x, y, and z are positive integers and x

Question

x, y, and z are positive integers and  x < y < z . If  x + y + z  is an even number less than 15, is x equal to 2?

(1) The product of any two numbers of x, y, and z is an even number.
(2) One of x, y, and z is twice the one of the other numbers.

AbbyJ · Answer

For the first statement Given that all the numbers must be even and in increasing order with an even sum less than 15, two combinations are possible: 2,4,6 (Sum=12) or 2,4,8 (sum=14). In both situations, smallest number has to be 2. Hence A is alone sufficient. For the second statement, two cases are possible for an even sum (o,o,e) or (e,e,e) eg, (3,6,5) (sum=14) and of course the previous two combinations identified in statement 1. Thus B alone is not sufficient.

chandnik3 · Answer

Bunuel  Could you please provide the solution?
For Statement 1:- Can't the solution be 3<4<6 ? The sum is also below 15. Also,The product of any two numbers of x, y, and z is an even number.

Bunuel · Answer

The stem says that  x + y + z  is an  even  number less than 15., whereas 3 + 4 + 6 = 13 = odd.

walterwhite756 · Answer

Stmt 1: if product of any two numbers is even then all 3 numbers have to be even and hence minimum can be 2, 4 and 6 which satisfy given constraints and hence  sufficient .

Stmt 2: take cases for example x = 1, y =2 and z = 5, in this case this statement is  insufficient .

Therefore  answer is A.

Navnit04 · Answer

According to question stem x+y+z= even number less than 15 and x<y<z

statement 1) The product of any two numbers of x, y, and z is an even number. For this statement to be true we need at-least two of then to be even but if 2 will even from x,y,and z then x+y+z will not be even number therefore all x,y and z must be even. Now, 
as x<y<z only possible set of number will be x= 2, y=4, and z=6 (x+y+z= 12 <15)

sufficient

Statement 2) One of x, y, and z is twice the one of the other numbers. let y=2x

therefore, 3x+z= even number<15

now consider x=2 and z=6 
therefore 3x+z= 12
let x=1 and z=3
3x+z= 6

not sufficient bz multiple value of x is possible.

x, y, and z are positive integers and x < y < z. If x + y + z

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