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If the 4 x 4 grid in the attached picture is filled with the

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If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 31 Jan 2012, 18:25
3
2
17
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

77% (02:20) correct 23% (02:26) wrong based on 198 sessions

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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

Any idea how to approach this question please?

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Re: Shaded part of the grid  [#permalink]

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New post 01 Feb 2012, 02:58
9
2
Sum of the numbers from 37 to 52 = (n/2)[2a + (n-1)d] = (16/2) [74+15] = 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).
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 08 Dec 2013, 07:01
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?
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 08 Dec 2013, 07:02
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?
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 10 Dec 2013, 08:16
This is just another great question.
The difficulty level of this question is high and it didn't strike me how to do this question when I tried it in the first attempt.
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 04 Jan 2016, 18:57
1
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
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If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 13 Jul 2016, 11:04
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enigma123 wrote:
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

Any idea how to approach this question please?


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

Attachment:
gmatclub.PNG
gmatclub.PNG [ 1.95 KiB | Viewed 3147 times ]


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.
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 13 Jul 2016, 21:31
2
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
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 29 Dec 2016, 05:46
manojach87 wrote:
enigma123 wrote:
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

Any idea how to approach this question please?


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

Attachment:
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.



It's a great solution.
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 29 Dec 2016, 19:16
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.
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 19 May 2017, 21:05
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
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 19 Jan 2018, 02:29
1
enigma123 wrote:
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

Any idea how to approach this question please?


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)
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 19 Jan 2018, 03:11
manojach87 wrote:
enigma123 wrote:
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

Any idea how to approach this question please?


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

Attachment:
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.


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

[#of items (first term+last term)] /2 = 16 (37+52)/2= 8*89= 712
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If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 30 Jun 2018, 10:26
enigma123 wrote:
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

Any idea how to approach this question please?

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
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Re: If the 4 x 4 grid in the attached picture is filled with the  [#permalink]

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New post 30 Jun 2018, 20:07
enigma123 wrote:
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

Any idea how to approach this question please?


The Sum will be two of the least number (37+38) + tow of the max (51 + 52) = 178
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Re: If the 4 x 4 grid in the attached picture is filled with the &nbs [#permalink] 30 Jun 2018, 20:07
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