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If a equals the sum of the even integers from 2 to 200, inclusive, and [#permalink]

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06 Jan 2013, 00:24

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If a equals the sum of the even integers from 2 to 200, inclusive, and b equals the sum of the odd integers from 1 to 199, inclusive, what is the value of a - b?

Re: If a equals the sum of the even integers from 2 to 200, inclusive, and [#permalink]

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06 Jan 2013, 01:09

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Drik wrote:

If a equals the sum of the even integers from 2 to 200, inclusive, and b equals the sum of the odd integers from 1 to 199, inclusive, what is the value of a - b? [*]10 [*]100 [*]190 [*]200 [*]210

Number of even integers from 2 to 200, inclusive is \((200-2)/2 + 1\) or \(198/2 + 1\) or \(100\). Mean of evenly spaced set that includes all the even numbers from 2 to 200 inclusive is \((200+2)/2\) or \(101\). Therefore a= \(101*100\)

Number of odd integers from 1 to 199, inclusive is \((199-1)/2 + 1\) or \(198/2 +1\) or \(100\). Mean of evenly spaced set that includes all the odd numbers from 1 to 199, inclusive is \((199+1)/2\) or \(100\). Therefore b=\(100*100\)

hence a-b=\((100 * (101-100))\) or \(100\) +1B
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If a equals the sum of the even integers from 2 to 200, inclusive, and b equals the sum of the odd integers from 1 to 199, inclusive, what is the value of a - b?

Re: If a equals the sum of the even integers from 2 to 200, inclusive, and [#permalink]

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24 Oct 2014, 02:00

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Re: If a equals the sum of the even integers from 2 to 200, inclusive, and [#permalink]

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13 Jan 2015, 19:28

PROBLEM: If a equals the sum of the even integers from 2 to 200, inclusive, and b equals the sum of the odd integers from 1 to 199, inclusive, what is the value of a - b?

A. 10 B. 100 C. 190 D. 200 E. 210

SOLUTION:

Use following formulae for such problems:

Sum of evenly spaced integers = (# of integers)*(mean of integers)

# of integers = [(last - first)/2] + 1 Mean of integers = (last + first)/2

In above problem:

# of integers = [(200 - 2)/2] + 1= 100 and [(199-1)/2]+ 1 = 100 Mean of integers = (200 + 2) = 101 and (199 + 1)/2 = 100

Sum of integers = (101*100) = 10100 and (100*100) = 10000

When a GMAT Quant question asks you to deal with a big group of numbers, there's almost always a built-in pattern that can help you to avoid a lengthy "math approach."

Here, we're given two groups:

Group A = the sum of all the EVEN integers from 2 to 200, inclusive Group B = the sum of all the ODD integers from 1 to 199, inclusive

We're asked for the value of A - B

Calculating any of these individual totals would take some work, but notice how the two groups follow a pattern:

(The first term in A) - (The first term in B) = 2 - 1 = 1

(The second term in A) - (The second term in B) = 4 - 3 = 1

This pattern will continue as you compare each successive term in each group

Since we're dealing with just the even numbers in Group A, 200/2 = 100 total terms. This means that we'll have 100 differences of "1" The Total difference will be 100(1) = 100

Re: If a equals the sum of the even integers from 2 to 200, inclusive, and [#permalink]

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30 Dec 2015, 06:52

if we have 100 even and 100 odd then the difference between the sum of even ones and odds one will be 100. logically thinking, each even number will be 1 more than preceded odd number. since we have 100 even numbers, then the difference is 100.

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