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GRE Prep Club Legend  Joined: 07 Jun 2014
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The sum of all the odd integers from 1 to 100, inclusive [#permalink]
Expert's post 00:00

Question Stats: 66% (00:32) correct 33% (01:58) wrong based on 18 sessions
 Quantity A Quantity B The sum of all the odd integers from 1 to 100, inclusive The sum of all the even integers from 1 to 100, inclusive

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Manager Joined: 29 Nov 2017
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Kudos [?]: 78 , given: 99

Re: The sum of all the odd integers from 1 to 100, inclusive [#permalink]
WE need to use the formula like this

sum of the series = number of pair(first+last term)

in this case choice B is 50(1+100) =5050
choice A 50(1+99)= 5000
hence option B is correct answer.
Director Joined: 20 Apr 2016
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Re: The sum of all the odd integers from 1 to 100, inclusive [#permalink]
IshanGre wrote:
in this case choice B is 50(1+100) =5050
.

Can you plz check it again? as 1 is not even number

and I think it should be = 25 * 102 =2550
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Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html Director Joined: 20 Apr 2016
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Re: The sum of all the odd integers from 1 to 100, inclusive [#permalink]
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sandy wrote:
 Quantity A Quantity B The sum of all the odd integers from 1 to 100, inclusive The sum of all the even integers from 1 to 100, inclusive

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Sum of even number =$$2 + 4 + 6 + 8 + ......100$$

it can be written as = $$2 * ( 1 + 2 + 3 + 4 + ....50) = 2 * \frac{{50 * 51}}{2} = 50 * 51 = 2550$$(sum of first n numbers = $$n* \frac{(n +1)}{2})$$

Now sum of odd numbers from 1 + 3 + ...100 = [ sum of n numbers ] - [ sum of even nos. from 1 to 100] = $$\frac{{100 * 101}}{2} - 2550 = 5050 - 2550 = 2500$$

Therefore

QTY A < QTY B

----------------------------------------------------------------------------------------------

One more option to solve this problem

Sum of even numbers from 1 ......100 i.e. = 2 + 4 + 6 + ....96 + 98 + 100

now add the first term + last term = 2 + 100 = 102

then add the second term + 2nd last term = 4 + 98 = 102

then add the third term + 3rd last term = 6 + 96 = 102

So as we move inward, adding each pair of numbers, we get 102 every time.

We know there are 50 numbers in this set, so we can use 25 pairs of numbers = 25 * 102 = 2550.

Similarly if we take the sum of odd nos i.e. 1 + 3 + 5 + .........95 + 97 + 99 (since 99 is the last odd number)

as we add the first and last numbers in the series, we get 1 + 99 = 100.

Adding the second and second-to-last numbers = 3 + 97 = 100

As we move inward, adding each pair of numbers, we get 100 every time.

We know there are 50 numbers in this set, so we can use 25 pairs of numbers.= 25 * 100 = 2500

Therefore QTY A < QTY B
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Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html Intern Joined: 21 Mar 2019
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Re: The sum of all the odd integers from 1 to 100, inclusive [#permalink]
1
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Quantity A
Number of odd integers:
[(99-1)/2]+1=50

Sum of odd integers:
[(99+1)/2]*50 =
50*50=
2500
-----
Quantity B
Number of even integers:
[(100-2)/2]+1=50

Sum of even integers:
[(100+2)/2]*50 =
51*50 =
2550

Answer B
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Diana

If you found this post useful, please let me know by pressing the Kudos Button Re: The sum of all the odd integers from 1 to 100, inclusive   [#permalink] 25 Mar 2019, 17:44
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