Data Sufficiency

[email protected] wrote:The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?

A. 5100
B. 7550
C. 10100
D. 15500
E. 20100

The sum of the first 50 positive even integers is 2550.
In other words, 2 + 4 + 6 + 8 + ...+ 98 + 100 = 2550

We want the sum: 102 + 104 + 106 + . . . 198 + 200
IMPORTANT: Notice that each term in this sum is 100 greater than each term in the first sum.
In other words, 102 + 104 + 106 + . . . 198 + 200 is the SAME AS...
(100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 98) + (100 + 100)
We can rearrange these terms to get: (100 + 100 + ... + 100 + 100) + (2 + 4 + 6 + 8 + ...+ 98 + 100)

IMPORTANT: There are 50 100's in the red sum, and we're told that the blue sum = 2550

So, our sum = 50(100) + 2550
= 5000 + 2550
= 7550
= B

Cheers,
Brent

[email protected] wrote:The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?

A. 5100
B. 7550
C. 10100
D. 15500
E. 20100

Another approach:

We want to evaluate: 102 + 104 + 106 + . . . 198 + 200
Since the terms in this series are EQUALLY SPACED, the sum = (average of first and last term)(# of terms)
So, sum = [(102 + 300)/2][50]
= [(302)/2][50]
= [151][50]
= 7550
= B

Cheers,
Brent

Hi shibsriz,

There's a variation on Brent's approach called "bunching", which can also be used to answer this question.

Since we're dealing with the EVEN integers from 102 to 200, we have a total of 50 numbers

Adding the biggest (200) and smallest (102) gives us 302
Adding the next biggest (198) and next smallest (104) also gives us 302

With 50 numbers, we have 25 groups of two. Each group of two = 302

25(302) is approximately 25(300), which = 7500

There's only one answer that's close to 7500

Final Answer: B

GMAT assassins aren't born, they're made,
Rich