Author Topic: Math paper4 Jun. 2009 help  (Read 7675 times)

Offline moon

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Math paper4 Jun. 2009 help
« on: October 20, 2009, 12:37:35 pm »
plz, can any body help me in math paper 4 Jun.09 question 11 d (i,ii) ??? ??? ???

Offline wassup

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Re: Math paper4 Jun. 2009 help
« Reply #1 on: October 20, 2009, 12:58:00 pm »
11(d)(i) (n(n+1))/2 + ((n+1)(n+2))/2
         =(n^2+n+n^2+2n+n+2)/2
         =(2n^2+4n+2)/2
         =(2(n^2+2n+1))/2
         =n^2+2n+1
         So, n^2+2n+1=(n+1)^2

       (ii) (n+1)^2=3481
             n^2+2n+1=3481
             n^2+2n-3480=0
             n=(-2+-underroot(2^2-4*1*-3480))/2
             n=(-2+-underroot(13924))/2
             n=(-2+-118)/2
             So, n=58, -60
             n=58 (It can't be negative)
             Two terms will be n, (n+1)
             =58, 59
             (58(58+1))/2=1711
             (59(59+1))/2=1770
             Answer=1711 and 1770

Offline moon

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Re: Math paper4 Jun. 2009 help
« Reply #2 on: October 20, 2009, 01:02:13 pm »
thanks a lot :) :)

Offline wassup

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Re: Math paper4 Jun. 2009 help
« Reply #3 on: October 20, 2009, 01:03:00 pm »
welcome

Offline moon

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Re: Math paper4 Jun. 2009 help
« Reply #4 on: November 02, 2009, 04:51:10 pm »
11(d)(i) (n(n+1))/2 + ((n+1)(n+2))/2
         =(n^2+n+n^2+2n+n+2)/2
         =(2n^2+4n+2)/2
         =(2(n^2+2n+1))/2
         =n^2+2n+1
         So, n^2+2n+1=(n+1)^2

       (ii) (n+1)^2=3481
             n^2+2n+1=3481
             n^2+2n-3480=0
             n=(-2+-underroot(2^2-4*1*-3480))/2
             n=(-2+-underroot(13924))/2
             n=(-2+-118)/2
             So, n=58, -60
             n=58 (It can't be negative)
             Two terms will be n, (n+1)
             =58, 59
             (58(58+1))/2=1711
             (59(59+1))/2=1770
             Answer=1711 and 1770

Please explain both parts still didn't get it ...

Offline moon

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Re: Math paper4 Jun. 2009 help
« Reply #5 on: November 02, 2009, 04:57:29 pm »
Please I need help in JUNE 2009 Q2(d)

Offline wassup

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Re: Math paper4 Jun. 2009 help
« Reply #6 on: November 02, 2009, 05:17:20 pm »
Please explain both parts still didn't get it ...

It's given that the nth term is n(n+1)/k where k comes out to be 2.

so, (n+1)th term will be (n+1)(n+2)/k where k is 2. (v'll add 1 because it's n+1 and not n)

so v'll do this now:
 
         (n(n+1))/2 + ((n+1)(n+2))/2
         =(n^2+n+n^2+2n+n+2)/2
         =(2n^2+4n+2)/2
         =(2(n^2+2n+1))/2
         =n^2+2n+1
         So, n^2+2n+1=(n+1)^2

Now v have to find 2 consecutive terms with sum of 3481.

v got frm d above ans that the sum of two consecutive nos. is (n+1)^2

now v'll use that and make it equal to 3481 as given below

             (n+1)^2=3481
             n^2+2n+1=3481
             n^2+2n-3480=0
             n=(-2+-underroot(2^2-4*1*-3480))/2
             n=(-2+-underroot(13924))/2
             n=(-2+-118)/2
             So, n=58, -60
             n=58 (It can't be negative)
             Two terms will be n, (n+1)
             =58, 59
             (58(58+1))/2=1711
             (59(59+1))/2=1770
             Answer=1711 and 1770

hope u get it now!! :)

nid404

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Re: Math paper4 Jun. 2009 help
« Reply #7 on: November 02, 2009, 05:18:53 pm »
Please I need help in JUNE 2009 Q2(d)

The total count was 148 for 50 rolls.

148=50X2.96

so 10 rolls.....count will be 10x...if x is considered to be the mean

together for sixty rolls

148+10x/60=2.95
therefore x=2.9

Offline @d!_†oX!©

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Re: Math paper4 Jun. 2009 help
« Reply #8 on: November 02, 2009, 05:20:04 pm »
see its quite simple:

mean for 50 rolls is 2.96(found in c part)
mean for 60 rolls is 2.95(given)

so to find the mean for the extra 10 throws....:

[(60*2.95)-(50*2.96)]/10
=>29/10
=>2.9

do u get it bro???
AAL IZZ WELL!!! ;)

Offline holychalice

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Re: Math paper4 Jun. 2009 help
« Reply #9 on: November 02, 2009, 05:20:17 pm »
c u get 2.96 as ur mean i guess in c part  outta 50 times
d mean after 50 +10=60 is 2.95

so (148+10x)/60 = 2.95
148+10x = 2.95*60
10x = 177-148
x=29/10
x=2.9
Shower Shortzzzz ..... For the MAN who has nothing to hide .... But still wants to !!!!

Offline cooldude

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Re: Math paper4 Jun. 2009 help
« Reply #10 on: November 02, 2009, 05:21:17 pm »
Please I need help in JUNE 2009 Q2(d)

im gonna use | for the summation (sigma) symbol
we knw that Mean=|fx/|f

therefore the sum of the 60 dice rolls is=Mean*|f
=> 2.95*60=177
the sum of the 50 dice rolls-Mean*|f
=> 2.96 (found in part c)*50=148
therefore the sum of the last 10 rolls=177-148=29
therefore the mean of the 10 rolls=|fx/|f
=>29/10=2.90
okay temme if u dnt get nething?

          

Offline SGVaibhav

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Re: Math paper4 Jun. 2009 help
« Reply #11 on: November 02, 2009, 05:23:11 pm »
why does it have to be june 2009?

Offline wassup

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Re: Math paper4 Jun. 2009 help
« Reply #12 on: November 02, 2009, 05:25:10 pm »
Please I need help in JUNE 2009 Q2(d)

c) Mean score for 50 rolls = ((15*1)+(10*2)+(7*3)+(5*4)+(6*5)+(7*6))/50
                                   = 2.96

d) Mean score for 60 rolls = 2.95

60*2.95 = 177
50*2.96 = 148

So, 177-148 = 29

Mean of 10 rolls = 29/10
                      = 2.9


  

nid404

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Re: Math paper4 Jun. 2009 help
« Reply #13 on: November 02, 2009, 05:26:22 pm »
lol 5 answers to that question........haha

Offline @d!_†oX!©

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Re: Math paper4 Jun. 2009 help
« Reply #14 on: November 02, 2009, 05:26:33 pm »
man!!!!!
after so many explanations anyone can understand it!!!
this shows how eager ppl at sf are to help others!!!
Thanks alot cooldude, nid, manbir, wassup and me too!!!

By the way moon....tll us if u didn't understand anything!!u'll have another 10 ppl tryina explain it to u!!
AAL IZZ WELL!!! ;)