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# Math Questions

Well like Metalfreek, I'm starting a topic with some "math questions". Not all of them require a lot of work, some of them well you just have to see those to solve them...

Question 1:
Well because I had to use a table for this one, I created a webpage to display the table properly.

Question is simple, what's the number that should replace the Questionmark.
Skills needed: + - * /

Question 2:
http://www.frihost.com/forums/vt-108646.html#905508

Question 3:
http://www.frihost.com/forums/vt-108646.html#905707

PS: Also check this topic: http://www.frihost.com/forums/vt-106519.html, some nice and tricky questions there too...

metalfreek
answer is 7 because (8 + 13) / 3 = 7
fx-trading-education got it right, you had to add up the first to numbers and divide them by the third number and then you get the fourth number.
 Code: 8+6=14/2=7 9+6=15/3=5 16+18= 34/17=2 10+10=20/5=4 8+13=21/3=7

New Question will be added soon...

Question 2:

My first customer bought half of the stock + half a bread. My Second customer, bought half of what was left + half a bread. The third customer asked for half of the stock that was still available + half a bread to feed the ducks. The last Customer, Mr Salomon, wanted to buy a bread, but there were no more breads.

Hogwarts
Pretty shonky baker if he only bakes 7 loaves o_o
Question 3:

Prove that:

• sin(a+b)*sin(a-b)=cos˛b-cos˛a
• cos3a = 4cosła-3cosa

*Those two have nothing to do with eachother, they are two seperate things.

_AVG_
The first one:

TO PROVE: sin(a+b)*sin(a-b)=cos^2b-cos^2a

sin(a+b)*sin(a-b)=(sinacosb+sinbcosa)(sinacosb-sinbcosa)
=sin^2acos^2b-sinasinbcosacosb+sinasinbcosacosb-sin^2bcos^2a
=cos^2b(1-cos^2a)-cos^2a(1-cos^2b)=cos^2b-cos^2acos^2b-cos^2a+cos^2acos^2b
=cos^2b-cos^2a

And the second one:

TO PROVE: cos3a = 4cos^3a-3cosa

cosa+isina=e^ia
cos3a+isin3a=e^3ia=(e^ia)^3=(cosa+isina)^3=cos^3a+3icos^2asina-3cosasin^2a-isin^3a
=cos^3a-3cosasin^2a+i(3cos^2asina-sin^3a)
Equating Real parts, cos3a=cos^3a-3cosasin^2a=cos^3a-3cosa(1-cos^2a)=cos^3a-3cosa+3cos^a
=4cos^3a-3cosa
metalfreek
 adri wrote: Question 3: Prove that: sin(a+b)*sin(a-b)=cos˛b-cos˛a cos3a = 4cosła-3cosa *Those two have nothing to do with eachother, they are two seperate things. Adri

#1
sin(a+b)*sin(a-b)
=(sina cosb + cosa sinb)*(sina cosb - cosa sinb)
=sin^2a * cos^2b - cos^2a * sin^2b
=(1 - cos^2a) * cos^2b - cos^2a * (1-cos^2b)
=cos^2b - cos^2a*cos^2b - cos^2a + cos^2a*cos^2b
=cos˛b-cos˛a
proved#

#2 cos3a=4cosła-3cosa
cos3a
=cos(a+2a)
=cosacos2a-sinasin2a
=cosa(2cos^2a-1)-sina(2sinacosa)
=2cos^3a-cosa-2sin^2acosa
=2cos^3a-cosa-2cosa(1-cos^2a)
=2cos^3a-cosa-2cosa+2cos^3a
=4cosła-3cosa
proved#

Ok, I know I'm bumping a topic, but I don't think there's a point of creating a new topic with the same purpose.

Question 4

The question is simple: How long is x?

Bikerman
square-root of 6
Incorrect, sorry bikerman.

Bikerman
Arrghhh....of course it is incorrect...I've just seen my rather elementary mistake
I'll have another bash later and this time no silly mistake
I must say, I didn't have the correct answer the first time too, but then with some good thinking (*really good thinking actually*), I found the answer.

Looking forward to see your calculations.

_AVG_
Before I took a crack at this, I just wanted to confirm - the angle between the line segment of length "x" and the line segment of length 6 units is 90 degrees, correct?
Correct.
_AVG_
Alright then, x=6?

I used similarity and congruency. Name the horizontal lines CD (above) and AB (below). The central point O. Then, Triangle DOC is similar to Triangle AOB (with ratio 2 : 1). But, Triangles ACD and BAC are congruent (as they are similar and have a common side). So, AC = BD = 9 = x + 0.5x => x = 6
Incorrect, AC isn't BD, if you watch the picture you see that AB isn't as long as CD.

_AVG_
Oh shoot - made a silly mistake (they aren't congruent because corresponding sides aren't equal - even though there is a common side).

Well, after proving the first two triangles similar (i.e. DOC and AOB) with ratio 2 : 1, we know OD = x, thus, AO = 0.5x. And, since OB = 3, by the pythagorean theorem, AB^2 = (0.5x)^2 + 9

Similarly, AC^2 = (0.5x)^2+36

And BC^2 = AB^2 + AC^2 = 2(0.5x)^2+45 = 81

Solving for x, x = 6 times the square root of 2

approximately : 8.5

Dam me always trying to solve simple problems with complicated methods
Correct, although there are easier ways of finding the answer.

I thought Bikerman would give another go but anyways...

Question 5

In my street, there's only one side with houses. The sum of all the housenumbers smaller than mine is equaly to the sum of all housenumbers bigger than mine. The 'connection/relation' between my housenumber x and all houses in my street is?

If it is too hard, where going to do multiple choice.

Bikerman

Damn - to be honest I forgot all about it....I'll have a bash at the next when I get a moment...
You can still solve Question four if you want.

PS: Congrats with your 7000 posts. :p

Bikerman
Thanks..

here's my starting point for the previous puzzle
From there is follows that area top right triangle = 1/2base*height = 1.5x*6/2 = 4.5x
alternatively, area = a^2*sin(B)*sin(C)/(2*sin(B+C)) = 1.5x^2*sin(90-phi)*sin(phi)/2*sin(90)

I think it solves relatively simply from there using simple substitution..

PS = 2x should be 1/2x = slip of the pen..

PPS - there is a simpler solution to this using just the 2 top right triangles, if I can get my brain in gear, based on one being half the area of the other....

God almighty...how did I not see it?....it is bleedin obvious...!

Tan(theta) = 12/x = x/6

12/x = x/6 x^2=72 x=+/-root(72)

or more elegantly

x=6*root2 (or -6*root2)

I apologise for being a complete numpty.
Bikerman
 adri wrote: I thought Bikerman would give another go but anyways... Question 5 In my street, there's only one side with houses. The sum of all the housenumbers smaller than mine is equaly to the sum of all housenumbers bigger than mine. The 'connection/relation' between my housenumber x and all houses in my street is? If it is too hard, where going to do multiple choice.

Bollox to multiple choice - I have to redeem myself after that last performance.

OK - so we have N houses in a series. The known house is x in the series (I'm going to assume they are numbered in 1s but hopefully any solution that drops out will work for other multiples).

The sum of any series from 1....X is given by X(X+1)/2
We are going from 1 to x-1
So the sum of the smaller numbers would be: x(x-1)/2
The larger numbers must sum to (N-x)(N+x)/2

Therefore 1/2*(x - 1)*x = 1/2*(N - x)*(N + x)
Multiply it out
1/2x^2-1/2x = 1/2 ( N^2-x^2)
1/2x(x-1) = 1/2 (N^2 - x^2)
1/2x(x-1) +1/2x^2 = 1/2N^2
x^2-1/2x = 1/2N^2
2x^2 - x = N^2
x(2x-1) = N^2

I think that is as far as I can take it....

Hang on....that second sum looks wrong......
Damn...I'm 1 short, it should be (N-x)(N+x+1)
OK - I'm gonna cheat and run it through Mathematica to quickly get me back (I've done the work myself once, I think I'm allowed a bit of mechanical aid this time - why have a dog and bark yourself? )

Mathematica simplifies it down to

N^2+N= 2x^2

So for N houses in your street, the relationship between your house number (x) and N is
x=ROOT(N^2+N)/2

(Mathematica chops that out to be x = +/- ROOT(N^2/2 + N/2)
_AVG_
 adri wrote: I thought Bikerman would give another go but anyways... Question 5 In my street, there's only one side with houses. The sum of all the housenumbers smaller than mine is equaly to the sum of all housenumbers bigger than mine. The 'connection/relation' between my housenumber x and all houses in my street is? If it is too hard, where going to do multiple choice. Adri

Ooh .. I think I need multiple choice for this one. Also, could you explain the question a bit more, I didn't exactly get what you mean by "the connection / relation" - are you looking for some equation that relates the housenumbers on the street to your house? Or is it some other kind of connection e.g. your number is the median or something; finally, could you comment on Bikerman's assumptions?
Bikerman
The only thing I have 'assumed' is the formula for a sum 1....X is given by X(X+1)/2. That is solid.
The rest follows from that first theorem.....
_AVG_
 Bikerman wrote: What assumptions have I made? The only thing I have 'assumed' is the formula for a sum 1....X is given by X(X+1)/2. That is solid. The rest follows from that first theorem.....

I agree your math is solid but ...

You've also assumed that the houses are numbered in 1s, that they are numbered regularly, that all are numbered and that they start from 1. adri, I think you should let us know - are the housenumbers all in a properly defined sequence - like 1, 2, 3, 4 ... OR 1, 3, 5, 7 etc. [In some cities, I've seen streets where one-side is odd-numbered and the other even-numbered]
Bikerman
The maths works if they are numbered in any regular integer (arithmetic, not geometric) manner I think...though I don't think I could prove that rigorously...
infinisa
Hi Bikerman

Of course, when you wrote:
x=ROOT(N^2+N)/2

you meant:
x=ROOT((N^2+N)/2)

(missing brackets!)