Set P consists of the first n positive multiples of 3 and set Q consis

Question

Set P consists of the first n positive multiples of 3 and set Q consists of the first m positive multiples of 5. The sum of all the numbers in set P is equal to R and the sum of all the numbers in set Q is equal to S. If m and n are positive integers, is the difference between R and S odd?

(1) m is odd and n is even

(2) m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x, where x is a positive integer

(1) m is odd and n is even

(2) m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x, where x is a positive integer

chetan2u · Answer

Let us check for pattern...  
Multiples of 3...3,6,9,12,15,18,21...
So n=1, sum=3=odd
n=2, sum =3+6=9=odd
n=3, sum =3+6+9=18=even
n=4, sum=18+12=30=even
n=5, sum is odd
 So O,O,E,E,O,O

same will be the case for multiple of 5 ...... O,O,E,E,O,O,E,E ...
  So we require to know n and m or ATLEAST in terms of multiples of 4 as pattern repeats after every 4th number.. that is 4k-1, 4k+3 etc..

Statement I nothing much
Statement II sum of m is surely E, as OOEEO.. but 2x can be anything..

Combined..
 S is even but R can be anything.. 
Examples 
 S is always EVEN ,
But if n is 2,  R will be O  (o,o, e,e,o,o..) but if n is 4,  R is E  (o,o,e,e,o,o..)
So answer will vary..

E

Raja1406 · Answer

Given that 
P={3,6,9,...….3n}
Q={5,10,......5m}
Sum of a series with k terms , first term=a and last term=b is
Sum= k/2*(a+b))
so Sum of P= n/2*(3+3n)= 3n/2*(n+1) here n or n+1 is even and may be 2,4,6,----
so Sum of P =odd or even will depend on value of n or n+1
ex- if n is even and n>2 then Sum will always be even and if n=even but n=2 the sum= odd always
Similarly for Sum of Q= m/2*(5+5m)= 5m/2*(m+1) will depend on actual value of m or m+1
Stat 1--m=odd & n=even
Sum of P may be even or odd depending on actual value of 'n'
Sum of Q may be even or odd depending on actual value of 'm+1'
So R-S may be odd or even ----- not sufficient
Stat 2----m=4x+3 so m+1 =4(x+1) a multiple of 4 so Sum of Q= always even
but as n=2x it may be multiple of 4 or not So sum of P= even/odd
So R-S= even/odd-even=may or may be odd---- not sufficient
Stat1+2--- also doesn't give actual value of n and m so not sufficient
Ans ---E

Archit3110 · Answer

took > 2mins 
the bottom line of this question is the learning of the fact that with an odd or even set of multiple the sum of multiples varies;
#1
m is odd and n is even
m= 5 
n=3,6=9
R-S= even
now m =5,10,15; 30
n=3,6,9,12,15 ; 45
R-S = 15 odd 
insufficient
#2
m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x, where x is a positive integer
with this we can say that irrespective of value of x 
m= will always be odd
and n also be even
but the count of m & n will vary the values of sum of R & S so we cannot conclusively say whether R-S will be odd or even
IMO E

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Set P consists of the first n positive multiples of 3 and set Q consis

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