Mathematics: Post your doubts here!

[hamksha]

i cant solve ..november 2011 version 21 (Paper 2) ..question no ,18 ..can any body show how can i get these answers..

another thing i need to ask ur help about is paper 2 november 2011 version 22 quesion no 1.....every time i get 34 ....i divide 8.5 hours(opening period for the port)bu 0.25hours (interval which us leaves .....the result is 34 ..while on ms they say 35!!!
any help??????????? plzzzzzzzzzzz.

 

[slayer]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s04_qp_2.pdf q 15 PLEASE

 

[GetSomeLife]

i cant solve ..november 2011 version 21 (Paper 2) ..question no ,18 ..can any body show how can i get these answers..

V 21
Q 18:
a) Just put '23' in the place of 'n' in the formula... use your calculator and you'll get the right answer = 4324

b) i) They already gave you the formula. Use it along with the formulas on the top of the page in your calculator and you'll get your answers.
ii) If you find the values of U1, U2 and U3... you'll notice they go like 4, 9 and 16 respectively. These numbers are all SQUARES. Furthermore, the first square is 4 when it should be 1 IF you use the formula n^2
Keeping that in mind, you have to make a slight modification to the formula and write it as (n+1)^2.

c) This one requires some brain:
Notice that:

V1 = 4*T1
V2 = 4*T2
V3 = 4*T3

So you can say

Vn = 4*Tn
Tn as given above is (1/6)n(n+1)(2n+1)

So replace it in the equation:
Vn = 4*Tn
Vn = 4* (1/6)n(n+1)(2n+1)
Vn = (4/6)n(n+1)(2n+1)
Simplify and you get your answer:

(2/3)n(n+1)(2n+1)

another thing i need to ask ur help about is paper 2 november 2011 version 22 quesion no 1.....every time i get 34 ....i divide 8.5 hours(opening period for the port)bu 0.25hours (interval which us leaves .....the result is 34 ..while on ms they say 35!!!
any help??????????? plzzzzzzzzzzz.

Hours = 8.5
Bus leaves every 0.25 hours
Number of times bus leaves = 8.5/0.25 = 34
In the end, it leaves ONCE AGAIN at 17.30

34 + 1 = 35

 

[GetSomeLife]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s04_qp_2.pdf q 15 PLEASE

It's a question of similarity:


Area of smaller triangle/area of larger triangle = [length of smaller triangle/length of larger triangle]^2
11/99 = (BC/12)^2
11/99 = BC^2/144
BC^2 = 1/9 * 144 = 16
BC = 16^(1/2)
BC = +- 4

Since length can't be negative
BC = 4 cm

 

[slayer]

It's a question of similarity:


Area of smaller triangle/area of larger triangle = [length of smaller triangle/length of larger triangle]^2
11/99 = (BC/12)^2
11/99 = BC^2/144
BC^2 = 1/9 * 144 = 16
BC = 16^(1/2)
BC = +- 4

Since length can't be negative
BC = 4 cm

Thanks for the reply really appreciate it but Iam not sure if the are of ABC is 11 shouldnt the rest be 88 :/

 

[GetSomeLife]

Thanks for the reply really appreciate it but Iam not sure if the are of ABC is 11 shouldnt the rest be 88 :/

Which two triangles are similar here?

ABC and APQ. So, we'll take the area of ABC an APQ! (the WHOLE of APQ) If we do what YOU say, we'll be talking the area of the quadrilateral BCQP. That would be wrong

 

[slayer]

Which two triangles are similar here?

ABC and APQ. So, we'll take the area of ABC an APQ! (the WHOLE of APQ) If we do what YOU say, we'll be talking the area of the quadrilateral BCQP. That would be wrong

ohhhhhhh Sorry I feel a bit stupid now Thank you

 

[natsume hyuga]

how do ya do the last question part b ii for this paper http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s11_qp_23.pdf
Also for question 5, did u do sum calculations or wat to get the answer?

 

[Mohammad Ahmad]

thxxx a bunch !!
i have another querstion .......
4^2n-3= 16
find n ?!

 

[suhaib05]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w08_qp_02.pdf

Can someone please help me with Q5? I do not understand how to answer the question once I rotate the square, and the mark scheme is not clear.

thxxx a bunch !!
i have another querstion .......
4^2n-3= 16
find n ?!

16 can be written as 4^2.
If 4^(2n-3) = 4^2, then the powers should be equal, i.e, 2n-3=2.
Add three on both sides. You'll end up with 2n=5.
Divide 2 on both sides of the equation.
n=2.5

To double check your answer, substitute 2.5 into the original equation: 4^[(2x2.5)-3] = 16.

 

[Mohammad Ahmad]

oh thxx btw ur answer is quite to that quetion is quite simple .... u should only rotate the the figure keeping the point B Unchanged

 

[Addoozz]

can anyone tell me the steps for http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s11_qp_41.pdf question 7(all of it)

 

[suhaib05]

oh thxx btw ur answer is quite to that quetion is quite simple .... u should only rotate the the figure keeping the point B Unchanged

Well, the rotation part is simple. It's the locus part which I'm not sure what to do with. And you're welcome.

 

[niassu]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_w04_qp_4.pdf
here Q 10 (d)


please guuuuuuyyys help with question is it possible that noone can do it here

I am waiting please do help urgently .

 

[bluefa1con]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w08_qp_02.pdf

Can someone please help me with Q5? I do not understand how to answer the question once I rotate the square, and the mark scheme is not clear.


16 can be written as 4^2.
If 4^(2n-3) = 4^2, then the powers should be equal, i.e, 2n-3=2.
Add three on both sides. You'll end up with 2n=5.
Divide 2 on both sides of the equation.
n=2.5

To double check your answer, substitute 2.5 into the original equation: 4^[(2x2.5)-3] = 16.

join BD this will be the radius
now draw 90 degree angle from BD
place one end of compass at B open till D
draw an arc to cut the 90 degree angle u made earlier
the point of intersection will be D'

 

[GlitteringLights]

this is w11 qp 21 # 11(b)
how do we solve this (without a calculator)
actually i have no idea how to do it at ALL

 

[Zeerik]

http://www.xtremepapers.com/papers/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_s04_qp_2.pdf
Question Number 18,15 and I have my paper 4 on the 14th What Chapters Should i Revise For My paper 2 is on 8th im almost Ready for it but not Paper 4 Its Tooo Tough!

 

[bluefa1con]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s04_qp_2.pdf
Question Number 18,15 and I have my paper 4 on the 14th What Chapters Should i Revise For My paper 2 is on 8th im almost Ready for it but not Paper 4 Its Tooo Tough!

comeon man
question 15 is simple as (ratio of sides)^2=(ratio of area)
question 18 is just a right angled triangle with bearings part question
for part a use Pythagoras theorem

for part b
90 - (tan(78/83)

good luck for your exam and pray for me i am trying for world ranking
keep these questions coming

 

[bluefa1con]

this is w11 qp 21 # 11(b)
how do we solve this (without a calculator)
actually i have no idea how to do it at ALL

they just want the inverse of the matrix
the given syntax is just to trick you

the determinant is (3*2)-(4*1)=2
final answer will be to simplyfy 1/2
(3 -1
-4 2)

so you will have
(1.5 -0.5
-2 1)

 

[GlitteringLights]

they just want the inverse of the matrix
the given syntax is just to trick you

the determinant is (3*2)-(4*1)=2
final answer will be to simplyfy 1/2
(3 -1
-4 2)

so you will have
(1.5 -0.5
-2 1)

*faceplam*
it was just an inverse!
i feel dumb