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Author Topic: VCE Methods Question Thread!  (Read 4802587 times)  Share 

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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17535 on: January 03, 2019, 11:37:15 am »
+1
Hi. I am having really big problems with methods and it's making me hate it. I am currently self learning the content. Right now, I'm doing circular functions. It's taking me 30 mins to do one question because it's so hard I have to take a lot of time to draw the graphs.
Does anyone have any tips,advice or things they did that helped them with this self learning stuff?
Anything would be appreciated
Don't worry, I'm also currently pre-learning methods 1/2 and also spending at the least 30 minutes to complete the harder questions.
What I did is just do the ones I know how to complete and if I do not know some certain things, for e.g Logarithms, I would search it up and have a look.

lyristis

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Re: VCE Methods Question Thread!
« Reply #17536 on: January 09, 2019, 07:56:34 pm »
0
hi, can anyone help with these? i would try them first but i honestly have no idea where to start

1: For g(x) = 3x − 2, find: { x : g(2x) = 4 }

2: Find the value of k for each of the following if f(3) = 3, where: f(x) = kx − 1

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Re: VCE Methods Question Thread!
« Reply #17537 on: January 09, 2019, 08:46:44 pm »
+1
1: For g(x) = 3x − 2, find: { x : g(2x) = 4 }

\[\begin{align*}  g(2x) =& \: 4 \\ 3(2x)-2 =& \: 4 \\ 6x =& \: 6 \\ x =& \: 1 \end{align*}\]

2: Find the value of k for each of the following if f(3) = 3, where: f(x) = kx − 1

\[\begin{align*}  f(3) =& \: 3 \\ 3k-1 =& \: 3 \\ 3k =& \: 4 \\ k =& \: \frac{4}{3} \end{align*}\]

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17538 on: January 09, 2019, 10:01:40 pm »
0
In four years time a mother will be three times as old as her son. Four years ago she was five times as old as her son. Find their present ages.

Two children had 110 marbles between them. After one child had lost half her marbles and the other had lost 20 they had an equal number. How many marbles did each child start with and how many did they finish with?

One hundred and fifty tickets were sold for a basketball match and $560 was the total amount collected. Adult tickets were sold at $4.00 each and child tickets were sold at $1.50 each. How many adults tickets and how many child tickets were sold?

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?

Sorry just wanted to get these questions across. :( I feel like my goal of achieving a 40+ in Math Methods has shattered.

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #17539 on: January 09, 2019, 10:15:44 pm »
+2
Just gonna drop a few words here to help you advance with these questions :).

Algebra is super handy here. Instead of considering two variables in two scenarios ie. four variables, try to use the information given to manipulate a single variable by adding or subtracting, or multiplying our dividing numbers. e.g. in the first example let the son's age be x, then in four years, he will be (x+4), then let the mothers age be 3(x+4) in four years. From there I think you get the gist. Solving for this single variable then substituting back into your work will yield you everything you ever needed to know.

The fourth one is a bit of a curveball, and it bucks the trend a little.You make it fit the trend by considering both numbers ie. xy and yx and that x+y=8. Adding them up will give you 88 from that same theory. See where that gets you using the paragraph above.

EDIT: Noting the below post, probably should've made myself clearer. Since they add up to 88, and have a difference of 36, you do get a pair of easily solvable simultaneous equations, which is slightly different from the below post but has the same end process. :)

Hope this helps :)
« Last Edit: January 09, 2019, 10:29:43 pm by fun_jirachi »
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Re: VCE Methods Question Thread!
« Reply #17540 on: January 09, 2019, 10:16:11 pm »
+2
Sorry just wanted to get these questions across. :( I feel like my goal of achieving a 40+ in Math Methods has shattered.

I'll start with this. Please don't get down on yourself. From your signature you seem to be doing Methods 1/2 this year. That is heaps of time to get the basics down and then to work hard in methods 3/4 and get a great score. From your other posts that I have seen, you will be more than capable of getting 40+ if you work efficiently, in particular, on the areas you struggle with most.

In four years time a mother will be three times as old as her son. Four years ago she was five times as old as her son. Find their present ages.
\[\text{First step should be to create simultaneous equations of some description. Let the current of the mum be "m" and the current age of her son to be "s".} \\ \therefore m+4=3(s+4) \\ and \\ m-4=5(s-4) \\ \therefore m=44 \quad \text{and} \quad s=12\]

Try to do the other questions by assigning a variable to each unknown and creating two simultaneous equations. I have skipped the steps needed to solve them, but if you need help doing them it may be helpful to find the textbook chapter on solving these or search up a Youtube video on the possible techniques.

EDIT:
The fourth one is a bit of a curveball, and it bucks the trend a little.You make it fit the trend by considering both numbers ie. xy and yx and that x+y=8.
I disagree a little with this, while I agree that it is a bit more challenging and requires another step/recognition, I think it can still be easily solved using simultaneous equations by considering the place value of each digit.
Eg. x+y=8 and 10x+y+36=10y+x.
« Last Edit: January 09, 2019, 10:20:57 pm by FelixHarvey »

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17541 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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Re: VCE Methods Question Thread!
« Reply #17542 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.

Sine

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Re: VCE Methods Question Thread!
« Reply #17543 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 #17544 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!

lyristis

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Re: VCE Methods Question Thread!
« Reply #17545 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

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Re: VCE Methods Question Thread!
« Reply #17546 on: January 10, 2019, 04:50:46 pm »
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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 »

Sine

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Re: VCE Methods Question Thread!
« Reply #17547 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 #17548 on: January 10, 2019, 10:57:00 pm »
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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




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Re: VCE Methods Question Thread!
« Reply #17549 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
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