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OG - Quant Review DS - #83

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Carla
 Post subject: OG - Quant Review DS - #83
 Post Posted: Mon May 14, 2007 8:26 pm 
Question:

If k and n are integers, is n divisible by 7?

(1) n-3 = 2k

(2) 2k-4 is divisible by 7.

The answer is C.

From statement (1) I was able to determine that n is not divisible by 2 or any multiple of 2 because

n = 2k + 3

If for example n = 2k the we would know that n is divisible by 2. In this case we can see that n is not divisible by 2 since there is the +3 which is the remainder.

So just because n is not even we still can not determine if n is divisible by 7. I was able to eliminate A and D.

From statement (2) I got a bit stuck because we know that (2k - 4) / 7 = integer

I was not sure where to go from here....

Any feedback would be great!! Thanks,
Carla
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Carla
 Post subject: Question follow up
 Post Posted: Mon May 14, 2007 8:44 pm 
I was going over this with a friend and he showed me the following technique.. but I just wanted to make sure that this is actually correct...

n - 3 = 2k

(2k - 4) /7 = b --> 2k-4 = 7b

where b is some integer

then we combined these two equations by adding the left to the left and the right to the right -->


n-3 + 2k - 4 = 2k + 7b ---> (can we just do this? what is this based on?)

then we eliminate the 2k on both sides and simplify:

n-7 = 7b

n = 7b + 7

n = 7(b+1)

and so we can see that n must be divisible by 7...

I just wanted to check this answer so any feedback would be great.. also this is different from the solution given in the book which I did not really understand...

Thanks!!

Carla
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Saurabh Malpani
 Post subject: Re: Question follow up
 Post Posted: Tue May 15, 2007 12:22 am 
Carla,

The technique is correct----

The other way to look at it is

n - 3 = 2k --> n=2k+3---(1)

(2k - 4) /7 = b --> 2k-4 = 7b ----> 2k=7b+4 --(2)

Now replace 2k from eq 2 in eq 1

You get n=7b+4+3 which is nothing but 7b+7 --> 7(b+1)

Well this is how I apporached the problem!

Saurabh Malpani


where b is some integer

then we combined these two equations by adding the left to the left and the right to the right -->


n-3 + 2k - 4 = 2k + 7b ---> (can we just do this? what is this based on?)



Carla wrote:
I was going over this with a friend and he showed me the following technique.. but I just wanted to make sure that this is actually correct...

n - 3 = 2k

(2k - 4) /7 = b --> 2k-4 = 7b

where b is some integer

then we combined these two equations by adding the left to the left and the right to the right -->


n-3 + 2k - 4 = 2k + 7b ---> (can we just do this? what is this based on?)

then we eliminate the 2k on both sides and simplify:

n-7 = 7b

n = 7b + 7

n = 7(b+1)

and so we can see that n must be divisible by 7...

I just wanted to check this answer so any feedback would be great.. also this is different from the solution given in the book which I did not really understand...

Thanks!!

Carla
Top 
esledge
 Post subject:
 Post Posted: Tue May 15, 2007 6:16 pm 
Offline
ManhattanGMAT Staff


Posts: 901
Location: St. Louis, MO
Quick citation note: This is from the Official Guide for GMAT Quantitative Review. For copyright reasons, we must cite full source name.

Hi Carla,

I think Saurabh pretty much covered it (thanks!), but I wanted to chime in about one thing. You asked "can we just do this?" about adding the equations from statements (1) and (2). Yes, you definitely can. It's just as acceptable as combining any two equations to solve. Consider this simpler example:

(1) x + y = 5
(2) 2x - y = 4

combine to give:
3x + 0y = 9
x = 3

Your process here was exactly the same. The only difference is that your equations for this divisibility problem had a variable that you made up (the "b is any integer" step). That's probably what made it look weird to you.

_________________
Emily Sledge
Instructor
ManhattanGMAT
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None
 Post subject:
 Post Posted: Sun Jun 10, 2007 4:11 pm 
I actually did it slightly differently. Is this still "Kosher"? I took n-3 = 2k and made it k= (n/2) - (3/2). Then subbed it into the 2nd equation as:

2( (n-3)/2 ) + 3

This boils down to just "n:, and since we no that 2k-4 is divisible by 7 we now know that n is divisible by 7 as well.

Is this correct as well?
Top 
dbernst
 Post subject:
 Post Posted: Mon Jun 11, 2007 1:48 pm 
Offline
ManhattanGMAT Staff


Posts: 304
Koser like bacon and sausage :) If I am correctly interpreting your work, you have just simplified the first statement to n = n, which doesn't really help you with the remainder of the problem.

My suggestion is to really focus on you DS process:
Step 1: Consider rephrase of question
-no good rephrase in this case

Step 2: Choose easier statement
-Statement (2) 2k-4 is divisible by 7 is easier, since there is no mention of n. Because statement (2) does not mention n, eliminate BD from BD/ACE grid.

Step 3: Rephrase statement (1)
-Statement (1) n-3 = 2k, so n=2k+3
-Since n depends on k we do not know whether n is divisible by 7. Eliminate A from ACE grid

Step 4: Combine statements
-since 2k=n-3 (1), and 2k-4 is div by 7 (2), n-3-4 is divisible by 7 (I simply substituted n-3 from (1) for 2k (2). Thus n-7 is divisible by 7, so n must be a multiple of 7.

The correct answer is C

Hope that helps!
-dan

Quote:
I actually did it slightly differently. Is this still "Kosher"? I took n-3 = 2k and made it k= (n/2) - (3/2). Then subbed it into the 2nd equation as:

2( (n-3)/2 ) + 3

This boils down to just "n:, and since we no that 2k-4 is divisible by 7 we now know that n is divisible by 7 as well.

Is this correct as well?


Last edited by dbernst on Tue Jun 12, 2007 6:00 pm, edited 1 time in total.

Top 
GMATPaduan
 Post subject: Another alternative
 Post Posted: Tue Jun 12, 2007 3:26 pm 
Another way to solve the problem: Plugging in Numbers

Is n divisible by 7?
(1) n-3 = 2k

n= 2k+3
if k = 2 yes, if k =3 -no --Not Sufficient

(2) 2k-4 is divisible by 7
Says nothing about N -- not sufficient

Combination:
If 2k-4 is divisible by 7-- k = 9, 16, 23, etc
n=2k+3
2 * any of the above (9, 16, 23) - 4 is divisible by 7

C is sufficient
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dbernst
 Post subject: Plugging In
 Post Posted: Wed Jun 13, 2007 2:25 pm 
Offline
ManhattanGMAT Staff


Posts: 304
GMATPaduan,

You are correct. One of the primary mistakes that students make on algebraic problems throughout the Quant section is to search only for the algebraic solutions. Often, students do not readily "see" an algebraic approach, so they then waste several minutes searching for one. Your "brute force" method of plugging in numbers can be just as effective on many difficult algebraic problems. Remember, your goal is simply to answer the question correctly in approximately 2 minutes. The computer (and thus your GMAT score) doesn't care how you arrived at your answer, just that you got there!
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GMATPaduan
 Post subject: Correction
 Post Posted: Wed Jun 13, 2007 3:05 pm 
I have a mis-print: The last part of my plugging in Numbers Solution should be:

Combination:
If 2k-4 is divisible by 7-- k = 9, 16, 23, etc
n=2k+3
2 * any of the above (9, 16, 23) +3 is divisible by 7

C is sufficient
Top 
Kris
 Post subject: No Algebra
 Post Posted: Wed Sep 19, 2007 2:17 am 
combining I & II

From statement II , 2K- 4 is divisible by 7

for statement I n=2k + 3=(2k- 4) + 7

So n will also be divisible by 7.



in simple terms,

if x is divisible by 7, then x+ 7, x + 14 , x+ 21 ..... are all divisible by 7;
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