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Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
11 Sep 2012, 04:21

5

This post received KUDOS

Expert's post

nishtil wrote:

If X is the sum of first 50 positive even integers and Y is the sum of first 50 positive odd integers, what is the value of x-y?

A. 0 B. 25 C. 50 D. 75 E. 100

----------------------------------------------- Please try to explain your answers

We can solve this question even if we don't know any formula for such sums:

First even minus first odd = 2-1 = 1; The sum of first 2 even integers minus the sum of first 2 odd integers = (2+4)-(1+3) = 2; The sum of first 3 even integers minus the sum of first 3 odd integers = (2+4+6)-(1+3+5) = 3; ... We can see the patterns here, so the sum of first 50 positive even integers minus the sum of first 50 positive odd integers will be 50.

Answer: C.

OR: each even minus its preceding odd is one, so x-y=50 (x-y=(even1+even2+...+even50)-(odd1+odd2+..+odd50)=(even1-odd1)+(even2-odd2)+...+(even50-odd50)=1+1+...+1=50).

Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
27 Mar 2013, 03:33

1

This post received KUDOS

Expert's post

jmuduke08 wrote:

fameatop wrote:

Shortcut way without even a single calculation:-

Sum of first 'n' even integers is given by - n(n+1) Sum of first 'n' odd integers is given by - n^2

x = n(n+1) = 50 x 51 y= n^2 = 50 x 50 x-y = 50 (51-50) = 50 (1) = 50 Answer C

Hope It helps

do these above formulas always hold true? And what is the formula for the first 100 positive integers?

Sum of n first positive integers: \(1+2+...+n=\frac{1+n}{2}*n\). So, the sum of 100 first positive integers is (1+100)/2*100.

Sum of n first positive odd numbers: \(a_1+a_2+...+a_n=1+3+...+a_n=n^2\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n-1\). Given \(n=5\) first odd positive integers, then their sum equals to \(1+3+5+7+9=5^2=25\).

Sum of n first positive even numbers: \(a_1+a_2+...+a_n=2+4+...+a_n\)\(=n(n+1)\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n\). Given \(n=4\) first positive even integers, then their sum equals to \(2+4+6+8=4(4+1)=20\).

Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
25 Apr 2013, 02:06

1

This post received KUDOS

I'll post two ways to solve this question : The formal way (in order for you guys to better understand the theory behind it) and the GMAT way (within the 2-minute scope).

Let's start.

1st method : The GMAT way

Bunuel actually proposed the easiest and fastest method to solve this question. That is you should look for patterns through examples.

First 2 even numbers : 2, 4 => summed up : (2+4) = 6 First 2 odd numbers : 1, 3 => summed up : (1+3) = 4

Their difference will be 2.

First 3 even numbers : 2, 4, 6 => summed up : (2+4+6) = 12 First 3 odd numbers : 1, 3, 5 => summed up : (1+3+5) = 9

Their difference will be 3.

And so forth. So eventually the difference between the sum of the first 50 even integers and the first 50 odd integers will be 50. Which is answer choice C.

2nd method : The formal way

To use this method you should be familiar and comfortable with :

- The general form of an even integer which is 2n ; - The general form of an odd integer which is 2n+1 ; - Counting the number of consecutive integers within a list which is given by the following formula : (Last number - First number) + 1 ; - The sum operator and manipulating it.

Therefore the difference between the sum of the 50 first even integers and the 50 first odd integers is written as such :

Attachment:

pic3.jpg [ 6.93 KiB | Viewed 4691 times ]

You'll notice two things :

- I chose to index the sums from 0 to 49 since the first even integer is 0 and the first odd integer is 1 ; - I inverted the difference since I've written Y-X instead of X-Y. This is due to the general form of the odd integer which if left as the original question stem suggests would leave me with a (-1) instead of 1.

If we develop the difference above we get :

Attachment:

pic4.jpg [ 10.55 KiB | Viewed 4690 times ]

Which, unsuprisingly, yields 50 which is answer choice C.

Note that you can combine two sums if and only if they have the same index range (0 to 49 in both cases).

Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
11 Sep 2012, 05:15

Nice way bunuel..!!

and i did it in this way..

first even and last even ..2 and 100 respectively, 100+2=102/2=51..51 is average of first 50 even integers=num of terms*average =50*51=2550..so sum of first 50 integers is 2550

first odd and last odd of first 50 intergers is =99 and 1..99+1=100/2=50.. so 50*50=2500..sum of first 50 odd integers is 2500..

2550-2500=50... ans c. _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
30 Mar 2013, 12:06

Bunuel wrote:

jmuduke08 wrote:

fameatop wrote:

Shortcut way without even a single calculation:-

Sum of first 'n' even integers is given by - n(n+1) Sum of first 'n' odd integers is given by - n^2

x = n(n+1) = 50 x 51 y= n^2 = 50 x 50 x-y = 50 (51-50) = 50 (1) = 50 Answer C

Hope It helps

do these above formulas always hold true? And what is the formula for the first 100 positive integers?

Sum of n first positive integers: \(1+2+...+n=\frac{1+n}{2}*n\). So, the sum of 100 first positive integers is (1+100)/2*100.

Sum of n first positive odd numbers: \(a_1+a_2+...+a_n=1+3+...+a_n=n^2\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n-1\). Given \(n=5\) first odd positive integers, then their sum equals to \(1+3+5+7+9=5^2=25\).

Sum of n first positive even numbers: \(a_1+a_2+...+a_n=2+4+...+a_n\)\(=n(n+1)\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n\). Given \(n=4\) first positive even integers, then their sum equals to \(2+4+6+8=4(4+1)=20\).

Bunuel, if I wanted to find the sum of the even integers between 26 and 62, inclusive, then the formula above for even integers n(n+1) would not work, because this formula is only for the first n even positive integers meaning we would need to start at 2. Is this the correct way to think about it?

Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
16 Apr 2013, 10:21

[quote= Bunuel, if I wanted to find the sum of the even integers between 26 and 62, inclusive, then the formula above for even integers n(n+1) would not work, because this formula is only for the first n even positive integers meaning we would need to start at 2. Is this the correct way to think about it?[/quote]

Sum of x consecutive even integers = 2xn + x(x+1)/2 (n = (First term of series/2) - 1)

Sum of x consecutive odd integers = (x+n)^2 - (n)^2 (n=(first term of series - 1)/2) _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

Re: If X is the sum of first 50 positive even integers and Y is [#permalink]
09 Dec 2013, 00:23

Expert's post

iyersu wrote:

Question: Is the number 0 even or odd? Answer: 0/2 = 0. The result is an integer, so the number 0 is divisible by 2. As a result, the number 0 is even

i read this in one of the manhatttan forums...hence the doubt..... could you throw some light based on the above info

As written above: yes, zero is an even number but the questions talks about positive even numbers and since zero is neither positive nor negative, we do not consider 0 for this question.

THEORY: 1. EVEN/ODD

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.

An odd number is an integer that is not evenly divisible by 2.

According to the above both negative and positive integers can be even or odd.

2. ZERO

Zero is an even integer. Zero is nether positive nor negative, but zero is definitely 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 (in fact zero is divisible by every integer except zero itself).

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