Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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
_________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

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]

Show Tags

21 Dec 2012, 08:20

2

This post received KUDOS

1

This post was BOOKMARKED

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]

Show Tags

27 Jun 2013, 12:10

1

This post was BOOKMARKED

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]

Show Tags

26 Nov 2013, 09:21

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]

Show Tags

01 Aug 2014, 02:55

1

This post received KUDOS

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
_________________

......................................................................... +1 Kudos please, if you like my post

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

Show Tags

02 Aug 2014, 10:52

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
_________________

I'm on 680... 20 days to reach 700 +

gmatclubot

Re: If, in the addition problem above, a, b, c, d, e, f, x, y, a
[#permalink]
02 Aug 2014, 10:52

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...