formular123triol:
Firstly,
Logic is a subset of mathematics
Mathematics is a fundamental of reasoning
Reasoning is the vital aspect of Intelligent Quotient(I.Q)

In logic/reasoning questions,sometimes you need not make use of all the information given,as this might be a tentative way of deceit.You know,they want to test how intelligent you are by confusing you.

For this particular question,the fact that they said use any/all of the digits doesn't mean they may appear in the box,so first approach is to do away with the last statement on the question.
Secondly,if you observe those digits,they form a SERIES in an ARITHMETIC PROGRESSION(A.P) with a Common Difference(d) of 2.Similarly,the three digits required must also form an A.P with common difference of 2.From knowledge of Series,Tn=a+(n-1)d and Sn=n/2[2a+(n-1)d],where Tn is the nth term,Sn is the sum of the first n-terms,a is the first term and d is the common difference.From the series given,a=1,d=2,Sn=30(given),n=3 because we need three unknown digits.Furthermore,fixing these values in the formular Sn=n/2[2a+(n-1)d] we get 30=3/2[2a+(3-1)2] which gives a=8(i.e the first digit of the second series to be formed is .To get the remaining two digits,add 2 consecutively in this order:8,8+2=10,10+2=12;which gives the required series adding 8,10,12 together,we have 30.

NB:The reason why they asked us to use all the numbers in the given series 1,3,5,7,9,11,13,15 is for us to deduce the parameters required to derive the second series 8,10,12 which we did unravelled above.I hope we all got it,thanks!!!

I think u tried in convincing me a little

But the useless thing about this type of questions is that the person asking d question doesn't know the answer so as to give us a satisfactory answer

Kudos with this

.

(3+7) + (9+11) = 30. abeg make una clap 4 me

[(3-1)+13] + [(11-9)-(7-5)] + [15]
[2+13] + [2-2] + [15]
15+0+15 = 30

emmyrichie:
[ ] + [ ] + [ ] = 30

Fill the boxes using

(1, 3, 5, 7, 9, 11, 13, 15)

You can also repeat the nos.

Wrong question. Addition of three old numbers can not give an even number.

kingphilip:
op solve it n stop shining ur teeth

Somebody already gave the answer, check the thread (page 4).

Olucheye:
use another operand within the boxes,

It's as simple as,

[15] + [11] + [{9-5}] = 30

PS: The answer lies within your knowledge of how brackets work.

This is the answer. Plain and simple. Someone z talkin about base 9

CHM11:

This is good...thinking out of the box

Thinking outside the box pls.

three odds will never give an even

mozeybee:
To solve this problem, you must understand algebraic operations in different number bases.
1) you must note that the normal operation will perform daily is in "base 10" ( which is an even number).
2) The sum of any three even numbers in any "even base" will always give an even number.
3) Similarly, the sum of any three odd numbers will only give an even result when operated in an "odd base"
4) Since the question did not specify which "base" the summation is operated in, one is at liberty to fix any arbitrary base for solution.
5) Since the three boxes must be filled with three odd numbers ( as provided), an "odd base" must be used.
6) The unique rule for number base operation is that " no digit in a number must be = or > the base of the number".
7) Hence, the suggestions of 9 base 9 is mathematically wrong although you will get a correct answer by 27/9= 30 (in base 9).
To solve this problem, numbers must be chosen in "base 7".
9) The basic idea of this algebraic operation is that "once the since the summation is = or > the base, the result of the summation is divided by the base and the result is the summation in that base.
10) For the solution, I pick 1, 11 and 15 (all in base 7) Note that all the digits in each number are < 7. Hence, mathematically valid.
11) Now
T U
1
+ 1 1
+ 1 5
.......................
2 7
12) Since the unit result is = to the base(7), we divide by 7 and round the whole whole answer to the Tens column, while the remainder is left in the Unit column.. Hence, 7/7 = 1 remainder 0. Hence the final result is
T U
1
+ 1 1
+ 1 5
.......................
3 0
..............,......
Q.E.S..

Very smart answer. Reasoning wide. However, when numbers are written and no base is specified, then it authomatically means it is in base ten. If we are at liberty to choose any base we want then every question will have at least 10 diff answers. That is chaos.

formular123triol:
You are a bloody liar,for the fact that you don't know it,doesn't mean you should commit falacy of instinct generalization,ok?

Oga whats your problem? Why should call somebody a blody liar over something you are ignorant of. Oya prove him wrong. Mumu. Tell me three odd numbers you can add to get an even number?

ganiyheart:
(5+5)+(5+5)+(5+5)=30

this manipulation of yours got me jumping from my 3 storey floor ha ha ha ah ah aha !! A lady for that matter ! Ha ha ha ha

Perochelsea:




Thinking outside the box pls.

Thank you....but I said 'out of the box' on purpose.
Box as regards the question.

pkjag:
It's simple base addition let's use base ₉:
So 9₉+9₉+9₉=30
Those who have done computer science or electrical engineering will get it, for those who don't, we operate using base₁₀, so for base₉ let's all calculations such as 4+4=8 are the same as base₁₀, but when it comes to calculations like
9+9= 20, it's just like base₂ where
1₂+1₂=10.

You see in base₉ the number 9 is the last number, so there's no 10, so when you add the last numbers together you get 20 for arithmetic convenience, how? well in base₉ there are 10 numbers: 0,1,2,3,4,5,6,7,8,9 so it'd expected that if you add the tenth number to itself it should give the twentieth number then if you add again it should give the thirtieth number.

You just complicated it...

[11]+[13]+[6]=30

note: if you turn 9 upside down, you'll get 6.

emmyrichie:


Nobody's got time for all these... Base or no base, top or no top.

a no does not exist in its own base my friend

pkjag:
It's simple base addition let's use base ₉:
So 9₉+9₉+9₉=30
Those who have done computer science or electrical engineering will get it, for those who don't, we operate using base₁₀, so for base₉ let's all calculations such as 4+4=8 are the same as base₁₀, but when it comes to calculations like
9+9= 20, it's just like base₂ where
1₂+1₂=10.

You see in base₉ the number 9 is the last number, so there's no 10, so when you add the last numbers together you get 20 for arithmetic convenience, how? well in base₉ there are 10 numbers: 0,1,2,3,4,5,6,7,8,9 so it'd expected that if you add the tenth number to itself it should give the twentieth number then if you add again it should give the thirtieth number.

a no does not exist in its own base

emmyrichie:
[ ] + [ ] + [ ] = 30

Fill the boxes using

(1, 3, 5, 7, 9, 11, 13, 15)

You can also repeat the nos.

Funny hw pple just use their brains to over think, when all they need is just understand d simplicity of life.
The question is simple:
"Fill the boxes using (1, 3, 5, 7, 9, 11, 13, 15)"
Ans:
[1, 3, 5, 7, 9, 11, 13, 15 ] + [1, 3, 5, 7, 9, 11, 13, 15 ] + [1, 3, 5, 7, 9, 11, 13, 15 ] = 30
NB: at times our too much brain, is what kills us.
-Life is simple

pkjag:

Base 5 doesn't have an 11 or 13, neither does base 7 or 9

It is read 'One' 'One' in base 5 not 'Eleven'
also 'One' 'Three' in base 5 not 'Thirteen'

so it has

emmyrichie:
[ ] + [ ] + [ ] = 30

Fill the boxes using

(1, 3, 5, 7, 9, 11, 13, 15)

You can also repeat the nos.

[15] +[13]+ [-3]=30

(0) + (15) + (15) = 30

Zero is a null number, doesn't have to be there, Anyone else reason like me?

usmanktg2:

It is read 'One' 'One' in base 5 not 'Eleven'
also 'One' 'Three' in base 5 not 'Thirteen'
so it has

Btw, you're right as long as it's not 15 or 16, so it becomes
11₅+13₅+1₅=30₅.
Yes people that is as correct as it comes, nobody said what base you should use, so these types of aptitude questions will usually have different methods of answering, but if you can show them how you came up with the answer and if it's authentic then you are correct. Like here, the largest number in base₅ is 4, so when you add 11₅ and 13₅ you get 24₅ then because 4₅ is the last number just like 9₁₀ in base₁₀, when you add 1₅ you get a 0₅ then you carry a 1₅ then you should get 30₅.

emmyrichie:
[ ] + [ ] + [ ] = 30

Fill the boxes using

(1, 3, 5, 7, 9, 11, 13, 15)

You can also repeat the nos.

(15-5) + (13-3) + (11-1)

Equal 30

(5+7) + (11+1) +(5+1)

Equal 30

pkjag:
Btw, you're right as long as it's not 15 or 16, so it becomes
11₅+13₅+1₅=30₅.
Yes people that is as correct as it comes, nobody said what base you should use, so these types of aptitude questions will usually have different methods of answering, but if you can show them how you came up with the answer and if it's authentic then you are correct. Like here, the largest number in base₅ is 4, so when you add 11₅ and 13₅ you get 24₅ then because 4₅ is the last number just like 9₁₀ in base₁₀, when you add 1₅ you get a 0₅ then you carry a 1₅ then you should get 30₅.

Yes! I argued with my friends for long about that the base is not stated, As an electrical engineer, I made them understand 'addition' is a redundant word. It could be binary or oct addition.

Correction the largest digit and not number in base₅ is 4₅ just like there are numbers like 9₁₀ and the larger ones like 11₁₀ are combinations of previous digits cause in base₁₀ we don't have an independent symbol that has a larger value than 9₁₀ and that doesn't use the other lesser digits. But in base₁₆ we have from 0₁₆ to 9₁₆ then instead of 10₁₆ we have A₁₆,B₁₆,C₁₆,D₁₆,E₁₆,F₁₆ so here F₁₆ is the largest digit
hope you now understand

[ 15] + [ 7] + [ 9-1] = 30

[9] + [11] + [15-5] =30

The type of bracket used in the question suggest a matrix,therefore use the numbers to for a 2by2 matrix in each matrix bracket given,add the three matix to become one matrix,hence the only way combinat of matrix can give a single number is by finding its determinant.im using a phone that is not easy to type i would have solve it here

The simple answer is you Cannot add 3 odd numbers to get an even number. That's why they give u less than a minute to answer... It's simple logic

jackpot:
[11]+[13]+[6]=30

note: if you turn 9 upside down, you'll get 6.

I doff my hat for you but mathematics is basically about manipulation

pkjag:
It's simple base addition let's use base ₉:
So 9₉+9₉+9₉=30
Those who have done computer science or electrical engineering will get it, for those who don't, we operate using base₁₀, so for base₉ let's all calculations such as 4+4=8 are the same as base₁₀, but when it comes to calculations like
9+9= 20, it's just like base₂ where
1₂+1₂=10.

You see in base₉ the number 9 is the last number, so there's no 10, so when you add the last numbers together you get 20 for arithmetic convenience, how? well in base₉ there are 10 numbers: 0,1,2,3,4,5,6,7,8,9 so it'd expected that if you add the tenth number to itself it should give the twentieth number then if you add again it should give the thirtieth number.

I would have supported you but 9₁₀ = 10₉.

theAtheist101:


Wrong question. Addition of three old numbers can not give an even number.

Definitely, the addition of three "young" numbers can give us an even number.