ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: thatisanote on November 07, 2011, 03:43:09 pm

Title: My proposed solutions for exam 2
Post by: thatisanote on November 07, 2011, 03:43:09 pm
Core
25.0
28.2
1.1
28.25 lower boundary, .:. outlier
65.5%
comment
comment
3-median line
2.39+5.89...
27.7 years

number patterns
plot the point
50
150
$3.20
$26.40
20GB
11GB
$33
9th month
$47.50
$11.31
$460
198100
Cn+1=0.99Cn+100, c1=200000
3781

Geometry & Trig
150m
0.2
152 degrees
78 degrees
208 degrees T
79 degrees T
show that...
1.4m^2
1.2m
21m^2
85.2 degrees
42m
438m^3

Graphs and relations
236g
200g
0.1x+0.04y>=16
shade...
plot line...
1000g
125g
2.3km/h
a=3/2 b=11/2
2 hours
1.38 hours
Title: Re: My proposed solutions for exam 2
Post by: Haras_Etak on November 07, 2011, 03:44:22 pm
How do you remember all these?
I can only remember the hard questions!
Title: Re: My proposed solutions for exam 2
Post by: Maths1234 on November 07, 2011, 03:46:43 pm
These are copied from iTute i'm pretty sure, no way you could remember without the exam.
Title: Re: My proposed solutions for exam 2
Post by: thushan on November 07, 2011, 03:52:37 pm
unless he copied his answers onto his formula sheet
Title: Re: My proposed solutions for exam 2
Post by: 123david321 on November 07, 2011, 03:54:18 pm
How did you get 1.38 for the last graphs and relations question ?
Title: Re: My proposed solutions for exam 2
Post by: Maths1234 on November 07, 2011, 03:55:10 pm
Why would you write answers down when you could check :|
Title: Re: My proposed solutions for exam 2
Post by: seretide on November 07, 2011, 03:56:51 pm
Whats the lowest average? How do u get that?
Title: Re: My proposed solutions for exam 2
Post by: M2Char on November 07, 2011, 03:57:23 pm
How did you get 1.38 for the last graphs and relations question ?

Find the first boundary by making the girl's equation equal the equation of the first part of the guy's hike plus 3.
Solve for t.
t = 1.625

Find the second boundary by making the girl's equation equal the equation of the second part of the guy's hike minus 3.
Solve for t.
t = 3

The time between those boundaries is the time that they are within 3km of each other.

3 - 1.625 = 1.375 ~= 1.38 hours
Title: Re: My proposed solutions for exam 2
Post by: Haras_Etak on November 07, 2011, 03:58:07 pm
Yeah, I still don't get that 1.38 graphs and relation questions. It was the only thing I got wrong!!!!!!!!!
Title: Re: My proposed solutions for exam 2
Post by: thatisanote on November 07, 2011, 03:59:46 pm
These are copied from iTute i'm pretty sure, no way you could remember without the exam.
Big accusation mate :P Actually it was written down by my mate and I had exactly the same for all of them.
Also confirmed by going through the exam with my teacher.
Title: Re: My proposed solutions for exam 2
Post by: thatisanote on November 07, 2011, 04:00:42 pm
Whats the lowest average? How do u get that?

Look at the stem and leaf and check the lowest number.
Title: Re: My proposed solutions for exam 2
Post by: seretide on November 07, 2011, 04:14:48 pm
Was that it? Lmao ..... first question, zero marks.
Title: Re: My proposed solutions for exam 2
Post by: Namit on November 07, 2011, 04:56:42 pm
Got a copy of the actual exam? Seems like you got some errors...
Title: Re: My proposed solutions for exam 2
Post by: thatisanote on November 07, 2011, 05:46:40 pm
Got a copy of the actual exam? Seems like you got some errors...


Nope, but do tell which things you think are errors.
These are confirmed by 2 people including my teacher.
Title: Re: My proposed solutions for exam 2
Post by: NO_HEART on November 07, 2011, 05:49:22 pm
How did you get $11.31 for that question in number patterns?
Title: Re: My proposed solutions for exam 2
Post by: woolworth on November 07, 2011, 06:19:23 pm
i think the solns are right, except for that maximum almond one where i got 120 >.> in graphs & relns
Title: Re: My proposed solutions for exam 2
Post by: Namit on November 07, 2011, 06:24:38 pm
Got a copy of the actual exam? Seems like you got some errors...


Nope, but do tell which things you think are errors.
These are confirmed by 2 people including my teacher.

Wasn't the last question in number patterns asking for what value gives 200000 in march?  I think you gave 3781,  i thought it may be 2000?
Title: Re: My proposed solutions for exam 2
Post by: davidle_10 on November 07, 2011, 06:26:34 pm
i think the solns are right, except for that maximum almond one where i got 120 >.> in graphs & relns
I got 125g..
You had to find inequation x+y >(greater or equal) 500
then rearrange to find y so y=500-x
Sub this into the equation 0.2x+0.04y=40
0.2x+0.04(500-x)=40
x=125g
Title: Re: My proposed solutions for exam 2
Post by: thatisanote on November 07, 2011, 09:09:41 pm
Got a copy of the actual exam? Seems like you got some errors...


Nope, but do tell which things you think are errors.
These are confirmed by 2 people including my teacher.

Wasn't the last question in number patterns asking for what value gives 200000 in march?  I think you gave 3781,  i thought it may be 2000?

Sorry this one was confirmed by my teacher, I watched him do it.
Work out the value for t3 (march), then go 200000-t3.
Title: Re: My proposed solutions for exam 2
Post by: Namit on November 07, 2011, 09:27:59 pm
Okayyy, damn
Title: Re: My proposed solutions for exam 2
Post by: some dude on November 07, 2011, 11:13:40 pm
for the geo answer of 1.4m^2 , didnt it ask for the area of the octagon. so
.5*2*2*sin(45)=1.4 *8 gives you 11.something. did u forget to times the 8, or am i missing something here?
Title: Re: My proposed solutions for exam 2
Post by: woolworth on November 07, 2011, 11:34:51 pm
i think the solns are right, except for that maximum almond one where i got 120 >.> in graphs & relns
I got 125g..
You had to find inequation x+y >(greater or equal) 500
then rearrange to find y so y=500-x
Sub this into the equation 0.2x+0.04y=40
0.2x+0.04(500-x)=40
x=125g

yeh i know i got it wrong, i did the same thing, but somehow got 120g
Title: Re: My proposed solutions for exam 2
Post by: thatisanote on November 08, 2011, 11:11:49 am
for the geo answer of 1.4m^2 , didnt it ask for the area of the octagon. so
.5*2*2*sin(45)=1.4 *8 gives you 11.something. did u forget to times the 8, or am i missing something here?

Nah just the triangle (OAB or something)
THen the next question you have to use PIr^2-area of octagon (which is 1.4*8)
Title: Re: My proposed solutions for exam 2
Post by: some dude on November 08, 2011, 11:54:45 am
for the geo answer of 1.4m^2 , didnt it ask for the area of the octagon. so
.5*2*2*sin(45)=1.4 *8 gives you 11.something. did u forget to times the 8, or am i missing something here?

Nah just the triangle (OAB or something)
THen the next question you have to use PIr^2-area of octagon (which is 1.4*8)

ahh so thats 2 silly mistakes already...:/