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Section I - Multiple Choice (1 Viewer)

Rafy

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Share your multiple choice responses here
 

RafH

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badabdacbc

guessed 7 and 10
 

bunnypacman

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b
a
a (i had no idea)
b (got through half my working out and ran outta time so had to guess)
b
d
c
c
c (another guess)
? (cant remember what i put for 10 cause it was a guess :/ )
 

panda15

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badabdccba

Completely guessed 10.
 

Siddy123

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i guessed d for 10
what was the answer to that revs per minute q?
 

Locus

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i guess d for q10 as well - looked like a nice number
 

BlugyBlug

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1b 2a 3d 4a 5b 6(?) 7c 8c 9b 10d (guess)

Forgot what i put for 6, pretty sure it's D though but i just did it now.

revs per minute = 100pi
v=rw
r=5, (period = 2pi/w) period was given so w =20pi

First time I've seen a pure circular motion q in a HSC paper lols.
 

pHyRe

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D was defs wrong for q 10 i reckon, i saw it and realised it was 2^12, or 4GB = 4096 MB so it was there to trick us im pretty sure. I think i guessed C.... whatevs. fuck that
 

hyphea

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D was defs wrong for q 10 i reckon, i saw it and realised it was 2^12, or 4GB = 4096 MB so it was there to trick us im pretty sure. I think i guessed C.... whatevs. fuck that
yep same here
 

Trebla

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I think the answer to Q10 is (C).
 

Trebla

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First consider the unrestricted case. We have six people going into four rooms and each person has four choices which gives 46.

Now subtract the cases where the restriction of 4 people per room is broken.

Case 1 - We have 5 people + 1 person. If we fix 5 people to be in the first room, the 1 person can go into the three other rooms. Similarly, if we fix the 5 people across the other rooms etc leading to 4x3 possibilities. However there are six possible people who can be that 1 person so there are 6x4x3 possibilities altogether.

Case 2 - We have 6 people and there are four possibile rooms they can be in.

Altogether we get 46 - 6x4x3 - 4 = 4020
 
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psychotropic

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First consider the unrestricted case. We have six people going into four rooms and each person has four choices which gives 46.

Now subtract the cases where the restriction of 4 people per room is broken.

Case 1 - We have 5 people + 1 person. If we fix 5 people to be in the first room, the 1 person can go into the three other rooms. Similarly, if we fix the 5 people across the other rooms etc leading to 4x3 possibilities.

Case 2 - We have 6 people and there are four possibile rooms they can be in.

Altogether we get 46 - 4x3 - 4 = 4080
Case I is incorrect as you have to consider which person is excluded. this means that you have to multiple case 1 by 6, yielding 4020
 

Trebla

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Case I is incorrect as you have to consider which person is excluded. this means that you have to multiple case 1 by 6, yielding 4020
Ah overlooked that. Corrected.
 
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