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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)
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Re: What is the sum of the integers in the table above? [#permalink]
17 May 2012, 09:23

1

This post received KUDOS

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

Re: What is the sum of the integers in the table above? [#permalink]
02 Jul 2013, 13:28

1

This post received KUDOS

Baten80 wrote:

Attachment:

Table.jpg

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

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

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
_________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Re: What is the sum of the integers in the table above? [#permalink]
02 Jul 2013, 15:45

1

This post received KUDOS

B is correct. Here is my solution:

Attachments

Sum of integers.png [ 25.72 KiB | Viewed 1401 times ]

_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

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.

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)

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

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)

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

Each row represents an evenly spaced set (aka arithmetic progression). In any evenly spaced set the arithmetic mean (average) is equal to the median and the sum of the terms in any evenly spaced set is the mean (average) multiplied by the number of terms.

The median of each row is the middle number and each row has 7 numbers in it so the sum of the table is 7*4+7*(-8)+7*12+7*(-16)+7*20+7*(-24)+7*28=7(4-8+12-16+20-24+28)=112.

Answer: B.

keiraria wrote:

subhashghosh wrote:

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

hello why do you have added all the average plz explain

thanks for your help

best regards

Those are not averages, but the sums of the numbers in each column.
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Re: What is the sum of the integers in the table above? [#permalink]
08 Oct 2013, 10:23

Baten80 wrote:

Attachment:

Table.jpg

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

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

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

gmatclubot

Re: What is the sum of the integers in the table above?
[#permalink]
08 Oct 2013, 10:23