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

Question

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

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amanvermagmat · Accepted Answer

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

eliaslatour · Answer

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).

EgmatQuantExpert · Answer

Reserving this space to post the official solution.

Luckisnoexcuse · Answer

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)

mehrotrayashraj · Answer

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

sjavvadi · Answer

Answer, I think, is k-30 i.e. choice B.

Solution Attached.

pushpitkc · Answer

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

ladyrenee95 · Answer

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).

hellosanthosh2k2 · Answer

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

stne · Answer

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 !

ChuHoaiNam · Answer

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.

Thanks and best regards

BrentGMATPrepNow · Answer

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

juliahamm24 · Answer

Hi Brent, 
I'm wondering how we know to start with 0 instead of 2 in this scenario?

BrentGMATPrepNow · Answer

The even integers look like this: .......-8, -6, -4, -2, 0, 2, 4, 6, 8, 10, ....
The key here is the word "non-negative" (as in "...what is the sum of first 30  non-negative  even integers?")
If we remove all of the negative even numbers from the list, the remaining numbers are: 0, 2, 4, 6, 8, 10, ....

ASIDE: A lot of students see the word "non-negative" and misinterpreted as meaning positive.

9962882832 · Answer

Why are we classifying 0 as an even number . It is non-negative integer but its not even right?

Bunuel · Answer

Zero is neither negative nor positive but it's an even number. An even number is an   integer   that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

ScottTargetTestPrep · Answer

Solution:

The first 30 positive odd integers are: 1, 3, 5, …, 59.

The first 30 non-negative even integers are: 0, 2, 4, …, 58.

We see that each of the first 30 non-negative even integers is 1 less than its counterpart in the first 30 positive odd integers. Therefore, the sum of the first 30 non-negative even integers will be 30 x 1 = 30 less than the sum of the first 30 positive odd integers. Since the sum of the first 30 positive odd integers is given to be k, the sum of the first 30 non-negative even integers is therefore k - 30.

Answer: B

Bambi2021 · Answer

Sum of positive odd integers starting from 1 is x^2 where x is the number of odd integers.
Sum of positive consecutive even integers starting from 2 is x(x+1) where x is the number of even integers.

k = 30*30

The even sequence starts from 0 and becomes x(x-1) = 30*29 = k-30

Shohinee · Answer

They say not to be dependent on formulae while solving GMAT questions. But in solving a bunch of problems like these I noticed that the sum of the first 'n' non-negative even nos. (which starts from 0 as opposed to 2 as in the case of a set of n numbers of 'positive' even nos.), is  -2n.

so, sum of first 30 positive odd integers = n^2 = 30^2 = k = 900
sum of first 30 non-negative even integers =  -2n = 900+30-60 =900-30=k-30, and that's option B.

But I do think a better way is to substitute smaller values and find the answer instead of trying to memorize formulae.

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

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

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