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Maths questions thread

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post maths questions here that you don't understand(no matter how simple as long you don't understand :) ) or that you thought were very difficult so others can also test themselves and learn.

i'll prolly be asking alot of questions ;) :p
Paper 1 Qs) Find the smallest possible integer value of n for which 99n is a multiple of 24. [1]
 
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i also got 8 but after trial and error of multiplying 99 by integers until i got something that was completely divisible by 24. it was very long so i thought i might be doing it the wrong way. can you explain the working please?
 

Nibz

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See!
First write down the factors of 24 => 1 x 2 x 2 x 2 x 3;
Then thos of 99 => 1 x 3 x 3 x 11;
Leave the '3 and 1' that are common between the two. You end up with => 2x2x2 = 8!
 
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For multiple of 24 that number needs to have all the prime factors of 24

24=2x2x2x3

99=3x3x11

Now 99 has that 3 but not any of the remaining numbers, those three 2.... So n=2^3
 
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U know what I did refresh the page just to look that if anyone has answered... And Nibz ur reply was not there... Just as I posted it found yours as you did it a fraction of second earlier.... :p
 
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We should have a questions thread like this for other subjects too and have them stickied so that people don't have to make a topic for every question.
 
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Yeah ill make it next year before the exams way before but its up too u guys to do it now take the step just like Math angel has a thread on Maths questions in IGCSE section.
 
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You can post Additional Mathematics and Statistics questions over here, too.
 
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Yeah, good point.

I'd appreciate it if someone could help me solve question 10 of this paper (Add Maths June 2008), it's extremely confusing. :\
 
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leosco1995 said:
Yeah, good point.

I'd appreciate it if someone could help me solve question 10 of this paper (Add Maths June 2008), it's extremely confusing. :\


Not with Add Maths yet.... Help him someone else?
 
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@leosco: Look at the pictures below for the solution. If you don't understand anything tell me. :)
 

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Thanks man.. you're a genius. :) I understood the first 4 parts very well but I'm still having problems with (v), it's quite confusing. A little explanation on that would be nice. :)
 

Nibz

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Zabardast Hamidali391! :)
Look lesco: If the both the ship and the submarine are to meet at a point! On that particular point they would have same Final Position! Agree?
And we have => starting position + (distance) = final position.. put distance = velocity x time
as the final position of both would be same :
=> final position of ship = final position of submarine
=> starting position of ship + (ship's velocity) x time = starting position of submarine + (submarine's velocity x time)
put the values
=> ( 2 3) + ( 15 15)t = (47 -27 ) + (0 25)t
=> (2+15t 3+15t) = (47 -27+25t) =====> (S)
=> 2+15t = 47 and 3+15t = -27+25t
solve this for t and u'l get t=3 for both like Hamidali's answer! which means that they meet at t= 3hrz
to find this postion ..simply substitue t=3 in =====> (S) either of them
=> (2 + 45 3 + 45)
=> (47 48)
=> 47i + 48j

Hope this clears the confusion!
 
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Oh... I see. Yeah, after your explanation and me going through hamidali's solution again I think I got it. It's made my concepts about this very clear and I just managed to do a similar question (June 2009 P1 Q9) without any problems. :)
 

Nibz

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It's simple factorisation!
6p^2 -4p + 3p - 2
(2p + 1) (3p - 2) Ans/.
 
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i kno but neither 4 or 3 are factors of 2 that multiply together to make 2. atleast thats how i thought factorization by splitting the middle value was done.

kinda simple and embarrassing but i have a doubt here so... :(
can u explain please?
 
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