like this, please resort to editing your post instead of posting consecutively.

olr1999 · « **Reply #251 on:** May 15, 2017, 11:15:58 am »

Quote from: jamonwindeyer on April 23, 2017, 05:15:56 pm

Yeah I'm thinking we are misinterpreting the question somehow, let us know how you go!

My maths teacher was also stumped! I guess we will just assume the answer given was a mistake...

Jess.martinuzzo · « **Reply #252 on:** May 15, 2017, 08:01:40 pm »

Hey guys,
Hope someone can help me, just doing a bit of maths homework and I'm struggling to figure out how to do this probability question.

'In Poker, a 'royal flush' is getting 10, J, Q, K and A of the same suit. Show that there is 1 chance in 649 740 of achieving this.'

please help!!

RuiAce · « **Reply #253 on:** May 15, 2017, 08:13:21 pm »

Quote from: tahliamag on May 15, 2017, 10:35:34 am

hey guys!
This is more a general question but does anyone have any tips on how to revise for prelim and hsc content at the same time? Thanks in advance!

Hi guys,
Does anyone have any tips on how to approach studying for multipe choice questions for exams? It's something I struggle with, thankyou!

Moderator action: Posts merged. At times like this, please resort to editing your post instead of posting consecutively.

20% of the final exam is based off preliminary content. That should give you a rough indication on how to study for it.

The main focus until the trials will still be HSC content so you should be well aware of what to focus on. Come HSC time, you may employ various strategies to ensure you have preliminary content covered, e.g. getting it back in your head first before everything else, adding bits and pieces of in as you study HSC content etc. It is all about time management at the end of the day.

You can only study by practicing. If there are specific areas of concern in multiple choice you may bring them up, but there is no better advice than keep practicing.

Quote from: olr1999 on May 15, 2017, 11:15:58 am

My maths teacher was also stumped! I guess we will just assume the answer given was a mistake...

It has to be a mistake then for sure.

Quote from: Jess.martinuzzo on May 15, 2017, 08:01:40 pm

Hey guys,
Hope someone can help me, just doing a bit of maths homework and I'm struggling to figure out how to do this probability question.

'In Poker, a 'royal flush' is getting 10, J, Q, K and A of the same suit. Show that there is 1 chance in 649 740 of achieving this.'

please help!!

$\text{Jokers are conventionally excluded in Poker.}$
$\text{There are only 4 favourable outcomes}\\ \text{This is quite easy to count; we know that the ranks must be 10 J Q K A}\\ \text{and there are only 4 suits}$
$\text{The total possible outcomes is the number of 5 card combinations drawn from a deck of 52 cards}\\ \text{To draw the cards, we have }52\times 51\times 50\times 49\times 48\\ \text{This part should not be too confusing}$
$\text{However there is repetition}\\ \text{Note that this is a situation where the }\textit{order does not matter}\\ \text{e.g. 2 3 4 5 6 is the same as 5 6 2 4 3}$
$\text{Since there are 5 cards, the repetition we must address is }5\times 4\times 3\times 2\times 1\\ \text{So the total possible outcomes is }\frac{52\times 51\times 50\times 49\times 48}{5\times4\times3\times2\times1} = 2598960$
$\text{So our required probability is }\frac{4}{2598960}=\frac{1}{649740}$

Jess.martinuzzo · « **Reply #254 on:** May 15, 2017, 08:22:04 pm »

{Jokers are conventionally excluded in Poker.}[/tex]
$\text{There are only 4 favourable outcomes}\\ \text{This is quite easy to count; we know that the ranks must be 10 J Q K A}\\ \text{and there are only 4 suits}$
$\text{The total possible outcomes is the number of 5 card combinations drawn from a deck of 52 cards}\\ \text{To draw the cards, we have }52\times 51\times 50\times 49\times 48\\ \text{This part should not be too confusing}$
$\text{However there is repetition}\\ \text{Note that this is a situation where the }\textit{order does not matter}\\ \text{e.g. 2 3 4 5 6 is the same as 5 6 2 4 3}$
$\text{Since there are 5 cards, the repetition we must address is }5\times 4\times 3\times 2\times 1\\ \text{So the total possible outcomes is }\frac{52\times 51\times 50\times 49\times 48}{5\times4\times3\times2\times1} = 2598960$
$\text{So our required probability is }\frac{4}{2598960}=\frac{1}{649740}$
[/quote]

Ahh thankyou so much! So essentially its: 52 x 51 x 50 x 49 x 48 over 5 factorial? Then with the 4 favourable outcomes over 2 598 960, which simplifies to the answer, 1/649 740?

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