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Mathematics paper 4 helper
- Thread starter Keeemoman
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hey where did u find 2011 specimen paper 4 2011...?? can u give the link...?/
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Link
Here you go bro !!!
http://www.xtremepapers.me/CIE/Cambridge IGCSE/0580 - Mathematics/0580_y11_sp_4a.pdf
Cheers !!! :good:
Here you go bro !!!
http://www.xtremepapers.me/CIE/Cambridge IGCSE/0580 - Mathematics/0580_y11_sp_4a.pdf
Cheers !!! :good:
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bt its actually summer 2007 paper...?? rite??
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I dunno
Haven't really been through both papers so can't give you a reply to that
Haven't really been through both papers so can't give you a reply to that
Here you go IGCSE28
a)i)You just need to multiply the 20 000 000 into 12 to find the actual distance
Next mark, you get for converting your answer to km as the answer you found in first step is in cm(divide it by 100 000, 100 for cm to m and 1000 for m to km)
a)ii)Here there are three steps
First convert the map scale to area map(1^220 000 000)^2)
Next find the answer by multiplying(4*10^14)*13 to get the actual distance
Lastly convert your answer to km2 as your answer is in cm2(divide it by 100 000*100 000)
b)i)the first mark you get is to convert one of numbers to the other unit to get the same units(1580km 158 000 000)
The second mark is for finding the ratio in the simplest form 31.6:158 000 000 to 1:5 000 000(n=5 000 000)
b)ii)First mark for the knowledge of the formula(distance/time)
Second mark for answering get the time(14 03 - 11 55 = 2 08)
Next mark for writing the time in numbers as shown 2+8/(60min)=2.13333333333333333hour
last mark for finding the answer 1580/2.1333333333=740.625km/hour
2)f)deals with a shear, you can multiply the transformation matrix into the points A,B,C
(1 0)(2 3 5)
(-1 1)(1 3 1)
A shear will always be of this format (1 0) or this one (1 k)(depends on which line is invarient)
(k 1) (0 1)
ii) Since 0 is the y axis part in the shear and it doesnt change so it is the y axis which is invarient
1
k is the scale factor so the scale factor is the -1
There the discription could be Shear with scale factor -1 and yaxis is the invarient line.
4f)i) p is vertical line so there is no any x-coordinate whereas p is half the value of DB as DB is 2p
Therfor the coordinate would be ( 0 )
(24)
F)ii)here x-coordinate is 7 as you need to move 7 units to your right and the y-coordinate is -24 as it goes down opposite to p and only
half the value of p
Therefore the matrix will be( 7 )
(-24)
g)this moves up to B as the value is 48 and then right to E by 14 units
There you could find that it is a right angle triangle in which you will have to find the hypotenuse using the Pythogoras theorm
square root of 14^2+48^2.
5)d)You just need to write the values of x-coordinates in which they make a line that is of negative gradient
(Note: A negative gradient line will make an obtuse angle with the Xaxis line on the upper side)
So the values would be from 0-6
6)C)I am writing it to you in steps
105/x-105/(x+4)=0.8
then make this fraction as single fraction
(105(x+4)-105x)/x(x+4)=0.8
Then simlify it by shifting the 0.8 to the denominator and x(x+4) to the other side
Hope this helped.
a)i)You just need to multiply the 20 000 000 into 12 to find the actual distance
Next mark, you get for converting your answer to km as the answer you found in first step is in cm(divide it by 100 000, 100 for cm to m and 1000 for m to km)
a)ii)Here there are three steps
First convert the map scale to area map(1^2
20 000 000)^2)Next find the answer by multiplying(4*10^14)*13 to get the actual distance
Lastly convert your answer to km2 as your answer is in cm2(divide it by 100 000*100 000)
b)i)the first mark you get is to convert one of numbers to the other unit to get the same units(1580km 158 000 000)
The second mark is for finding the ratio in the simplest form 31.6:158 000 000 to 1:5 000 000(n=5 000 000)
b)ii)First mark for the knowledge of the formula(distance/time)
Second mark for answering get the time(14 03 - 11 55 = 2 08)
Next mark for writing the time in numbers as shown 2+8/(60min)=2.13333333333333333hour
last mark for finding the answer 1580/2.1333333333=740.625km/hour
2)f)deals with a shear, you can multiply the transformation matrix into the points A,B,C
(1 0)(2 3 5)
(-1 1)(1 3 1)
A shear will always be of this format (1 0) or this one (1 k)(depends on which line is invarient)
(k 1) (0 1)
ii) Since 0 is the y axis part in the shear and it doesnt change so it is the y axis which is invarient
1
k is the scale factor so the scale factor is the -1
There the discription could be Shear with scale factor -1 and yaxis is the invarient line.
4f)i) p is vertical line so there is no any x-coordinate whereas p is half the value of DB as DB is 2p
Therfor the coordinate would be ( 0 )
(24)
F)ii)here x-coordinate is 7 as you need to move 7 units to your right and the y-coordinate is -24 as it goes down opposite to p and only
half the value of p
Therefore the matrix will be( 7 )
(-24)
g)this moves up to B as the value is 48 and then right to E by 14 units
There you could find that it is a right angle triangle in which you will have to find the hypotenuse using the Pythogoras theorm
square root of 14^2+48^2.
5)d)You just need to write the values of x-coordinates in which they make a line that is of negative gradient
(Note: A negative gradient line will make an obtuse angle with the Xaxis line on the upper side)
So the values would be from 0-6
6)C)I am writing it to you in steps
105/x-105/(x+4)=0.8
then make this fraction as single fraction
(105(x+4)-105x)/x(x+4)=0.8
Then simlify it by shifting the 0.8 to the denominator and x(x+4) to the other side
Hope this helped.
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i need help in specimen 2011
Q8 (c) & (d)
Q8 (c) & (d)
Here you go test128
c)First convert the 1.3mm to m you will get it as 0.0013
you already have height of the segment and the total area of the segment
so you need to find the radius
0.0013*pie*r^2=2 and then you make r as the subject of the formula and find the radius
d)First you need to get the total area that is 50.4/0.12=420
The next step is to find the perimeter, so as you in the diagram in the previous page that you 2sides of length 0.8
and the below one only is 1.2 and so the total will be 0.8+0.8+1.2=2.8
If you multiply the length which you have to find into the total perimeter, you will get the area
so you have the area and the perimeter you need to find the length
Here you go 420/2.8=150(this is the answer)
Hope this helped.(Good luck for your tommorow's examination)
c)First convert the 1.3mm to m you will get it as 0.0013
you already have height of the segment and the total area of the segment
so you need to find the radius
0.0013*pie*r^2=2 and then you make r as the subject of the formula and find the radius
d)First you need to get the total area that is 50.4/0.12=420
The next step is to find the perimeter, so as you in the diagram in the previous page that you 2sides of length 0.8
and the below one only is 1.2 and so the total will be 0.8+0.8+1.2=2.8
If you multiply the length which you have to find into the total perimeter, you will get the area
so you have the area and the perimeter you need to find the length
Here you go 420/2.8=150(this is the answer)
Hope this helped.(Good luck for your tommorow's examination)
umm , whats the difference between " shear " and " stretch" and how do you know if its a horizontal shear/stretch or vertical shear/stretch ??
+ do u know any easy method to memorize all the volume and area formulas ?
+ in cumulative frequency qs, when its 1<x<=6 and the cumulative frequency of that interval is 2, do we plot the point (1,2) or (6,2) ??
PLEASE ANSWER ASAP !! i need to pass xD
+ do u know any easy method to memorize all the volume and area formulas ?
+ in cumulative frequency qs, when its 1<x<=6 and the cumulative frequency of that interval is 2, do we plot the point (1,2) or (6,2) ??
PLEASE ANSWER ASAP !! i need to pass xD
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Help's here
And the point you need to plot is (6,2)
Hope that helped !!!
Cheers !!! :good:
Go to: http://www.xtremepapers.me/forums/viewtopic.php?f=24&t=6947Candi00 said:
And the point you need to plot is (6,2)
Hope that helped !!!
Cheers !!! :good:
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hey ok im damn scared, what exactly is specimen 2011?any relationship to the exercises in paper 4 2011? and is there a markscheme for it? if not PLEASE help me in question 9, pls pls pls and HUGE thanks to anybody that'll help.
cheers, waiting :/
cheers, waiting :/
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Mark scheme
Here you go: http://www.xtremepapers.me/CIE/Cambridge IGCSE/0580 - Mathematics/0580_y11_sm_4a.pdf
DeeaAlice13 said:
Here you go: http://www.xtremepapers.me/CIE/Cambridge IGCSE/0580 - Mathematics/0580_y11_sm_4a.pdf
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thanks for the markscheme.. A LOT...but...i still dont get how they did 9 c)....help?pls? , onli i) nd ii) i cant get the 10/3 nd the 32/3 .....
, onli i) nd ii) i cant get the 10/3 nd the 32/3 .....
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Solution
The question states: ‘The total number of one centimetre lines in the first n diagrams is given by the expression’.
2/3 x n^3 + f x n^2 + g x n
So, when n = 1, you have to read through the table in Part 9 (b) and find the number of 1cm lines in the first diagram, which is 4.
Now equate and replace n with 1 at the same time equaling the expression to 4 since this is what it should give.
2/3 x 1^3 + f x 1^2 + g x 1 = 4
2/3 + f + g = 4
f + g = 4 – 2/3
f + g = 10/3
Good Luck !!!
Cheers !!! :good:
The question states: ‘The total number of one centimetre lines in the first n diagrams is given by the expression’.
2/3 x n^3 + f x n^2 + g x n
So, when n = 1, you have to read through the table in Part 9 (b) and find the number of 1cm lines in the first diagram, which is 4.
Now equate and replace n with 1 at the same time equaling the expression to 4 since this is what it should give.
2/3 x 1^3 + f x 1^2 + g x 1 = 4
2/3 + f + g = 4
f + g = 4 – 2/3
f + g = 10/3
Good Luck !!!
Cheers !!! :good: