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

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What is the sum of the integers in the table above? [#permalink]

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06 Apr 2011, 14:21
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What is the sum of the integers in the table above?

(A) 28
(B) 112
(C) 336
(D) 448
(E) 784
[Reveal] Spoiler: OA

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06 Apr 2011, 14:26
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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

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"
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06 Apr 2011, 14:36
fluke wrote:
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

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"

Why will we multiply the sum of column by 7?
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06 Apr 2011, 14:43
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Baten80 wrote:
Why will we multiply the sum of column by 7?

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

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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06 Apr 2011, 18:11
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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

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06 Apr 2011, 18:37
=(1+2+3+4+5+6+7)(1-2+3-4+5-6+7)
=112

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07 Apr 2011, 21:55
Spidy001 wrote:
=(1+2+3+4+5+6+7)(1-2+3-4+5-6+7)
=112

What is the explanation of your approach?
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07 Apr 2011, 23:25
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.
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11 May 2012, 01:00
Guys,
what do you thing is the level of a difficulty of the such question? Anyone knows?
Thanks
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11 May 2012, 04:05
YuraK wrote:
Guys,
what do you thing is the level of a difficulty of the such question? Anyone knows?
Thanks

I'd say it's 600 level question.

Check similar one to practice: sum-in-the-chart-126125.html
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17 May 2012, 00:43
fluke wrote:
Baten80 wrote:
Why will we multiply the sum of column by 7?

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

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
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17 May 2012, 00:52
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

hello
why do you have added all the average plz explain

best regards
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17 May 2012, 01:43
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keiraria wrote:
fluke wrote:
Baten80 wrote:
Why will we multiply the sum of column by 7?

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

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.

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

hello
why do you have added all the average plz explain

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]

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17 May 2012, 10:23
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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
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Re: What is the sum of the integers in the table above? [#permalink]

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02 Jul 2013, 01:16
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Re: What is the sum of the integers in the table above? [#permalink]

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02 Jul 2013, 14:28
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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
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Re: What is the sum of the integers in the table above? [#permalink]

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02 Jul 2013, 16:45
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B is correct. Here is my solution:
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Sum of integers.png [ 25.72 KiB | Viewed 7761 times ]

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Re: What is the sum of the integers in the table above? [#permalink]

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08 Oct 2013, 11:23
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
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Re: What is the sum of the integers in the table above? [#permalink]

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Re: What is the sum of the integers in the table above? [#permalink]

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29 Jan 2015, 16:46
One more approach:

Look at each column starting from last.

The difference between each pair of numbers (beyond the first number) results same.

For example, in the last column:

-14+21 = 7. -28+35 = 7. -42+49 = 7. The first entry of that column is 7.
So we have 4*7 = 28 as sum of the last column.

Similarly, sum of preceding column = 4*6 = 24
Ones before that results in:

4*5
4*4
4*3
4*2
4*1

So result is 4* (1+2+3+4+5+6+7) = 112.
Re: What is the sum of the integers in the table above?   [#permalink] 29 Jan 2015, 16:46

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