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What's the raw mark needed for state ranking? (2 Viewers)

bossleymaths

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i got 101 but i dont seem to have a solution to the circle geometry, could anyone please show me how to do question 6 b) ii
 

Kikkoman

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Our maths teacher once professed to us that there are (and have been) people who get (got) full marks. I'm like........... WAZZUPWIDDAT!?!?!?!?
Keep in mind this guy's been marking 4U HSC for quite a few years now
 
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khorne

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I believe there has only been one person to get the full 120/120 in the 4U HSC (1993)
 

untouchablecuz

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i got 101 but i dont seem to have a solution to the circle geometry, could anyone please show me how to do question 6 b) ii
PTQ+TQP+QBP+BPT=360 (angle sum of a quad) (*)

but from i) TPB+TQB=α+β+2θ and PBQ=180-PBS=180-θ (angles on a straight line)

sub this into (*)

PTQ=180-(α+β+θ) (**)

DAB=2α+θ (exterior angle of triangle equal to the sum of interior, opposite angles)

DAB=BAC=2α+θ (interior angle of cyclic quad is equal to opposite, exterior angle)

CBR=SBP=θ (vert. oppo. angles)

so, now the angle sum of ΔBQR is,

CBR+BRC+RCB=180

subbing in CBR=θ, BRC=2β and DAB=2α+θ

2(α+β+θ)=180

α+β+θ=90 (***)

sub (***) into (**)

PTQ=180-(α+β+θ)=180-90=90

.'. ST is perpendicular to RT
 

hermand

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At least 118/120. Keep in mind, that this is only an educated estimate. It is highly likely that someone (or many people) may get full marks in this year's paper, in which case a raw mark of 118/120 would be lucky to get a state ranking.
we had this conference with a mx2 marker and he said noone's ever got full marks in an mx2 paper?
 

gurmies

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we had this conference with a mx2 marker and he said noone's ever got full marks in an mx2 paper?
Really? I thought there was some guy who now lectures @ Usyd or UNSW? Could be mistaken though (it appears that I am).
 

hermand

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Really? I thought there was some guy who now lectures @ Usyd or UNSW? Could be mistaken though (it appears that I am).
i dunno. i walked in halfway through the conference so i may not have been listening properly but i thought that's what he said?

eh.

116+ is close enough to full marks for my liking haha.
 

gurmies

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i dunno. i walked in halfway through the conference so i may not have been listening properly but i thought that's what he said?

eh.

116+ is close enough to full marks for my liking haha.
Lol. In my opinion, 110 and 120 imply the same level of skill.
 
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khorne

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Nah, someones got full marks...Anthony Henderson is the name, I believe.
 

bossleymaths

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PTQ+TQP+QBP+BPT=360 (angle sum of a quad) (*)

but from i) TPB+TQB=α+β+2θ and PBQ=180-PBS=180-θ (angles on a straight line)

sub this into (*)

PTQ=180-(α+β+θ) (**)

DAB=2α+θ (exterior angle of triangle equal to the sum of interior, opposite angles)

DAB=BAC=2α+θ (interior angle of cyclic quad is equal to opposite, exterior angle)

CBR=SBP=θ (vert. oppo. angles)

so, now the angle sum of ΔBQR is,

CBR+BRC+RCB=180

subbing in CBR=θ, BRC=2β and DAB=2α+θ

2(α+β+θ)=180

α+β+θ=90 (***)

sub (***) into (**)

PTQ=180-(α+β+θ)=180-90=90

.'. ST is perpendicular to RT
NICE thanks man now i get it
 

shaon0

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Nah, someones got full marks...Anthony Henderson is the name, I believe.
I think he equalled with some other person. I think there's been more than one person to get 120 for MX2 previous to 1993.
 

tommykins

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terrence doesn't lecture at usyd or unsw
 

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