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January 24, 2019, 09:32:13 pm

Author Topic: VCE Methods Question Thread!  (Read 2114852 times)  Share 

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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17580 on: January 09, 2019, 10:31:35 pm »
0
-
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Ok. Thank you both! I know I am struggling right now, but I'm pretty sure people in my (2019) class already pre-learnt Methods (1/2 and 3/4).........
Will I be at any disadvantages? Or should I just take every single topic seriously? :/
Thanks for the answers to my questions, I will post again if I still have another problem!
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FelixHarvey

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Re: VCE Methods Question Thread!
« Reply #17581 on: January 09, 2019, 10:47:37 pm »
0
Ok. Thank you both! I know I am struggling right now, but I'm pretty sure people in my (2019) class already pre-learnt Methods (1/2 and 3/4).........
Will I be at any disadvantages? Or should I just take every single topic seriously? :/
Thanks for the answers to my questions, I will post again if I still have another problem!

Depends on a heap of factors to determine whether you are at a disadvantage. It is hard to tell how well they learnt the course, how well you will learn the course, how much work they will do, what your base aptitude is etc. If you just focus on the topic in class and consistently making sure you understand everything and why you do processes or use certain methods/strategies, you will put yourself in a great position in year 12. Just be consistent for Y11, my one tip would be to try to understand the fundamentals as well as you can.

For example it will prove much more beneficial to understand why when the discriminant is less than 0 you will have 0 real solutions to a quadratic equation. Rather than just memorise that as a fact. Try to understand where the discriminant comes from and in what cases it can be useful. I'm sure many people on this years Methods Exam 1 would have thought to use it in one of the questions because its the first thing that can come to mind when you want a certain number of solutions. But having a deeper understand that is comes from the quadratic formula and that it only applies to quadratics, would have quickly dispelled any want to use the discriminant as a viable method.
2017: Business Management 44
2018: Methods 45 | Specialist 41 | English 41 | Physics 44 | Software Development 48

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Tutoring! PM me for details.

Sine

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Re: VCE Methods Question Thread!
« Reply #17582 on: January 10, 2019, 01:01:47 am »
0
Ok. Thank you both! I know I am struggling right now, but I'm pretty sure people in my (2019) class already pre-learnt Methods (1/2 and 3/4).........
Will I be at any disadvantages? Or should I just take every single topic seriously? :/
Thanks for the answers to my questions, I will post again if I still have another problem!

Disadvantage? Well technically yes since they are at an advantage of already learning the content

Take each topic seriously? yes all the topics in the study design are examinable

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17583 on: January 10, 2019, 10:11:10 am »
0
Depends on a heap of factors to determine whether you are at a disadvantage. It is hard to tell how well they learnt the course, how well you will learn the course, how much work they will do, what your base aptitude is etc. If you just focus on the topic in class and consistently making sure you understand everything and why you do processes or use certain methods/strategies, you will put yourself in a great position in year 12. Just be consistent for Y11, my one tip would be to try to understand the fundamentals as well as you can.

For example it will prove much more beneficial to understand why when the discriminant is less than 0 you will have 0 real solutions to a quadratic equation. Rather than just memorise that as a fact. Try to understand where the discriminant comes from and in what cases it can be useful. I'm sure many people on this years Methods Exam 1 would have thought to use it in one of the questions because its the first thing that can come to mind when you want a certain number of solutions. But having a deeper understand that is comes from the quadratic formula and that it only applies to quadratics, would have quickly dispelled any want to use the discriminant as a viable method.
Disadvantage? Well technically yes since they are at an advantage of already learning the content

Take each topic seriously? yes all the topics in the study design are examinable
Ok thanks both!! Will do it cause I'm only good at memorising formula's but not the entire meaning behind it!
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lyristis

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Re: VCE Methods Question Thread!
« Reply #17584 on: January 10, 2019, 04:34:27 pm »
0
An aircraft, used for fire spotting, flies from its base to locate a fire at an unknown
distance, x km away. It travels straight to the fire and back, averaging 240 km/h for the
outward trip and 320 km/h for the return trip. If the plane was away for 35 minutes, find
the distance, x km

i tried using the d = s x t formula but got it wrong
here's my working out:
d = s x t
   = ((320+240)/2) x 35 = 9800 km
ANS: 80 km.

i did the same thing for this next one too, and it didn't work.
 
A group of hikers is to travel x km by bus at an average speed of 48 km/h to an
unknown destination. They then plan to walk back along the same route at an average
speed of 4.8 km/h and to arrive back 24 hours after setting out in the bus. If they allow
2 hours for lunch and rest, how far must the bus take them?

d = s x t
   = ((48+4.8)/2) x 22 = 580.8
ANS: 96 km

FelixHarvey

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Re: VCE Methods Question Thread!
« Reply #17585 on: January 10, 2019, 04:50:46 pm »
0
An aircraft, used for fire spotting, flies from its base to locate a fire at an unknown
distance, x km away. It travels straight to the fire and back, averaging 240 km/h for the
outward trip and 320 km/h for the return trip. If the plane was away for 35 minutes, find
the distance, x km

i tried using the d = s x t formula but got it wrong
here's my working out:
d = s x t
   = ((320+240)/2) x 35 = 9800 km
ANS: 80 km.

i did the same thing for this next one too, and it didn't work.
 
A group of hikers is to travel x km by bus at an average speed of 48 km/h to an
unknown destination. They then plan to walk back along the same route at an average
speed of 4.8 km/h and to arrive back 24 hours after setting out in the bus. If they allow
2 hours for lunch and rest, how far must the bus take them?

d = s x t
   = ((48+4.8)/2) x 22 = 580.8
ANS: 96 km

Good atttempt. However you have a misconception. Using the formula:
\[Distance=Speed \: \times\: Time\]
Is correct, however the plane does not travel at the respective speeds for the same amount of time. Thus, the average speed is not the same as the median speed. Try to create another equation with this fact. It may help to split the journey up into two parts. Also, remember unit conversions, the current working that you have suggests that the plane flies for 35 hours...
« Last Edit: January 10, 2019, 04:54:19 pm by FelixHarvey »
2017: Business Management 44
2018: Methods 45 | Specialist 41 | English 41 | Physics 44 | Software Development 48

ATAR: 99.40

Tutoring! PM me for details.

Sine

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Re: VCE Methods Question Thread!
« Reply #17586 on: January 10, 2019, 04:58:45 pm »
+1
An aircraft, used for fire spotting, flies from its base to locate a fire at an unknown
distance, x km away. It travels straight to the fire and back, averaging 240 km/h for the
outward trip and 320 km/h for the return trip. If the plane was away for 35 minutes, find
the distance, x km

i tried using the d = s x t formula but got it wrong
here's my working out:
d = s x t
   = ((320+240)/2) x 35 = 9800 km
ANS: 80 km.

i did the same thing for this next one too, and it didn't work.
 
A group of hikers is to travel x km by bus at an average speed of 48 km/h to an
unknown destination. They then plan to walk back along the same route at an average
speed of 4.8 km/h and to arrive back 24 hours after setting out in the bus. If they allow
2 hours for lunch and rest, how far must the bus take them?

d = s x t
   = ((48+4.8)/2) x 22 = 580.8
ANS: 96 km

distance = x (km)
speed = 240 (km/h) out and 320 (km/h) in
t = 35

It's important to realise that since the speeds are not the same thus the time spend going in and out will also not be the same.
Also the total distance travelled by the plane is 2x (km)

First i'll define a couple of things, distance (d) = x
Time out = t
Time in = 35 -t

d = st

x = 320 (t)       (1)

x = 240 (35 -t)   (2)

Equate (1) and (2)

320t = 240(35 - t)
320t = 8400 - 240t
560t = 8400
t = 15 = 0.25 hours

Sub this back in

d = 320 x (0.25) = 80km



lyristis

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Re: VCE Methods Question Thread!
« Reply #17587 on: January 10, 2019, 10:57:00 pm »
0
thanks guys, i understood your explanations and got the other one correct!

here are 2 more questions i need help with. sorry for being annoying and asking questions that are probably really simple  :(

Solve each of the following pairs of simultaneous equations for x and y:
(a + b)x + cy = bc
(b + c)y + ax = −ab

here's what i tried:
(a + b)x + cy = bc
(b + c)y + ax = −ab

ax + bx + cy = bc
by + cy + ax = -ab

bx - by = bc + ab
b(x-y) = b(c+a)

x = (b(c+a)/b) + y
x = c + a + y
i got stuck after this and stopped bc the answer is x = c but i don't know how to get rid of the a and y
 
q2: For the simultaneous equations x/a + y/b = 1 and x/b + y/a = 1, show that x = y = ab/a+b

here's what i tried for the first equation:
x/a + y/b = 1
abx + aby = ab
ab(x+y) = ab
x + y = 1
there's no answer given but i stopped here bc it already doesn't look right




MB_

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Re: VCE Methods Question Thread!
« Reply #17588 on: January 11, 2019, 10:40:35 am »
+1
thanks guys, i understood your explanations and got the other one correct!

here are 2 more questions i need help with. sorry for being annoying and asking questions that are probably really simple  :(

Solve each of the following pairs of simultaneous equations for x and y:
(a + b)x + cy = bc
(b + c)y + ax = −ab

here's what i tried:
(a + b)x + cy = bc
(b + c)y + ax = −ab

ax + bx + cy = bc
by + cy + ax = -ab

bx - by = bc + ab
b(x-y) = b(c+a)

x = (b(c+a)/b) + y
x = c + a + y
i got stuck after this and stopped bc the answer is x = c but i don't know how to get rid of the a and y
Try isolating x in the first equation and subbing it into the second equation
2015-16: VCE
2017-: BSci UoM - Maths & Psych

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17589 on: January 11, 2019, 01:19:45 pm »
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Sorry for reposting this question but I still cannot understand because it is a little weird.
I successfully completed the other ones but:

Linda thinks of a two-digit number. The sum of the digits is 8. If she reverse the digits, the new number is 36 greater than her original number. What was Linda's original number?

xy = 2 digit number
x+y = 8
yx - xy = 36

y = 8 -x
x(8-x) - x(8-x) = 36
8x - x^2 - 8x + x^2 = 36
hmmm?
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Sine

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Re: VCE Methods Question Thread!
« Reply #17590 on: January 11, 2019, 01:29:38 pm »
+1
Sorry for reposting this question but I still cannot understand because it is a little weird.
I successfully completed the other ones but:

Linda thinks of a two-digit number. The sum of the digits is 8. If she reverse the digits, the new number is 36 greater than her original number. What was Linda's original number?
Good effort but there may be a couple problems with the assumptions you have made.

Two digit number: let the "tens" place be x and the "units" place be y
thus two digit number is xy
but is expressed as 10x+y
If we reverse the digits we get 10y + x

So,
x + y = 8 (i)
and
(10y + x) - (10x + y) = 36 (ii)

From (i) =>   y = 8 -x
Sub into (ii)

(10(8-x) + x) - (10x + 8 -x) = 36
(80 - 10x + x) - (9x +8) = 36
72 - 18x = 36
36 = 18x
x = 2

sub into (i)
y = 8 - x = 8 -2 = 6

So the two digit number is 26

S_R_K

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Re: VCE Methods Question Thread!
« Reply #17591 on: January 11, 2019, 01:52:10 pm »
0
thanks guys, i understood your explanations and got the other one correct!

here are 2 more questions i need help with. sorry for being annoying and asking questions that are probably really simple  :(
 
q2: For the simultaneous equations x/a + y/b = 1 and x/b + y/a = 1, show that x = y = ab/a+b

here's what i tried for the first equation:
x/a + y/b = 1
abx + aby = ab
ab(x+y) = ab
x + y = 1
there's no answer given but i stopped here bc it already doesn't look right

If x/a + y/b = 1 and x/b + y/a = 1, then x/a + y/b = x/b + y/a. Multiplying through by ba gives bx + ay = ax + by, hence x = y. Then just substitute into one of the original equations to solve for x, y in terms of a, b.

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17592 on: January 11, 2019, 03:07:21 pm »
0
Good effort but there may be a couple problems with the assumptions you have made.

Two digit number: let the "tens" place be x and the "units" place be y
thus two digit number is xy
but is expressed as 10x+y
If we reverse the digits we get 10y + x

So,
x + y = 8 (i)
and
(10y + x) - (10x + y) = 36 (ii)

From (i) =>   y = 8 -x
Sub into (ii)

(10(8-x) + x) - (10x + 8 -x) = 36
(80 - 10x + x) - (9x +8) = 36
72 - 18x = 36
36 = 18x
x = 2

sub into (i)
y = 8 - x = 8 -2 = 6

So the two digit number is 26
  o yeahhh >.<
missed that.
Thanks Sine!
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mihir999

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Re: VCE Methods Question Thread!
« Reply #17593 on: January 12, 2019, 05:44:24 pm »
0
Solve the following system of simultaneous equations in terms of a.
2x − y + az = 4
(a + 2) x + y − z = 2
6x + (a + 1) y − 2z = 4

DBA-144

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Re: VCE Methods Question Thread!
« Reply #17594 on: January 12, 2019, 06:04:23 pm »
+1
Solve the following system of simultaneous equations in terms of a.
2x − y + az = 4
(a + 2) x + y − z = 2
6x + (a + 1) y − 2z = 4

Why dont you have a go first? What have you got? Maybe then we can help you better.

Hint: follow same, typical process you would for any set of equations.