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
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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 =
2500Therefore QTY A < QTY B