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

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

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.

Simply look at the formulas
sum of n odd integers = n^2
sum of n even integers = n^2+n