What is the fastest way to tackle the question attached?
I started by figuring out the sum from top to bottom (i.e. 1-2+3-4+5-6+7 = 4)
Then I noticed a consecutive pattern = 4 + 8 + 12 + 16 + 20 + 24 + 28
Therefore (middle number x number of terms)
= 16 x 7
= 112
Answer: B
Is this the most efficient way? Please advise, thanks.
GMAT Prep 1_ PS Number Properties #15
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I used the same logic
if u take a close look, u will see that every row is just a set of consecutive integers.
since the sum of the set of consecutive integers is equal to median*number of integers , we have the following
4*7+(-8)*7+12*7+(-16)*7+20*7+(-24)*7+28*7 =
=7(4-8+12-16+20-24+28)=7*16=112
if u take a close look, u will see that every row is just a set of consecutive integers.
since the sum of the set of consecutive integers is equal to median*number of integers , we have the following
4*7+(-8)*7+12*7+(-16)*7+20*7+(-24)*7+28*7 =
=7(4-8+12-16+20-24+28)=7*16=112
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i remember seeing a response for this question from one of the experts and i think it was Mitch Hunt. Anyway, here is the approach given:
if you notice the table shows a multiplication table, i.e row 1 * column 1
therefore sum = (sum of numbers in row 1) * (sum of numbers in col 1)
sum = (1+2+3+4+5+6+7) * (1-2+3-4+5-6+7)
sum = 28 * 4
sum = 112
if you notice the table shows a multiplication table, i.e row 1 * column 1
therefore sum = (sum of numbers in row 1) * (sum of numbers in col 1)
sum = (1+2+3+4+5+6+7) * (1-2+3-4+5-6+7)
sum = 28 * 4
sum = 112