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I am posting this approach though it takes more than 2 min. Hope it is helpful to you.

From 1, we have 3a = f = 6y. The only possible scenario is y=1, f=6, a=2 Hence: 2bc + de6 ------ x1z

From above, we have: c+6=z or c+6=z+10 b+e=1(this is not possible since repetition is not allowed) or b+e=11 if c+6=z+10, then b+1+e=11 2+1+d=x

Thus the possibilities are: c+6=z, b+e=11, 2+1+d=x (i) or c+6=z+10, b+e=10,2+1+d=x (ii)

From i, c+6=z, then c=0,1,2,3. c cannot be 0,1,2 (no repetition) hence c=3, z=9 b+e=11, the only solution is b and e are 4,7 or 7,4. 3+d=x, the only solution is 5 and 8.

From ii. c+6=z+10 The possibilities are c z 4 0 5 1 etc If c=4 and z=0, b+e=10 thus b and e are 3 and 7 3+d = x thus d and x are 5 and 8.

So we have 2 solutions for Z z= 9 and z = 0 Thus 1 is not sufficient.

Now from 2, we have f – c = 3 f c 3 0 4 1 5 2 6 3 7 4 ... etc 3 0 cannot be the case since z will be 3 and we will have a repetition. Considering 4 1, z will be equal to 5 and 291+384=675 is valid solution. Considering 5 2, z will be equal to 7 and 395+412=807 is valid solution. Hence we have 2 values of z and therefore it is not sufficient.

Taking 1 and 2 together, we have c=3, z=9 and f=6 and therefore taking 1 and 2 together it is sufficient to answer the question.

Sorry for the mess-up and I hope this was valuable to you.

now c+f=z----> for z to be a positive single unit c has to be 0 or 3 ( 1 and 2 already ocuupied) and for c>3 z= two digit no. and hence violate the given condition

now if c=0 z=f which is not possible since f and z are diff

therefore c is 3 and z is 9

suff

s2) insuff
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The Reasoning is wrong because if C>3 the +1 would be transfer to the next column and z would be single unit. The correct way is Step 1- From 1st statement we can fix the values of a=2,f=6 and y=1,After that the remaining numbers are 3,4,5,7,8,9.We check these values for C.

Step-2 4,5 can easily removed because Z cant be equal to 0 or 1. If we put c=7 then z=3....... and 'b+e' should be equal to 10 which is not possible because remaining numbers are (4,5,8,9) If we put C=8 then z=4..........remaining numbers are (3,5,7,9),b+e=10 by choosing 3 and 7,..........remaining numbers (7,9) which are not satisfying d and x. if we put c=9 then z=5,..........remaining numbers are (3,4,7,8),b+e=10 by choosing 3 and 7,...........remaining numbers (4,8) which are not satisfying d and x.

But when we put c=3 then Z=9,remaining numbers are (4,5,7,8) and b+e=11 here so we can choose 7 and 4,remaining numbers are (5,8) which satisfies d and x.

I just came across this question in the MGMAT question banks and I think that ytarun is correct in that you can't rule out c=7,8,or 9 on the basis that z is a single positive digit, which the official explanation seems to do. (If the question didn't mention "positive," then c=4, z=0 is another possibility.)

Is there a faster way of ruling out c=7,8,9 other than plugging it in and seeing if the numbers work out? It seems like a rather slow way of doing it but I can't see any other option.

St1: 3a=f=6y => f must be a multiple of 2*3 but if we multiply 2*3 with any number other than 1, value of f will be > 9, which is not allowed. so, f=6; a=2 and y=1.

now, from the given condition c+f=z => (9-0) + 6 = z, from the set for c we can rule out values of 0, 1, 2 (as 1 and 2 are already taken and 0 will make z=6, which is not allowed as f=6) so, c can be any value from 9,8,7,5,4 => Not Sufficient

St2: f-c=3, alone this statement is Not sufficient.

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

How to solve this in under 2 mins ??

Took me >5 mins to prove "A's" sufficiency, yet I fear that I may be missing something. Hope this is not from GPrep!!!

a b c + d e f ----- x y z -----

(1) 3a = f = 6y

y=1; f=6; a=2

2 b c + d e 6 ----- x 1 z -----

Now, y=1 must be the unit's digit of 11 with 1 being carried over to the hundred's place.

So, 2+d+1=x; here "1" is the carried over 1 from y. 3+d=x x can't be 4 as d can't be 1. x can't be 5 as d can't be 2. x can't be 6. x can be 7 if d=4 x can be 8 if d=5 x can't be 9 as d can't be 6.

Let's try with x=8, d=5 2 b c + 5 e 6 ----- 8 1 z -----

1,2,5,6,8---used z can be 3,4,7,9 z=3, it got to be 3 of 13; So, c=7; "b+e" becomes 9+4=13 and 1 carried over from z=13; so 13+1=14; but y=1 NOT 4. Rule out. z=4, it got to be 4 of 14; c=8(Not possible). Rule out. z=7, c can't be 1. Rule out. z=9, c=3; b+e=11; perfect y's "1" properly fits and 1 carries over. ------- This is first case so far.

Let's try the other scenario:

Let's try with x=7, d=4 2 b c + 4 e 6 ----- 7 1 z -----

1,2,4,6,7 used up 3,5,8,9 left. z=3; it must be 3 of 13; c=7(Not possible). Rule out z=5; it must be 3 of 15; c=9. b+e+1(carried over)=8+3+1=12; but y=1. Rule out. z=8; c=2(Not possible). Rule out. z=9; c=3. b+e=13; but y=1. Rule out.

Thus, only one scenario fits that gave us z=9. Sufficient.

(2) f – c = 3 a b c + d e f ----- x y z -----

f=4, c=1, z=5 2 9 1 + 3 8 4 ----- 6 7 5 -----

And we already know from 1 that z=9 Not Sufficient.

z = 9 and c = 3 z = 8 and c = 2 ........ this cant be true as 'a' is already 2 z = 7 and c = 1 ........ this cant be true as 'y' is already 1.

hence Z =9 and X = 3... perfect.

As I said in my previous post, I don't think you can ignore that possibility that c=7,8, or 9 which would make z= 3, 4, or 5. Thinking of that values that only fit 6 = z - c is not accurate because c + 6 can be over 10 and the 1 would carry over to the tens digit.

Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]

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21 Dec 2012, 07:20

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The key is to keep in mind that the digits are DISTINCT

c + f = z ;; Constraint => z < 9

b + e = y ;; Constraint => y < 9

a + d = x ;; Constraint => x < 9

z = ?

1) 3a = f = 6y y cannot be greater than 1 as it would make f needs to be a single distinct digit. y cannot be zero as then a = f = y = 0 is NOT possible as digits are distinct.

Hence y = 1, f = 6 , a = 2

We know z<9 = c + f = c + 6 => c<3 Since digits are DISTINCT and y =1 & a = 2 are already taken values. c = 0 or c = 3

now c+f=z----> for z to be a positive single unit c has to be 0 or 3 ( 1 and 2 already ocuupied) and for c>3 z= two digit no. and hence violate the given condition

now if c=0 z=f which is not possible since f and z are diff

therefore c is 3 and z is 9

suff

s2) insuff

I was stuck as hell on this question. I crossed B out because it is clearly NS. And felt that there was a trap answer in C it makes problem to easy. Using your logic I just listed out numbers 1=>9 and ticked off the numbers that have already been used. But forgot that C could not be zero which I was confused about until I reread the problem.

Since the only available numbers left are 0,3,4,5,7,8,9 I crossed out 4 through 9 because they would produce a 2 digit integer. But, through my illogical thinking I forgot that c/=0 and fell for the trap answer C in haste.

The easiest way to answer this problem is to forget about complex math, and just write down all of the digits that satisfy the fact pattern.

What I did was write out the numbers that are available, cross out the numbers that have been used, assuming a=2, f=6, y=1 and then cross out all digits that would cause a 2 digit integer.
_________________

Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]

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27 Jun 2013, 11:10

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I reasoned it out, although it took me 3 min 5 seconds.

We know from A. that Y = 1, and F = 6.

There are TWO conditions: C + 6 = Z and C+6 = 10+Z.

Let's tackle the first condition: C+6 = Z.

We can't have C=1, because we already know that Y is 1. We can't have C = 2, because we already know that A is 2. We can have C = 3, which would make Z = 9.

When Z = 9, that means that B + E has to equal 11, because Y = 1 (and we can't use 0, or 1 again)

Are there still ways left to have B+E = 11? We could use 7 for B, and 4 for E (or vice versa) and that would give us 11. Neither of those numbers have been used yet, so that's valid.

Now let's look at the second condition:

C + 6 = 10 + Z We can't use C = 5, because that would make Z = 1, and 1 is already used by Y. We can't use C = 6, because 6 is already used by F (furthermore, it would make Z = 2, and 2 is already used by A)

We can test C = 7, which would make Z = 3... but we would have to carry over the 10 (from 13) to the tens digit. B + E + 1 (carried over from single digit 6+7=13) = 10+ 1 (which coincides with y = 1) Which means that B+E has to equal 10. Lets examine some of the ways that that would be possible. We can't use 5+5, because that would use 5 twice. We can't use 6+4, because 6 is already being used by F We can't use 7+3, because 7 is already being used in this hypothetical test by C. We can't use 8+2, because 2 is used by A. We can't use 9+1, because 1 is used by Y. We can't use 0 + 1, because we can't use 0...

Which means we can't use C=7.

The same logic applies to C=8. If C=8, Z = 4, and B+E has to equal 10 Can't use 5+5 for B+E, can't use 6+4 for B+E. It's possible to use 7+3 for B+E, and in this case we then have to go to the hundreds digits. If we're using 1,2,3,6,and 8 for the singles column as well as tens column, that means we would have to use 9 in the hundreds column. This is impossible though, because if 9 is in the hundreds column then the answer would have to have four digits rather than three. That same logic applies for the rest of the B+E answer choices if C+F (6) = 10+z

Therefore, the only possible choice left is C=3, which would make Z = 9.

A summary: The "9" option has to be in either the tens or the single digit C+6 < 10, otherwise, because we'd have to carry the 10 over, B+E would have to equal 10, and that's not possible without putting the 9 in the hundreds digit because if we use 9 for B or E, we'd have to use 1 for the other variable, which is impossible since 1 is already used by Y.

Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]

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26 Nov 2013, 08:21

I have a semi related question as to the digit "0" Many people are considering this as a possibility, but when I first read the question that they are all different POSITIVE single digits I excluded 0 from the options. It was my understanding that positive numbers are > 0 and that 0 is neither + or -.

I have a semi related question as to the digit "0" Many people are considering this as a possibility, but when I first read the question that they are all different POSITIVE single digits I excluded 0 from the options. It was my understanding that positive numbers are > 0 and that 0 is neither + or -.

Can someone shed light on this?

Yes, 0 is neither positive nor negative, thus neither of the digits can be zero.

Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]

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01 Aug 2014, 01:55

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mn2010 wrote:

a b c + d e f ----- x y z -----

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

How to solve this in under 2 mins ??

Got wrong answer after 1.5 minute .

Thanks for interesting question I think the most important part of the question is "different positive single digits" when we have y = 1, f = 6, a = 2 we have some options for (c, z) = (3,9); (7,3); (8,4) or (5,9) after trying all options, only (3,9) is correct.

answer A
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a [#permalink]

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02 Aug 2014, 09:52

Took me 2 mins and 22 seconds to solve it though this problem seemed tough to even solve at first.

stmt 2 looks the easiest..lets start with it

f - c = 3 , if f = 4 and c = 1, f - c = 3 and f + c = z = 4 if f = 5 and c = 2, f - c = 3 and f + c = z = 5, 2 values for z, not sufficient eliminate choice B and D

Stmt 1: 3a = f = 6y

now all are DISTINCT POSITIVE SINGLE DIGITS

3a = f => a = f / 3 6y = f => y = f / 6

a, y are positive digits ( 1 to 9) hence they're also single digit integers so f is divisible by both 3 and 6 The smallest positive single digit that is divisible by 6 is 6 ITSELF

so we have f = 6, a = f/3 = 2, y = f/6 = 1

so addition is :

2 b c + d e 6 ----------------- x 1 z

Now 1 , 2 are already used up, and c is distinct positive single digit, so is z

c cannot be greater than or equal to 4 because 4 + 6 would give 10 and we're told z is positve single digit

the only value that fits for c is 3

3 + 6 = 9 = z sufficient. answer is A
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Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a
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02 Aug 2014, 09:52

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