Is (p+r)(q+r) odd?
1. p is odd, q is odd.
2. q is odd, r is odd.
Which is correct?
a-1 only
b-2 only
c-Both 1 & 2
d-Neither 1 nor 2

• Neither 1 nor 2
• 7 years agoHelpfull: Yes(29) No(4)
• ANS: B
  
  2 only
  
  statement 1: if we assume
  CASE1: r is even
  (odd[p]+even[r])(odd[q]+even[r])
  (odd[p+r])(odd[q+r])
  =>odd*odd= odd
  CASE 2: r is odd
  (odd[p]+odd[r])(odd[q]+odd[r])
  =>even*even= even
  Statement 2:
  odd[q]+odd[r]=even
  even*any number (p+r)=even
  so, with only statement it can be determined that (p+r)(q+r) is not odd
  
  and is b-2 only
• 7 years agoHelpfull: Yes(12) No(6)
• Ans- 1 only
  p is odd, q is odd
  (odd + even)(odd + even)
  (odd)*(odd) = odd
  so answer is p is odd, q is odd
• 7 years agoHelpfull: Yes(10) No(15)
• st1=> p is odd , q is odd
  if r is even then
  P(odd)+r(even)=odd
  q(odd)+r(even)=odd
  result= odd*odd=odd
  else
  if r is odd then
  P(odd)+r(odd)=even
  q(odd)+r(odd)=even
  result even*even= even
  in stmt 1 dosn't give unique answer so stmt one is not ture
  in stmt 2
  q(odd)+r(odd)=even
  so
  (P+r) is odd or even doesn't mean the multiplication of two if one is even the result always be even
  so answer found no form ans: b-2
• 7 years agoHelpfull: Yes(10) No(17)
• I don't know you guys are thinking from which point of view
  But you need Both to ans the question
• 7 years agoHelpfull: Yes(3) No(1)
• stmt 1: take p=3(odd),q=5(odd) , first put r=2(even) we get (p+r) (q+r)=5*7=35(odd),
  now ,put r=9(odd) we get (p+r) (q+r)=12*14=168(even) so stmt 1 not satisfied condition
  stmt 2: take q=3(odd), r=5(odd) first put p=2(even), we get (p+r) (q+r)=7*5=35(odd)
  now put p=3(odd) we get (p+r) (q+r)=8*8=64(even) so stmt 2 also not satisfied
  ANSWER is: d-Neither 1 nor 2
• 6 years agoHelpfull: Yes(3) No(0)
• Answer is b-2
  Sol:
  First consider Statement 1: Let us suppose r is odd, then p(odd)+r(odd)--> even.
  q(odd)+r(odd)--->even.
  Now, let us suppose r is even, then p(odd)+r(even)--->odd.
  & q(odd)+r(even)-->odd.
  Hence nounique answer if we consider just statement 1.
  
  Similarly evaluate this for statement 2, You will find out that statement 2 gives unique answer.
  Hence, answer is b-2.
• 7 years agoHelpfull: Yes(2) No(1)
• p is odd and q is odd. then sum of odd and even gives odd number and final result is product of two odd numbers is odd
• 7 years agoHelpfull: Yes(1) No(2)
• statement 1
  (odd+even)(odd+even)=odd
  (odd+odd)(odd+odd) =even so we can see question is depend upon r we can not tell
  statement 2
  (even+odd)(odd+odd)
  so whatever the value of (p+r) ,(q+r) will be always even so
  it is sufficient to tell that NO
  its not odd
  so ans will be
  only statement 2 required to tell
  Note - for who dont know the concept of data sufficiency
  in question asking is it odd so if so u dont have to proof its odd
  u just have to tell yes or no by statement so if yes then its true if its no its allso true
  only will be wrong if its ans is ambiguos means when its give yes or no both
  u are not sure .
• 7 years agoHelpfull: Yes(1) No(1)
• (p+r).(q+r)= odd (acc. to given situation)
  RULE:- odd*odd=Odd
  Option 1- (3+2).(5+2)= odd
  (Let P and Q as an odd no and R as an even no.)
  5.7= odd
  35= odd
  Option 2- (2+5).(7+5)=odd
  (Let q and r as an odd no. And P as an even no)
  84 is not an odd no
  
  Hence option 1 (1only) satisfy the condition
• 6 years agoHelpfull: Yes(1) No(1)
• 2 only
  q(odd)+r(odd)=even
• 7 years agoHelpfull: Yes(0) No(5)
• 2 only since first have 2 answers with 2 diff types of value of r.But 2nd case
  If p=odd & r is odd then then p+r=even
  (Anything)×(even)=even
  Since this gives answer of our question that given term is not odd.
• 7 years agoHelpfull: Yes(0) No(1)
• p-odd, q-odd, r-odd
  
  (odd+odd)(odd+odd)
  (even)(even)
  even
  is (p+r)(q+r) odd ans d-Neither 1 nor 2
• 6 years agoHelpfull: Yes(0) No(0)
• Neither 1 nor 2
• 5 years agoHelpfull: Yes(0) No(0)