Post by: Will T on November 05, 2012, 03:44:34 pm
Core:
Question 1.
a.
i. 20 degrees Celsius.
ii. 23.3%.
b. 97.5%.
Question 2.
Sketch: the best points were probably (0,13) and (20, 26.4).
b. When the minimum temperature is 0, the maximum will be 13 degrees Celsius.
c. Direction: positive. Strength: moderate.
d. For every increase of 1 degrees Celsius in minimum temperature, there is a resultant increase of 0.67 degrees Celsius in maximum temperature.
e. 40%.
f. -8 degrees Celsius.
Question 3.
a. South-east, North-east.
b. 2,2,2,3,4,4,4,4. Note: this question was not ambiguous and to obtain this answer you had to reason through the five-figure summary provided (minimum: 2, quartile 1: 2, median: 3.5, quartile 3: 4, maximum: 4).
Question 4.
a. y= 3.4 + 6.6x.
b. 13 kilometres per hour.
Number patterns:
Question 1.
a. 156 - 162 = 162 - 168 = -6.
b. The sixth term of this arithmetic sequence is: 138.
c. The difference between the eighth term and the tenth term is 12.
d. The sum of the first 18 terms is 2,106.
ii. The amount of time taken for no more blocks of land to be sold was 29 months. Hence, the answer is the difference between the sum of the first 29 terms and the first 18 terms, which is 330.
Question 2.
a. The third term is 36.
b. The nth term is greater than 100 when n is approximately 5.52. Therefore, it happens in the 6th year.
c. The sum of the first five terms of this geometric sequence is 211.
d. The sum of this geometric series is greater than 1000 when n = 8.57. Therefore, in the 9th year.
e. a = 1.5, b = 0, c = 16.
Question 3. (For this, it is advised you solve for the nth term of the difference equation, see page. 329 of the Essentials textbook for details.
a. The third term of our difference equation is 1026.08, so 1026 to the nearest whole number.
b. The nth term is greater than 4000 when n = 10.35. Hence, in the 11th year.
c. I suggest to graph the nth term and you will see from its equation that 12,500 is an asymptote, and therefore the correct answer. Although other more rigorous methods are acceptable.
Graphs & relations:
Question 1.
a.
i. The gradient = 50.
ii. C = 20,000 + 50n.
b.
i. Sketch the line R = 150n. Label the origin and the intercept between the Cost-graph and the Revenue-graph (200,30,000) as well as labeling the equation on the graph. (I'm not sure how much of that is a requirement for the mark.
ii. 54,000/150 = 360 phones.
c. When R = C, the number of phones is 200.
Question 2.
a. When R = C, the number of laptops is 446.43. But you would need to sell 447 to be profiting, and at 446 you are still in a negative profit (only by a small amount).
b. When R = C, the number of laptops is now 400. Therefore, the new selling price is $682.50.
Question 3.
a. The time available to repair all the mobile phones and laptops in one day must be less than or equal to 1,750 minutes. (Just because there's only 1,440 minutes in a day does not make this question incorrect.) I would suppose that the company that repairs these devices only has a total of 1,750 minutes worth of time shared between all their workers.
b. At most 8 laptops can be repaired for every 10 phones.
c. Sketch the line: 35x + 50y = 1750, make sure to label the x-intercept; y-intercept; and the equation of the line on the graph.
d. The highest y-value that is possible based on the constraints is 18.67 (56/3 in exact value), this means 18 laptops is the maximum possible.
e. Sketch the line y = 9 on the graph, find the two points of intersection, the maximum will be the highest value, which is 37.14, and the minimum, which is 11.25. Now, this means 37 phones is the maximum that can occur.
f.
i. $P(x,y) = 60x + 100y. Corner-point principle tells us that the maximum value of this will occur either on (0,0); (70/3, 58/3); or (50,0). Now, this is where you must be careful. The laptops cannot exceed 18.67, and so cannot exceed 18. This means that if you manufacture 19 laptops, your co-ordinates will be outside of the feasible region. Therefore, the point that will be inside the feasible region will be (24,18), because we're dealing with a real-life, integer problem. Substituting all these values into the objective function yields: $0; $3,240; $3,000. Therefore, the answer to this question is 24 phones and 18 laptops.
ii. This question is technically answered by solving the question above (good one VCAA). Irrespective of that, the answer is P(24,18) = $3,240.
Matrices:
Question 1.
a. Anvil and Dantel.
b. Anvil, Berga, Dantel, then to Cantor.
c. [1 2 1 1].
d. The matrix G tells us the total amount of locations you can travel to from locations A, B, C and D respectively. (i.e., they've summed the columns).
Question 2.
a.
i. The matrix C is as follows (it is a 2 by 3 matrix).
[1 3 2]
[3 9 6]
ii. 133926.
b. A = B^{-1} x C. For this question, it would've been advisable to show full working of how you multiply matrix C by the inverse of B, and in the order shown above, because matrices are not commutative. For example, saying A = C x B^{-1} would've been incorrect. I think showing how to calculate the inverse of B, stating the above equation and then the result should be sufficient for this question.
Question 3.
a. 0.7 x 100 + 0.8 x 200 + 0.9 x 50 = 275. So 275 won't rotate to a different position.
b. The meaning of a_{4},{4} is that, 100% of the people who leave in this year are divorced from the company permanently. I.e., it is a one-way ticket out of the company if you leave.
c.
i. (This is a 4 by 1 matrix)
[70]
[170]
[65]
[45]
ii. Entry_{2},{1} in T^{2} x Initial State is 143. Therefore, 143 operators are expected to be there.
iii. This question is asking when will Entry_{2},{1} be less than 30. A pragmatic approach would be to continually multiply matrix S by T until the desired result is found. Doing this yields T^{10} x Initial State to be the first matrix where the number of operators is less than 30. Therefore, this is 10 years after 2011. And so the answer to this question is 2021.
iv. Evaluating T^{n} x Initial State for large values of n should reveal that, eventually, there will be nobody left at the company.
d. Evaluating S_{2013} using the difference equation yields a value of 182 for Entry_{2},{1}, and so there shall be 182 operators who are expected to be working there.
Post by: Jezza on November 05, 2012, 03:53:45 pm
Post by: Stick on November 05, 2012, 03:54:41 pm
b) When the minimum temperature is zero degrees Celcius, the maximum temperature is expected to be 13 degrees Celcius.
d) For every increase in minimum temperature by one degree Celcius, the maximum temperature is expected to increase by 0.67 degrees Celcius.
Post by: Will T on November 05, 2012, 04:02:26 pm
OK, so you might not lose marks for this, but you probably should've been more careful with your wording in core question 2. For example, my answers for these were:On my actual exam I think I said 'on average', or 'is predicted to be' or something along those lines.
b) When the minimum temperature is zero degrees Celcius, the maximum temperature is expected to be 13 degrees Celcius.
d) For every increase in minimum temperature by one degree Celcius, the maximum temperature is expected to increase by 0.67 degrees Celcius.
Post by: Nima2703 on November 05, 2012, 04:16:04 pm
Post by: Yendall on November 05, 2012, 04:24:04 pm
Post by: Will T on November 05, 2012, 04:44:07 pm
i dont understand the number patterns for houses not sold. It said how many were not sold? so hows that 29th - 18th? was that worth 2 marks?It was worth 2 marks. Think of the sequence like this, by the time it gets to the 19th month, the amount of houses sold in that month would've been 60. The next month there will be 54, 48, 42 and so on. They're saying that after the 18th month, if they were to stop selling houses, how many will there be that should have been sold had it continued but will now not be sold. Essentially you have to add up all the numbers between 60, 54...... 12, 6, 0. And this gets you 330.
Post by: kimk2kr on November 05, 2012, 04:59:00 pm
So angry i got that wrong...
Post by: fisa2001 on November 05, 2012, 05:00:18 pm
Post by: ~T on November 05, 2012, 05:03:23 pm
"The land sales continued in this pattern until all the blocks of land were sold."
However, this doesn't necessarily mean that they continue up to the point where one term equals zero. The moment where "all the blocks of land were sold" when it supposedly stops isn't necessarily at the last term, it could stop earlier. Had to presume that is what the question meant, which I'm sure everyone did anyway, but still a little dodgy.
Post by: caluuu on November 05, 2012, 05:03:55 pm
For question 1 b in Core, do you reckon it will be wrong if you have used Binomial CDF in your CAS to find it? Cause I got 97.7%. only .2% away from the answer....I did that too!
And another much easier way of doing part 3c of number patterns was to use
solve(a=0.96(a)+500,a)
a=12500
Post by: StumbleBum on November 05, 2012, 05:05:14 pm
For question 1 b in Core, do you reckon it will be wrong if you have used Binomial CDF in your CAS to find it? Cause I got 97.7%. only .2% away from the answer....
I doubt it, because they were more looking for students to use the 68-95-99.7 rule. So i doubt a more accurate methods approach will yield you the mark when its more of an approximation.
Post by: Jajsjsjsjsjs on November 05, 2012, 05:10:32 pm
Hey will. Just got a question for ya, for graphs and relation question 3 part (f) if i got part (i) wrong (for some reason got 23 phomes and 18 laptops) would i get consequentials for part (ii)??
So angry i got that wrong...
he is incorrect, 24 is outside the feasible reagion thus the point is (23,18)
Post by: Will T on November 05, 2012, 05:12:30 pm
Hey will. Just got a question for ya, for graphs and relation question 3 part (f) if i got part (i) wrong (for some reason got 23 phomes and 18 laptops) would i get consequentials for part (ii)??Funny you should mention that. I have a similar predicament as I wrote down something similar. I can't be sure of what they'll do with consequential marks, but I hope it will be sufficient because otherwise I'm thoroughly screwed. I think for the last question, if you showed working out, at least 1 mark will be awarded. I guess the question is, how sufficient was your working out?
So angry i got that wrong...
Post by: oppalovesme on November 05, 2012, 05:13:49 pm
No, (24,18) is within the feasible region. It satisfies both 35x + 50y <(=) 1750 and y<(=)(4/5)xHey will. Just got a question for ya, for graphs and relation question 3 part (f) if i got part (i) wrong (for some reason got 23 phomes and 18 laptops) would i get consequentials for part (ii)??
So angry i got that wrong...
he is incorrect, 24 is outside the feasible reagion thus the point is (23,18)
Post by: xcookieFD on November 05, 2012, 05:15:11 pm
Post by: Will T on November 05, 2012, 05:15:37 pm
I don't agree that (24,18) is outside the feasible region. Substituting those points into the inequality that they appear to be breaching yields: 35 x 24 + 50 x 19 < or = 1750. Which gives 1740 < = 1750 which is true. So those points should in fact lie in the feasible region.Hey will. Just got a question for ya, for graphs and relation question 3 part (f) if i got part (i) wrong (for some reason got 23 phomes and 18 laptops) would i get consequentials for part (ii)??
So angry i got that wrong...
he is incorrect, 24 is outside the feasible reagion thus the point is (23,18)
Post by: Will T on November 05, 2012, 05:16:43 pm
if i just wrote "the 10th year" for the matrices question about in which year will the number of operators be less than 30...they'll probably take a mark off right? :(Unfortunately for your sake, I suspect that they will. I saw something very similar in a Number patterns question on last year's exam where they wanted you to add a result to the starting year. So sorry but, yes I think you will be penalised for not writing the year.
Post by: Jordzs on November 05, 2012, 05:20:46 pm
if i just wrote "the 10th year" for the matrices question about in which year will the number of operators be less than 30...they'll probably take a mark off right? :(Unfortunately for your sake, I suspect that they will. I saw something very similar in a Number patterns question on last year's exam where they wanted you to add a result to the starting year. So sorry but, yes I think you will be penalised for not writing the year.
Omg, I think I did the same thing!!
Post by: kimk2kr on November 05, 2012, 05:21:12 pm
Hey will. Just got a question for ya, for graphs and relation question 3 part (f) if i got part (i) wrong (for some reason got 23 phomes and 18 laptops) would i get consequentials for part (ii)??Funny you should mention that. I have a similar predicament as I wrote down 23 laptops and 19 phones. I can't be sure of what they'll do with consequential marks, but I hope it will be sufficient because otherwise I'm thoroughly screwed. I think for the last question, if you showed working out, at least 1 mark will be awarded. I guess the question is, how sufficient was your working out?
So angry i got that wrong...
i think i did show sufficient working out. so might pull off a mark...
one thing i dont understand though, is that how is it 24? when i find the intersection between 35x + 50y <(=) 1750 and y<(=)(4/5)x i get x=23.3333333 and y=18.666667
so wouldnt it be 23,18 not 24 as its outside the given region?
Post by: Will T on November 05, 2012, 05:23:01 pm
Hey will. Just got a question for ya, for graphs and relation question 3 part (f) if i got part (i) wrong (for some reason got 23 phomes and 18 laptops) would i get consequentials for part (ii)??Funny you should mention that. I have a similar predicament as I wrote down 23 laptops and 19 phones. I can't be sure of what they'll do with consequential marks, but I hope it will be sufficient because otherwise I'm thoroughly screwed. I think for the last question, if you showed working out, at least 1 mark will be awarded. I guess the question is, how sufficient was your working out?
So angry i got that wrong...
i think i did show sufficient working out. so might pull off a mark...
one thing i dont understand though, is that how is it 24? when i find the intersection between 35x + 50y <(=) 1750 and y<(=)(4/5)x i get x=23.3333333 and y=18.666667
so wouldnt it be 23,18 not 24 as its outside the given region?
(24,18) isn't outside of the feasible region, nor is (23,18). But in order to maximize it's better to produce that one extra phone. And so the answer will be 24 phones and 18 laptops. I suggest drawing this problem graphically and plotting the points (24,18) and (23,18) and you will have a better picture of what is going on in the question.
Post by: kimk2kr on November 05, 2012, 05:26:27 pm
Hey will. Just got a question for ya, for graphs and relation question 3 part (f) if i got part (i) wrong (for some reason got 23 phomes and 18 laptops) would i get consequentials for part (ii)??Funny you should mention that. I have a similar predicament as I wrote down 23 laptops and 19 phones. I can't be sure of what they'll do with consequential marks, but I hope it will be sufficient because otherwise I'm thoroughly screwed. I think for the last question, if you showed working out, at least 1 mark will be awarded. I guess the question is, how sufficient was your working out?
So angry i got that wrong...
i think i did show sufficient working out. so might pull off a mark...
one thing i dont understand though, is that how is it 24? when i find the intersection between 35x + 50y <(=) 1750 and y<(=)(4/5)x i get x=23.3333333 and y=18.666667
so wouldnt it be 23,18 not 24 as its outside the given region?
(24,18) isn't outside of the feasible region, nor is (23,18). But in order to maximize it's better to produce that one extra phone. And so the answer will be 24 phones and 18 laptops. I suggest drawing this problem graphically and plotting the points (24,18) and (23,18) and you will have a better picture of what is going on in the question.
na i cant be bothered drawing the graph, haha
and because its a 3mark question (overall, both part i and ii), thinking that i gave enough working out, would i get 2/3? or would it still be 1/3...
Post by: StumbleBum on November 05, 2012, 05:37:22 pm
Post by: Nima2703 on November 05, 2012, 05:52:11 pm
would i get 1 for working out?i dont understand the number patterns for houses not sold. It said how many were not sold? so hows that 29th - 18th? was that worth 2 marks?It was worth 2 marks. Think of the sequence like this, by the time it gets to the 19th month, the amount of houses sold in that month would've been 60. The next month there will be 54, 48, 42 and so on. They're saying that after the 18th month, if they were to stop selling houses, how many will there be that should have been sold had it continued but will now not be sold. Essentially you have to add up all the numbers between 60, 54...... 12, 6, 0. And this gets you 330.
Post by: Will T on November 05, 2012, 05:59:25 pm
You really need to learn how to useMaybe some day. Originally when I was writing this I tried to, but it became too difficult.;)
Post by: Will T on November 05, 2012, 06:03:03 pm
Quite possibly, I guess it would depend on where you went wrong?would i get 1 for working out?i dont understand the number patterns for houses not sold. It said how many were not sold? so hows that 29th - 18th? was that worth 2 marks?It was worth 2 marks. Think of the sequence like this, by the time it gets to the 19th month, the amount of houses sold in that month would've been 60. The next month there will be 54, 48, 42 and so on. They're saying that after the 18th month, if they were to stop selling houses, how many will there be that should have been sold had it continued but will now not be sold. Essentially you have to add up all the numbers between 60, 54...... 12, 6, 0. And this gets you 330.
Post by: melvinpig on November 05, 2012, 06:40:42 pm
Post by: kimk2kr on November 05, 2012, 06:42:40 pm
Post by: Will T on November 05, 2012, 07:01:02 pm
will you have not yet answered my question! :P
So you said the company should produce 23 phones and 18 laptops for part i, correct?
Well, that question is worth 2 marks, so I'm wondering if it will be 1 mark for the correct amount of phones and I mark for the correct amount of laptops. At any rate, 18 laptops is correct, but 23 phones is incorrect. So, if my prognosis of how they are going to score this question is correct you should get 1 mark for part i.
Now, for part ii. It's only worth 1 mark, so I'm wondering if they're only going to give out the mark if you have the correct answer. But, I think if you have correctly input your values for part i. into part ii. then you're showing that you know how to perform the computation that the question is asking you to do, and that your answer is only incorrect consequently because of a previous mistake.
If they're kind, then I think a lot of people will get the 1 mark for the last question, as showing how to use an objective function isn't too hard, and it would be a challenge to make a mistake whilst computing the answer (although I'm sure I'd find a way).
But that would make me think that they're going to be strict, and only award the mark for the correct answer.
At any rate, I think you've scored at least 1 mark, and quite possibly 2 marks for that section, provided they're going to be a bit generous with consequential marking.
Post by: kimk2kr on November 05, 2012, 07:18:22 pm
will you have not yet answered my question! :P
So you said the company should produce 23 phones and 18 laptops for part i, correct?
Well, that question is worth 2 marks, so I'm wondering if it will be 1 mark for the correct amount of phones and I mark for the correct amount of laptops. At any rate, 18 laptops is correct, but 23 phones is incorrect. So, if my prognosis of how they are going to score this question is correct you should get 1 mark for part i.
Now, for part ii. It's only worth 1 mark, so I'm wondering if they're only going to give out the mark if you have the correct answer. But, I think if you have correctly input your values for part i. into part ii. then you're showing that you know how to perform the computation that the question is asking you to do, and that your answer is only incorrect consequently because of a previous mistake.
If they're kind, then I think a lot of people will get the 1 mark for the last question, as showing how to use an objective function isn't too hard, and it would be a challenge to make a mistake whilst computing the answer (although I'm sure I'd find a way).
But that would make me think that they're going to be strict, and only award the mark for the correct answer.
At any rate, I think you've scored at least 1 mark, and quite possibly 2 marks for that section, provide they're going to be a bit generous with consequential marking.
thanks will! youre a champ! <3
hopefully theyre generous enough to give out a mark haha
like last year, for bearing questions they gave a mark for writing 79 or something rather than having to write 079T which I personally think would be the only correct answer haha
Post by: Nima2703 on November 06, 2012, 10:30:06 am
what do you mean add a result? youre saying if i just put the 11th year in number patterns answer its wrong??if i just wrote "the 10th year" for the matrices question about in which year will the number of operators be less than 30...they'll probably take a mark off right? :(Unfortunately for your sake, I suspect that they will. I saw something very similar in a Number patterns question on last year's exam where they wanted you to add a result to the starting year. So sorry but, yes I think you will be penalised for not writing the year.
Post by: shmootz on November 06, 2012, 01:54:30 pm
Mine started with a 0 and ended with a 7 but the Q1, Median and Q3 were all correct.
Thanks!
Post by: ~T on November 06, 2012, 02:08:00 pm
For Core Q3b: (You got: 2,2,2,3,4,4,4,4) would they accept anything else that worked?You'll see on the box plot that there are no "whiskers." Whiskers indicate the maximum and minimum values, so when there aren't any (or more accurately, that they don't protrude from the box at all), the maximum is the same as Q3, and the minimum is the same as Q1. Hope that makes sense?
Mine started with a 0 and ended with a 7 but the Q1, Median and Q3 were all correct.
Thanks!
Post by: Jordzs on November 06, 2012, 02:33:02 pm
For Core Q3b: (You got: 2,2,2,3,4,4,4,4) would they accept anything else that worked?
Mine started with a 0 and ended with a 7 but the Q1, Median and Q3 were all correct.
Thanks!
There was nothing else that worked as Q1 is the min and Q3 was the max as there was no whiskers.
I believe 2,2,2,3,4,4,4,4 would be the only answer they will accept
Post by: shmootz on November 06, 2012, 05:01:02 pm