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If x is the sum of the first 50 positive even integers and y is the

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nigina93
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink] New post   12 Jun 2018, 05:55
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I used the following method.
The sum of first 50 even integers is nu of terms* av or 50/2+1=26* 25 (is av)
The sum of first 50 odd integers is nu of terms*av 50/2-1=24*25.
So we have 26*25-24*25=25(26-24)=25*2=50
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oskarw93s
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink] New post   04 Aug 2018, 04:04
First all the answer-choices are non-negative, so figured x would must be equal or bigger to y. Helped in my thinking.

Started with taking 2 (first even integer) + 100 (50th even integer)/2, average=51. Then multiple the number by the amount of even integers=50
51*50.

Then first odd and 50th odd--> 1+99/2=50, 50*50(amount of odd integer)

So (51*50)-(50*50), you dont have to do the calculation, you simply see its gonna be 50 left when (50*50) is subtracted.

Hence, we got our answer :D
Thumbs up if you thought it was a good explanation :-) :thumbup:
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Probus
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink] New post   31 Aug 2018, 23:07
Hi,

well there are several super explanations given by experts which we must internalize to better our qunat score. However, just an alternate approach.

If we look at set of numbers which we get in x { 2,4,6,8,10,12.......100} and similarly set of numbers y {1,3,5,7,9...99)

now x-y would be {2 -1 + 4-3 + 6-5+ 8-7+....... 100-99}
if we just see the set performs for first 5 even odd we can then infer about the reaming set
So in first 5 even -odd it would be { 2 -1 + 4-3 + 6-5+ 8-7 +10-1}= 5

Now since we have for first 5 (even -odd) the result is 5 , so there are 10 such sets , so the answer must be 10 *5= 50


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shreyasawhney
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink] New post   20 Sep 2018, 22:17
Sum of first 50 odd numbers: 1+3+....+97+99=100x25=2500
(bunch the numbers together, 99+1=100, 97+3=100 and there are 25 such pairs)

Sum of first 50 even numbers: 2+4+...+98+100=102x25=2550
(bunch the numbers together, 100+2=102, 98+4=102 and there are 25 such pairs)

so 2550-2500=50
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west95
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink] New post   16 Oct 2019, 09:19
I dealt with d first 10 integers (even and odd)
Even
2+4+6+8+10=30
Odd
1+3+5+7+9=25

Even-odd ; 30-25=5

We have 10 more integers in 9 places to go and their differences would always be 5

5 in 10 places makes 50

Hope this helps

Posted from my mobile device
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink] New post   22 Feb 2022, 22:10
Sum of first 'n' positive even integers = \(n(n+1)\)
Sum of first 'n' positive odd integers = \(n^2\)

Given:
\(n=50\)
\(X = n(n+1) \)
\(Y = n^2\)

\(X-Y\) = \(n^2 + n - n^2\) = \(n\) = 50 (C)
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adityasuresh
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink] New post   11 Dec 2022, 06:50
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

Sum of terms in a AP =\(N/2\) *(First + Last term)
Sum of first 50 positive even integers = \(50/2\)(2 + 100) = 50 * 51 = X
Sum of first 50 positive odd integers = \(50/2\)(1 + 99) = 50 * 50 = Y
Difference(X -Y) = 50
C
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink] New post   04 Jan 2023, 11:32
What is the formula for the quantity of even and odd integer pairs? Thank you.
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink] New post   15 Jan 2024, 07:44
nishtil wrote:
If x is the sum of the first 50 positive even integers and y is the sum of the first 50 positive odd integers, what is the value of x – y ?

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

PS00605


Sum of numbers formula: n/2 (a1 + aN)
Sum of first 50 even numbers: 50/2 (2+100)
Sum of first 50 odd numbers: 50/2 (1+99)
Therefore :
50/2(102-100)
50/2 x 2
50
Hence C
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Re: If x is the sum of the first 50 positive even integers and y is the [#permalink]
15 Jan 2024, 07:44
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