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20 Mar 2020, 05:09
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C
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Difficulty:
55%
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Question Stats:
62% (02:14) correct
38%
(02:26)
wrong
based on 117
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Set S contains 5 integers, labeled A, B, C, D, and E. If the sum of all of the elements of set S is odd, how many of the elements of set S are even?
(1) The sum of A and B is odd.
(2) The product of B, C, and D is odd.
(1) The sum of A and B is odd.
(2) The product of B, C, and D is odd.
20 Mar 2020, 05:49
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Set S contains 5 integers, labeled A, B, C, D, and E. If the sum of all of the elements of set S is odd, how many of the elements of set S are even?
Constraint: A+B+C+D+E=Odd
Asked : Number of even in
S={A,B,C,D,E} ??
(1) The sum of A and B is odd.
(A+B)+C+D+E=Odd
(A+B)=odd so either A/B =Even/Odd
(even+odd)+even+odd+odd=odd
S={even,odd,even,odd,odd} or
S={even,odd,odd,odd,even}
Here set has two(2) evens
Or
(even+odd)+even+even+even=odd
S={even,odd,even,even,even}
Here set has four(4) evens
(Not sufficient)
(2) The product of B, C, and D is odd.
B•C•D = odd —> odd•odd•odd=odd
A+(B+C+D)+E = odd
Odd+(odd+odd+odd)+odd=odd
S={odd,odd,odd,odd,odd}
The set has no even
Or
Even+(odd+odd+odd)+even=odd
S= {even,odd,odd,odd,even}
Here set has two evens
(1+2) set has two evens in
St.(1) S={even,odd,odd,odd,even}
A and E are even and in
St.(2) S={even,odd,odd,odd,even}
A and E are even ,
so we are certain that S has two evens which is element A and E
(sufficient)
Tricky one lolx
Hence,C like Carnivores
Posted from my mobile device
Constraint: A+B+C+D+E=Odd
Asked : Number of even in
S={A,B,C,D,E} ??
(1) The sum of A and B is odd.
(A+B)+C+D+E=Odd
(A+B)=odd so either A/B =Even/Odd
(even+odd)+even+odd+odd=odd
S={even,odd,even,odd,odd} or
S={even,odd,odd,odd,even}
Here set has two(2) evens
Or
(even+odd)+even+even+even=odd
S={even,odd,even,even,even}
Here set has four(4) evens
(Not sufficient)
(2) The product of B, C, and D is odd.
B•C•D = odd —> odd•odd•odd=odd
A+(B+C+D)+E = odd
Odd+(odd+odd+odd)+odd=odd
S={odd,odd,odd,odd,odd}
The set has no even
Or
Even+(odd+odd+odd)+even=odd
S= {even,odd,odd,odd,even}
Here set has two evens
(1+2) set has two evens in
St.(1) S={even,odd,odd,odd,even}
A and E are even and in
St.(2) S={even,odd,odd,odd,even}
A and E are even ,
so we are certain that S has two evens which is element A and E
(sufficient)
Tricky one lolx
Hence,C like Carnivores
Posted from my mobile device
20 Mar 2020, 07:32
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A+B+C+D+E =odd
--> how many of the elements of set S are even?
(Statement1): The sum of A and B is odd.
A+B =Odd
Odd+C+D+E =Odd
--> C+D+E = Odd-Odd = Even
If two of C,D and E are odd, there will be 2 Even
if all of C,D and E are even, there will be 4 Even.
Insufficient
(Statement2): The product of B, C, and D is odd.
BCD= odd
In order BCD to be odd, all of B,C and D must be odd.
--> A+odd+odd+odd+ E= odd --> A+E= odd-(odd+odd+odd) =Even
A+E =even
--> A and E could be EVEN (2 Even)
--> A and E could be ODD (0 Even)
Insufficient
Taken together 1&2,
A+B =Odd
BCD= Odd
--> B is ODD. --> A is Even
Also, A+E =even
--> E is Even too.
There are Two EVEN numbers.
Sufficient .
Answer (C)
--> how many of the elements of set S are even?
(Statement1): The sum of A and B is odd.
A+B =Odd
Odd+C+D+E =Odd
--> C+D+E = Odd-Odd = Even
If two of C,D and E are odd, there will be 2 Even
if all of C,D and E are even, there will be 4 Even.
Insufficient
(Statement2): The product of B, C, and D is odd.
BCD= odd
In order BCD to be odd, all of B,C and D must be odd.
--> A+odd+odd+odd+ E= odd --> A+E= odd-(odd+odd+odd) =Even
A+E =even
--> A and E could be EVEN (2 Even)
--> A and E could be ODD (0 Even)
Insufficient
Taken together 1&2,
A+B =Odd
BCD= Odd
--> B is ODD. --> A is Even
Also, A+E =even
--> E is Even too.
There are Two EVEN numbers.
Sufficient .
Answer (C)
