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June 20, 2019, 03:08:37 pm

Author Topic: VCE Methods Question Thread!  (Read 2321436 times)  Share 

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lyristis

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Re: VCE Methods Question Thread!
« Reply #17550 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 #17551 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 »
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Sine

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Re: VCE Methods Question Thread!
« Reply #17552 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 #17553 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 #17554 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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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17555 on: January 11, 2019, 01:19:45 pm »
0
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 #17556 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 #17557 on: January 11, 2019, 01:52:10 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  :(
 
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 #17558 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 #17559 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 #17560 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.

matthewzz

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Re: VCE Methods Question Thread!
« Reply #17561 on: January 12, 2019, 07:19:35 pm »
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I've just finished chapter 1 of the year 12 Cambridge textbook in prep for 3/4 methods, and I ran into a few problems that need some light shed if anyone can help!

The question attached is on composite functions, and I attempted to solve for a so that the range of g fit into the domain of f and got 2 (so that the graph is translated up 2 to satisfy the range of g=[2,infinity]), even though the answer is [2,3]. I'm not sure how to find this out no matter how many times I try. Thanks in advance!

lzxnl

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Re: VCE Methods Question Thread!
« Reply #17562 on: January 12, 2019, 07:33:25 pm »
+2
For f(g(x)) to be defined, the range of g must be in [2,inf). Now the range of g, quite obviously, is [a,inf] so for this to fit inside the domain of f, we have a >= 2 (if a = 3, for instance, the range of g is then a subset of the domain of f, which is fine)

For g(f(x)) to be defined, the range of f must be in (-inf, 1], and the range of f is (-inf, a-2]. For (-inf, a-2] to be in (-inf, 1], we require a-2 <= 1, or a <= 3. Combining the two gives you 2<=a<=3.

You need to remember that f(g(x)) is defined even when the domain of f isn't equal to the range of g; the domain of f just has to include the range of g.
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Gogurt

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Re: VCE Methods Question Thread!
« Reply #17563 on: January 13, 2019, 02:38:50 pm »
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Not sure on how to approach this probability question:

David has either a sandwich or fruit salad for lunch. If he has a sandwich for lunch one day, the probability he has a sandwich the next day is 0.4. If he has fruit salad for lunch one day, the probability he has a fruit salad for lunch the next day is 0.3. Suppose he has a sandwich for lunch on a Monday. What is the probability that he has fruit salad for lunch the following Wednesday.

Any thoughts?

Quinapalus

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Re: VCE Methods Question Thread!
« Reply #17564 on: January 13, 2019, 03:54:59 pm »
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Not sure on how to approach this probability question:

David has either a sandwich or fruit salad for lunch. If he has a sandwich for lunch one day, the probability he has a sandwich the next day is 0.4. If he has fruit salad for lunch one day, the probability he has a fruit salad for lunch the next day is 0.3. Suppose he has a sandwich for lunch on a Monday. What is the probability that he has fruit salad for lunch the following Wednesday.

Any thoughts?

Best way is to use a tree diagram.

Split up into possibilities i.e What is probability that he has/does not have a sandwich on Tuesday? If he has a sandwich on Tuesday, what is the probability that he has a sandwich on Wednesday/does not? etc.


This is testing conditional probabilities.
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