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Math paper4 Jun. 2009 help

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moon:
Please I need help in JUNE 2009 Q2(d)

wassup:

--- Quote from: moon on November 02, 2009, 04:51:10 pm ---Please explain both parts still didn't get it ...

--- End quote ---

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:

--- Quote from: moon on November 02, 2009, 04:57:29 pm ---Please I need help in JUNE 2009 Q2(d)

--- End quote ---

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

@d!_†oX!©:
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???

holychalice:
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

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