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You guys should keep it up!!!!
This time around the questions would be much more difficult and technical to that of yesterday..You all have got what it takes to reason critically..Kudos future ALBERT ENSTEINs

JungleboyT: D sum of 6th and 8th term of an Ap is 20, while d ratio of 3rd and 5th term is 10. Find d product of d 13th term, add it up by twice d common difference then divide by 3. Leave ur answer in 3 significant figure.

let the T1=a
and common difference=d
T6=a+5d
T8=a+7d
but T6+T8=2a+12d=20
2a+12d=20
a+6d=10---eqn1
T3/T5=a+2d/a+4d=10
a+2d=10a+40d
9a+38d=0---eqn2
solve simultaneously to obtain
a= -95/4
d=45/8
T13=a+12d
T13 +2d/3
=a+14d/3= -95/4+14(45//3
=55/3=18.33...
=18.3(3s.f)

jaryeh: To Maximus' question.

let the age of the man be m,and his niece' be n.

therefore,from the 1st statement
m = 2n+38 .....eqn(i)
2nd statement
n^2-m =42 ....eqn(ii)

To solve eqns(i) and (ii) simultaneously,substitute 2n+38 for m in eqn (ii),
hence, n^2 -(2n+38)=42

n^2 -2n-38=42

n^2 -2n-80=0

(n+ (n-10)=0
therefore,n=-8 or n=10,but age can't be negative,hence n=10

by substituting 10 for n in eqn (i)

m=2(10) +38
m=58.

therefore the man's present age is 58,while his niece' is 10.

then,to the first part of the question,the closest perfect square to 58 is 64,that will mean an addition of extra 6yrs,which if added to the niece' age will give 16 which is also a perfect square. Hence,their ages would simultaneously be perfect squares in the next 6 yrs

SOLUTION to the second part of the quest
let the no of y be p
n+p=(m+p)/2 + 3
n+p=(2n+38+p)/2 + 3
2n+2p=2n+38+p+6
p=44
therefore in 44yrs the niece age will satisfy the statement that it is 3 more than half his uncle's age

The sum of the interior angles of a regular polygon=(n-2)180
where n=no of sides
there4 (n-2)180=1440
n-2=1440/180
n-2=8
n=10.
Ans 10 sides(DECAGON)

Using integratin by substitution
let 2x be u
u=2x
du/dx=2
dx=du/2
$sinu.du/2
1/2$sinu.du
-1/2cosu=-1/2cos2x
ans -1/2cos2x + c

Multiples of an even no can never be odd.Hence if Q is even it implies that 3Q is even.
CASE1
Q+3
Since Q is even and 3 is odd hence Q+3 is odd
CASE3
Q^2-1=(Q-1)(Q+1)
since 1 is odd Q-1 and Q+1 are odd.hence their product are odd.
AnsB= 3Q

Pls i hope you guyz are scoring me..I got the 2 questns right...Its just that i am having problems with my email so i had to paste my solutions publicly.I will b glad if i am accepted as a participant..

Let the 3 consecutive no b x,x+2,x+4
but x+x+2+x+4=x(x+2)/x/3
=3x+6=105
=x=99/3=33
from the 2nd statement
x(x+2)(x+4)/11(x+2)/7
=7x(x+4)/11
substitute for x=33
=7x33(33+4)/11
=231x37/11=777
ans 777

Let sinx be = y
substitute for y
2y^2+3y+1=0
factorize
(2y-1)(y-1)=0
y=1/2 or 1
hence sinx=1/2 or 1
x=sin^-1(1/2)=30 or 150
OR
X=Sin^-1(1)=90
x=30,90,150 degrees
range of x is
30_<x_<150
pls i am a newcomer here i dont knw the particular email to message my solution to.PARDON

It's not the matter of going to BRAZIL instead it's the question of what can we deliver?After all we showcased at the 2012 S.A WC with the likes of ayegbeni,haruna,yobo,taiwo etc only to disgrace ourselves despite being placed in a favourable group.

It's not the matter of going to BRAZIL instead it's the question of what can we deliver?After all we showcased at the 2012 S.A WC with the likes of ayegbeni,haruna,yobo,taiwo etc only to disgrace ourselves despite being placed in a favourable group.