Maths, Addmaths and Statistics: Post your doubts here!

[Hassan Kazmi]

Compound Interest is in the CIE O Level Mathematics D (Calculator Version) syllabus. But that is a relatively simple compared to the questions above. For compund interest, you just have to use the I = PRT formula again and again using the modified P (principal) which means adding the interest to the principal to calculate next year's interest. I'm not sure how compound interest is tested in the O Level Statistics exam.

pikachu, if you could post a question?

For invariant lines, the O Level Mathematics syllabus will include only lines which are parallel to the x-axis or y-axis. These can be identified by looking at the shape. The formula to calculate the exact equation of the invariant line (whether horizontal, vertical or diagonal) is a bit complex. It would be easier to explain if you could post a question.

 

[Young Stunner]

Can anyone please help! For the first part

 

[Saad Mughal]

Can anyone please help! For the first part

This year,
Ada = x years , Bill = y years,
Last year,
y - 1 = 6 (x - 1)
y - 1 = 6x - 6
y = 6x - 5

 

[Young Stunner]

Oh

This year,
Ada = x years , Bill = y years,
Last year,
y - 1 = 6 (x - 1)
y - 1 = 6x - 6
y = 6x - 5

OK Jazakallah u Khair bro...

 

[Young Stunner]

Please Help

 

[Ishaq]

I have questions of Permutations in Admaths,
(1)Find the number of 4-letter code-words that can be made from the letters of the word ADVANCE
(i)using both of the A's.

(2)Eight people go the theatre and sit in a particular group of eight adjacent reserved seats in the front row. three of the eight belong to one family and sit together.
(i)If the other five people donot mind where they sit, find the number of possible seating arrangements for all eight people.(I've done this one, but I thought you should know this as it relates to the second part).
(ii)If the other five people donot mind where they sit, except two of them refuse to sit together, find the number of possible seating arrangements for all eight people.
^Please help me with this. Thanks

 

[Hassan Kazmi]

1) Simply doing 7! gives us all combinations, but, half of them are repetitive due to the 2 As. So, divide by 2! to remove those repetitions. = 7!/2!

2) Do this by first thinking about the possible combinations when the two people sit together, then subtract it from the total.
So, when there are two people out of those five who always want to sit together, it is 3! x 4! (considering the two as one block). The combinations without any restrictions, as in part (i), are 3! x 5!. So, the answer is (3! x 5!) - (3! x 4!) = 576.

 

[Ishaq]

1) Simply doing 7! gives us all combinations, but, half of them are repetitive due to the 2 As. So, divide by 2! to remove those repetitions. = 7!/2!

2) Do this by first thinking about the possible combinations when the two people sit together, then subtract it from the total.
So, when there are two people out of those five who always want to sit together, it is 3! x 4! (considering the two as one block). The combinations without any restrictions, as in part (i), are 3! x 5!. So, the answer is (3! x 5!) - (3! x 4!) = 576.

1) In doing that, the answer although will contain those codes consisting of both A's but will also consist those excluding A's, so what i am thinking about this is to find the total then subtracting from it the chances of there being either no A's or either one A.
2) In part (i), I have further multiplied 720 by 6 as it is the no. of ways the eight people can sit. and in part two I don't get how the chances of two people sitting together are (3!x4!). Can you please explain??

 

[Hassan Kazmi]

No, wait, I just overlooked the fact that we have to make 4-digit codes! Look at how this is done:

All possible 4-letter codes: 7p4 (fair enough? although this contains those repetitive codes) = 840
4-letter codes containing both As: 2p2 x 5p2 (to make sure the 2 As are selected) = 40 (divide by 2 for repetitive cases) = 20
So, 840 - 20 = 820 is the answer.

My solution to the second question is also flawed. Let me take a look again.

 

[Hassan Kazmi]

Yes, you are right for the second question.

For the first part the answer is 3! x 6! because there are six blocks of people, like this, where one block contains 3 people: [1][1][1][1][1][3] = 4320

For the second part, again, it will be 3! x 2! x 5! because there are five blocks now: [1][1][1][2][3] = 1440

So the answer to the second part is 4320 - 1440 = 2880.

 

[AhsanAfzal]

explain part (b) pls

 

[Hassan Kazmi]

Since HCF of N and 500 has 5^2 in it, N must have 5^2 so q=2.

Since LCM of N and 500 has 2^3 in it, N must have 2^3 because 500 only has 2^2, so p=3.

Since LCM of N and 500 has 7 in it, N must also have a 7 as 500 does not have a 7, so r=1.

 

[Mohammad Farzanullah]

from where can we get maths marking scheme of 90s?

 

[Hassan Kazmi]

If you have the questions, post them here and maybe we can help solve them.

I'm not sure where to get the mark schemes.

 

[AhsanAfzal]

Since HCF of N and 500 has 5^2 in it, N must have 5^2 so q=2.

Since LCM of N and 500 has 2^3 in it, N must have 2^3 because 500 only has 2^2, so p=3.

Since LCM of N and 500 has 7 in it, N must also have a 7 as 500 does not have a 7, so r=1.

thnks. got it

 

[AhsanAfzal]

No, wait, I just overlooked the fact that we have to make 4-digit codes! Look at how this is done:

All possible 4-letter codes: 7p4 (fair enough? although this contains those repetitive codes) = 840
4-letter codes containing both As: 2p2 x 5p2 (to make sure the 2 As are selected) = 40 (divide by 2 for repetitive cases) = 20
So, 840 - 20 = 820 is the answer.

My solution to the second question is also flawed. Let me take a look again.

i dont get this one.. why isnt the answer 40?? the question asks for having both A's in the code

 

[Ishaq]

No, wait, I just overlooked the fact that we have to make 4-digit codes! Look at how this is done:

All possible 4-letter codes: 7p4 (fair enough? although this contains those repetitive codes) = 840
4-letter codes containing both As: 2p2 x 5p2 (to make sure the 2 As are selected) = 40 (divide by 2 for repetitive cases) = 20
So, 840 - 20 = 820 is the answer.

My solution to the second question is also flawed. Let me take a look again.

Well actually you've again misread the question, it's asking for the no of chances WITH BOTH THE A's. By subtracting what we just did, we actually removed it.
What I'm thinking is that we are surely gonna select the 2 A's. so can't it be like, 2C2(cause 2 A's is necessary) x 5C2 (the other two) x 4! (need random 4-letter code) . Then we divide the answer 240 by 2! to remove repetitions, so the answer would become 120, no ?

 

[Ishaq]

Yes, you are right for the second question.

For the first part the answer is 3! x 6! because there are six blocks of people, like this, where one block contains 3 people: [1][1][1][1][1][3] = 4320

For the second part, again, it will be 3! x 2! x 5! because there are five blocks now: [1][1][1][2][3] = 1440

So the answer to the second part is 4320 - 1440 = 2880.

Thanks for this one

 

[Hassan Kazmi]

Well actually you've again misread the question, it's asking for the no of chances WITH BOTH THE A's. By subtracting what we just did, we actually removed it.
What I'm thinking is that we are surely gonna select the 2 A's. so can't it be like, 2C2(cause 2 A's is necessary) x 5C2 (the other two) x 4! (need random 4-letter code) . Then we divide the answer 240 by 2! to remove repetitions, so the answer would become 120, no ?

Ah, damn it! You're right. If we want only codes that contain both As, your method is fine. Another way of doing it is using the permutation:

2P2 x 5P2 x 3! (just like the blocks in the other question, it will be like [A|A][?][?] where these letters need to randomise). Then divide by 2! for repetitions. =120.

 

[Ishaq]

Ah, damn it! You're right. If we want only codes that contain both As, your method is fine. Another way of doing it is using the permutation:

2P2 x 5P2 x 3! (just like the blocks in the other question, it will be like [A|A][?][?] where these letters need to randomise). Then divide by 2! for repetitions. =120.

Yeah Thanks