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What is the difference between the sum of all even integers

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What is the difference between the sum of all even integers

by Max@Math Revolution » Wed Jan 09, 2019 12:05 am

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[Math Revolution GMAT math practice question]

What is the difference between the sum of all even integers between 51 and 100 (inclusive) and the sum of all even integers between 1 and 50 (inclusive)?

A. 1,250
B. 2,000
C. 2,500
D. 3,750
E. 5,000
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by fskilnik@GMATH » Wed Jan 09, 2019 8:13 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

What is the difference between the sum of all even integers between 51 and 100 (inclusive) and the sum of all even integers between 1 and 50 (inclusive)?

A. 1,250
B. 2,000
C. 2,500
D. 3,750
E. 5,000
$${S_A}\,\,\, = \,\,\,2 + 4 + 6 + \ldots + 50$$
$${S_B}\,\,\, = \,\,\,\left( {50 + 2} \right) + \left( {50 + 4} \right) + \left( {50 + 6} \right) + \ldots + \left( {50 + 50} \right)\,\,\, = \,\,\,{S_A} + 25 \cdot 50$$
$$?\,\,\, = \,\,\,{S_B} - {S_A}\,\,\, = \,\,\,25 \cdot 50\,\,\, = \,\,\,1250\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {\rm{A}} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by swerve » Wed Jan 09, 2019 9:26 am
Each set has 25 even consecutive numbers, and the difference between each element is 50,
(2, 4, 6, . . . , 50)
and
(52, 54, 56, . . . , 100).

So, the difference between the two sets is 25*50 = 1250.
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by Max@Math Revolution » Fri Jan 11, 2019 3:32 am
=>

( 52 + 54 + ... + 100 ) - ( 2 + 4 + ... + 50 ) = ( 52 - 2 ) + ( 54 - 4 ) + ... + ( 100 - 50 ) = 50 + 50 + ... + 50 = 50 * 25 = 1250.

Therefore, the answer is A.
Answer: A
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by Scott@TargetTestPrep » Mon Jan 21, 2019 5:36 pm
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

What is the difference between the sum of all even integers between 51 and 100 (inclusive) and the sum of all even integers between 1 and 50 (inclusive)?

A. 1,250
B. 2,000
C. 2,500
D. 3,750
E. 5,000
To find the sum of a set of numbers, we first find the arithmetic average of the set and then multiply by the number of items in the set.

The sum of the even integers from 51 to 100, which is really from 52 to 100, is:

sum = average x number of items = (100 + 52)/2 x [(100 - 52)/2 + 1]

sum = 76 x 25 = 1,900

The sum of the even integers from 1 to 50, which is really from 2 to 50 is:

sum = average x number of items = (2 + 50)/2 x [(50 - 2)/2 + 1]

26 x 25 = 650

Thus, the difference is 1,900 - 650 = 1,250.

Alternate solution:

(52 + 54 + ... + 98 + 100) - (2 + 4 + ... + 48 + 50)

= (52 - 2) + (54 - 4) + ... + (98 - 48) + (100 - 50)

= 50 + 50 + ... + 50 + 50 (Note: Since there are there are twenty-five pairings, there are twenty-five 50s.)

= 50 x 25

= 1,250

Answer: A

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