X + Y + Z= 10.

(1) X < Y < Z

0 < 4 < 6
2 < 3 < 5
-2 < 4 < 8

in all the above cases x + y + z = 10

Not sufficient.


(2) The three-digit number selected was even.

0 < 4 < 6
-2 < 4 < 8

Not Sufficient.

Both 1 and 2 taken together will also gives multiple even number choices.

Hence Ans is E

Please let me know if my assumption is wrong.

ssgmatter wrote:A child selected a three-digit number, XYZ, where X, Y and Z denote the digits
of the number and X + Y + Z= 10. If no two of the three digits were equal,
what was the three-digit number?
(1) X < Y < Z
(2) The three-digit number selected was even.

(1) x< y< z can be 1 +2+7, 2+3+5 different sets of three nos. Not suff
(2) 3 even nos can be only 2+2+6 or 2+4+4 = 10
but 2 nos are repeated.
Unless it is 0+4+6 or 0+2+8
but zero is neither even nor odd.
Isn't the Q flawed.

Ans should be C

Taking both together the only number that satisfies the 2 conditions is 136

kstv wrote:

ssgmatter wrote:A child selected a three-digit number, XYZ, where X, Y and Z denote the digits
of the number and X + Y + Z= 10. If no two of the three digits were equal,
what was the three-digit number?
(1) X < Y < Z
(2) The three-digit number selected was even.

(1) x< y< z can be 1 +2+7, 2+3+5 different sets of three nos. Not suff
(2) 3 even nos can be only 2+2+6 or 2+4+4 = 10
but 2 nos are repeated.
Unless it is 0+4+6 or 0+2+8
but zero is neither even nor odd.
Isn't the Q flawed.

E is correct !

0 is even....I think u made the same mistake in April...and stuart corrected u bhai

STMT 1:

X < Y < Z

1 < 3 < 6
1 < 2< 7

in all the above cases x + y + z = 10

Not sufficient

STMT 2:


Number z is even so

z can be 0,2,4,6,8

2 0 8
4 0 6

in all the above cases x + y + z = 10

Not sufficient


Taking both together


Number z is even so

z can be 0,2,4,6,8

when z= 8 x,y should be 1,1(all the three are not distinct) or 0 and 1 (x<y<z so X should be 0 but then it can't be three digit number) Wrong

z= 6

possible values for x,y = 1,3 or 0, 4 (wrong ) above stated reason

z= 4 (wrong ) above stated reason

possible values for x,y = 2,4 or 3,3



so ans is c

Hi

I checked to see if there are any answers to this question on other forums and found one that was discussed on manhattan gmat, the ans is E , however, the ques is slightly different from the one discussed here:
"A child selected a three-digit number, XYZ, where X, Y, and Z denote the digits of the number. If no two of the three digits were equal, what was the three-digit number?
(1) The sum of the digits was 10.
(2) X < Y < Z"

But the ques posted here mentions in (2) that the number is even and combining post statements, the only three digit number that satisfies both conditions is 136 (already noted by a previous poster). IMO answer should be C.

HTH

Answer should be C - 136 will be the 3 digit number.

One more mistake I observed in previous posts is that they have taken x as 0
x cannnot be 0 bcos xyz is a 3 digit number.

(1) alone is insufficient as XYZ can be 127 or 136 (for example).

(2) alone is insufficient as XYZ can be 712 or 136.

(1) + (2):

As kvcpk points out, X can't be 0 because then XYZ would be a 2-digit number. So, the smallest x can be is 1. We can't make X too large (in fact, X must be 1), because we need each of Y and Z to be larger, and the sum is only 10. You can quickly pick numbers or use odd/even sum rules to conclude that the only permutation that works is 136.

Choose C.

1) It can be 127 or 235 for example - insuff.
2) _ _ 2/4/6/8
Now the last digit can't be 8 since x+y cannot be 2 (since they are distinct digits). It can be 136 or 316 or 514 etc etc. - insuff.

1&2: no. can't be anything other than 136

Hence, C