Mathematics: Post your doubts here!

[drewrock]

can someone provide me pure mathematics as level paper held on 7 may...

 

[Alice123]

Okay, this is a tricky one.
View attachment 25529

See the picture. The red seats highlight the ones that Mrs. Lin can occupy. She has to sit directly behind a student, so if she occupies those seats then it's possible for a student to be placed in front of her. The reason I kept one seat in the second row uncolored is because one of the seats in the front row is occupied my Mrs. Brown. A student can only take two seats in the front row.
So for Mrs. Lin the possible combinations are 10P1
Mrs. Brown has to sit in one of the front seats so 3P1
She can sit behind any of the 5 students, so it's 5P1, the reason you don't count the possible seats for the student is because it doesn't matter where they are placed as long as one of them is in front of Mrs. Lin
And finally after Mrs. Lin, a student and Mrs. brown has been placed there are 9 remaining passengers and 11 remaining seats. There is no restrictions. So it's 11P9

Total no. of ways = 10P1 * 3P1* 5P1 * 11P9

If all 12 passangers are seated randomly = 14P12
so probability = 10P1 * 3P1* 5P1 * 11P9/ 14P12 =.0687

Hope you get this!

very well explained.... thankyou sooooo much littlecloud11
!!! syed1995 A star Rutzaba ............U can have a look at this

 

[syed1995]

Okay, this is a tricky one.
View attachment 25529

See the picture. The red seats highlight the ones that Mrs. Lin can occupy. She has to sit directly behind a student, so if she occupies those seats then it's possible for a student to be placed in front of her. The reason I kept one seat in the second row uncolored is because one of the seats in the front row is occupied my Mrs. Brown. A student can only take two seats in the front row.
So for Mrs. Lin the possible combinations are 10P1
She can sit behind any of the 5 students, so it's 5P1, the reason you don't count the possible seats for the student is because it doesn't matter where they are placed as long as one of them is in front of Mrs. Lin
Mrs. Brown has to sit in one of the front seats so 3P1
And finally after Mrs. Lin, a student and Mrs. brown has been placed there are 9 remaining passengers and 11 remaining seats. There is no restrictions. So it's 11P9

Total no. of ways = 10P1 * 3P1* 5P1 * 11P9

If all 12 passangers are seated randomly = 14P12
so probability = 10P1 * 3P1* 5P1 * 11P9/ 14P12 =.0687

Hope you get this!

Some help with Part ii as well.. Since I don't understand why the mark scheme is ignoring the other window seats... and just doing a 3P3*4P4*5P5 ..

Would really appreciate it ...

Thanks

 

[syed1995]

ok guys heres the deal
6 ii 5 students sit anywhere by the window... 5! =120
3 business men sit anywhere= 3!= 6
now for the married couples... there are 2 couples =2 two seats each =2 and 3 seats remaining for each couple =3p2
3p2 *2*2= 24
5!=120
3!=6

24*6*120 = 17280

There are 9 windows.. why are you ignoring the other windows and doing a 5! instead of 9P5?

Btw love your avatar

 

[littlecloud11]

Some help with Part ii as well.. Since I don't understand why the mark scheme is ignoring the other window seats... and just doing a 3P3*4P4*5P5 ..

Would really appreciate it ...

Thanks

The students cannot take the window seats in the first row because the 3 business people have to be seated there. Each married couple must be seated in a different row, but they have to be on the same side of the isle. So one of the couples have to take two window seats so that both couples are in different rows. So you see, only 5 window seats remain, and the students have to take these 5 seats in any order. So it's 5! and not 9P5.

 

[Alice123]

View attachment 25598

The students cannot take the window seats in the first row because the 3 business people have to be seated there. Each married couple must be seated in a different row, but they have to be on the same side of the isle. So one of the couples have to take two window seats so that both couples are in different rows. So you see, only 5 window seats remain, and the students have to take these 5 seats in any order. So it's 5! and not 9P5.

Thanks for solving all our problems....

 

[littlecloud11]

Thanks for solving all our problems....

Glad to help =)

 

[Farru]

can anyone please further solve it, making Ysq the subject please..
1/2 ln(Ysq + 4)=ln x + 1/2 ln 4

PS I know it simple, but not getting the right ans. :$

 

[Pie-man]

Sorry to bother you but can you please solve it for me ? I didn't get you..

bother? you aren't bothering me

so first thing is writing the forces so that you wouldn't get confused by them, as the 5.5 N and the W N are pulling the particle upwards.
then draw a line in the middle of the right angle to make 2 letter Z (one from AP1X to the line and the other from CP2X to the line) and label the angles AP1X with a symbol and CP2X with another symbol, so now you have the angle AP1X equal to the left side of the line you drew and CP2X is the same as the right side of the line.

so lami says that 7.3/sin(90) = 5.5/sin(AP1X) = W/sin(CP2X)

so using cross-multiplication (5.5 x sin(90))/7.3 = 48.9 (angle CP2X)

angle AP1X = 90 - 48.9 = 41.1

so angle AP1X= 41.1


then to get W you do the same but you use the angle 41.1 so
(sin(41.1) x 7.3)/sin(90) = 4.8

so W= 4.8

did you get it now? if you want me to scan the paper I solved it on i don't mind

 

[syed1995]

View attachment 25598

The students cannot take the window seats in the first row because the 3 business people have to be seated there. Each married couple must be seated in a different row, but they have to be on the same side of the isle. So one of the couples have to take two window seats so that both couples are in different rows. So you see, only 5 window seats remain, and the students have to take these 5 seats in any order. So it's 5! and not 9P5.

Ah that explains it.. I wasn't doing it on paper.. so that's why I was getting confused.. Thanks!

 

[Pie-man]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_4.pdf

@Obsidian Fl1ght
@Pie-man

its question 6(iii) ?

JazakAllah

well this was a fun question

the main thing is to know that air resistance is present the whole time P is in the air. (well that was my mistake then i started thinking about my next physics exam and drifted away thinking about the units i want to revise which made me notice) i'm so awesome xD

so what i did was get the total work done against the particle while moving upwards making my Initial Energy from point O and final energy to point of max height.

which gave me 45= 45.312 - Wr
so Wr = 0.312

then made the initial energy from the max height and final to ground.

so i got 43.2= 45 - Wr

so Wr = 1.8

then i just added them to get the total resistance while particle P was in the air so i got 1.8 + 0.312 = 2.11 J

 

[A star]

computing people join the thread https://www.xtremepapers.com/commun...9691-post-your-doubts-here.25670/#post-530873

 

[Farru]

anyone please??

 

[Ebby_SaLeem]

Can someone explain me union and intersection of the compliements of sets, A and B in this case.
PLEASE! With proper and clear diagrams... thanks!

 

[soul]

can someone please explain to me the link between work done by driving force, work one by resistance, k.e, p.e

 

[Ebby_SaLeem]

can someone please explain to me the link between work done by driving force, work one by resistance, k.e, p.e

isn't this a MATHEMATICS thread?

 

[Farru]

isn't this a MATHEMATICS thread?

Its Mathematics mechanics

 

[Ebby_SaLeem]

Omg! I just scrolled up and found out this is for A levels! MY BAD! HAHA!
LMAO! Next year I'll be posting on this page!

 

[Pie-man]

can someone please explain to me the link between work done by driving force, work one by resistance, k.e, p.e

the link from the rule "Final Energy = Initial Energy + Driving force - Resistance force" ? or the one "Driving force - Resisting force = mass x acceleration"?

so you get the final and initial energy using "K.E. [1/2 mass x velocity(squared) ] + P.E. (mass x gravity [10] x height)" and substituting the mass, height and velocity as given at each point.

if given a driving force or resistant force you put them in the equation "F.E. = I.E. + DF - RF"

 

[Pie-man]

Omg! I just scrolled up and found out this is for A levels! MY BAD! HAHA!
LMAO! Next year I'll be posting on this page!

good luck with that O-Level seems like a piece of cake after taking the AS

anyone please??

well i'm doing the AS-Level but i'll ask one of my friends ... when any of them wake up [i woke up early to study]