May be rushed in to it. I am going with A
Stmt II

f-c = 3

f ->9 c->6
f->8 c->5

INSUFF

Stmt I

3a = f = 6y

a can be 1 ,2 or 3 with f being 3,6,9

But since we are given f = 6y therefore y has to be 1. a=2 f=6

SUFF

273
546

819

Dmateer,
Whats the OA?

cramya wrote:Dmateer,
Whats the OA?

I added the OA with a spoiler to the original post.

I meant to say A) since I did stmt I second mixed up my choice.

I have edited it. Crap got to watch out for this on the GMAT i.e if I do stmt I second need to make sure I mark A and not B

I got to this point

2bc
de6

x1z

Then I got stuck.

2bc
de6

x1z

Then I got stuck.

Let me rethink how I did and try to explain the best I can.

This is how I proceeded from here

since y is 1 the sum of b+e has to be 11. 7+4 = 11 or 8+3 =11 (6+5 is also 11 but we have used 6 already in a different place)


2bc
de6

x1z

Case 1

b->8 e->3

Digits available : 0,1,2,3,4,5,6,7,8,9

Digits used if I took b->8 e->3 would be

1,2,3,6,8

Digits remaining-> 0,4,5,7,9

There will be a 1 carryover from b+e addition so carryover 1+ 2(a) + some one digit number(d) has to be x. Only 4 and 7 possible
but the rest of the addition process given in the problem will fall apart with 0,5,9 remaining.

Eliminate. Not possible


Case 2

This has to be it
b->7 e->4


P.S:

I remembered Stuart's golden words that in a DS question there will be something common between the 2 choices for the variables involved (i.e they will never contradict each other). For eg: If one choice said z is negative the other choice will not say z is positive

Since f-c=3 also matched my case 2 I confidently went with A)


May be there is a better way to decipher this! Good problem!

cramya wrote:This is how I proceeded from here

since y is 1 the sum of b+e has to be 11. 7+4 = 11 or 8+3 =11 (6+5 is also 11 but we have used 6 already in a different place)


2bc
de6

x1z

Case 1

b->8 e->3

Digits available : 0,1,2,3,4,5,6,7,8,9

Digits used if I took b->8 e->3 would be

1,2,3,6,8

Digits remaining-> 0,4,5,7,9

There will be a 1 carryover from b+e addition so carryover 1+ 2(a) + some one digit number(d) has to be x. Only 4 and 7 possible
but the rest of the addition process given in the problem will fall apart with 0,5,9 remaining.

Eliminate. Not possible


Case 2

This has to be it
b->7 e->4


P.S:

I remembered Stuart's golden words that in a DS question there will be something common between the 2 choices for the variables involved (i.e they will never contradict each other). For eg: If one choice said z is negative the other choice will not say z is positive

Since f-c=3 also matched my case 2 I confidently went with A)


May be there is a better way to decipher this! Good problem!

ahhh it says DIFFERENT positive integers! Now I see, thanks cramya!

dmateer25 wrote:This is a pretty good one.




If, in the addition problem above, a, b, c, d, e, f, x, y, and z each represent different positive single digits, what is the value of z ?

(1) 3a = f = 6y

(2) f – c = 3

OA is A

3a = f = 6y

so y =1
a = 2
f =6

i am using hit and trial to get to this,,,

IMO A

cramya wrote: since y is 1 the sum of b+e has to be 11.

Hi Cramya,

Don't you think b+e can also be 10? What am I missing that says there is no carryover from the unit digits' sum?

Mozartain wrote:

cramya wrote: since y is 1 the sum of b+e has to be 11.

Hi Cramya,

Don't you think b+e can also be 10, or 9 or 8 or some other number for that matter? What am I missing that says there is no carryover from the unit digits' sum?

It can't be any of those because we know the tens digit is a a 1. In order for it to be a 1 it has to be 11. If it was 10 the tens digit would be 0.

dmateer25 wrote:

Mozartain wrote:

cramya wrote: since y is 1 the sum of b+e has to be 11.

Hi Cramya,

Don't you think b+e can also be 10, or 9 or 8 or some other number for that matter? What am I missing that says there is no carryover from the unit digits' sum?

It can't be any of those because we know the tens digit is a a 1. In order for it to be a 1 it has to be 11. If it was 10 the tens digit would be 0.

probably i didn't frame the question very well. Let's think of it this way -
if c is, say, 4, or 5 or 7 or 8 or 9, then you need to add carryover 1 to b+e. In such a case, b+e=10 for y to be 1. (However, b+e cannot be less than 10, as that would require a carryover of more than 1.)