quantskillsgmat wrote:in a grid of 4x4, consecutive integers from 38 to 52 inclusive have to be arranged in such a way so that every column, row and major diagonals sum to same value. which of the following is a possible value of sum of four central cells.
a)124 b)153 c)178 d)192 d)214
Let the grid look as follows:
ABCD
EFGH
IJKL
MNOP
The grid must contain 16 integers.
Let's assume that the intended range is from 37 to 52.
Given consecutive integers:
The number of integers = biggest-smallest+1 = 52-37+1 = 16.
The average of the integers = (biggest+smallest)/2 = (57+32)/2 = 89/2.
The sum of the integers = number*average = 16*(89/2) = 712.
Thus, the sum of the entire grid must be 712.
Since the sum of the integers in each row must be the same, each row = 712/4 = 178.
Thus, the sum of the integers in any row, column, or main diagonal = 178.
We can plug in the answers, which represent the sum of the four central cells.
It seems VERY likely that the correct answer will be
C:
F+G+J+K=178.
Since the sum of rows 2 and 3 must be 2*178, E+H+I+L=178.
Since the sum of columns 2 and 3 must be 2*178, B+C+N+O =178.
Since the sum of the two main diagonals must be 2*178, A+D+M+P=178.
These 4 sums represent the entire grid.
Thus, the sum of the entire grid = 4*178=712.
Success!
If we plug in any other answer choice, the sum of the entire grid will not be 712.
The correct answer is
C.
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