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gmatprep - sum question

Expert replies
Source: — Problem Solving |

by sethids » Wed Sep 17, 2008 1:36 pm
Observe closely the first 2 rows.
We need to sum these two and the sum of the rest of the rows will follow. the rows are positive and negative alternately.
Also, the numbers in rows 3, 5, and 7 are the multiples of the numbers in row 1. Sum row 1 and to find the sum of rows 3, 5, and 7, multiply the sum of row 1 which is 28 by the corresponding number (which happens to be equal to the row number for the positive rows in this case).
Sum row 3 = 28*2
Sum row 5 = 28*5
Sum row 7 = 28*7

Similarly sum row 2 = -56
Sum Row 4 = -56*2
Sum row 6 = -56*3

Adding all these you get 28
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by cramya » Wed Sep 17, 2008 1:43 pm
I think the sum is 112.

Each row is a sequence in which terms differ by a constant and are arranged in ascending order.

For example for row 1 the sum is = number of terms*mean(middle term here since th e median is the same as mean)

7*8 = 56

If u do it for all we get -336 and +448

The difference is 112. The answer should be 112 and not 28

Are u saying u picked 28 or the correct answer is 28?
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by lunarpower » Thu Sep 18, 2008 1:14 am
there's a whole bunch of ways to do this problem.
here are two.

(1) use COLUMNS (probably the easiest way - definitely the smallest numbers):
the first column adds to -1 - 1 - 1 + 7 = 4.
the second column is exactly twice the first column, so it's 2(4).
the third column is exactly 3 times the first column, so it's 3(4).
etc.
so the total of all the columns is 1(4) + 2(4) + 3(4) + ... 7(4)
= (1 + 2 + 3 + ... + 7)(4)
= 28 x 4
= 112

(2) use ROWS
the first row adds to 28.
the second row is -2 times the first row, so it's -2(28).
the third row is 3 times the first row, so it's +3(28).
etc.
so the total of all the rows is 28 - 2(28) + 3(28) - ... + 7(28)
= (1 - 2 + 3 - 4 + 5 - 6 + 7)(28)
= (4)(28)
= 112

done
mm hmm
oh yeah.
Ron has been teaching various standardized tests for 20 years.

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by sethids » Mon Sep 22, 2008 1:53 pm
Apologies... the answer should be 112 and not 28 (as mentioned in my earlier post)
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