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If the sum of the first 30 positive odd integers is k, what

Expert replies
by BTGmoderatorDC » Fri Sep 27, 2019 4:15 pm
If the sum of the first 30 positive odd integers is k, what is the sum of first 30 non-negative even integers?

A. k-29
B. k-30
C. k
D. k+29
E. k+30

OA B

Source: e-GMAT
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Source: — Problem Solving |

by Jay@ManhattanReview » Mon Sep 30, 2019 6:09 am
BTGmoderatorDC wrote:If the sum of the first 30 positive odd integers is k, what is the sum of first 30 non-negative even integers?

A. k-29
B. k-30
C. k
D. k+29
E. k+30

OA B

Source: e-GMAT
So, we know that k = 1 + 2 + 3 + ... (30 terms)

The sum of the first 30 non-negative even integers = 0 + 2 + 4 + .. (30 terms)

Note that each of the first 30 non-negative even integers is 1 less than that of the first 30 odd even integers. Since there are 30 terms, the sum of of the first 30 non-negative even integers would be 30 less than k = k – 30

The correct answer: B

Hope this helps!

-Jay
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by swerve » Mon Sep 30, 2019 7:23 am
BTGmoderatorDC wrote:If the sum of the first 30 positive odd integers is k, what is the sum of first 30 non-negative even integers?

A. k-29
B. k-30
C. k
D. k+29
E. k+30

OA B

Source: e-GMAT
Sum of First \(n +\)ve odd integers is \(n^2\)

Sum of First \(n +\)ve even integers is \(n(n+1)\)

If we take the scenario for first 5 even/odd numbers

First \(5 +\)ve odd integers would be 25 which is \(K\) here

First \(5 +\)ve even integers would be \(5\cdot 6 = 30\) (Here "0" is not counted)

"0" is considered as a non-positive and non-negative even integer. and hence will go with \(K-N\).
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by Brent@GMATPrepNow » Mon Sep 30, 2019 7:49 am
BTGmoderatorDC wrote:If the sum of the first 30 positive odd integers is k, what is the sum of first 30 non-negative even integers?

A. k-29
B. k-30
C. k
D. k+29
E. k+30
k = 1 + 3 + 5 + 7 + . . . . . . + 57 + 59

Sum of the first 30 non-negative even integers = 0 + 2 + 4 + 6 + . . . . . . . . + 56 + 58

Notice the following: 0 + 2 + 4 + 6 + . . . . . . . . + 56 + 58 = (1 - 1) + (3 - 1) + (5 - 1) + (7 - 1) + . . . . . . . + (57 - 1) + (59 - 1)
= (1 + 3 + 5 + 7 + . . . . . . + 57 + 59) - (1 + 1 + 1 + 1 + . . . . . + 1 + 1)

ASIDE: since we're finding the sum of 30 integers, we know there are 30 1's in the sum of 1's
So, we can keep going....
= (1 + 3 + 5 + 7 + . . . . . . + 57 + 59) - (30)
= (k) - (30)

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Thu Oct 03, 2019 11:04 am
BTGmoderatorDC wrote:If the sum of the first 30 positive odd integers is k, what is the sum of first 30 non-negative even integers?

A. k-29
B. k-30
C. k
D. k+29
E. k+30

OA B

Source: e-GMAT
If we were to pair up the first 30 non-negative even numbers with their counterparts in the first 30 positive odd numbers, we’d have:

0, 1
2, 3
4, 5
etc.

Notice that in each pair, the even number is one less than its odd counterpart. Since there are 30 such pairings, the sum of the first non-negative even integers is k - 30.

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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