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

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


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All rows are evenly spaced.
The numbers in 4th column are the averages of respective rows.
Every row has 7 elements.

Thus,
Total = (4-8+12-16+20-24+28)*7 = (64-48)*7= 16*7= 112

Ans: "B"

Take 1 row at a time

1st row: Total = Average*Number of elements = 4*7
2nd row: Total = Average*Number of elements = -8*7
3rd row: Total = Average*Number of elements = 12*7
4th row: Total = Average*Number of elements = -16*7
5th row: Total = Average*Number of elements = 20*7
6th row: Total = Average*Number of elements = -24*7
7th row: Total = Average*Number of elements = 28*7

Add them up:
4*7+(-8)*7+12*7+(-16)*7+20*7+(-24)*7+28*7
Take 7 common:
7(4-8+12-16+20-24+28)

=(1+2+3+4+5+6+7)(1-2+3-4+5-6+7)
=112

Answer B.

Guys,
what do you thing is the level of a difficulty of the such question? Anyone knows?
Thanks

Hello Fluke
just to make sure I understood
as thoses questions look tricky and time consuming

first each line is a AP Progression
you use the average formula in reverse to find the sum
since in an AP Progression the mean or average is the same as median number
and then you factorise by 7 to complete the sum and multiply it by 7 as each line is a multiple of 7

right ?

Thanks for 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
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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