# HSC 2016 General Maths Marathon (1 Viewer)

Status
Not open for further replies.

#### davidgoes4wce

##### Well-Known Member
My initial think was 0.25% as well

$\bg_white \frac{0.5}{200} \times 100 \%= \pm 0.25 \%$

#### InteGrand

##### Well-Known Member
My initial think was 0.25% as well

$\bg_white \frac{0.5}{200} \times 100 \%= \pm 0.25 \%$
How reliable are the answers of the question's source?

#### davidgoes4wce

##### Well-Known Member
There are no answers, just ticks and crosses.

No solutions given to student.

#### InteGrand

##### Well-Known Member
There are no answers, just ticks and crosses.

No solutions given to student.
But you said the answer was D) – is this reliable?

#### davidgoes4wce

##### Well-Known Member
No its not reliable. if you read the first post on Page 3 of this forum, they got the radial survey question wrong as well. I'll post up the exam page online now.

#### davidgoes4wce

##### Well-Known Member
The teacher who marked it on the MCQ gave Q14 D (was deemed wrong), Q15 D (was deemed correct)

Im skeptical because when I did Q 14 :

$\bg_white A=\frac{1}{2} ab sin C$

$\bg_white = \frac{1}{2} \times 6 \times 4 \times sin 95^\circ = 11.95m^{2}$

$\bg_white Which is 12 m^{2} to the nearest metre$

Which was marked wrong.

#### InteGrand

##### Well-Known Member
In that case, yeah, answer should surely be +/- 0.25%.

#### davidgoes4wce

##### Well-Known Member
I also won't name the school paper this is from, for privacy reasons.

#### InteGrand

##### Well-Known Member
The teacher who marked it on the MCQ gave Q14 D (was deemed wrong), Q15 D (was deemed correct)

Im skeptical because when I did Q 14 :

$\bg_white A=\frac{1}{2} ab sin C$

$\bg_white = \frac{1}{2} \times 6 \times 4 \times sin 95^\circ = 11.95m^{2}$

$\bg_white Which is 12 m^{2} to the nearest metre$

Which was marked wrong.
Yeah it appears that these markings are quite unreliable.

#### davidgoes4wce

##### Well-Known Member
This school has fckued up their students , they have done Question 16 wrong as well.

#### InteGrand

##### Well-Known Member
This school has fckued up their students , they have done Question 16 wrong as well.
Wouldn't the students have complained if there were so many blatant errors?

Question 16:

#### davidgoes4wce

##### Well-Known Member
Wouldn't the students have complained if there were so many blatant errors?
The student that I am tutoring wasn't good enough , he failed the General Maths exam. I don't think he would have been able to spot the errors out and he left half of the short answer section.

#### davidgoes4wce

##### Well-Known Member

Theory behind that question

#### davidgoes4wce

##### Well-Known Member
$\bg_white A=\frac{h}{3}(d_f+4d_m+d_l)$

$\bg_white h=11, d_f=12,d_m=7,d_l=10$

$\bg_white A=\frac{11}{3}[12+4(7)+10]$

$\bg_white A=\frac{11}{3}(12+28+10)$

My answer would have been B, the answer that was marked correct was D

#### InteGrand

##### Well-Known Member
$\bg_white A=\frac{h}{3}(d_f+4d_m+d_l)$

$\bg_white h=11, d_f=12,d_m=7,d_l=10$

$\bg_white A=\frac{11}{3}[12+4(7)+10]$

$\bg_white A=\frac{11}{3}(12+28+10)$

My answer would have been B, the answer that was marked correct was D
Yeah it should be B) (Simpson's Rule).

#### InteGrand

##### Well-Known Member
Is the reason the teachers marked that wrong that they didn't understand the theory themselves?

If you have access to other years' papers from that school, you could maybe check how correctly those ones were marked, to see if this is a recurring problem.

#### davidgoes4wce

##### Well-Known Member
I won't show the name of the student or the school for privacy reasons but this is how they mark the MCQ's:

#### davidgoes4wce

##### Well-Known Member
Just wondering if there has been a mistake made in the Year 11 Oxford General Maths textbook for Chapter 2 Q 11.

Saw an example of this nature from the text book:

$\bg_white Oxford Year 11 Chapter 2 Review Set$

$\bg_white Q11. The mass of a can of soup was 250 grams to the nearest 10 grams. The percentage error in this measurement is:$

$\bg_white A \ \pm 4 \% \ B \ \pm 2 \% \ C \ \pm 0.4 \% \ D \ \pm 0.2 \%$

My initial thinking with this question was to find the Limit of Reading value which in this case, I thought it was 1 gram. (As I speak I've realised they have used the Limit of Reading value of 10 grams as stated in the question)

$\bg_white Absolute Error =\frac{1}{2} \times 1 \ gram=0.50 \ grams$
$\bg_white \frac{0.50}{250} \times 100 \% = \pm 0.20 \%$

After having a deeper thought about it, I think the answer is (but I really don't like the logic behind it):

$\bg_white Absolute Error= 10 grams \times \frac{1}{2}= 5 grams$

$\bg_white \frac{5}{250} \times 100 \% =\pm 2 \%$

Status
Not open for further replies.