og quant review #160 REally Interesting!

There are probably more efficient ways to do this problem. Here's my take:

You'll see different arithmetic progressions (AP) in every row.

Sum of all numbers in AP = (n/2)*(first term + last term)

Therefore sum of all numbers in the grid:

Sum = (7/2)(1+7) + (7/2)(-2 - 14) + (7/2)(3+21) + (7/2)(-4 - 28) + (7/2)(5+35) + (7/2)(-6 - 42) + (7 + 49)
Sum = (7/2)(8 - 16 + 24 -32 + 40 - 48 + 56)
Sum =(7/2)(-8 -8 -8 +56)
Sum = (7/2)(32)
Sum = (7)(16) = 112

Choose B.

-BM-

Actually, a way more efficient approach is to look at the grid column wise:

1. Look at the numbers in the first column by first looking at the last 2 numbers. Their sum = 7 + (-6) = 1. Notice that the next last two numbers in this column also has the sum of 1. So sum of all numbers in column 1 is 1+1+1+1 = 4.

2. The same is done in the 2nd column: Sum of all numbers = 2+2+2+2=8

3. The same is done for the 3rd column and so forth. You should get the sum to be 3+3+3+3 = 12 here.

So the sum of the entire grid is:

(1+1+1+1)+(2+2+2+2)+....+(7+7+7+7)
=4(1+2+3+4+5+6+7)
=4(28)
=112

-BM-