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
What is z
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Source: Beat The GMAT — Data Sufficiency |
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cramya
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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!
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!-
dmateer25
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ahhh it says DIFFERENT positive integers! Now I see, thanks cramya!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!
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ronniecoleman
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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
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dmateer25
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Mozartain wrote:Hi Cramya,cramya wrote: since y is 1 the sum of b+e has to be 11.
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.
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Mozartain
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probably i didn't frame the question very well. Let's think of it this way -dmateer25 wrote:Mozartain wrote:Hi Cramya,cramya wrote: since y is 1 the sum of b+e has to be 11.
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.
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.)
