Sum of even -Sum of odd

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moron: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Tue Oct 06, 2015 9:04 pm; Thanked: 1 times

by moron » Mon Nov 02, 2015 10:44 pm

This question has tedious calculation but I am 100% positive that my calculation is correct (verified using Excel spread sheet and small computer program) . So my concept is wrong and I will present my logic vs the correct logic . An explanation will be appreciated. So here goes the question..

Q If x is the sum of first 50 positive even integers 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

My solution
Using formula n(n+1)/2 for sum of first n numbers.
Sum of all 50 numbers comes to 25(51) --> Lets call it as t (for total)
Sum of first 50 positive even integers 25(26) which is x
So sum of first 50 positive odd integers will be y=t-x=25(51)-25(26)=25(25)
Now x-y
=25(26)-25(25)=25 My answer is B

GMAT Solution
x-y
(2+4+6+8....100)-(1+3+5...+99)
(2-1)+(4-3)+(6-5)+....(100-99)
50(1) So answer is 50 Correct answer is C

What I don't understand why is it x is (2+4+6+8....100) and not (2+4+6+8....50) or y is (1+3+5...+99) and not (1+3+5...+49)
Even if i use GMAT solution and use (2+4+6+8....50) similarly calculate y I get answer B

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Nov 03, 2015 1:22 am

moron wrote:If x is the sum of first 50 positive even integers 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 evenly spaced integers = (number of integers) * (average of biggest and smallest)

x is the sum of first 50 positive even integers.
Smallest = 2.
Biggest = 100.
Average of biggest and smallest = (100+2)/2 = 51.
Thus:
x = (number of integers)(average of biggest and smallest) = (50)(51).

y is the sum of first 50 positive odd integers.
Smallest = 1.
Biggest = 99.
Average of biggest and smallest = (99+1)/2 = 50.
Thus:
y = (number of integers)(average of biggest and smallest) = (50)(50).

Thus:
x-y = (50)(51) - (50)(50) = 50(51-50) = 50.

The correct answer is C.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Nov 03, 2015 1:39 am

To count EVENLY SPACED INTEGERS:

Number of integers = (biggest - smallest)/interval + 1

The INTERVAL is the distance between one term and the next.

moron wrote: What I don't understand why is it x is (2+4+6+8....100) and not (2+4+6+8....50) or y is (1+3+5...+99) and not (1+3+5...+49)

The sets in red are composed of only THE FIRST 25 positive even integers and THE FIRST 25 positive odd integers.

(2+4+6+8....50):
Since only the EVEN integers are to be counted here, the interval = 2.
Thus:
Number of integers = (biggest - smallest)/interval + 1 = (50-2)/2 + 1 = 25.

(1+3+5...+49):
Since only the ODD integers are to be counted here, the interval = 2.
Thus:
Number of integers = (biggest - smallest)/interval + 1 = (49-1)/2 + 1 = 25.

Since x is the sum of THE FIRST 50 positive even integers and y is the sum of THE FIRST 50 positive odd integers, the correct sets for x and y are {2, 4, 6...96, 98, 100} and {1, 3, 5...95, 97, 99}.

{2, 4, 6...96, 98, 100}:
Since only the EVEN integers are to be counted here, the interval = 2.
Thus:
Number of integers = (biggest - smallest)/interval + 1 = (100-2)/2 + 1 = 50.

{1, 3, 5...95, 97, 99}:
Since only the ODD integers are to be counted here, the interval = 2.
Thus:
Number of integers = (biggest - smallest)/interval + 1 = (99-1)/2 + 1 = 50.

More practice:
https://www.beatthegmat.com/greatest-pri ... 76728.html

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Nov 03, 2015 6:25 am

moron wrote:
Q If x is the sum of first 50 positive even integers 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 first 50 even integers) - (sum of first 50 odd integers)
= (2 + 4 + 6 + 8 + . . . + 98 + 100) - (1 + 3 + 5 + 7 + . . . 97 + 99)
= 2 - 1 + 4 - 3 + 6 - 5 + . . . + 98 - 97 + 100 - 99
= (2 - 1 ) + (4 - 3 ) + ( 6 - 5) + . . . + (98 - 97 ) + ( 100 - 99)
= (1) + (1) + (1) + . . . + (1) + (1)
= 50
= C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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ceilidh.erickson: GMAT Instructor; Posts: 2095; Joined: Tue Dec 04, 2012 3:22 pmFollowed by: 247 members; Thanked: 1443 times

by ceilidh.erickson » Tue Nov 03, 2015 3:55 pm

To extrapolate on Brent's approach... often when you're asked to find differences between the sums of consecutive sets, it's easiest not to calculate, but to think: can I just compare the 1st term in one set to the 1st term in the other, the 2nd to the 2nd, and so on.

As Brent demonstrated, each term in the even set is +1 more than the corresponding term in the odd set. If there are 50 terms, then the even set must be (1)(50) greater.

Try this technique with OG 13/2015 PS 90: https://www.beatthegmat.com/consecutive- ... 09968.html

Here's more on sums of consecutive integers:
https://www.beatthegmat.com/92-x-is-the- ... tml#539349

Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

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Jim@StratusPrep: MBA Admissions Consultant; Posts: 2279; Joined: Fri Nov 11, 2011 7:51 amFollowed by: 266 members; Location: New York; Thanked: 660 times; GMAT Score:770

by Jim@StratusPrep » Tue Nov 03, 2015 5:42 pm

Just think that there are 50 different sets of 2 numbers:

2 - 1 = 1
4 - 3 = 1
6 - 5 = 1
...etc

50 x 1 = 50

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akash singhal: Master | Next Rank: 500 Posts; Posts: 152; Joined: Fri Apr 24, 2015 1:39 amFollowed by: 3 members; Location: Rourkela Odisha India; Thanked: 2 times; GMAT Score:650

by akash singhal » Tue Nov 03, 2015 11:25 pm

moron wrote:This question has tedious calculation but I am 100% positive that my calculation is correct (verified using Excel spread sheet and small computer program) . So my concept is wrong and I will present my logic vs the correct logic . An explanation will be appreciated. So here goes the question..

Q If x is the sum of first 50 positive even integers 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

My solution
Using formula n(n+1)/2 for sum of first n numbers.
Sum of all 50 numbers comes to 25(51) --> Lets call it as t (for total)
Sum of first 50 positive even integers 25(26) which is x
So sum of first 50 positive odd integers will be y=t-x=25(51)-25(26)=25(25)
Now x-y
=25(26)-25(25)=25 My answer is B

GMAT Solution
x-y
(2+4+6+8....100)-(1+3+5...+99)
(2-1)+(4-3)+(6-5)+....(100-99)
50(1) So answer is 50 Correct answer is C

What I don't understand why is it x is (2+4+6+8....100) and not (2+4+6+8....50) or y is (1+3+5...+99) and not (1+3+5...+49)
Even if i use GMAT solution and use (2+4+6+8....50) similarly calculate y I get answer B

X =(2+4+6+8........94+96+98+100)
We can solve this by taking pairs..
As you can see (2+100)=(4+98)=(6+96)=(8+94)......=102
thus, X = 102*25 (as 25 such pairs exists)

Similarly, Y=(1+3+5......95+97+99)
Y=100*25 (similar process involved above)

X-Y= 102*25 - 100*25 = 25*2 = 50 HenceC

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