Set M is composed of the positive even integers up to 100. Set N is co

Question

Set M is composed of the positive even integers up to 100. Set N is composed of the odd integers from –1 to 99. What is the value of (the sum of Set M) – (the sum of Set N)?

A) 49 
B) 50
C) 51
D) 100
E) 101

push12345 · Accepted Answer

Positive even integers=2,4,6,8.........100
Odd integers= -1,1,3,5,7.......99

Keep -1 aside for now

We have 50 pairs of
(2-1)+(4-3)+(6-5).....and so on

So 50*1=50

Including -1 now
We get

50-(-1)=51

Give kudos if it helps

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saraheja · Answer

M = Positive Even Integers = (2, 4 ,6 ...100 )

N = Odd Integers = ( -1 , 1 , 3 , .... 99 )

M will have total of 50 Integers = Sum will be n^2 + n

N will have total of 51 integers , lets inlcude -1 from the set and only consider the integers from 1 to 99 , so sum of odd integers will be n^2

Now M - N will be n^2 + n - n^ = n ( n=50 )

now considering -1 ( which we exclued earlier ) so result will be 50 - (-1) = 51

souvonik2k · Answer

Set M = 2+4+6+....+100
Set N = -1+1+3+5+....+99
Let us first neglect -1 in set N, Then M-N = (2-1)+(4-3)+(6-5).....(100-99) = 1+1+1....+1=50
Now since N also contains -1,subtracting -1 from the above = 50-(-1)=51
Answer C.

EgmatQuantExpert · Answer

2 cents:

We can look into different ways of finding out the individual sums of the given series.

Set M = {2, 4, 6, …, 100}  
Sum of the elements of M = 2 + 4 + 6 + … + 100

Method 1:  
 2 + 4 + 6 + … + 100
= 2 (1 + 2 + 3 + … + 50)
=  2 * \frac{(50*51)}{2} = 50 * 51

Method 2: 
 2 + 4 + 6 + … + 100
This is an arithmetic progression with 50 terms
Hence sum =  \frac{50}{2}   = 25 (4 + 49 * 2) = 25 * 2 (2 + 49) = 50 * 51

Method 3: 
 2 + 4 + 6 + … + 100
This expression denotes sum of first 50 even natural numbers
We know, sum of first n even natural numbers is calculated by  n^2  + n = n (n + 1)
Hence, sum of this series = 50 (50 + 1) = 50 * 51

Set N = {-1, 1, 3, 5, …, 99}  
Here we will consider the values barring -1 initially, and then add -1 with the computed result
Initial sum = 1 + 3 + 5 + … + 99

Method 1: 
 1 + 3 + 5 + … + 99
This is an arithmetic progression with 50 terms
Hence, sum of this series =  \frac{50}{2}   = 25 (2 + 49 * 2) = 25 * 2 (1 + 49) = 50 * 50 = 50^2

Method 2: 
 1 + 3 + 5 + … + 99
This expression denotes sum of first 50 odd natural numbers
We know, sum of first n odd natural numbers is calculated by  n^2  
Hence, sum of this series =  50^2

Method 3: 
 1 + 3 + 5 + … + 99
= (1 + 2 + 3 + … + 100) – (2 + 4 + 6 + … + 100)

With the result obtained, we need to add -1 to get the final sum.

BrentGMATPrepNow · Answer

Set M is composed of the positive even integers up to 100.   
Set M = { 2, 4, 6, 8, . . . . 96, 98, 100 }

Set N is composed of the odd integers from –1 to 99  
Set N = { -1, 1, 3, 5, . . . 95, 97, 99 }

What is the value of (the sum of Set M) – (the sum of Set N)?  
SUM of set M =  2 + 4 + 6 + 8 + . . . .+ 96 + 98 + 100 
SUM of set N =  -1 + 1 + 3 + 5 + . . . 95 + 97 + 99

ASIDE: Notice that there are 100 POSITIVE integers from 1 to 100 inclusive
HALF of them are EVEN and HALF are ODD

So,  set M  consists of 50 integers
and  set N  consists of 51 integers (since set N also has one NEGATIVE odd number)

(the sum of Set M) – (the sum of Set N) = ( 2 + 4 + 6 + 8 + . . . .+ 96 + 98 + 100 ) - ( -1 + 1 + 3 + 5 + . . . 95 + 97 + 99 )
= ( 2 + 4 + 6 + 8 + . . . .+ 96 + 98 + 100 ) - ( 1 + 3 + 5 + . . . 95 + 97 + 99 )   + 1      
= ( 2   -  1 ) + ( 4   -  3 ) + ( 6   -  5 ) + . . . + ( 98   -  97 ) + ( 100   -  99 )   + 1  
= (1) + (1) + (1) + . . . + (1) + (1)   + 1  
= 50   + 1      
= 51

Answer: C

Cheers,
Brent

exc4libur · Answer

num.terms=last(multiple)-first(multiple)/multiple+1
sum.terms=n/2*(2first+multiple(n-1))

M=positive.even
M: n=100-2/2+1=50
M: sum=50/2(2*2+2(50-1))=50/2*(102)

N=odds
N: n=99-(-1)/2+1=51
N: sum=51/2(2*2+2(51-1))=51/2*(98)

M-N sums
50/2*(102)-51/2*(98)=50*51-51*49=51(1)=51

Ans (C)

Crytiocanalyst · Answer

Set M = 2+4+6+....+100

Set N = -1+1+3+5+....+99

T M-N = (2-1)+(4-3)+(6-5).....(100-99) = 1+1+1....+1=50-(-1) = 51

Therefore IMO C

HarshavardhanR · Answer

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