HSC 2015 MX1 Marathon (archive) (5 Viewers)

Kaido · Mar 7, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
My teacher probably puts on a troll face when putting the permutation questions in

bruh, my teacher's like: "i'll make sure you don't get 100%, so I'll find an extra hard perm q this year" LOL.

Drsoccerball · Mar 7, 2015

Re: HSC 2015 3U Marathon

Kaido said:
bruh, my teacher's like: "i'll make sure you don't get 100%, so I'll find an extra hard perm q this year" LOL.

i threatened my teacher sayind id get 100 he trolls everyone and makes the hardest test everyone gets 50 ahhaha

photastic · Mar 7, 2015

Re: HSC 2015 3U Marathon

Kaido said:
bruh, my teacher's like: "i'll make sure you don't get 100%, so I'll find an extra hard perm q this year" LOL.

Dw, my students fail my exams I give them

braintic · Mar 7, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
according to the answers iii and iv are wrong. can you explain your method for i and ii

I answered the wrong question for (IV). It should be 1-(1/2)⁵ = 31/32.
But again ... I don't know why people insist on using perms and combs for a straightforward 2 unit question.
The only way of getting an odd product is if every number is odd. That is 1/2 for each of 5 rolls.

But I'm not sure why (III) is wrong. Your answer is exactly double mine. Can you see where I might have missed a factor of 2?

Edit: Silly error ... the first of the 5C2s should have been 5P2 ... I was talking of choosing positions for 2 and 3, so the order is important.

FrankXie · Mar 9, 2015

Re: HSC 2015 3U Marathon

new question

$Find the volume of the solid generated when the region enclosed by the curves $ y=\frac{\sqrt{x}}{x+1}, y=\frac{1}{2\sqrt{x}} $ and the line $ x=4 $ is rotated about the $ x-$axis.$

Drsoccerball · Mar 12, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
new question

$Find the volume of the solid generated when the region enclosed by the curves $ y=\frac{\sqrt{x}}{x+1}, y=\frac{1}{2\sqrt{x}} $ and the line $ x=4 $ is rotated about the $ x-$axis.$

Intersection at x=1
$\pi \int_1^4(\frac{x}{(x+1)^2})dx - \pi \int_1^4(\frac{1}{4x})dx$

$$ Let $ u =x+1$

$du=dx$

$\pi \int_2^5\frac{u-1}{u^2}du$

$\pi [(ln(u) + \frac{1}{u})]_2^5 -\frac{\pi}{4}[ln(x)]_1^4$

$\therefore $ volume $ = \pi(ln(\frac{5}{8})+\frac{7}{10})$

$$ volume $ = 0.72 u^3 $ (2.dp)$

FrankXie · Mar 12, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Intersection at x=1
$\pi \int_1^4(\frac{x}{(x+1)^2})dx - \pi \int_1^4(\frac{1}{4x})dx$

$$ Let $ u =x+1$

$du=dx$

$\pi \int_2^5\frac{u-1}{u^2}du$

$\pi [(ln(u) + \frac{1}{u})]_2^5 -\frac{\pi}{4}[ln(x)]_1^4$

$\therefore $ volume $ = \pi(ln(\frac{5}{8})+\frac{7}{10})$

$$ volume $ = 0.72 u^3 $ (2.dp)$

the integral was correct anyway, not sure about the answer, the evaluation of integral looks wrong

Ekman · Mar 13, 2015

Re: HSC 2015 3U Marathon

If 4 digits (1,2,3,4) are used and arranged in an order to create a 4-digit number, what is the value of the sum of all the possible 4-digit numbers formed.

What would be a quick way of going about for this question besides calculating the sum of all 24 cases....

VBN2470 · Mar 13, 2015

Re: HSC 2015 3U Marathon

So you will have 6 numbers beginning with 4, 6 with 3, 6 with 2, and 6 with 1, so the sum of all of these is 24000 + 18000 + 12000 + 6000 = 60000 (considering only the thousands place in each number). Then you have the same thing for the hundreds, tens and ones place for each number. Therefore the sum is 6000 for the digits in the hundreds place, 600 for the digits tens place and 60 for the digits in the ones place, so the total sum would be 60000 + 6000 + 600 + 60 = 66660.

Carrotsticks · Mar 13, 2015

Re: HSC 2015 3U Marathon

Ekman said:
If 4 digits (1,2,3,4) are used and arranged in an order to create a 4-digit number, what is the value of the sum of all the possible 4-digit numbers formed.

What would be a quick way of going about for this question besides calculating the sum of all 24 cases....

Notice that each number appears in each digit 6 times.

ie: 1234, 1243, 1324, 1342, 1423, 1432

Observe how the '1' appears in the first digit place 6 times, so we have 6x(1000).

Apply a similar technique to the other digit places and the other numbers...

Ekman · Mar 13, 2015

Re: HSC 2015 3U Marathon

Carrotsticks said:
Notice that each number appears in each digit 6 times.

ie: 1234, 1243, 1324, 1342, 1423, 1432

Observe how the '1' appears in the first digit place 6 times, so we have 6x(1000).

Apply a similar technique to the other digit places and the other numbers...

This was a MC question in my ext 2 exam and I used the same technique to estimate 60000 as a result. Turns out the actual answer is 66660, I just wanted to know if there was a quicker way to getting the exact answer.

VBN2470 · Mar 13, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
So you will have 6 numbers beginning with 4, 6 with 3, 6 with 2, and 6 with 1, so the sum of all of these is 24000 + 18000 + 12000 + 6000 = 60000 (considering only the thousands place in each number). Then you have the same thing for the hundreds, tens and ones place for each number. Therefore the sum is 6000 for the digits in the hundreds place, 600 for the digits tens place and 60 for the digits in the ones place, so the total sum would be 60000 + 6000 + 600 + 60 = 66660.

.

Ekman · Mar 13, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
.

Whoops sorry didn't see your post, yeah I understand it now.

Carrotsticks · Mar 13, 2015

Re: HSC 2015 3U Marathon

Ekman said:
This was a MC question in my ext 2 exam and I used the same technique to estimate 60000 as a result. Turns out the actual answer is 66660, I just wanted to know if there was a quicker way to getting the exact answer.

Well since the number is only a 4 digit number, doing it 'by hand' isn't too long.

You could ever so slightly increase the speed by using the sum of an arithmetic series (as the addition is 1+2+3+4+...) but as I said since the number is small, there's little point.

VBN2470 · Mar 15, 2015

Re: HSC 2015 3U Marathon

NEXT QUESTION:

$Find the number of ways in which the letters of the word PERMUTATION are arranged such that the vowels occupy themselves in the order AEIOU from left to right, but not necessarily together.$

braintic · Mar 15, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
NEXT QUESTION:

$Find the number of ways in which the letters of the word PERMUTATION are arranged such that the vowels occupy themselves in the order AEIOU from left to right, but not necessarily together.$

11! ÷ 2 ÷ 5!

VBN2470 · Mar 16, 2015

Re: HSC 2015 3U Marathon

NEW QUESTION:

$Six diners in a restaurant choose randomly from a menu featuring six main courses. Find the probability that exactly one of the main courses is NOT chosen by any of the diners.$

Drsoccerball · Mar 19, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
NEW QUESTION:

$Six diners in a restaurant choose randomly from a menu featuring six main courses. Find the probability that exactly one of the main courses is NOT chosen by any of the diners.$

$(\frac{1}{6})^6$

Carrotsticks · Mar 19, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
$(\frac{1}{6})^6$

You just found the probability of all six people choosing the same particular main course.

Ekman · Mar 19, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
NEW QUESTION:

$Six diners in a restaurant choose randomly from a menu featuring six main courses. Find the probability that exactly one of the main courses is NOT chosen by any of the diners.$

$\frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} \times \frac{5}{6} = 0.0129$

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