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[quote author=Dr.Gboy]i sent 778. You said consecutive numbers, in which mine was 34, 35, 36. I don't think 33, 35 and 37 are really consecutive. My opinion. Thanks[/quote]This is why we implore you all to study the questions painstakingly before solving. I typed consecutive ODD/EVEN numbers the series of either having a common difference of 2. I.e: 1,3, 5...(Odd) or 2,4,6(even)

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

.... This is an elimination round where half of you will be evicted at the end of the day! I can't post simple questions. Besides the solution was quite easy, I haven't even started setting hard questions. Goodluck

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

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

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??

Pls don't give any clues.

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

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

NB:My question has been modified.

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?

Next question loading..

Hmmmm, great day to solve Mathematics

benbuks: 0X =9
find x.

....x=infinity

LogoDWhiz: tomorrow is alright by me...

The votes have already been cast and majority opted for Monday

Alright the NAYS have it!! (Since 6/10 participants have already opposed) Case closed. The quiz will be moved to Monday and the time will be disclosed later.

Due to some unforseen network impediments being faced by Richiez, it was concluded that the quiz would be postponed till 10.00am tomorrow. Normally,Doubledx and I would have carried on the torch and proceeded without Richiez but this would not have been effective due to the fact that all the work would have been concentrated on me due to Doubledx's absence and Richiez internet issue. If tomorrow isn't a convenient time for majority of the competitors, then we the co-ordinators will postpone the event till next week Monday. If you are in support of having the quiz tomorrow, post 'AYE', if you oppose, post 'NAY'. Thank you.

smurfy: Here goes...

(a) Sample space = 6^3 = 216

Set of combinations which gives a sum of 11 include:
164, 146, 461, 416, 641, 614, 155, 515, 551, 245, 254, 425, 452, 524, 542, 236, 263, 326, 362, 623, 632, 344, 434, 443, 335, 353, 533.

That's 27 ways of getting the sum 11. So answer to question (a) is 27/216 = 3/24

(b) Set of combination which gives the sum 12 include:
156, 165, 516, 561, 615, 651, 246, 264, 426, 462, 624, 642, 255, 525, 552, 336, 363, 633, 345, 354, 435, 453, 534, 543, 444.

That's 25 ways of getting the sum of 12. That gives 25/216.

Hmmm, nice approach bruv but here's a shorter approach:
(I) Pr(sum of 11) : for this to occur the first two numbers must have a sum of at least 5 and at most 10(11-6 and the 2 dice can't sum up to 11 since there's no die bearing 0) and from sample space there, are 27 outcomes out of 36
Thus a probability of 27/36, then the third number must make the previous sum add up to 11 and there is a 1/6 probability of this occurence since the last number is from the last die(singular of dice)
Thus,pr(sum of 11 from 3 dice)=27/36 * 1/6=27/216(it wasn't meant to be this lengthy if not for explanation's sake)
(II) Pr(sum of 12 from a single throw of 3 dice)
Same method above: the first 2 dice must have a sum of at least 6(12-6) and at most 11, from sample space there are 25 outcomes out of 36 possible outcomes satisfying this designation...then the last die must show a number which makes the sum 12 and there is a 1/6 chance for this to occur, therefore:
Pr(sum of 12 from single throw of 3 dice)=25/36 * 1/6 =25/216...this method saves mathematicians fro the strenuous approach of having to mentally try out all possible combinations such as 1,4,6...4,1,6 ...etc, like you did in your solution

Alright , this was a simple case of word problems leading to simultaneous equations. Mathematically transforming the aforementioned question into equations, we have that a=humans and b=animals, since humans, lions and Thomson's gazelle each have a single head, we have that
A+b=7678....(I)
Since humans have two legs each and the animals have four, we have an equational interpretion of :
2a+4b=28490
Using the elimination method and multiplying (I) by 2 for the purpose of the extrication of the unknown 'a', we have
2b=13134
B=6567 animals, inputting 'b' into (I), we establish that a=1111humans
We were given info that the gazelles constitute 8796 legs so we arrive at the following
(4*6567)-8769=17472legs owned by lions
We divide by four since lions have four appendages. Thus
17472/4=4368 lions

Hold on for solution...typing....

The last question has already been posted. Comrade Richiez please commence countdown!!

Final Question By Quizmaster Alpha Maximus
Question: An American billionaire who had been pondering ways to create jobs finally decided to build a zoo. A year later, his zoo was already on the verge of completion. On one particular day, he instructed his Head-zookeeper to do a count of visitors to the zoo and to obtain the population of the animals. The Head-zookeeper who had a rather unorthodox means of counting recorded 7,678 heads and 28,490 legs and claimed he still knew the number of humans and animals. The only animals the zoo had at that time were lions and Thomson's gazelle. Of the animals, the Thomson's gazelle constituted 8,796 legs. How many lions are in the zoo?

Alright guyz hold on for the last question

smurfy: No Maximus. You don't understand. When Richiez first posted the question, the multiple choice I only, II and III only etc. wasn't included. Ask Richiez. Only the answers I, II and III were there. Then option III had a typo. That's I didn't pick it. I submitted my answer, refreshed the page only to see A, B, C, etc. and correction to option III

Alright...next time please do not send any answer before cross checking questions and spotting irregularities

smurfy: What!!!!
I got the question! All the options are correct. The initial question I opened on my phone had error for option c

option I only,option I and II only and option I, II and III are three entirely different choices. When either of the first two options are chosen by a participant, the consequence is that he/she is implying that that option II or III is wrong and the chosen are the only correct options , whichever the case may be

Please post your answers to both Doubledx and me. And no double-answering please; cross-check before sending.

Idenyijoshua: Hope u got my answer before the time was up. Network hassles

Got it