Author Topic: Mathematics Question Thread (Read 1309793 times) Tweet Share

Mathew587 · « **Reply #1185 on:** February 15, 2017, 09:36:16 pm »

Hiya everyone,
Can someone pls help me with q 15 b ii) iii) iv) from the 2unit 2014 paper. I dont understand how the marking guidelines got the answers like what specifically they used.
Ty peeps

RuiAce · « **Reply #1186 on:** February 15, 2017, 10:11:07 pm »

Quote from: Mathew587 on February 15, 2017, 09:36:16 pm

Hiya everyone,
Can someone pls help me with q 15 b ii) iii) iv) from the 2unit 2014 paper. I dont understand how the marking guidelines got the answers like what specifically they used.
Ty peeps

Since you've presented the BOSTES answers (thanks - makes life easier), I'll just explain what they did a bit more slowly.

I'll assume you got part (i) out. In part (ii), they commence by using part (i). Since you have a pair of similar triangles, you know that the ratios of their sides are in proportion to each other. Hence the original ratio DR/DF = DS/DE

Then, if you look at the diagram, the side DE is really the side DS AND the side SE joined together.
Hence DE = DS+SE. This is simply substituted into the above ratio.

And lastly, sub in your x's and y's
______________________________

Solving part (iii) needs a bit of thought and/or a tad looking ahead.

The question's interested in areas. Well, since we have triangles involved, we can use A=1/2 bh or A=1/2 ab sinC. But we have no height, so the latter gets picked.

A₁ is the area of triangle DSR, which they tell you in the question. What do they do? Apply the formula 1/2 ab sin C straight into it.
And then A is the area of your large triangle. They apply the same formula here. This is what gets you the first mark.

The question wants you to find sqrt(A₁/A). So we start by finding A₁/A without the square root. That's just substitution.
But the reason 1/2 ab sin C gets used is that they realised "oh wow, sin D gets cancelled out!" And in fact, so does the 1/2. Which yields the result beautifully.
______________________________

You need to notice how in this part, triangle DSR is replaced by triangle SEQ. You need to somehow "see" that this is another pair of similar triangles. Hence, in a SIMILAR way to the above, they deduce that expression for sqrt(A₂/A), which you obviously need because sqrt(A₂) somehow appears in the final result.

Being able to see that if you repeat this whole process, except on a different triangle, is what earns you the first mark.

Lastly, direct substitution.

bananna · « **Reply #1187 on:** February 17, 2017, 06:32:09 am »

Hi I'm having a bit of trouble with these questions (cambridge textbook)

thank you!

RuiAce · « **Reply #1188 on:** February 17, 2017, 08:26:11 am »

Quote from: bananna on February 17, 2017, 06:32:09 am

Hi I'm having a bit of trouble with these questions (cambridge textbook)

thank you!

$\text{The first layer of tinting cuts out }\frac15\text{ of all the UV}\\ \text{So }\frac45\text{ of the UV penetrates through}$
$\text{The second layer cuts out }\frac15\text{ what penetrates through the first layer}\\ \text{So a further }\frac45\times \frac15=\frac4{25}\text{ gets cut out}$
$\text{In total, we cut out }\frac15 + \frac45\times \frac15\text{ of the UV}$
________________________
$\text{Following the pattern above, a third layer would cut out this much UV:}\\ \frac15 + \frac45\times \frac15 + \left(\frac45\right)^2\times \frac15$
$\text{So in general, the }n\text{-th layer will cut out this much UV:}\\ \frac15+\frac45\times \frac15 +\left(\frac45\right)^2\times \frac15+\dots + \left(\frac45\right)^{n-1}\times \frac15\\ \text{which is a geometric series, so by formula is equal to}\\ \frac{\frac15\left(1-\left(\frac45\right)^n\right)}{1-\frac45}$
$\text{We need to cut out at LEAST }\frac9{10}\text{ of the UV}\\ \text{Hence, we are solving:}\\ \begin{align*}\frac{\frac15\left(1-\left(\frac45\right)^n\right)}{1-\frac45}&\ge\frac9{10}\\ 1-\left(\frac45\right)^n &\ge \frac{9}{10}\\\left(\frac45\right)^n &\le \frac1{10} \\ \ln \left(\frac45\right)^n&\le \ln \frac1{10}\\ n \ln \frac45&\le \ln \frac1{10}\\ n &\ge \frac{\ln \frac1{10}}{\ln \frac45}\tag{*}\\ &=10.31\dots \end{align*}\\ \text{Hence 11 layers}\\ (*)-\text{Note, sign flips as }\ln \frac45\text{ is negative}$

RuiAce · « **Reply #1189 on:** February 17, 2017, 08:57:54 am »

Quote from: bananna on February 17, 2017, 06:32:09 am

Hi I'm having a bit of trouble with these questions (cambridge textbook)

thank you!

This whole concept of retelling, especially for just a 2U level, is exactly why the extension Cambridge in the questions do not reflect the scope of the HSC, and purely exist for self interest. In general I will not do these questions. However, the previous one was not so difficult - it was only tricky because of the wording.
$\text{Suppose he chooses his socks for day 1 first}\\ \text{This is important, as we will adjust this assumption for the next parts.}$
$\text{The first sock he picks can be any out of the 10: }\frac{10}{10}\\ \text{But the second sock must be matching to the first.}\\ \text{There is only ONE sock that matches the first, so the probability he picks it is }\frac{1}{9}\\ \therefore \frac{10}{10}\times \frac19= \frac19\\ \textit{Note - Once a sock is picked it can't be put back, hence NO replacement.}$
________________________________________
$\text{Problem: There's FIVE different pairs of socks}\\ \text{He can choose certain socks for day 1 first}\\ \text{But then we have to factor in day 2, 3, 4 AND 5 as well}\\ \text{He can not get matching pairs of socks}\\ \text{and he can ALSO get matching pairs, but then the order of which pair gets muddled up}$
$\text{There is simply way too much to count}\\ \text{Hence, suppose instead he chooses socks for }\textbf{day 5}\text{ first}$
$\text{But if does this, then it just becomes the same thing as part i)}\\ \text{Hence also }\frac19$
________________________________________
$\text{Now choose for day 3 first instead}\\ \text{Same thing, }\frac19$
________________________________________
$\text{Note that the pairs of socks will ALWAYS be MATCHING}\\ \text{So we start analysing similar to a)}$
$\text{We know that in the first morning, the probability of matching socks is }\frac19\\ \text{Counting successively, for the second day he has 8 socks to start with}\\ \text{And he can pick any of them with probability }\frac88$
$\text{Then, only ONE of the 7 socks remaining matches the one he picked already}\\ \text{The probability he gets it is }\frac17\\ \text{So the probability of getting matching socks on the second day}\\ \textbf{having factored in the FIRST day}\\ \text{is }\frac88 \times \frac17 = \frac17$
$\text{So the probability for both days is }\frac19 \times \frac17$
________________________________________
$\text{Repeating the above analysis for days 3, 4 and 5 gives us }\\ \frac19 \times \frac17 \times \frac15 \times \frac13\times \frac 11$
________________________________________
$\text{This is nonsense}\\ \text{If you got matching socks for 4 days in a row}\\ \text{On the 5th day, you only have 2 socks left, and they're GUARANTEED to be matching}\\ \text{So the probability of this is effectively }0$

Hplovers · « **Reply #1190 on:** February 17, 2017, 05:58:56 pm »

Hi!
I am having trouble with these two questions:

1. Find the area bounded by the curve x = y^2 - 2y - 3 and the y-axis.

2. Find the area bounded by the curve x = -y^2 - 5y - 6

Thanks!

Kle123 · « **Reply #1191 on:** February 17, 2017, 07:52:51 pm »

Quote from: Hplovers on February 17, 2017, 05:58:56 pm

Hi!
I am having trouble with these two questions:

1. Find the area bounded by the curve x = y^2 - 2y - 3 and the y-axis.

2. Find the area bounded by the curve x = -y^2 - 5y - 6

Thanks!

Since i don't exactly know what is troubling you (may it be that it is the area with y-axis instead of x?), i just assumed you had the background knowledge to understand the following working.

jamonwindeyer · « **Reply #1192 on:** February 18, 2017, 12:15:19 am »

Quote from: Kle123 on February 17, 2017, 07:52:51 pm

Since i don't exactly know what is troubling you (may it be that it is the area with y-axis instead of x?), i just assumed you had the background knowledge to understand the following working.

Legend! Thanks for the awesome answer Kle123

bananna · « **Reply #1193 on:** February 18, 2017, 04:22:26 pm »

hello!
had some trouble with this q:

A jar contains red buttons and white buttons in the ratio of 3:2. If three buttons are chosen at random from the jar, find the probability that:

a) exactly two are red

b) not more than one is white

thank you!!

RuiAce · « **Reply #1194 on:** February 18, 2017, 04:35:15 pm »

Quote from: bananna on February 18, 2017, 04:22:26 pm

hello!
had some trouble with this q:

A jar contains red buttons and white buttons in the ratio of 3:2. If three buttons are chosen at random from the jar, find the probability that:

a) exactly two are red

b) not more than one is white

thank you!!

$\text{Note that we treat ratio-based probability questions}\\ \textbf{as though there IS REPLACEMENT}$
$\text{Here is the reasoning:}\\ \text{Suppose you only have 5 buttons.}\\ \text{Then, ok, only 3 are red and 2 are white}\\ \text{So if we only had 5 buttons, we should treat it without replacement}\\ \text{and for a) we'd get }\frac{3}5\times \frac24$
$\text{But suppose instead you have 5000 buttons}\\ \text{Then in fact, 3000 are red and 2000 are white}\\ \text{So without replacement, the answer would now be }\frac{3000}{5000}\times \frac{2999}{4999}\\ \text{Which is very very close to just }\frac{3}{5}\times \frac{3}{5}\\ \text{But COMPLETELY different to }\frac35\times \frac24$
$\text{In general, if ratios are used}\\ \text{We assume that the total quantity is so large, like above}\\ \text{To the point a small change (take one out)}\\ \textbf{ does virtually NOTHING to the ratio}$
$\textit{i.e. the disturbance to the ratio is NEGLIGIBLE}\\ \text{Hence, we approximate by assuming there WAS replacement}$
______________________________________________
$\text{With replacement, the answer to a) is just}\\ \frac35\times \frac35$
$\text{And for b), we use the complement event}\\ \text{that is, both are white.}\\ \text{Answer: }1-\frac25\times \frac25$

Kle123 · « **Reply #1195 on:** February 18, 2017, 05:08:18 pm »

Quote from: RuiAce on February 18, 2017, 04:35:15 pm

$\text{Note that we treat ratio-based probability questions}\\ \textbf{as though there IS REPLACEMENT}$
$\text{Here is the reasoning:}\\ \text{Suppose you only have 5 buttons.}\\ \text{Then, ok, only 3 are red and 2 are white}\\ \text{So if we only had 5 buttons, we should treat it without replacement}\\ \text{and for a) we'd get }\frac{3}5\times \frac24$
$\text{But suppose instead you have 5000 buttons}\\ \text{Then in fact, 3000 are red and 2000 are white}\\ \text{So without replacement, the answer would now be }\frac{3000}{5000}\times \frac{2999}{4999}\\ \text{Which is very very close to just }\frac{3}{5}\times \frac{3}{5}\\ \text{But COMPLETELY different to }\frac35\times \frac24$
$\text{In general, if ratios are used}\\ \text{We assume that the total quantity is so large, like above}\\ \text{To the point a small change (take one out)}\\ \textbf{ does virtually NOTHING to the ratio}$
$\textit{i.e. the disturbance to the ratio is NEGLIGIBLE}\\ \text{Hence, we approximate by assuming there WAS replacement}$
______________________________________________
$\text{With replacement, the answer to a) is just}\\ \frac35\times \frac35$
$\text{And for b), we use the complement event}\\ \text{that is, both are white.}\\ \text{Answer: }1-\frac25\times \frac25$

Hey Rui, great explanation. thanks for teaching me that in ratio questions we should treat it as replacement. But aren't three buttons being chosen? I haven't revised probability since prelims, but shouldn't the answer to part (a) be (3/5)*(3/5)*(2/5)*3?(times 3 as there are 3 combination)

and part (b) 1-(P(3white) +P(2white,1red)=== 1-((2/5)^3+((2/5)^2*(3/5)*3)
Or
P(2red,1white)+P(3red)=((3/5)^2*(2/5)*3)+(3/5)^3

RuiAce · « **Reply #1196 on:** February 18, 2017, 05:12:09 pm »

Quote from: Kle123 on February 18, 2017, 05:08:18 pm

Hey Rui, great explanation. thanks for teaching me that in ratio questions we should treat it as replacement. But aren't three buttons being chosen? I haven't revised probability since prelims, but shouldn't the answer to part (a) be (3/5)*(3/5)*(2/5)*3?(times 3 as there are 3 combination)

and part (b) 1-(P(3white) +P(2white,1red)=== 1-((2/5)^3+((2/5)^2*(3/5)*3)
Or
P(2red,1white)+P(3red)=((3/5)^2*(2/5)*3)+(3/5)^3

I might've misread the question. If 3 buttons are being picked then a) is definitely correct and at least the second answer to b) is correct (didn't check the complement)

Edit: Yep, misread. Looks great to me.

Mathew587 · « **Reply #1197 on:** February 19, 2017, 05:09:17 pm »

can someone tell me why i'm wrong...
$ a) P(2 reds)= P(RWR or RRW or WRR) = (\frac{3}{5} * \frac{2}{5} * \frac{3}{5})+(\frac{3}{5}*\frac{3}{5}*\frac{2}{5}) + (\frac{2}{5}*\frac{3}{5}*\frac{3}{5})=\frac{54}{125} \\ b) P(no more than one white) = P(2 reds) + P(RRR) = \frac{54}{125} + (\frac{3}{5}*\frac{3}{5}*\frac{3}{5}) = \frac{81}{125} $

jamonwindeyer · « **Reply #1198 on:** February 19, 2017, 05:24:38 pm »

Quote from: Mathew587 on February 19, 2017, 05:09:17 pm

can someone tell me why i'm wrong...
$ a) P(2 reds)= P(RWR or RRW or WRR) = (\frac{3}{5} * \frac{2}{5} * \frac{3}{5})+(\frac{3}{5}*\frac{3}{5}*\frac{2}{5}) + (\frac{2}{5}*\frac{3}{5}*\frac{3}{5})=\frac{54}{125} \\ b) P(no more than one white) = P(2 reds) + P(RRR) = \frac{54}{125} + (\frac{3}{5}*\frac{3}{5}*\frac{3}{5}) = \frac{81}{125} $

Hey mate, I'm not 100% sure since I didn't see the debate on this question above, but it looks correct to me and Rui agrees? Why do you say it is incorrect?

Mathew587 · « **Reply #1199 on:** February 19, 2017, 07:00:16 pm »

Oh I just realised that I misread part a) solution that Rui answered. When checking I kept typing (3/5)^3*3 instead of what it should have been. However didn't I do part b) wrong as (3/5)^2 * (2/5)^3 + (3/5)^3 = 747/3125 instead of it being my answer of 81/125?

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