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

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

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Updated on: 24 Oct 2014, 03:15
1
9
00:00

Difficulty:

25% (medium)

Question Stats:

75% (01:40) correct 25% (01:39) wrong based on 209 sessions

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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?

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

Originally posted by Drik on 06 Jan 2013, 01:24.
Last edited by Bunuel on 24 Oct 2014, 03:15, edited 2 times in total.
Edited the question.
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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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06 Jan 2013, 02:09
2
2
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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Re: If a equals the sum of the even integers from 2 to 200, inclusive, and  [#permalink]

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07 Jan 2013, 02:05
1
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?

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

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Hope it helps.
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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, 20:28
2
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

Thus their difference (a - b) = 100

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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, 21:07
1
Hi All,

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

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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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23 Mar 2015, 02:59
A: 2,4,6,8,10,12,14

B: 1,3,5,7, 9, 11,13

see that every number in A is exactly 1 more than such number in B.

difference is exactly the number of term in A and B and it is 100

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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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30 Dec 2015, 07:52
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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Re: If a equals the sum of the even integers from 2 to 200, inclusive, and  [#permalink]

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15 Dec 2016, 06:11
Here is my solution to this one =>
a= 100/1[2+200] = 50*202
b= 100/2[1+199]=50*200

Hence a-b= 50[202-200] = 50*2 = 100

Hence B

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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 Mar 2019, 06:04
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Re: If a equals the sum of the even integers from 2 to 200, inclusive, and   [#permalink] 13 Mar 2019, 06:04
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