# Easiest Paper Ever (1 Viewer)

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#### Bliske

##### New Member
Asking to verify the calculation sounds well within the scope of the syllabus. I'm pretty sure I got taught how to do verify it too.
I thought so as well. My teacher assured me that you would never have to calculate the correlation coefficient by calculator as it would have to be given to you, but I learnt how to do it just in case. I'm glad I did.

#### FootballHooligan

##### New Member
The financial expectation question...
Yeah it's $0. #### FootballHooligan ##### New Member Yeah it's$0.
I'm seriously expecting to get 80+ for this paper
But being ranked 32/50 is not the best

#### BigBusty

##### New Member
Did any1 get 8:30am Saturday landing time for Dubai ?

#### FootballHooligan

##### New Member
Did any1 get 8:30am Saturday landing time for Dubai ?

#### BigBusty

##### New Member
and for the chocolate volume, it was around 192 ?

#### Mysterious_girl

##### New Member
If we are talking about question 30C) (The one with the school playground) I agree that we definitely can't use Pythagoras and should use a different formula.
this is how i calculated the triangle value; 1/2 X 45 X 45 X sin100

#### InteGrand

##### Well-Known Member
and for the chocolate volume, it was around 192 ?
It is 193.1417... (cubic cm), which is roughly 193 to nearest whole number.

#### FootballHooligan

##### New Member
I worked it out like this

1) Found the rectangular area surrounding the marshmallows (20 x 15)= 300
2) The depth of the area (volume of the rectangular area) = 300 x 6 (height of two marshmallows) = 1800
3) Space covered by marshmallows = volume of marshmallows (58.9) x 24 = 1413.6
4) Subtract from volume of rectangular area = (1800- 1413.6)= 386.4

Pretty sure I'm wrong

#### BigBusty

##### New Member
I think its a rectangle with 15 x 10 (had to draw it on the diagram and then work out the radius distances) and then subtract 6 circles area from that rectangle to find the remaining area and then multiply the area by 6cm to find the volume of the chocolate
I subtracted 6 circles because inside the rectangle there are 4 1/4 of a circle (which equals 1 circle) + 6 semi circles (3 circles) + 2 normal circles. Sorry if im confusing.

#### BigBusty

##### New Member
BTW question 19 = A and 21 = D ? does any1 know? and isn't question 25 = B?

#### InteGrand

##### Well-Known Member
$\bg_white \noindent The area of each shaded region in the `top view' is 5^{2} - \pi \times (2.5)^{2} (proved below). There are six such shaded regions and the depth of each is 6 cm (since there are two levels of 3 cm high marshmallows). So the volume of chocolate used is: 6\times \left(5^{2} - \pi \times (2.5)^{2}\right)\times 6 = 193.1417\ldots.$

$\bg_white \noindent Proof of area of shaded region: In general, for circles of radius r and diameter d = 2r, the shaded area of a single region like that is 4r^{2} - \pi r^{2} = d^{2} - \pi r^{2}. To see this, join up the centres of the four touching circles. The area of the square formed by this is 4r^{2}, since each side of this square has length 2r. From this, we subtract the parts of this square that belong to the circles, which are just four circular quadrants, which thus have area equal to one of the circles, i.e. \pi r^{2}. So we subtract this from 4r^{2}, meaning that the remaining area (which is the shaded region) is 4r^{2}-\pi r^{2} = d^{2} - \pi r^{2}. In the chocolate question, we had r = 2.5, d = 5.$

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#### BigBusty

##### New Member
For the memory speed question did any1 get 61 mbps (to nearest whole number)

#### InteGrand

##### Well-Known Member
I worked it out like this

1) Found the rectangular area surrounding the marshmallows (20 x 15)= 300
2) The depth of the area (volume of the rectangular area) = 300 x 6 (height of two marshmallows) = 1800
3) Space covered by marshmallows = volume of marshmallows (58.9) x 24 = 1413.6
4) Subtract from volume of rectangular area = (1800- 1413.6)= 386.4

Pretty sure I'm wrong
This method would unfortunately fail to exclude volume from the parts of the rectangle's "edges" where there is no chocolate. These should be excluded (since there's no chocolate there).

#### whatarethoose

##### Member
lmfao fuck my life i got 2:30am friday because youre going 21 hours back in time from 11:30pm

#### Bliske

##### New Member
For the memory speed question did any1 get 61 mbps (to nearest whole number)
Yes.

#### InteGrand

##### Well-Known Member
For the memory speed question did any1 get 61 mbps (to nearest whole number)
(Assuming you're talking about Question 30 (b) (ii)) How did people get 61? (Caveat, I don't do General Maths though, so I might have missed something.) Note that 1 byte = 8 bit. So 3072 MB = 3072 megabyte = 3072*8 mb (megabit).

Therefore, the speed in megabit per second required (noting 7 min. = 420 s) is

(3072*8 mb)/(420 s) = 58.51429... mbps, which is 59 to the nearest whole number.

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