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If the sum of the first 30 positive odd inte

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If the sum of the first 30 positive odd inte  [#permalink]

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New post 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 non-negative even integers?


Answer Choices



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

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Originally posted by EgmatQuantExpert on 26 May 2017, 05:09.
Last edited by EgmatQuantExpert on 07 Aug 2018, 06:54, edited 1 time in total.
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Re: If the sum of the first 30 positive odd inte  [#permalink]

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New post 26 May 2017, 05:10
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Re: If the sum of the first 30 positive odd inte  [#permalink]

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New post 26 May 2017, 06:20
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EgmatQuantExpert wrote:
Q.

If the sum of the first 30 positive odd integers is k, what is the sum of first 30 non-negative even integers?


Answer Choices



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

Thanks,
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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 non-negative 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 k-30.
I'll pick answer choice (B).
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Re: If the sum of the first 30 positive odd inte  [#permalink]

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New post 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 non-negative even integers?


Answer Choices



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

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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 non-positive and non-negative even integer. and hence will go with K-N i.e. (B)
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Re: If the sum of the first 30 positive odd inte  [#permalink]

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New post 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 non-negative even integers?


Answer Choices



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

Thanks,
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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 non-positive and non-negative even integer. and hence will go with K-N 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  [#permalink]

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New post Updated on: 26 May 2017, 21:10
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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 = k-30

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  [#permalink]

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New post 26 May 2017, 13:47
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Answer, I think, is k-30 i.e. choice B.

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Re: If the sum of the first 30 positive odd inte  [#permalink]

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New post 26 May 2017, 14:00
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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*n-1
Also, the sum of the first 30 non-negative 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 k-30 to get the value
k-30 = 30^2 - 30 = 30(30-1) = 30*29
Hence, Option B is the correct answer
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Re: If the sum of the first 30 positive odd inte  [#permalink]

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New post 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(n-1) 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 non-negative even integers (Non-Negatives 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

900-30=870.....

Which is k-30..... (B).
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Re: If the sum of the first 30 positive odd inte  [#permalink]

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New post 17 Dec 2017, 02:54
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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

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Re: If the sum of the first 30 positive odd inte  [#permalink]

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New post 13 Feb 2018, 05:14
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EgmatQuantExpert wrote:
Q.

If the sum of the first 30 positive odd integers is k, what is the sum of first 30 non-negative even integers?


Answer Choices



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

Thanks,
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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 = K-30
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  [#permalink]

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New post 15 Aug 2019, 09:15
The first 30 positive odd numbers: 1 3 5 7 ....
The first 30 non-negative even numbers: 0 2 4 6
Now, you see: 1-0 = 1; 3-2=1; 5-4=1, 7-6=1; .... 59-58=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  [#permalink]

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New post 30 Sep 2019, 09:03
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EgmatQuantExpert 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

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Re: If the sum of the first 30 positive odd inte   [#permalink] 30 Sep 2019, 09:03
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