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If the sum of the first 30 positive odd inte
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Updated on: 07 Aug 2018, 06:54
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Q. If the sum of the first 30 positive odd integers is k, what is the sum of first 30 nonnegative even integers? Answer Choices A. k29 B. k30 C. k D. k+29 E. k+30 To read all our articles:Must Read articles to reach Q51To solve question of the week:Question of the WeekThanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: If the sum of the first 30 positive odd inte
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26 May 2017, 05:10
Reserving this space to post the official solution.
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Re: If the sum of the first 30 positive odd inte
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26 May 2017, 06:20
EgmatQuantExpert wrote: Q. If the sum of the first 30 positive odd integers is k, what is the sum of first 30 nonnegative even integers? Answer Choices A. k29 B. k30 C. k D. k+29 E. k+30 Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts Well, I'm going to start by running a scenario with the first 3 odd integers. Those would be 1, 3, and 5 for a total of 9. The first 3 nonnegative integers would be 0, 2, and 4 for a total of 6. It seems that each even integer is 1 less than its odd counterpart. So my sum is 3 less for the evens because I had 3 terms. So if I use 5 terms, it should be 5 less. Let's test that. 1 + 3 + 5 + 7 + 9 = 15 0 + 2 + 4 + 6 + 8 = 20 Sure enough, each even integer is 1 less than its counterpart. So the answer must be k30. I'll pick answer choice (B).
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Re: If the sum of the first 30 positive odd inte
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26 May 2017, 06:34
EgmatQuantExpert wrote: Q. If the sum of the first 30 positive odd integers is k, what is the sum of first 30 nonnegative even integers? Answer Choices A. k29 B. k30 C. k D. k+29 E. k+30 Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts 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(6) = 30 (Here "0" is not counted) "0" is considered as a nonpositive and nonnegative even integer. and hence will go with KN i.e. (B)
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Re: If the sum of the first 30 positive odd inte
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26 May 2017, 09:24
mynamegoeson wrote: EgmatQuantExpert wrote: Q. If the sum of the first 30 positive odd integers is k, what is the sum of first 30 nonnegative even integers? Answer Choices A. k29 B. k30 C. k D. k+29 E. k+30 Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts 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(6) = 30 (Here "0" is not counted)"0" is considered as a nonpositive and nonnegative even integer. and hence will go with KN i.e. (B) Odd sum is 25 > This means k = 25 Even sum is 30. > This is 5 more than k. n = 5 (No. of elements) > This means Sum for even is k+5 > k+n When n is 30 > k+30



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Re: If the sum of the first 30 positive odd inte
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Updated on: 26 May 2017, 21:10
We should know that in an evenly spaced set (Arithmetic Progression), Mean or average is always = Median Also, for a series having EVEN number of terms, Median = average of two middle terms
First 30 positive odd integers: 1,3,5,7,...59 This is an evenly spaced set hence mean = median = average of two middle terms = average of 15th and 16th terms = (29+31)/2 = 30 Mean=30, So sum = 30*30 (Sum = mean*n) Sum = 900, this is given as K.
Now, first 30 non negative even integers: 0,2,4,6,8,...58 This is again an evenly spaced set hence mean = median = average of 15th and 16th terms = (28+30)/2 = 29 Mean=29, So Sum = 29*30 Sum = 870
If K=900, we can clearly see that 870 = k30
Hence B answer
Originally posted by amanvermagmat on 26 May 2017, 11:06.
Last edited by amanvermagmat on 26 May 2017, 21:10, edited 1 time in total.



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Re: If the sum of the first 30 positive odd inte
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26 May 2017, 13:47
Answer, I think, is k30 i.e. choice B. Solution Attached.
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Re: If the sum of the first 30 positive odd inte
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26 May 2017, 14:00
Since the sum of n positive integers(odd) can be got by simple formula Sum(Odd positive numbers) = n^2, which is equal to k. We can deduce that k = 30^2 as we are asked the sum of 30 odd integers. Coming to the second part of the question, we have a formula Sum(Even positive numbers) = n*n1Also, the sum of the first 30 nonnegative even numbers is 30*29 Since we need to find it in form of k & we already know that k = 30^2, we can use k30 to get the value k30 = 30^2  30 = 30(301) = 30*29 Hence, Option B is the correct answer
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Re: If the sum of the first 30 positive odd inte
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11 Oct 2017, 19:14
Can someone tell me if this is another way of doing these kind of problems? I saw pushpitkc use n(n1) and was wondering if that was just a shortcut from what I did on 2 below.
N^2 = sum of odd numbers N(N+1) = sum of even numbers. To find n = (First Even + Last Even)/2
1. The sum of the first 30 positive odd integers (0 is not included).... N^2=k=30^2=900
2. The sum of first 30 nonnegative even integers (NonNegatives include 0)... 0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58... (58 + 0)/2 = 29 = n 29(29+1)= 870
3. 900  x = 870
90030=870.....
Which is k30..... (B).



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Re: If the sum of the first 30 positive odd inte
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17 Dec 2017, 02:54
first 30 odd sequence: 1,3,5,7..........................................59 => sum = k subtracting one from 1 from each term we get first 30 even sequence: 0,2,4,6.............................58 => sum = k 30 kudos if you like my approach, i need them badly to unlock GMAT club tests. Thanks



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Re: If the sum of the first 30 positive odd inte
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13 Feb 2018, 05:14
EgmatQuantExpert wrote: Q. If the sum of the first 30 positive odd integers is k, what is the sum of first 30 nonnegative even integers? Answer Choices A. k29 B. k30 C. k D. k+29 E. k+30 Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts If one remembers the formula's here's another way: Sum of first n positive odd integers = \(n^2\) =\(30^2\) =K Sum of first n positive even integers = n(n+1) = 29 (30) > (30 1)30 > \(30^2\) 30 = K30 Answer :B Sum of the first 30 positive even integers =n(n+1) Please note this formula was derived taking 2 as the first even positive integer. Now if we are to include 0 as the first term then the 29th even integer is actually the 30th term in this question Hence we need the sum of the first 29 positive even integers = 29(30) ( which is actually the sum of first 30 non negative even integers,adding zero does not change the total.) Hope this helps !
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Re: If the sum of the first 30 positive odd inte
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15 Aug 2019, 09:15
The first 30 positive odd numbers: 1 3 5 7 .... The first 30 nonnegative even numbers: 0 2 4 6 Now, you see: 10 = 1; 32=1; 54=1, 76=1; .... 5958=1. It means: if we have 30 numbers and the sum of the first 30 positive odd number = k , then, the sum of the first 30 non negative even = k 30.
It will takes you less than 30 seconds if you do this way.
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Re: If the sum of the first 30 positive odd inte
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30 Sep 2019, 09:03
EgmatQuantExpert wrote: If the sum of the first 30 positive odd integers is k, what is the sum of first 30 nonnegative even integers?
A. k29 B. k30 C. k D. k+29 E. k+30
k = 1 + 3 + 5 + 7 + . . . . . . + 57 + 59Sum of the first 30 nonnegative even integers = 0 + 2 + 4 + 6 + . . . . . . . . + 56 + 58Notice 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
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Re: If the sum of the first 30 positive odd inte
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