Mathematics: Post your doubts here!

[Stefan Salvatore]

Hi, in dire need of some urgent help:

Q: Find how many terms there are in the geometric sequence using logarithms:
-1, 2, -4, 8, ....., -16777216

a= -1

r= -2

nth term (last term) = ar^n-1

 

[Saad the Paki]

Does Anyone the w 2015 examiner report for maths?

 

[Rizwan Javed]

Does Anyone the w 2015 examiner report for maths?

http://papers.gceguide.com/A Levels/Mathematics (9709)/9709_w15_er.pdf

 

[Saad the Paki]

http://papers.gceguide.com/A Levels/Mathematics (9709)/9709_w15_er.pdf

Oh wow I searched the website for so long, must have missed it. Thanks

 

[TariqBhai]

Spiceangel The Sarcastic Retard farhan141 Awesome12 *Le me* Akshajistari mistique_bee Copy Cat muhammadali233 qwertypoiu

 

[farhan141]

Spiceangel The Sarcastic Retard farhan141 Awesome12 *Le me* Akshajistari mistique_bee Copy Cat muhammadali233 qwertypoiu

 

[Copy Cat]

(iii)

 

[funky brat]

Anyone please try to solve these parts and explain.
ON13 Q6iv V62
MJ12Q5iii V62
ON11Q3b V62
ON11Q2ii V62
MJ11Q4iii V62
MJ09Q4iii V62

 

[Rizwan Javed]

Anyone please try to solve these parts and explain.
ON13 Q6iv V62

REMEMBRANCE

Here you are told that the selection does not contain Rs and Ms, so eliminate these letters from the word; you'll be left with:
EEE BNCA

Now the selection can contain either 3Es OR 2 Es. So consider these possibilities separately:
1) Selection contains 3 Es.

*EEE
^the asterisk here can be replaced by any of the other 4 letters, so no. of ways this can be done is 4C1

2) Selection contains 2Es

**EE
^again the asterisks can be replaced by any two letters from B,N,C or A. So no. of possible selections are: 4C2

Add these two results, you'll get the total no. of possible selections, which contain atleast 2 Es: 4C1 + 4C2 = 10 Ans.

 

[Rizwan Javed]

MJ12Q5iii V62

There are 8 questions in Section A, and 3 questions in Section B.

First consider the situation, that both Q1 & Q2 are included. Then you'll have to select 4 more questions from the remaining 9 questions. This can be done in 9C4 ways.

Now, consider the situation that neither Q1 is included nor Q2 is included, so you'll have to choose all of the 6 questions from the remaining 9 Questions. This will be done in 9C6 ways.

Sum these two, you'll get the answer: 9C4 + 9C6 = 210 selections.

 

[funky brat]

REMEMBRANCE

Here you are told that the selection does not contain Rs and Ms, so eliminate these letters from the word; you'll be left with:
EEE BNCA

Now the selection can contain either 3Es OR 2 Es. So consider these possibilities separately:
1) Selection contains 3 Es.

*EEE
^the asterisk here can be replaced by any of the other 4 letters, so no. of ways this can be done is 4C1

2) Selection contains 2Es

**EE
^again the asterisks can be replaced by any two letters from B,N,C or A. So no. of possible selections are: 4C2

Add these two results, you'll get the total no. of possible selections, which contain atleast 2 Es: 4C1 + 4C2 = 10 Ans.

Oh right. Thanks.

 

[funky brat]

for the part when both questions are to be chosen from part A, we have 4 options to choose right. Why aren't these the correct options?
4C4*3C0+ 4C3*3C1+ 4C2*3C2+ 4C1*3C3 since questions from part B are also chosen at the same time.

 

[Rizwan Javed]

There're two mistakes. Firstly, you aren't considering both of the scenarios i.e. when both questions are included and when neither of them is included. You are missing the scenario when none of those two questions are selected. Secondly, When two first questions are chose, you are left with 6 options to choose from part A, not 4.

So going with your method, the correct solution would have been like this:
(6C4*3C0+ 6C3*3C1+ 6C2*3C2+ 6C1*3C3) + (6C6 * 3C0 + 6C5 * 3C1 + 6C4 * 3C2 + 6C3 * 3C3)
= 210 Ans.

Get it?

 

[psychiatrist]

w11 qp 33 Q8 i and ii

 

[The Sarcastic Retard]

w11 qp 33 Q8 i and ii

qwertypoiu camscanner zindabad

 

[Copy Cat]

(i)

 

[funky brat]

There're two mistakes. Firstly, you aren't considering both of the scenarios i.e. when both questions are included and when neither of them is included. You are missing the scenario when none of those two questions are selected. Secondly, When two first questions are chose, you are left with 6 options to choose from part A, not 4.

So going with your method, the correct solution would have been like this:
(6C4*3C0+ 6C3*3C1+ 6C2*3C2+ 6C1*3C3) + (6C6 * 3C0 + 6C5 * 3C1 + 6C4 * 3C2 + 6C3 * 3C3)
= 210 Ans.

Get it?

Yeah I know that was incomplete, I was talking about the first scenario only and I think I misread it or like didn't pay attention and thought we had to choose 6. Thanks anyway.

 

[farhan141]

In part 1 i got the answer to number but the probability is confusing. Also why is variance being treated as p in part 2?

 

[farhan141]

View attachment 60531
(i)

 

[slisjunknown]

Can someone please try these parts and explain
ON 2011 p33 q 3 ii[I get two more values that are not in the ms]
ON 2011 p33 q6 ii
ON 2011 p33 8 ii and iii
ON 2011 p33 q9i[why do we use scalar product if its parallel don't we take ratios?] and q9 iii
ON 2011 p33 q10

that's almost the whole paper but can't help being dumb