Kendzyma: Chai...only 1 question since

factorial1: following though...no calculator, no pen(*coughs*) to solve, nothing nothing! Me Ah no du again jawe...

#teamleaveamfortheogas'

Kendzyma: Chai...only 1 question since

Kendzyma: abeg b4 u go ..let's swap brain

smurfy: @Richiez

Let the HOUSE provide the solutions. You endorse the correct one. That way, more people get to participate.

aysuccess99:
@kendzyma, easy my oga cos i knw dat u want 2 destroy all d qstn,i fear u o......... @factorial, u ke...? u want 2 leave....? if i hear...? bt seriously speaking u are one of our best ogas in d quiz, don't leave nw o...pls continue d quiz cos i knw dat dere's hope 4 u o...even without pen nd jotter with calc, u can stil solve cos u ar a maths guru....

Alpha Maximus: Quiz Question By Alpha Maximus
Question: The sum of three consecutive even / odd integers is the
equivalent of the result of the ratio of the product of the two smaller
integers to the ratio of the smallest integer to 3 which is 105. What
is the result of the ratio of the product of all three integers to 11
times the result of the ratio of the second smallest integer to
7?

Richiez:

5 mins more!!!

echibuzor: Maximus is a Spartacus o... I have sent the answer.. av u received it

smurfy: What do you mean by 'to 3'? Question not clear.

factorial1: boss, oya whisper the ans. to me ....

Alpha Maximus: ....I.e:ratio of x TO y

aysuccess99:
u funny sha...if he whisper it 2 u ...u to whisper it to me o...ceteris paribus.

smurfy: I hate your word problems, Max.

echibuzor:
Read the question very very well, Infact try reading it from the back.. You should see a key to work with..
I hope what I just said is allowed??

Alpha Maximus: Pls don't give any clues.

bitex2: 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

Alpha Maximus: No participant got this right! This is the correct and simple solution

factorial1:
...bro Alpha please...many of us here...are stil new to some qts(GMAT) u are used to setting, easy!

bitex2: 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

Alpha Maximus: No participant got this right! This is the correct and simple solution

Alpha Maximus: .... This is an elimination round where half of you will be evicted at the end of the day! I can't post simple questions. We aim to seperate the masters from the novices(no offence meant to anybody) . Besides the solution was quite easy, I haven't even started setting hard questions. Goodluck