If the 4 x 4 grid pictured at right is filled with the consecutive

Question

https://gmatclub.com/forum/download/file.php?id=72783 If the 4 x 4 grid pictured at right is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)? (A) 124 (B) 153 (C) 178 (D) 192 (E) 214 Grid.png

manojach87 · Accepted Answer

I approached it this way: there are 16 cells; 37,38,39,...,52 can be denoted as 36+1,36+2,36+3,...,36+16 the total of 37+38+...+52 is therefore =36*16+16*17/2 =16(36+8.5) =16*44.5 =8*89 =4*178 =2*356 =712 The average value of a cell is 712/16 = 44.5 gmatclub.PNG Also given : sum of values in column 2 = sum of values in column 3 = sum of values in row 2 = sum of values in row 3. To satisfy this, if we assigned the average value of a cell to each cell, the 4 cells will get a total of 44.5*4 = 178. Hence C is the answer.

GyanOne · Answer

Sum of the numbers from 37 to 52 = (n/2)  = (16/2)   = 8 * 89 = 712

Therefore the sum of all numbers in the grid = 712
If x is the sum of all numbers in a row/column/major diagonal, then 
4x = 712
=> x = 178 = sum of all numbers in any row = sum of all numbers in any diagonal = sum of all numbers in either major diagonal

Now, consider the grid as follows: 
1' 2' 3' 4'
5' 6' 7' 8'
9' 10' 11' 12'
13' 14' 15' 16'

We know that 1' + 6' + 11' + 16' = 178
4' + 7' + 10' + 13' = 178
=> 1' + 6' + 11' + 16' + 4' + 7' + 10' + 13' = 356
=> 6' + 7' + 10' + 11' + 1' + 13' + 4' + 16' = 356

Also 5' + 6' + 7' + 8' + 9' + 10' + 11' + 12' = 356
and 1' + 5' + 9' + 13' + 4' + 8' + 12' + 16' = 356
=> 6' + 7' + 10' + 11' - 1' - 13' - 4' - 16' = 0

Therefore 6' + 7' + 10' + 11' = 356/2 = 178 = sum of the middle four numbers

Option (C).

sumitaries · Answer

In the last step "=> 6' + 7' + 10' + 11' - 1' - 13' - 4' - 16' = 0"
how can we be sure that the +ve groupings and -ve groupings each add up to 178 or 6' + 7' + 10' + 11' = 178. It could be 180 for 6' + 7' + 10' + 11' and 176 for - 1' - 13' - 4' - 16'. Am i missing something here?

sumitaries · Answer

In the last step "=> 6' + 7' + 10' + 11' - 1' - 13' - 4' - 16' = 0"
how can we be sure that the +ve groupings and -ve groupings each add up to 178 or 6' + 7' + 10' + 11' = 178. It could be 180 for 6' + 7' + 10' + 11' and 176 for - 1' - 13' - 4' - 16'. Am i missing something here?

nitinmeharia009 · Answer

I followed a rather simple approach. Since all the four sides and the major diagonal has same value. The sum of four center cell should also be the same. and it should be (37+38+51+52) = 178

mbaprep2016 · Answer

mean of all 16 numbers is 44.5 
take two number on right of this mean and take two number left , but they all are in such a way that mean is 44.5.
so basically 44.5* 4 = 178

RMD007 · Answer

I solved it in this way:

Min number is 37 so Min Sum of 4 numbers will be 37*4 = 148. (Option A out).
Max Number is 52 so max sum of 4 numbers will be 52*4 = 208 (Option E out).

Because the numbers are uniformly distributed their sum should should lie in middle of 144 and 208 not towards any of these numbers.(Most likely B and D are out.)
148+208 = 356/2 = 178.

Hope this logic makes sense.

shravan2025 · Answer

hello .

i blv we can solve it using algebra.
sum of n numbers as we know 
Sn =n/2(2a +(n-1)d Here d=1, n=16, a=37

we will get that sum of all from 37-52 incl is 712.
but we have to find sum in each column & dignoal uniform so =712/4 =178 hence ans is C

NDND · Answer

Here's how I solved it

the number of integers= 52-37+1=16
The median (8th & 7th term)= (37+8)+(37+7)= 44&45 (No need to divide by 2) 
Add the previous number and the following number= 43+44+45+46= 178 (D)

NDND · Answer

I believe there's a simpler way to calculate the total:

/2 = 16 (37+52)/2= 8*89= 712

vivapopo · Answer

10-If the 4 x 4 grid in the attached picture is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)?

(A) 124
(B) 153 
(C) 178 
(D) 192 
(E) 214

37 38 39 40 178
44 43 42 41 178
45 46 47 48 178
52 51 50 49 178
178 178 178 178

C

rahul16singh28 · Answer

The Sum will be two of the least number (37+38) + tow of the max (51 + 52) = 178

puneetfitness · Answer

Hi please advice if the as per question stem each consecutive box will be filled with consecutive integer starting from 37-52

37,36,38,39,40
41,42,43,44,45 and so on

Or we will use consecutive integers from 37-52 but they can be in any box in 4*4 grid

Posted from my mobile device

nivivacious · Answer

If you want every row, column and diagonal to sum to same value, the middle needs to be in the centre.
Middle numbers of the set : 43,44,45,46

Unit digits of 3+4+5+6 sums of 8 in units place.
Hence C

Sonia2023 · Answer

KarishmaB  - Is this a GMAT like question? Can you kindly explain what really is the concept involved as random testing of numbers to make them equal took a lot of time?

Thank you

KarishmaB · Answer

GMAT could give some puzzle type of questions though they would have logic in them. It wouldn't give you a question in which you need to use hit and trial and solve for exact values. You may not be able to neatly bracket them into one topic. 
Here I would just find the average and take it 4 times. The number distribution is symmetrical - all rows, columns and diagonals have the same sum so likely the centre four numbers would have the same sum too.

mellowmev · Answer

I got stuck in this, but I scrolled to the answers to see a variety of approaches leading to the same answer.

Which of these would be the best to remember in such logic puzzles, that ensure minimum time used as well as accuracy across different variations of the question?

I thought the algebraic method would the most generic, however, quite time consuming.

bumpbot · Answer

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If the 4 x 4 grid pictured at right is filled with the consecutive

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