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If the 4 x 4 grid in the attached picture is filled with the [#permalink]
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31 Jan 2012, 18:25
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Question Stats:
84% (01:40) correct 16% (02:40) wrong based on 139 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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01 Feb 2012, 02:58
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Sum of the numbers from 37 to 52 = (n/2)[2a + (n1)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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15 Nov 2013, 21:23
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Re: If the 4 x 4 grid in the attached picture is filled with the [#permalink]
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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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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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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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04 Jan 2016, 18:57
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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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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 [ 1.95 KiB  Viewed 1985 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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13 Jul 2016, 21:31
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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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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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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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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 +(n1)d Here d=1, n=16, a=37
we will get that sum of all from 3752 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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19 Jan 2018, 02:29
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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? Here's how I solved it the number of integers= 5237+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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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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