DS Problem: ax + ay = 15

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DS Problem: ax + ay = 15

by Alpha800 » Sat Aug 02, 2008 2:29 pm
Hello All. Please be gentle as this is my first post in asking a problem question.

This is probably a very simple DS question from Princeton's Cracking the GMAT book, but I'm just not seeing the solution given by Princeton.
If ax + ay = 15, what is x + y + z?

1) x = 2
2) a = 5


Actual answer is suppose to be (B) but I don't see how one could obtain the value of z, and thus the ultimate answer to "what is x+y+z?" I want to answer (E).

Please expound on why the correct answer should be B.

Thanks!

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by Ankush2008 » Sat Aug 02, 2008 2:47 pm
Hi Alpha

Its a typo on Princeton's part. I saw two questions in OG that had a typo. So relax...no way in hell can the solution be known if Z is not given in the question.

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by Alpha800 » Sat Aug 02, 2008 3:30 pm
Ankush2008 wrote:Hi Alpha

Its a typo on Princeton's part. I saw two questions in OG that had a typo. So relax...no way in hell can the solution be known if Z is not given in the question.
Hi Ankush2008, well here is the official explanation from Princeton why the answer should be (B) but I don't see it. Take a look:
Statement (1) gives us a value of x, but we need x + y + d. Statement (1) is not sufficient. We're down to BCE. Statement (2) might not have seemed much more helpful, BUT using the distributive property, we can rewrite the original equation to read a (x+y+z) = 15. If a is 5, then x + y + z must equal 3. The correct answer is choice B.
This is an exact quote, including what I believe to be a typo highlighted in red above. I don't know why they suddenly introduced the variable "d" in their solution, but I believe they're actually referring to the variable "z" above. Nevertheless, does their solution make sense?

I'm having a very difficult time with DS problem sets in general because I've never had to solve DS problems before--like most people--and it's just very strange. I'm always trying to solve the whole problem.

Anyway, given their solution above, do you still think it's a typo? Or is their answer B correct?

Thanks.

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by newera » Sat Aug 02, 2008 5:40 pm
Based on the answer, it looks like the question should have been written as "ax+ay+az=15" because in the official answer, they write a(x+y+z)=15 and this is only possible if the question was written as shown above.

I don't know what they did in statement (1) in their official answer. Looks funky to me. I think its safe to assume that the question was just written wrong.

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by Alpha800 » Mon Aug 04, 2008 11:44 am
newera wrote:Based on the answer, it looks like the question should have been written as "ax+ay+az=15" because in the official answer, they write a(x+y+z)=15 and this is only possible if the question was written as shown above.

I don't know what they did in statement (1) in their official answer. Looks funky to me. I think its safe to assume that the question was just written wrong.
Thanks for your comments.

These books indeed have quite a few typos/errors. I found a couple of other wrong answers and typos. Bad bad bad! It makes an already confusing situation even more confusing. :(

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The question in this problem is the following:

If ax + ay = 15, what is x + y + z?

(1) x = 2
(2) a = 5

Given this information it is reasonable to assume that we do not need to simplify for all a, x, y, and z. Therefore, we may rewrite ax + ay = 15 as a(x + y) using the associative property of multiplication. Assuming that we can ‘tack on’ z in this equation (regardless of positive or negative assumed values) by using the commutative property of addition and/or multiplication. This leaves us with a(x + y + z) [to verify both properties you can google the associative and commutative properties or visit www.wikipedia.org for more information] . Now assuming we are solving for Data Sufficiency with the following possible answers as our premise:

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement alone is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.

we may use the pre-defined Process Of Elimination (POE) to find our answer. This shouldn’t be too hard as we simply attempt to solve our inequality (hopefully this is the correct terminology) with each statement one by one to try to solve for our data sufficiency using Boolean logic (https://en.wikipedia.org/wiki/Truth_table)

So using (1) as our first substitution method, we derive the following:

[please remember the associative and commutative properties]

ax + ay = 15
a(x + y + z) = 15
a(2 + y + z) = 15

As we can see we still have 3 variables that cannot clearly equate to a set of x + y + z = 15 because we still have ‘a’ as a variable that has not yet been initialized to a value. REMEMBER all we need to do is just find a way to set ‘x + y + z = 15’ and not initialize a value for all x, y, and z. The question NEVER asked us to initialize any values. Therefore with (1) defining x, we have only just done that: initialized x with a value and now have another variable a unsolved for in our inequality. We still need to meet our objective of setting ‘x + y + z = 15’.

Using the POE we can reasonably remove A and D. We know this simply because with statement (1) not allowing us to meet our objective of finding ‘x + y + z = 15’; both A and D cannot be met. If you cannot understand why please see the aforementioned wikipedia Boolean logic link for further information in regards to the truth table and the AND operator (meaning that both have to be true in order for the inequality to equate to true for more than one Boolean variable with the exclusion of two falses-E).

This leaves us with B, C, and E as our possible answer because we have logically excluded both A and D. Right away we can also eliminate C since we were unable to meet our objective of setting x + y + z = 15 with statement (1). Now we have eliminated all of A, C, and D. We now only have B and E and our potential set of answers.

Finally, if we exclude statement (1) and we use statement (2), we get the following:

[please remember the associative and commutative properties]

ax + ay = 15
a(x + y + z) = 15
5(x + y + z) = 15
Divide both sides by 5

[if we had distributed there would be a coefficient of 5 in front of all of our variables]

x + y + z = 3.

And we have now met our objective. We can now exclude E as we have found at least one sufficient source. Our answer is clear, B!

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by shargaur » Tue Mar 03, 2009 1:26 pm
Either the prob mentioned is wrong.

but i couldnt derive how you come to this conclusion

a(x+y)=15
a(x+y+z) =15....are you assuming z = 0..

...
Gaurav

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by quocbao » Tue Mar 03, 2009 8:01 pm
I think it should be E, or the problem is asking x + y

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Agreed it should be E

by carlosluis » Sat Sep 12, 2009 2:03 pm
Exentrick1, how does the z magically gets introduced after you factor a ?

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Re: Agreed it should be E

by Ian Stewart » Sat Sep 12, 2009 4:42 pm
carlosluis wrote:Exentrick1, how does the z magically gets introduced after you factor a ?
There are serious problems with the original question, and with the 'solution' posted above. Obviously you cannot 'magically introduce' a new letter into an equation, and the answer to the original question above has to be E. It's best to simply ignore the question and commentary above, and move on to problems from higher quality sources.
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