What number should replace the "?"
Source: A modern approach to verbal and non-verbal reasoning - RS Aggarwal
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I think Answer will be
Sum of all first 3 numbers (A+B+C) in each column divide by 4th number (D) is result of last row. (A+B+C)/D
(1+7+8)/4 = 4
(13+7+5)/5 = 5
(15+10+11)/? = 6
It seems in sequence next number might be 6 (4,5,6)
Number Six (6) will be replace question mark..
(15+10+11) is 36.. 36/6 = 6
I think the answer might be
If we, instead of regarding the numbers numerical value, pay attention to how many letters they contain, we can see a pattern. The third column is a product representation of the absolute difference between column 1 and column 2. ONE(3) THIRTEEN(8) --> 8-3 = 5 = 1x5 (15). SEVEN(5) SEVEN(5) --> 5-5 = 0 = 1x0 (10). EIGHT(5) FIVE(4) --> 5-4 = 1 = 1x1 (11). And lastely FOUR(4) FIVE(4) --> 4-4 = 0 = 1x0 (10)
Reasoning for 1st:
Number of distinct prime factors in each row: 3
Number of prime factors with multiplicity per row: 3,4,5,6
Why 11^3 and not some different prime number?
Product of distinct prime factors of columns 1 and 2 just below product of distinct prime factors in column 3 where largest prime factor in column 3 is as small as possible: 13 < 3x5, 7 < 2x5, 2x5 < 11, 2x5 < 11
Reasoning for 2nd:
Each number in column 3 is the sum of two numbers in the first two columns and not the same row as the sum, and each number is referenced exactly once.
I think the answer is
If you look at the number of letters in the number itself, and add them all up in the column (Except the number in row 4, which is at the bottom, so add the number itself.) Keep doing this until i got these numbers:
13 and 19. These numbers make it so that the sum is one away from a multiple of 4. Pretty cool huh?
You do the thing again and we still have the numbers: 13 and 19.
This 4-sided table is not entirely square, so i should just do the 7, 5 and with the ? square. So i did it again, and something happened. When i did it by 13, i got 22 which is not one away from a multiple of four. BUT... when i did it by 19. I got 28 WHICH IS A MULTIPLE OF FOUR.
The missing number is 14.
We observe all 3 columns have 8 odd numbers and 3 even. The sum of the odd numbers is
1+13+7+7+5+5+11+15=64. If we replace the question mark with the number 14 then the sum of
the even numbers is 4+8+10+14=36 and 64+36=100. The sum of the numbers from column 1 and 2
is 1+13+7+7=8+5+5+4=50 which is equal to the sum of the 3rd column 15+10+11+14=50
The sum of two numbers from columns 1 and 2 equals one number of column 3.