Topic, Math Whiz Quiz Game V2

  1. 250 (g of cream cheese)=2*125 = 2 pounds
200 (g of powdered sugar) = 1/5 kg = 0.5 pounds
total 2.5 pounds (I don't know the value of the pound in Euros, but I have the impression that the glazing could be more expensive that the cake itself!)

Extra question, for the fun :)
first boat: x=5t, y=4+12t
second boat: x=10-3u, y=4u (I have to use another variable for the parameter, because they are independent).

The intersection occurs when
5t=10-3u
4+12t=4u

5t+3u=10 ==> 20t+12u=40
12t-4u=-4 ==> 36t-12u=-12

Making the sum of both equations, I get 20t+36t=40-12
then 56t=28 ==> t=1/2
4u=4+12t=4+6=10 ==> u=5/2

The intersection is: x=5t=5/2 y=4+12t=10

  That's the same Ferrari
200/cbrt(2)+50*cbrt(4)= [200+50*cbrt(4)*cbrt(2)]/cbrt(2)=[200+50*cbrt*(8)]=300/cbrt(2)
woops I made a little fault. I forgot to multiple 50 with 2...

The first curve is given by r(t) = (5t)x + (4+12t)y and the second as s(u) = (10-3u)x + (4u)y. (the x and y used here are unit vector pointing along the x-axis and y-axis) Then we have to solve r(t) = s(u) to u and t, so we get this set of equations:
|5t = 10-3u
|4+12t = 4u

To solve this we can use a matrix:
( 5 3 | 10)
(12 -4 | -4)
An echelon form is:
(5 3 | 10)
(0 2 | 5)
So u = 5/2 and t = (10-3u)/5 = 1/2
Then fill in the t and u to get the intersection point(s).
I get the intersection is (5/2, 10).
Btw is this for homework or just interested? :)

  First question is correct, third question is cotrrect, but one the second question; how do you get 200/cbrt(2)+50*cbrt(4) to become 250/cbrt(2)? They aren't the same, are they?

200/cbrt(2)+50*cbrt(4)=238,1101 ... (Which is the correct answer)
250/cbrt(2)=198,4251 ...

Or is my calculator lying again? I will deduct no points either way. Even if there is a mistake there it is so small compared to the task in its eniterity, so I don't think it's fair to penalize it.

  Brain burst X_X

  First question: x=25 and y=10. First subsitute y=35-x in first equation. You'll get 35-x=2sqrt(x). Thus x = 25. So then y = 10.


Second question: Name x the width of the base and y the height of the box.
The volume is given by 4x^2*y=200 Thus y=50/(x^2).
The surface is given by 2*4x*y+2*x*y+2*x*4x=10xy+8x^2

Then subsitute y=50/(x^2) so you get 500/x + 8x^2.

Now differentiate this so you become -500/(x^2)+16x.

To get a extremum equalize it to 0. so -500/(x^2)+16x=0. So -500+16x^3=0 So x=cbrt(500/16)= (5/2)*cbrt(2).

Now the surface is 500/[(5/2)*cbrt(2)]+8*[(5/2)*cbrt(2)]^2=200/cbrt(2)+50*cbrt(4)=250/cbrt(2)

(where cbrt(x) is the cube root of x)

third question: Say OH = time from Oslso to Hamar and HM = time from Hamar to Molde and OM = time from Oslo to Molde. Then is 4*OH=HM. OM = 7,5 hour. Then is OH+HM=OM So 5*OH=OM. Thus OH= 1,5 hour which is 1 hour and thirthy minutes.

  Interesting. I have only solved it the way Nelson did!

Anyway, three new questions are up :)

  The proportion is similar to the construction of the golden ratio.

You could also prove the second question in a different way:

a/|AD|=cos(A) and x/|AD|=sin(A) so a/x=cot(A).
The same way we find x/|CD|=sin(C) and b/|CD|=cos(C) so x/b=tan(C).

Because the triangle ACD is rectangle we know C = 90°-A. So x/b = tan(C)= tan(90°-A) = cot(A) = a/x. Thus a/x = x/b.

edit: gave second solution for fun. :)

  1. The volume of a pyramid is equal to (base area)*height/3

V=230^2*146/3=52900*146/3=2574466.667 m3, about 2574467 m3

2. A triangle in a semi-circle is rectangle:
AD^2+DC^2=AC^2=(a+b)^2

Since BD is perpendicular to AC, the triangles ABD and DBC are rectangle:
AD^2=AB^2+BD^2=a^2+x^2
CD^2=DB^2+BC^2=x^2+b^2

From the three equations, we get:
a^2+x^2+x^2+b^2=(a+b)^2=a^2+2ab+b^2 ==> 2x^2=2ab ==> a/x=x/b

Sorry, I don't know the English term.

  Actually, I have made a typo in the question. Alternative E should have been 1043, and then 5011 would've been the only possible answer. Your reasoning is 100% correct, so I'm going to give you the points, of course :)

And, I guess my calculator was the problem then, tricking me to believe that the parametric equations were for two different lines. Bad Calculator! :/

  Again about the parametric equations, your graphic calculator seems to give strange results:
- [-3t-1, -t+2] and [-3t+2, -t+3] have the same derivative [-3, -1]
- the point [-1, +2] belongs to both curves, first curve for t=0, second curve for t=1

Concerning the suite 3938 - 7606 - 9170 - 3217 - 2234, I have remarked that the sums of the digits of each number are
3+9+3+8 = 23
7+6+0+6 = 19
9+1+7+0 = 17
3+2+1+7 = 13
2+2+3+4 = 11
These numbers are all prime numbers, the following being 7. The problem is that 5011 and 1042 have both 7 for sum. Could it be that your problem has 2 solutions? Maybe I am missing something or my solution is totally wrong.

  When I plotted them, they were the same, but for a certain value of t, you won't get the same point. But because for a parametric equation of a straight line the variable goes from -infinity to infinity. So basically you get the same collection of points, so it represents the same.

  PineappleDude gave the correct answer to task 1, and Nelson answered task 2 correctly. Well done!

Back to the parametres, I know there is an infinite number of parametric equations for the same line, but when I plotted my answer and your answer in on my graphic calculator, I got two different lines. I thought this meant that they don't represent the same object. If someone could explain why this is happening, then I would be grateful :) And just to be sure, I'll plot them in one more time, to make sure I didn't put in the worng number or something.

  About the parametric equations, I agree with Simon, I got exactly the same result at first try.

Concerning the suite 3456 - 9801 - 11781 - 22959, the following number is 115254. Explanation:
3456+6345=9801
9801+1980=11781
11781+11178=22959
22959+92295=115254
The number to add is obtained by circular permutation : just move the last digit at first position.

  Cocktail question: Onnly way this is possible is if all of the fanta is used. If all 1.5 liters of fanta is used, 1 liter of coke will be used. That is 2.5 liters in total, or 10 glasses :)

  Btw Ferrari there are hundreds of parametric equations of the same object. You can for example subsitute t by 2s or something and you would get a parametric equation in s.

  I'm not going to be very persistant here, so unless there are any protest I can happily give you your four points Simon :)

And Pineappledude gets one point as well!

  One icecube weighs 180/3=60/2= 30g ^^

  That's rather strange, it's the standard procedure. I also got the derivation right and the t0. And for the i, j, k stuff, that's the international convention.

edit:
my tangent and your tangent

I think they are just the same except I got another parameter representation. proof of the same line. So give me my points :).

  I actually get x=2-3t and y=3-t, and so does the book I got the task from. I also found a list of known flaws in the book, and there should be nothing wrong with that task.

Regarding the i, j, k ... thing, I'm presenting my questions the way I have been taught. Maybe they teach a different setup in other countries, I don't know :/

Psychomaster: I can make an easy question for you now, and I can make sure that whenever I add new questions, I can add an easy one too :)

  I wish there were a few easy questions, too.

  Let's say x(t)=t^2-t and y(t)=t^3-4t and let's say t0 is the value of t in point (2,3) which is -1. Then the tangent line is given by
| x=x(t0)+x'(t0)*(t-t0)=2-3*(t+1)
| y=y(t0)+y'(t0)*(t-t0)=3-1*(t+1)

(x'(t)=2t-1 and y'(t)=3t^2-4)

So the parameter equation of the tangent is given by [-3t-1, -t+2].

btw for vectors they mostly use i, j, k, ... unit vectors instead of [ ].

  That's correct!

  The speed is the derivative of the position versus time: [40, 20-10t]. At t=0, the speed vector is [40,20] ==> its module is sqrt(40^2+20^2) = sqrt(2000) = 20*sqrt(5) m/s, about 44.721 m/s or 160.997 km/h.

  Simon got the answer I was looking for, and since this obviously was a bit tougher than the previous one, I'll make it a 4p question instead.

  1; 3,5; 5; 6,5
2/2 7/2 10/2 13/2

Everyting points for 11 to me.

  This new sequence is pretty difficult.
Could 8 be the next number(probably not)

  I fully agree with both of you, I have found this question very easy. Don't hesitate to reduce my score. Anyway, I don't care about the score, my main pleasure is to try to solve the problems :)

  Maybe not, how about three points? I'm not especially good at these kind of questions, so I may have over-estimated its difficulty.

  Damnit I always see the question after Nelson solved it... But I don't think this question deserves 4 points.

  And the kind of question I am very fond of!

If we multiply each number by 2, we get 1 - 8 - 27 - 64, that is 1^3 - 2^3 - 3^3 - 4^3.

The next number is 5^3/2 = 125/2 = 62.5