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If a equals the sum of the even integers from 2 to 20, inclusive, and

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If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?

A. 1
B. 10
C. 19
D. 20
E. 21
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Nov 2014, 04:44, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 04 Apr 2007, 20:36
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The sum is 10.

The fastest way is to notice that since we're taking a-b, all we are going to end up with is addings ten '1' since each even number is one larger than the odd.

Like 2-1 = 1, then 4-3 = 1.... 20-19 = 1

That's just adding up a bunch of 1s.

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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 04 Apr 2007, 23:33
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faifai0714 wrote:
Is there a faster way to calculate this other than adding the even numbers in A and adding the odd numbers in B?


A = 2+4+6 ... +20
B = 1+3+5 ...+19

We want A-B
essentially 2-1 + 4-3 + 6-5 + ... + 20-19

thus ... 1 + 1 + 1 .. + 1 (ten times)

Ans: 10 B

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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 05 Apr 2007, 01:04
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Answer is 10

Yes! there is really a faster way to solve it.

Sum of consecutive odd or even integers = (no. of odd or even ints) * (first int + last int) / 2

Here A = sum of even ints from 2 to 20, inclusive
number of even ints = 10,
first int + last int = 2+20 = 22
A = 10*22 / 2 = 110

B = sum of odd ints from 1 to 19, inclusive
number of odd ints = 10,
first int + last int = 1+19 = 20
A = 10*20 / 2 = 100

A-B = 110 - 100 = 10

Hope it helps :-D

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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 05 Apr 2007, 02:17
Good explanation ywilfred....

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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 12 Jan 2016, 06:02
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This is a solution from Beatthegmat:

http://www.beatthegmat.com/if-a-equals- ... 38885.html

even numbers:

(20-2)/2 + 1 = 10 even integers.
(20+2)/2 = 11 is the average of the even set.
sum = avg*(#of elements) = 11*10 = 110 = a


odd numbers:
(19-1)/2 + 1 = 10 odd integers.
(19+1)/2 = 10 is the average of the odd set.
sum = avg*(#of elements) = 10*10 = 100 = b

a-b = 110 - 100 = 10. (B)
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?

A. 1
B. 10
C. 19
D. 20
E. 21



Hi,
Another way..
Without getting into the sum etc, the answer can be found easily...
2-20 contains 10 integers and similarily 1-19 also has 10 integers..
smallest integer of a is greater by 1 than the corresponding integer of b.. and these is same for all 10 integers..
so a-b=1*10=10
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 07 Feb 2016, 03:18
chetan2u wrote:
faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?

A. 1
B. 10
C. 19
D. 20
E. 21



Hi,
Another way..
Without getting into the sum etc, the answer can be found easily...
2-20 contains 10 integers and similarily 1-19 also has 10 integers..
smallest integer of a is greater by 1 than the corresponding integer of b.. and these is same for all 10 integers..
so a-b=1*10=10


how to count number of odd integers from 1 to n inclusive? Please advise
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If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 07 Feb 2016, 03:29
smartguy595 wrote:
chetan2u wrote:
faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?

A. 1
B. 10
C. 19
D. 20
E. 21



Hi,
Another way..
Without getting into the sum etc, the answer can be found easily...
2-20 contains 10 integers and similarily 1-19 also has 10 integers..
smallest integer of a is greater by 1 than the corresponding integer of b.. and these is same for all 10 integers..
so a-b=1*10=10


how to count number of odd integers from 1 to n inclusive? Please advise


Hi,
If n is odd, add 1to it and divide by 2...
If n is even, straight way divide by 2...
In this example, n is 19.. Add one,so 20..
20/2=10..

smartguy595, at times it might be that you are given some integer in between to another integer ahead..
example, it may not be 1 to n, but n to x..
few points to note in that scenario..

1) both x and n are even
total integers will be (x-n)+1...
even will be 1 more than odd..
even= (x-n)/2+1 and odd=(x-n)/2..

2) x is even and n is odd..
total integers = x-n +1..
even will be equal to odd..
even= odd={(x-n)+1}/2 ..

3)x is odd and n is even..
total integers = x-n +1..
even will be equal to odd..
even= odd={(x-n)+1}/2 ..

4) both x and n are odd
total integers will be (x-n)+1...
odd will be 1 more than even..
odd= (x-n)/2+1 and even=(x-n)/2..

Your query on 1 to 19 will fall under 3) and 4) above..
hope this helps..
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 26 Mar 2016, 18:21
Here's a visual that should help.
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?

A. 1
B. 10
C. 19
D. 20
E. 21


Solution :

a = Sum of all even integers between 2 to 20
Theory : Sum of n even integers = n(n+1)
Therefore a = n^2+n

b = sum of all odd integers between 1 to 19
Theory : Sum of n odd integers = n^2

Therefore a-b = n^2+n-n^2 = n

Here n = no of even integers between 2 to 20 = no of odd integers between 1 to 19 = 10

Therefore, a-b = n = 10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 29 Aug 2016, 01:26
faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?

A. 1
B. 10
C. 19
D. 20
E. 21


Option B
sum of even integers = 10*22/2= 110
sum of odd integers = 10*20/2 =100
a - b= 110-100 =10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and [#permalink]

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New post 02 Sep 2017, 05:14
We can list numbers as 2 4 6 .....20
and odd numbers 1 3 5 ....19
There are a total of 10 even 10 odd numbers so if we subtract each corresponding odd number from even numbers we will get 1 so total will be 10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and   [#permalink] 02 Sep 2017, 05:14
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