Author Topic: Mathematics Question Thread (Read 1626447 times) Tweet Share

RuiAce · « **Reply #2970 on:** October 21, 2017, 10:55:51 pm »

Quote from: blasonduo on October 21, 2017, 10:53:46 pm

Hey! Just a quick question

For questions such as these, I always normally answer the question with decimal places, however, the answers normally give it in exact form.

My question is if they don't specify the exact value, and I put an approximate answer (to 3 d.p for example) will I lose a mark?

The examiner would be extremely rude to deduct you marks for it, but personally I believe that whenever it can be given in exact form it should be. You're taking a gamble if you just obtain a rounded answer without having an exact answer before-hand.

asd987 · « **Reply #2971 on:** October 21, 2017, 11:47:17 pm »

Hey so for number iv) the answer is 45.433... does that mean this is the 46th year and so the year will be 1991+46 which is the year 2037, or will it be rounded down to 45 then 1991+46 which is 2036. the answers/markers notes say its 2036, but i always thought you had to round it up. Can anyone please clarify, thanks.

kemi · « **Reply #2972 on:** October 21, 2017, 11:58:26 pm »

Quote from: asd987 on October 21, 2017, 11:47:17 pm

Hey so for number iv) the answer is 45.433... does that mean this is the 46th year and so the year will be 1991+46 which is the year 2037, or will it be rounded down to 45 then 1991+46 which is 2036. the answers/markers notes say its 2036, but i always thought you had to round it up. Can anyone please clarify, thanks.

Hey

They way I'd think about it is, the count is from the beginning of 1991, and the question says during which year? So if you got 45.433... that would be from the beginning of 1991, then as it is only partially through the 45th year, it must be 2036, not 2037 because well, it really hasn't reached 2037 at all. It's still part way through the 45th year - so add 45 to 1991.

RuiAce · « **Reply #2973 on:** October 22, 2017, 12:05:29 am »

Quote from: asd987 on October 21, 2017, 11:47:17 pm

Hey so for number iv) the answer is 45.433... does that mean this is the 46th year and so the year will be 1991+46 which is the year 2037, or will it be rounded down to 45 then 1991+46 which is 2036. the answers/markers notes say its 2036, but i always thought you had to round it up. Can anyone please clarify, thanks.

$\text{The key word was }\textbf{during}.$
$\text{Because they wanted the year }\textbf{during which}\text{ the population will reach 30 million}\\ \text{and you arrived at 45.433}\\ \text{then it was }\textbf{during}\text{ the 45th year when that happened}$
i.e. in 2036.
$\text{If they wanted the first year where the population}\\ \text{was over 30 million }\textbf{at the start of the year}\\ \text{then you would say the 46th year (2037).}$

bsdfjnlkasn · « **Reply #2974 on:** October 22, 2017, 08:43:24 am »

Hey!

Can I get some help as to how they derived the equation for the total income in the last part?
I don't get why it's (n-1)

RuiAce · « **Reply #2975 on:** October 22, 2017, 09:07:15 am »

Quote from: bsdfjnlkasn on October 22, 2017, 08:43:24 am

Hey!

Can I get some help as to how they derived the equation for the total income in the last part?
I don't get why it's (n-1)

These dodgy questions are part of the reason why I generally don't vouch for questions before 2001 unless you really need it. The method they provide is correct but trying to interpret why can be hard. Here's the reasoning:
$\text{Under ordinary circumstances, the most the two parties care}\\ \text{is that }\textbf{exactly}\,\boxed{15400}.$
$\text{But the fish farmer will ultimately run out of fish.}$
That was basically the whole point of part iii), to find when this happens.
$\text{So the supplier offers the fish farmer the opportunity}\\ \text{to buy }\textbf{ALL}\text{ of the fish left over at the price of }\$15\\ \text{and we're interested in how the farmer can use it to his advantage.}$
__________________________________________________
$\text{The key words are in what we want to actually maximise:}\\ \text{the }\textbf{total}\text{ income}.$
$\text{Instead of focusing on how much he can get during just that year}\\ \text{(which, is kinda useless),}\\ \text{we are focused on how much he gains }\textbf{in total}\\ \text{i.e., over the course of the }\textbf{entire}\text{ duration}\\ \text{he is in the industry.}$
$\text{The }(n-1)\text{ bit is reflective of all the years}\\ \text{that have }\textbf{already passed}.\\ \text{In the years that have already gone by, he sold }15400\text{ every year}\\ \text{at }\$10\text{, for }(n-1)\text{ years}.\\ \text{That is the first part: }\boxed{15400\times 10(n-1)}$
$\text{To maximise the profits, we want to find }\textbf{the exact year}\\ \text{in which he should sell the entire stock.}\\ \text{Basically, we make this the }n\text{-th year, and thus in this year}\\\text{he will actually make }\$15 \times\text{ all the fish he has} = \boxed{15F_n}$
$\text{Combine those together and there's our total income}\\ \text{That's why we maximise }\boxed{I = 154000(n-1) + 15F_n}$

bsdfjnlkasn · « **Reply #2976 on:** October 22, 2017, 10:05:08 am »

Hey there!

Just need some help with the last probability, thought it was 1/3 just by reading off the branch because that's what I thought the probability was... answer is 1/5

Nuwanda_g · « **Reply #2977 on:** October 22, 2017, 10:09:28 am »

Are we ought to mentally find the equation of velocity to integrate then find maximum displacement. Or is there an easier way? Thanks in advance

RuiAce · « **Reply #2978 on:** October 22, 2017, 10:15:58 am »

Quote from: Nuwanda_g on October 22, 2017, 10:09:28 am

Are we ought to mentally find the equation of velocity to integrate then find maximum displacement. Or is there an easier way? Thanks in advance

Already addressed in the compilation. The equation of the velocity isn't actually too hard to find here, because we have a straight line.

Mental computations are one possible way as well but that requires quite some good memory

K9810 · « **Reply #2979 on:** October 22, 2017, 10:16:55 am »

Hey, could you please help me with part iv)
Thanks

RuiAce · « **Reply #2980 on:** October 22, 2017, 10:17:11 am »

Quote from: K9810 on October 22, 2017, 10:16:55 am

Hey, could you please help me with part iv)
Thanks

Also already addressed in the compilation

RuiAce · « **Reply #2981 on:** October 22, 2017, 10:22:00 am »

Quote from: bsdfjnlkasn on October 22, 2017, 10:05:08 am

Hey there!

Just need some help with the last probability, thought it was 1/3 just by reading off the branch because that's what I thought the probability was... answer is 1/5

$\text{Because of how the probability works.}\\ \text{The first ball didn't need to be red. It could've been the second one.}$
$\text{The probability of at least one red is actually}\\ \mathbb{P}(\textbf{at least one red}) = \mathbb{P}(RR)+\mathbb{P}(RB)+\mathbb{P}(RW)+\mathbb{P}(BR)+\mathbb{P}(WR) = \frac56$
$\text{And of course, }\mathbb{P}(RR) = \frac16$
__________________
$\text{We actually require the }\textbf{conditional probability}\text{ of }RR\\ \text{given at least one }R.\\ \text{This will be }\frac{1/6}{5/6} = \frac15$

bsdfjnlkasn · « **Reply #2982 on:** October 22, 2017, 11:47:02 am »

How often do we have to use the product of roots in questions? Like is it something we should keep in mind when dealing with intersection points?

teapancakes08 · « **Reply #2983 on:** October 22, 2017, 11:53:34 am »

Part 1/2 List of SOS 2010 2U HSC Asks – 5(b)(iii), 8(d), 9(b)

...help.

teapancakes08 · « **Reply #2984 on:** October 22, 2017, 11:55:25 am »

Part 2/2 2010 2U HSC Asks – 10(a)(iv), 10(b)(ii)

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