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Sum of integers in the above table..

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by welcome » Sun Mar 01, 2009 5:40 pm
If you will observe and Name first row as A, then your question is to take some common in all rown and convert them in form of A.

So if A = 1+2+3+4+5+6+7 = 28

Than you have to find A-2A+3A-4A+5A-6A+7A = 4 A = 4*28 = 112 - Answer.
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by vineethyoung » Sun Dec 04, 2011 10:39 pm
try this link

https://www.beatthegmat.com/sum-of-integ ... 43127.html
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by GMATGuruNY » Mon Dec 05, 2011 5:13 am
Image

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

A 28

B 112

C 336

D 448

E 784
Look for a pattern.
The table shows the results of multiplying every integer in the top row by every integer in the leftmost column.

Thus, we need to determine the sum of all these products:
(Top row)*(Leftmost column) = (1+2+3+4+5+6+7)(1-2+3-4+5-6+7) = 28*4 = 112.

The correct answer is B.
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by SwatiAgarwal » Tue Dec 06, 2011 5:59 am
GMATGuruNY wrote:
Image

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

A 28

B 112

C 336

D 448

E 784
Look for a pattern.
The table shows the results of multiplying every integer in the top row by every integer in the leftmost column.

Thus, we need to determine the sum of all these products:
(Top row)*(Leftmost column) = (1+2+3+4+5+6+7)(1-2+3-4+5-6+7) = 28*4 = 112.

The correct answer is B.
Can you please share the rationale for concluding the statement --
sum of all number = (Top row)*(Leftmost column)?
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by GMATGuruNY » Tue Dec 06, 2011 7:18 am
SwatiAgarwal wrote:
GMATGuruNY wrote:
Image

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

A 28

B 112

C 336

D 448

E 784
Look for a pattern.
The table shows the results of multiplying every integer in the top row by every integer in the leftmost column.

Thus, we need to determine the sum of all these products:
(Top row)*(Leftmost column) = (1+2+3+4+5+6+7)(1-2+3-4+5-6+7) = 28*4 = 112.

The correct answer is B.
Can you please share the rationale for concluding the statement --
sum of all number = (Top row)*(Leftmost column)?
To illustrate:
Multiply 4 in the top row by every number in the leftmost column:
4*(1...-2...3...-4...5...-6...7) = 4...-8...12...-16...20...-24...28.
The resulting products form the fourth column of the table.

Multiply 5 in the top row by every number in the leftmost column:
5*(1...-2...3...-4...5...-6...7) = 5...-10...15...-20...25...-30...35.
The resulting products form the fifth column of the table.

Extending this logic, we can see that the table shows the results of multiplying every integer in the top row by every integer in the leftmost column.
Thus, the sum of all these products = the sum of all the integers in the table:
(Top row)*(Leftmost column) = (1+2+3+4+5+6+7)(1-2+3-4+5-6+7) = 28*4 = 112.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
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