VCE Methods Question Thread!

[aspiringantelope]

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?

[Sine]

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]

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]

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!

[mihir999]

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]

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]

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]

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.

[Gogurt]

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]

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.

[darkz]

eh.. I'm still not getting the answer I want. I've tried a few things but I'm still not getting it

Well at least you had a try! Please refer to the attachment

[Quinapalus]

eh.. I'm still not getting the answer I want. I've tried a few things but I'm still not getting it

Investigate the following possibilities and each of their probabilities (ordered Monday, Tuesday, Wednesday)

Sandwich, fruit salad (0.6), Fruit salad (0.3), so total probability for this event is 0.6 x 0.3 = 0.18
Sandwich, sandwich (0.4), fruit salad (0.6), so total probability for this event is 0.4 x 0.6 = 0.24


Sum up probabilities for total probability that he has fruit salad for lunch on Wednesday.

Hope that helps

[lyristis]

1) A student has the choice of three mathematics subjects and four science subjects. In
how many ways can she choose to study one mathematics and two science subjects?

i tried doing nPr(3,1) = 3 + nPr(4,2) = 12; 3 + 12 = 15 but it's wrong and should be 18.

i also tried using combinations which didn't work out either

could anyone explain the difference between combinations and permutations? i still don't know when to use which. i know that for permutations order matters, but how do you know when it matters or not?

2) A survey is to be conducted, and eight people are to be chosen from a group of 30.
If the group contains 10 men and 20 women, how many groups of eight people
containing exactly two men are possible?

3) From a standard 52-card deck, how many seven-card hands have exactly five spades and
two hearts?

i really have no idea where to begin with these 2. can someone give me some steps to begin with?

[addict]

1) A student has the choice of three mathematics subjects and four science subjects. In
how many ways can she choose to study one mathematics and two science subjects?

i tried doing nPr(3,1) = 3 + nPr(4,2) = 12; 3 + 12 = 15 but it's wrong and should be 18.

i also tried using combinations which didn't work out either

could anyone explain the difference between combinations and permutations? i still don't know when to use which. i know that for permutations order matters, but how do you know when it matters or not?

2) A survey is to be conducted, and eight people are to be chosen from a group of 30.
If the group contains 10 men and 20 women, how many groups of eight people
containing exactly two men are possible?

3) From a standard 52-card deck, how many seven-card hands have exactly five spades and
two hearts?

i really have no idea where to begin with these 2. can someone give me some steps to begin with?

For 1) You should be using nCr not nPr since we're looking at combinations (where the order in which items are selected don't matter), not permutations (where the order does matter). Moreover, you multiply nCr(3,1) and nCr(4,2), not add them up. This is because choosing maths subjects and choosing science subjects are two separate tasks you're looking at.

[MB_]

2) A survey is to be conducted, and eight people are to be chosen from a group of 30.
If the group contains 10 men and 20 women, how many groups of eight people
containing exactly two men are possible?

3) From a standard 52-card deck, how many seven-card hands have exactly five spades and
two hearts?

i really have no idea where to begin with these 2. can someone give me some steps to begin with?

These two follow a similar solution to question 1 using nCr