Further Maths Exam 2: Answers + Discussion!

[Dan1995]

Same deal as Friday, hope everyone went well today, the answers will be up as each module is completed, I will only be doing solutions for Core, Trig, Networks and Matrices and the others will be added in from other people's solutions. If you find any errors then let us know, stating why you think that answer should be changed and it will be discussed.

Core
1a. 19%
1.b 29 440 000 people
1.c No association as all three percentages are similar, with Australias population being approximately 67% aged 15-64 and India and Japan's populations both being 64% aged 15-64, this suggests that that age group does not vary much in relation to the country in which they live.

2a. Population.
2b. (0,5330) and (9,29450) or two other points on the plot, labelled and then connected with a straight line.
2.c On average, it is predicted that for every 1km² increase in Area, population will increase by 2680 people.
2di. -9360
2dii. 44.6%

3a. population=7.7 + 7.7x log₁₀(area)
b. 23000 people (to the nearest thousand)

4a. weak, negative, linear
4bi. -0.8
   ii. 1 suburb
   iii. 2 suburbs

Module 1: Number Patterns

1a. 20 000
1b. 1%
1c. L2015 = 0.99L2014
      L2015 = 0.99L*20000
      L2015 = 19800
1.d L2016 - L2015 = 19800 - 19602 = 198km2

2a. 0.68/0.8 = 0.578/0.68 = 0.85 = r
2b. 0.35km2
2c. S5 = 2.97km2
2d. Sn > 4, n = 8.53, in year 2022

3a. H2015 = 0.85H2014 + 500
      H2015 = 0.85*14000 + 500
3b. E2016 = 5618
      H2016 = 11040
      11040/5618 = 1.97km2 for each elephant in 2016
      Since 1.97 < 2, in 2016 the elephant habitat will be overpopulated.
3c. 5000 = 1.06*5000 - k, k = 300 elephants
3d. 4900 = 1.06*5000 - k, k = 400 elephants
3e. H2016 = 11040
      H2016/P2016 >= 2
      11040/P2016 >= 2
      0 < P2016 <= 5520
      Let P2016 = 5520 for minimum.
      P2016 = 1.06P2015 - k
      P2015 = 1.06*5000 - k = 5300 - k
      5520 = 1.06(5400 - k) - k
      5520 = 5618 - 2.06k
      k = 47.57
      For minimum, k = 48 elephants.


Module 2: Geometry and Trigonometry
1a. 8m²
b. 12.8m

2a. 75 degrees
2b. AX/sin (45)= 3.16/sin(75)
2c. 2.31m
2d. 3.2m²
2e. 17m²

3a. 157cm²
3b. 1440cm³
3c. 761cm²

4. 228 degrees

Module 3: Graphs and relations
Got some tentative suggested solutions for Number Patterns and Graphs and Relations.

1a. 0.04kg
1b. 25kg
1c. 0.06x + 0.04y >= 180
1di. 0.02x + 0.06y = 120
1dii. Space in the top right hand corner of the graph. Graph Here
1ei. 4000kg
1eii. all integer values along the line 0.05x + 0.05y = 120 from x = 1000 to x = 3000. Graph Here
2a. Straight line from (0,0) to (40000, 140000). Graph Here.
2b. n = 20000.
      P = R - C = 3.5n - (1.25n +36000)
      P = 2.25n - 36000
      P = 2.25(20000) - 36000
      P = $9000

3a. 10.8+4(8-2) = $34.80
3b. at n = 10, 10.8+4(10-2) = a + 2(n-10).
      10.8+4(10-1) = 42.8 = a + (10-10)
      Therefore, a = 42.8
3c. C = 3.50n
      C = R
      3.5n = 42.8 + 2(n-10)
      n = 15.2kg
   
Module 4: Business-related Mathematics
1a. 20%
1b. $330
1c. $15
1di. $17000
1dii. 3.5% p.a
2a. 3.75% p.a
2b. $20000 (uh this question was on MC last year...)
2ci. $772.50
2cii. $558
3a. $14450
3b. 6.9%
3ci. [0.006] and [885]
3cii. $75443
4. 78%
(Let PV=143585.44, PMT=-2500, N=1, I=4.5 and PpY=12, the FV=141623.77 which means 1961.56 has been paid off the principal, which is 78% of the 2500 payment)

Module 5: Networks
1a. 2
1b. miniature trains

2a.  0
      0
      1
b. Tasks can only be assigned where a zero exists, Brianna has no zeros in her respective column, therefore she cannot be assigned a task and the optimal task allocation can not be made.

ci. Equipment
cii. 36 hours

3ai. Bower and Eden
3aii. 910km

3b. 270km
3c. Bower and Derrin

4a. 7 Hours
4b. 18 hours
4c. 2 hours
4d 4 hours
4e. $270

Module 6: Matrices
1a. 4x2
b. 1850
c. The total female population of the city.
d.The product is defined because the number of columns in matrix V (2 columns) is equal to the number of rows in matrix P (2 rows).
e. w=1360x0.45 + 1460x0.55= 1415
f. 6021 votes

2ai. 20%
aii. 25%
b. 1164 votes
cu. 4900
    4634
    2466
ii. The number of votes, for each candidate, in March.
d. 5303 votes

3a. 50%
b. 6451 votes.

[Lucho23]

Got some tentative suggested solutions for Number Patterns and Graphs and Relations.

Module 1: Number Patterns

1a. 20 000
1b. 1%
1c. L₂₀₁₅ = 0.99L₂₀₁₄
      L₂₀₁₅ = 0.99L*20000
      L₂₀₁₅ = 19800
1.d L₂₀₁₆ - L₂₀₁₅ = 19800 - 19602 = 198km²

2a. 0.68/0.8 = 0.578/0.68 = 0.85 = r
2b. 0.35km²
2c. S₅ = 2.97km²
2d. S_n > 4, n = 8.53, in year 2022

3a. H₂₀₁₅ = 0.85H₂₀₁₄ + 500
      H₂₀₁₅ = 0.85*14000 + 500
3b. E₂₀₁₆ = 5618
      H₂₀₁₆ = 11040
      11040/5618 = 1.97km² for each elephant in 2016
      Since 1.97 < 2, in 2016 the elephant habitat will be overpopulated.
3c. 5000 = 1.06*5000 - k, k = 300 elephants
3d. 4900 = 1.06*5000 - k, k = 400 elephants
3e. H₂₀₁₆ = 11040
      H₂₀₁₆/P₂₀₁₆ >= 2
      11040/P₂₀₁₆ >= 2
      0 < P₂₀₁₆ <= 5520
      Let P₂₀₁₆ = 5520 for minimum.
      P₂₀₁₆ = 1.06P₂₀₁₅ - k
      P₂₀₁₅ = 1.06*5000 - k = 5300 - k
      5520 = 1.06(5400 - k) - k
      5520 = 5618 - 2.06k
      k = 47.57
      For minimum, k = 48 elephants.

Module 3: Graphs and relations

1a. 0.04kg
1b. 25kg
1c. 0.06x + 0.04y >= 180
1di. 0.02x + 0.06y = 120
1dii. Space in the top right hand corner of the graph. Graph Here
1ei. 4000kg
1eii. all values (integer and non-integer) along the line 0.05x + 0.05y = 120 from x = 1000 to x = 3000.
2a. Straight line from (0,0) to (40000, 140000). Graph Here.
2b. n = 20000.
      P = R - C = 3.5n - (1.25n +36000)
      P = 2.25n - 36000
      P = 2.25(20000) - 36000
      P = $9000

3a. 10.8+4(8-2) = $34.80
3b. at n = 10, 10.8+4(10-2) = a + 2(n-10).
      10.8+4(10-1) = 42.8 = a + (10-10)
      Therefore, a = 42.8
3c. C = 3.50n
      C = R
      3.5n = 42.8 + 2(n-10)
      n = 15.2kg

[silverpixeli]

      k = 47.57
      For minimum, k = 47 elephants.

Sorry for commenting on a solutions thread, but isn't it rounded the other way to k=48?

[Lucho23]

Sorry for commenting on a solutions thread, but isn't it rounded the other way to k=48?

My bad, changing it now

[DanielJ]

Is 4a in core really linear? A reciprocal transformation would fit it slightly better I thought?

Also, 4bii) Was it not 2 standard deviations greater than or equal to the mean? 3.4 + 2*(1.6) = 6.6. There were 5 points on or above y = 6.6

[brenden]

Is 4a in core really linear? A reciprocal transformation would fit it slightly better I thought?

Yeah it looks that way to me.

[Lucho23]

Is 4a in core really linear? A reciprocal transformation would fit it slightly better I thought?

It's a weak relationship, hence why it doesn't look so linear. But it is, in theory, a linear association. Just a weak one.
That's my take on it

[Dan1995]

Is 4a in core really linear? A reciprocal transformation would fit it slightly better I thought?

Also, 4bii) Was it not 2 standard deviations greater than or equal to the mean? 3.4 + 2*(1.6) = 6.6. There were 5 points on or above y = 6.6

It appears to be linear to me, it doesnt quite have enough of a curve,

Also as it was talking about the mean and standard deviation of Area, you need to look at the x values. There are only 2 values above x=6.6

[brenden]

Is anyone doing solutions for Business?

[Dan1995]

Is anyone doing solutions for Business?

I didnt do the business module for my VCE but I am attempting to do the solutions, if someone else would like to do them to compare that would be great.

[jt33]

surely graphs and relations 3c is 15.2 kg?

[DanielJ]

It appears to be linear to me, it doesnt quite have enough of a curve,

Also as it was talking about the mean and standard deviation of Area, you need to look at the x values. There are only 2 values above x=6.6

Hmm, I agree that it is weak, but it does seem to have a bit of curvature, the right end would need to be lower for a linear model.

Hmm, I can't remember exactly which was on what axis. Will check it once I see the exam again.

[Dan1995]

Hmm, I agree that it is weak, but it does seem to have a bit of curvature, the right end would need to be lower for a linear model.

Hmm, I can't remember exactly which was on what axis. Will check it once I see the exam again.

Area was on the x axis, I'm looking at the exam now, only two points were to the right of x=6.6

[Lucho23]

surely graphs and relations 3c is 15.2 kg?

Can you explain how you got this?
I think you're right

[brenden]

I didnt do the business module for my VCE but I am attempting to do the solutions, if someone else would like to do them to compare that would be great.

Currently attempting to do it with a computer calculator. Lol. Gonna go try find my CAS.