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I've got one! I encountered this problem while reviewing for the MTAP Sectorals. . .which is two days from now scaredplz.gif

@Radium Gonna compete as well? See ya there!!~

 

 

Insert two arithmetic means between x + 3y and x - 3y.

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Yep! You're right! 

 

Well, the way I solved it is a bit different from yours. But you still got it right!

 

To get the common difference, first subtract the first term x + 3y from the last term x - 3y. 

 

= x - 3y - (x + 3y)

= x - 3y - x - 3y         distributing the negative sign~

= -6y                        collecting like terms

 

Now, since two terms are given, x + 3y and x - 3y, and two arithmetic means are to be inserted, we'll have a total of 4 terms. The formula for getting the common difference is

 

 

       an - a1

d =  ------------

           n - 1

 

where d is the common difference, an is the last term, a1 is the first term, and n is the number of terms.

 

Since we now have an - a1 = -6y, all that's left is to subtract 1 from the number of terms (4).  4 - 1 = 3

 

So dividing:

 

- 6y

-----  =   -2y

  3

 

Adding -2y to the first term:

 

x + 3y - 2y = x + y

 

Adding -2y to the next term:

 

x + y - 2y = x - y

 

Here we have the two arithmetic means,

 

x + y and x - y

 

 

Missed ya as well! Are ya competing in the Regionals too? :D

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I'm a bit lazy memorizing formulas and do it my way sometimes :P

I'll try and find ya there!

 

Well, that's really what Mathematics teaches to people. Finding your own way of solving problems, the formulas are just there to guide you~

 

(Which Region do you belong to? We might not even be in the same region :/)

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NCR! The National Comfort Room! Jk.

What division are you, if you're NCR? And what grade also?

 

MUHAHAHAHAHAHAAH. SEE YA ON TUESDAY, RADIUM-CHAN :D :D :D

 

Yours first :P Are you on Category A or B?

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@Radium-chan Category B as well. As for the grade, I think I'll be more comfortable disclosing it in a private message since disclosing my grade level would mean disclosing my age~ And I'm very private about my age.

 

 

Problem:

 

What is the area of a quadrilateral whose vertices are (0, 0), (4, 0), (3, 2), and (0, 4)?

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^I'd answer, but spoiler tags don't work for me >.<

 

Hmm MTAP-MMC contest by chance? I remember when I competed in the PMO once. I fell short of winning in the nationals  <_<  Care to share any problem you're having difficulty with? I forgot what it felt like when a competition is looming near; maybe seeing some of those problems will jog my memory lol.

 

Anyway, what's a Math thread without a math problem?

 

 

You have two boxes. Each has a square base, and is half as tall as it is wide. If the larger box is two inches wider than the smaller box, and is 244cu.in greater in volume, what is the width of the smaller box?

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Yep! MTAP-MMC~ Actually, in all my years of studying, it's the first time I've ever qualified for the MTAP-MMC. My previous school didn't let us join those kinds of contests, but when I got to study in a Science High School, I could finally join. Last year I didn't qualify because my score didn't make it for our school eliminations, but this year I got qualified (yay!) I highly doubt we'll make it to the Nationals, though :/ Our competitors are Chinese-Filipino schools and other Science High Schools, and I don't know if our team stands a chance against those Chinese students who  probably have been studying Maths ever since birth XD

 

The width of the smaller box is 8 inches.

 

Solution:

 

Let            x = the width of the smaller box

             x/2 = the height of the smaller box

          x + 2 = width of the larger box

     x + 2 /2 = the height of the larger box

 

Since the larger box is two inches wider than the smaller box (x), its width will be x + 2. 

 

So. . .forming an equation~

 

          x                                     x + 2

x2 *   ---  + 244  =  (x + 2)2 *  ----------

         2                                       2

 

The left side is the volume of the smaller box, wherein when you add 244 to it, becomes equal to the volume of the larger box. The right side of the equation is the volume of the larger box.

 

Solving:

 

 

          x 3                  (x2 + 4x + 4)(x + 2)

         ---  + 244  =    ------------------------

         2                                   2

 

 

         x 3                   x3 + 6x2 + 12x + 8

         ---  + 244  =    ------------------------

         2                                   2

 

 

         x 3                   x3 + 6x2 + 12x + 8

   (2) ---  + 244  =    ------------------------ (2)           Multiplying both sides by 2 to get rid of the denominator...

         2                                   2

 

        x 3  + 488  =   x3 + 6x2 + 12x + 8

 

6x2 + 12x - 480 = 0                                             Collecting like terms...and let the quadratic fun begin.

 

We can divide both sides of the equation by 6 in order to make things more simple.

 

Here we have:

 

     x2 + 2x - 80 = 0

(x + 10)(x - 8)   = 0

 

There are two possible roots for this quadratic equation:

 

 x + 10 = 0                                            x - 8 = 0

       x  = -10                OR                         x = 8

 

Since we are talking about measurements, only the positive root is possible, hence x = 8 is the root we are looking for.

 

The width of the smaller box is 8 inches.

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@Rye:

Is the answer 10 square units? I'm not sure because I suck at the Cartesian Coordinate Plane >_<

Oh, and do you take MTG?

@scudzilla:

Let the width of the smaller box be 2x, so its height is x.

Its volume is (2x)(2x)(x)=4x^3

And the width of the larger box is 2x+2, so its height is x+1.

(2x+2)(2x+2)(x+1)= 4x^3+12x^2+12x+4

So 244=12x^2+12x+4

240= 12(x+1)(x)

20=(x+1)(x)

X=4

2x=8

8cm

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@Rye:

Is the answer 10 square units? I'm not sure because I suck at the Cartesian Coordinate Plane >_<
Oh, and do you take MTG?


@scudzilla:

Let the width of the smaller box be 2x, so its height is x.
Its volume is (2x)(2x)(x)=4x^3
And the width of the larger box is 2x+2, so its height is x+1.
(2x+2)(2x+2)(x+1)= 4x^3+12x^2+12x+4
So 244=12x^2+12x+4
240= 12(x+1)(x)
20=(x+1)(x)
X=4
2x=8
8cm

 

 

@Radium Yes, it is. And no, I don't :/ I want to, though.

 

Oh. . . and @Radium

The unit for scudzilla-san's problem is inches, not cm. Just a friendly note, be careful with those units! I lost a point there once. I got the correct numerical value but I forgot to put a unit :/ Good luck for Tuesday!  

 

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Those tricky units! DX<

A bulk of documents is for photocopying. Using Machine A, it will take 11 hours. With Machine B, it will take 13 hours. If they work together, a total of 28 copies less will be printed. When both are used, the job will be done in 6 1/4 hours. How many pages are for photocopying?

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Those tricky units! DX<

A bulk of documents is for photocopying. Using Machine A, it will take 11 hours. With Machine B, it will take 13 hours. If they work together, a total of 28 copies less will be printed. When both are used, the job will be done in 6 1/4 hours. How many pages are for photocopying?

 

If I understood the problem correctly, 336 pages in all?

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Nope

EDIT: Hello! This is the creator of this thread. Since I'll be leaving soon, I need some volunteers to contribute to this thread when I'm gone. Please post here if you want :)

And maybe some people are confused with some of the problems we've posted. Here's an easier one:

What is the smallest number you can multiply to 126 to get a perfect square?

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Those tricky units! DX<

A bulk of documents is for photocopying. Using Machine A, it will take 11 hours. With Machine B, it will take 13 hours. If they work together, a total of 28 copies less will be printed. When both are used, the job will be done in 6 1/4 hours. How many pages are for photocopying?

I really don't understand the bold part -.-"

But if my understanding is right, the equation can be like this:

6 1/4 * (X-28) * (1/11 + 1/13) = X

X (the pages) = 600

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