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Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by mozeybee: 12:37pm On Apr 16, 2015
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).
cool 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..
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Omowalksola(f): 12:41pm On Apr 16, 2015
(11-1)+(13-3)+(15-5)
(10)+(10)+(10)
=(30)

2 Likes

Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by logica(m): 12:47pm On Apr 16, 2015
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.
Excuse me, when the base is not specified, it is base 10. Sorry.
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by 500samuel(m): 12:49pm On Apr 16, 2015
obum88:
[15]+[13]+[7]= 30

obum's theory.. grin grin grin
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by logica(m): 12:49pm On Apr 16, 2015
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).
cool 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..
There also is no digit 9 in Base 7.

1 Like

Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by 2naira50kobo: 12:50pm On Apr 16, 2015
Biko! Ki re?
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.
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by 500samuel(m): 12:51pm On Apr 16, 2015
Niyoung:


Simply: [3] + [7] + [9] + [11] = 30

bro u were given three brackets cool cool cool

1 Like

Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by 500samuel(m): 12:57pm On Apr 16, 2015
Emmanuelemman3:
[8] + [10] + [12] = 30

u just dey manufacture figure... u try no be small
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by anakenyi: 12:58pm On Apr 16, 2015
This does not need not need much grammar just a simple logic

[1day(24hours)] + [5hrs] + [1hr] = 30hours

this is the answer.
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Flameneo(m): 1:01pm On Apr 16, 2015
[15-9] + [13-7] + [7-1] + [9-1] + [13-9] =30
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by frisky2good(m): 1:04pm On Apr 16, 2015
In that case, it is not meant to be solved with "ordinary eye".

emmyrichie:


It's an aptitude test question from Apple and Microsoft.
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by iyemizi: 1:04pm On Apr 16, 2015
1+5+9+15=30. FIN!
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by frisky2good(m): 1:05pm On Apr 16, 2015
I may be wrong but adding 2 odd numbers will give an even number as answer and adding an odd number to an even number will always give you an odd number as answer. Long story short, we are expected to arrive at 30 (an even number) when from a logical point of view the only possibility is an odd number. If job no dey why dem dey disturb people head na?

Meanwhile, OP please don't come and give us a stupid answer like:

[5+5] + [5+5] + [5+5] = 30

1 Like

Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Nobody: 1:05pm On Apr 16, 2015
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).
cool 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..
That's not how base addition is done, when you have a larger number than 7 you subtract it not divide it, and for the record the number 7 is not in base 7, the last number is 6! I've already tried base addition from 2 up to 20 and none of any of three of the given numbers adds up to 30.
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Nobody: 1:05pm On Apr 16, 2015
emmyrichie:
[ ] + [ ] + [ ] = 30

Fill the boxes using

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

You can also repeat the nos.

Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Dotunwa(m): 1:06pm On Apr 16, 2015
Since i can repeat the nos ill just stroll over d question and go like this:
[15] + [15] + [ ] = 30.

Do i have to explain or you garrit?
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Phillfree(m): 1:07pm On Apr 16, 2015
(11-1) +(13-3)+(15-5)=30
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by affigurl5(f): 1:10pm On Apr 16, 2015
(7)+(13)+(15)=30
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by otikobaba: 1:10pm On Apr 16, 2015
Let us us 2 to multiply each bracket

2(1)+2(3)+2(11)=30
2+6+22=30

shikeennahhhhhhhhhhhh!!!!!!!!!!!!!
FTC straightttttttttttttttttttttttttttttttttt!!!!!!!!!!!!!
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by cassidy1996(m): 1:14pm On Apr 16, 2015
[5+5]+[9]+[11= 30]
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by kblaze111: 1:18pm On Apr 16, 2015
it is mathematically impossible for the addition of any set of 3 odd numbers to give an even number
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by joescase: 1:24pm On Apr 16, 2015
Making sence pls more of dat
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Nobody: 1:31pm On Apr 16, 2015
9 = 15 = 24, 5=1 =6
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by M4gunners: 1:38pm On Apr 16, 2015
[1]+[3]+[11]+[15]=30.
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Sleekydee(m): 1:38pm On Apr 16, 2015
affigurl5:
(7)+(13)+(15)=30

You joking right.
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Sleekydee(m): 1:40pm On Apr 16, 2015
[1+3]+[5+7]+[9+5]=30

1 Like

Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by M4gunners: 1:41pm On Apr 16, 2015
cassidy1996:
[5+5]+[9]+[11= 30]
Hw many 5 u see for dia na?
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Nobody: 1:42pm On Apr 16, 2015
Guys it isn't a correct question.
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by M4gunners: 1:46pm On Apr 16, 2015
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.
I disagree with you. Op only use + and no - you failed the test.
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by usmanktg2(m): 1:47pm On Apr 16, 2015
In base 5:
1 + 11 + 13 = 30
In base 7:
3 + 11 + 13 = 30
In base 9:
5 + 11 + 13 = 30

(Usman, 2015)
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by PezzoNovante(m): 1:53pm On Apr 16, 2015
[1+1]+[13]+[15] = 30
Re: Who Can Solve This Simple, Yet Hard Aptitude Test Question by Jtmanager(m): 1:57pm On Apr 16, 2015
shoro niyen

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