What is the sum of the integers in the table above?

(A) 28
(B) 112
(C) 336
(D) 448
(E) 784


_________________

Why will we multiply the sum of column by 7?
_________________

I went for a different approach, the sum of every coulmn came as a multiple of 4, starting from 4, and there are 7 columns.

So sum = 4 + 8 + 12 + 16 + 20 + 24 + 28

= 112

Answer - B
_________________

Each row - ( 1+2+3+4+5+6+7) = 28
Let that be a constant A = 28.
Now, visual analysis shows that each row is a multiple of A,
as
row 1 = A
row 2 = -2(A)
row 3 = 3(A)

and so on.

Hence, we get A- 2A + 3A-4A+5A-6A+7A = 4A => 4* 28
= 112.
Thanks.

I'd say it's 600 level question.

Check similar one to practice: https://gmatclub.com/forum/what-is-the- ... 26125.html
_________________

hello
why do you have added all the average plz explain

thanks for your help

best regards

Let X be the sum of 1 to 7 = 28
1st row = x
2nd row=-2x
3rd row=3x
4th row = -4x
5th row = 5x
6th row = -6x
7th row = 7x
Total = [x-2x+3x-4x+5x-6x+7x]
= 4x=4*28=112

from first row take out 1 in common==>sum =1((7*8)/2)
from 2nd row take out 2 in common==>sum =-2((7*8/2)
...
.
.
.from 7th row take out 7 in common==>sum= 7((7*8)/2)
now we have to add all ...each row has (7*8)/2=28 in common
take 28 common from all===>28(1-2+3-4+5-6+7)=28*4=112
_________________

I think the best way to solve this type of problems is to try to cancel out some of the numbers before operating since you have positives and negatives.
I quickly realized that lines 1,3 and 4 could cancel out. Then by combining lines 5,6 and 7 you get the same numbers in line 6 but with different signs. So you can actually then combine that one with line 2 and just get a series of multiples by 4 starting with 4+8+12....+28.

So 7 terms average 16 = 112

It seems more complicated than it really is. I guess one could find other patterns and even cancel something else out, but you should not overthink it.
At least cancel 2/3 lines and then you can start adding/subtracting to get another evenly spaced set of numbers.

Hope it helps

What is the sum of the integers in the table above?

(A) 28
(B) 112
(C) 336
(D) 448
(E) 784

It can be solved as following (1+2+3+4+5+6+7)*4=112. How we can come to it?

We have seven rows, right? We can add corresponding numbers in row 7 to row 6 (positive number to negative number) for instance: 49+(-42)=7; 42+(-36)=6; 35+(-30)=5; 28+(-24)=4; 21+(-18)=3; 14+(-12)=2; 7+(-6)=1.

We can do the same calculations for the corresponding numbers of rows 5 and 4 as well as of rows 3 and 2 and all the time we will have the same results (1+2+3+4+5+6+7). Since we have four such calculations, including 1st row, we just proceed with: (1+2+3+4+5+6+7)*4=112