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The sum of all the odd integers from 1 to 100, inclusive

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The sum of all the odd integers from 1 to 100, inclusive [#permalink] New post 30 Aug 2018, 16:28
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Question Stats:

59% (00:46) correct 40% (01:14) wrong based on 64 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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Re: The sum of all the odd integers from 1 to 100, inclusive [#permalink] New post 01 Sep 2018, 07:59
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.
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Re: The sum of all the odd integers from 1 to 100, inclusive [#permalink] New post 01 Sep 2018, 10:33
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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Re: The sum of all the odd integers from 1 to 100, inclusive [#permalink] New post 01 Sep 2018, 10:44
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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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Re: The sum of all the odd integers from 1 to 100, inclusive [#permalink] New post 25 Mar 2019, 17:44
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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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Re: The sum of all the odd integers from 1 to 100, inclusive [#permalink] New post 24 Sep 2019, 21:41
IshanGre wrote:
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.



IshanGre is fully wrong, incorrect.

Solution:

case 1: for QA: B >> (50/2) * (2+100) = 2550
case 2: for QA: A >> (50/2) * (1+99) = 2500

SO, B is greater;
Answer : B.
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Re: The sum of all the odd integers from 1 to 100, inclusive   [#permalink] 24 Sep 2019, 21:41
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