Mathematics: Post your doubts here!

[MNU]

can anyone explain the 6 b ii question in this paper http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w08_qp_04.pdf

 

[always-smile :)]

Thanks

no need .. u'r welcome

 

[сᴏᴏʟ сяуѕтᴀʟѕ]

Thanks alot! Really helped me

its okie..i am glad if i helped

 

[Yousif Mukkhtar]

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we have 2 cards with no. 1
the probability of it is 2/6 the we solve bii) the first not 1 is 4/6 and the second is 1 is 2/5 ((without replacement)) so the probability that one card is 1 from 2 cards is 4/6 x 2/5
then he wants that 1 card chosed is 1 before the third try ---so we have the probability that 1 from 2 cards has the no.1 . add this to the probability of getting 1 which is 2/6-------->2/6+ 4/6 x 2/5 = 3/5

But, why do we add 2/6?

 

[always-smile :)]

But, why do we add 2/6?

cuz we have 2 card and the probability that e multiplied are for 1 card the first not 1 and the second is 1
so we add the prob. of 1 from all the six cards to the prob. of getting one card of two wih the no. 1

 

[kboss]

can anyone explain the 6 b ii question in this paper http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w08_qp_04.pdf

the height for 22-24 is straight forward just 26/10= 2.6
(the 10 comes from 115/11.5)
then for 10-20... the gap between 10 and 20 = 10
so we divide 10 by 2 which was the gap between 20 and 22
then multiply the ans=5 by 10
that = 50
thus we divide 35/50 = 0.7
the last one is similar...
the gap is 6 this time
so 6/2 = 3
then 10*3= 30
therefore 24/30 = 0.8
hope it helps!!!

 

[Shaaaayan]

can anyone please help me with Q10 ii) on http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w10_qp_42.pdf and the steps on working it out please.

Appreciate it

 

[always-smile :)]

can anyone please help me with Q10 ii) on http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w10_qp_42.pdf and the steps on working it out please.

Appreciate it

can u tell me how did u find i first?

 

[Shaaaayan]

can u tell me how did u find i first?

Simultaneous Equations

 

[always-smile :)]

Simultaneous Equations

with the nth term formula right?

 

[Shaaaayan]

with the nth term formula right?

i didnt use any nth term formula just derived a simultaneous equation from the difference between the terms in c i) but wasnt able to do so in c ii)

 

[always-smile :)]

i make the same steps with the same numbers then the answer is completely different from the mark scheme
what's wrooong ???????// is it my calculator????

 

[kboss]

can anyone please help me with Q10 ii) on http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w10_qp_42.pdf and the steps on working it out please.

Appreciate it

its quite simple:
1) make 3 eqns. (since there are 3 variables)
2) they are:
a) -33+x=y
b) x+y=z
c) y+z=18
3) then replace y with -33 + x
i.e. x + -33 + x= z
and -33 + x + z =18
4) reorder and make it into a simultaneous eqn.
i.e. 2x-z=33 and x +z = 51
5) solve the eqn.
6) u will get the value of x, = 28
7) then put the value of x (28) in the first eqn. (a)
8) u will get y as -5
9) then put the value of y (-5) in the last eqn. (c)
10) u will get z as 23

so there u have it
hope it helps and best of luck!!!

 

[MNU]

the height for 22-24 is straight forward just 26/10= 2.6
(the 10 comes from 115/11.5)
then for 10-20... the gap between 10 and 20 = 10
so we divide 10 by 2 which was the gap between 20 and 22
then multiply the ans=5 by 10
that = 50
thus we divide 35/50 = 0.7
the last one is similar...
the gap is 6 this time
so 6/2 = 3
then 10*3= 30
therefore 24/30 = 0.8
hope it helps!!!

thank u sooo much!!! =D

 

[Shaaaayan]

th

its quite simple:
1) make 3 eqns. (since there are 3 variables)
2) they are:
a) -33+x=y
b) x+y=z
c) y+z=18
3) then replace y with -33 + x
i.e. x + -33 + x= z
and -33 + x + z =18
4) reorder and make it into a simultaneous eqn.
i.e. 2x-z=33 and x +z = 51
5) solve the eqn.
6) u will get the value of x, = 28
7) then put the value of x (28) in the first eqn. (a)
8) u will get y as -5
9) then put the value of y (-5) in the last eqn. (c)
10) u will get z as 23

so there u have it
hope it helps and best of luck!!!

yes it did. thank you!

 

[kboss]

i make the same steps with the same numbers then the answer is completely different from the mark scheme
what's wrooong ???????// is it my calculator????

LOL u seem to be going over the top XD
just do this:
1) leave ur calculator aside (its unnecessary for the question)
2) create the 2 eqns.
a) 2 + d=e
b) d + e =10
3) reorder eqn. (a)
i.e. d-e = -2
4) solve the simultaneous eqn.
5) 2d = 8 is what u should get
6) so d = 8/2 which is 4
7) put the value of d in eqn. (b)
8) therefore 4-e=-2
9) so e should be 4 + 2 = 6
10) answers: d = 4, e = 6

now u can calm down
good luck!!

 

[Yousif Mukkhtar]

Thanks always smile.

I have some other doubt. How to answer question 6 b) iii:
http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w09_qp_4.pdf

 

[always-smile :)]

LOL u seem to be going over the top XD
just do this:
1) leave ur calculator aside (its unnecessary for the question)
2) create the 2 eqns.
a) 2 + d=e
b) d + e =10
3) reorder eqn. (a)
i.e. d-e = -2
4) solve the simultaneous eqn.
5) 2d = 8 is what u should get
6) so d = 8/2 which is 4
7) put the value of d in eqn. (b)
8) therefore 4-e=-2
9) so e should be 4 + 2 = 6
10) answers: d = 4, e = 6

now u can calm down
good luck!!

LOOOL i don't mean this Q i mean with other questions using sine and cosine and things like that
IT IS DRIVING ME CRAAAAAAAAAAAAZY

 

[kboss]

Thanks always smile.

I have some other doubt. How to answer question 6 b) iii:
http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w09_qp_4.pdf

if u hv the paper printed out it would be easier for u.
1) the way to do it is to draw a line parallel to the y - axis from where both the graphs intersect, till the line meets the x axis
2) one should meet the x axis at around -4.5 and the other at around 1.3
3) make an inequality from those 2 points
4) the inequality would look like this : -4.5<x<1.3
5) thats because between -4.5 and 1.3 the graph of y=f(x) is under the graph of y = g(x)
hope it clears any doubts

best of luck !!!

 

[kboss]

LOOOL i don't mean this Q i mean with other questions using sine and cosine and things like that
IT IS DRIVING ME CRAAAAAAAAAAAAZY

oh ok....my bad