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LCD problem

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by biker317 » Mon Dec 14, 2009 3:37 am
Given the fraction p/q, if p and q are positive integers and p<=6, what is the value of q?

1) The lowest common denominator of p/q and 1/8 is 24.
2) The sum of p/q and 1/8 is p/6 .


(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. your answer correct
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.



I need a real quick way of solving this..Can someone take a shot?
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Source: — Data Sufficiency |
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by biker317 » Mon Dec 14, 2009 3:38 am
The OA is C
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by mehravikas » Mon Dec 14, 2009 7:26 pm
Tough one..not able to nail this one...!! Statement 1 and 2 are not sufficient individually. No clue after that.
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by Testluv » Mon Dec 14, 2009 8:43 pm
Choose C.
Last edited by Testluv on Mon Dec 14, 2009 9:20 pm, edited 1 time in total.
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by viju9162 » Mon Dec 14, 2009 8:54 pm
I was able to solve it .. but took lot lot lot of my sweet time :). I'm not sure about my solution.. please do comment on it..

By the facts, P and Q are +ve integers and P can be either 1,2,3,4,5,6...

From (1) The lowest common denominator of p/q and 1/8 is 24.

LCD means the LCM of denominators. Hence, LCM of q and 8 is 24. Let us deduce this:

8 = 2*2*2
Q = can be any value ( such as 3,6,12,24)

From (1), it cannot be determined.

From (2), The sum of p/q and 1/8 is p/6

I took lot of time deducing it.. but this can be reduced..

p/q + 1/8 = p/6
=> p/q - p/6 = -1/8
=> p ( 1/q-1/6) = -1/8

I started substituting values for q here .. with p as 1.

with q=24, and p=1.. on the left hand side I get the value as -1/8..

1*( 1/24-1/6) = -1/8

Therefore LHS=RHS (-1/8 = -1/8).

From both (1) and (2), value of q = 24.

Regards,
Viju
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by Testluv » Mon Dec 14, 2009 9:31 pm
As viju points out, from 1, q is either 3, 6, 12, or 24.

Insufficient.

From 2, there is only one equation but two unknowns. Insufficient.

1 + 2:

We know from the stem that p and q are positive integers.

If q = 3, then, subbing into 2:

p/3 + 1/8 = p/6

or

2p/6 + 1/8 = p/6 or p/6 = -1/8, which will result in p being negative. So, q cannot be 3.

If q = 6, then:

p/6 + 1/8 = p/6 or 1/8 = 0, which is clearly impossible. So, q cannot be 6.

If q = 12, then:

p/12 + 1/8 = p/6 or p/12 + 1/8 = 2p/12 or p/12 = 1/8, which results in p being a non-integer.

Therefore, q must equal 24.

When checking the different values of q, the process is sped up if you are actively checking to see whether p will end up being negative or fractional; as soon as you see that (or that the equation is nonsensical as when q =6), then you don't need to complete the algebra on that check.

Viju, I think you could have saved some time by not manipulating the equation from 2. Since you know you are going to test values for q anyways, it is quicker to just start plugging into the equation. Also, we can't assume any value for p. You know that q has only four values, so just plug those directly into 2, checking to see if any of the information about p (ie, that it is a pos integer) gets violated.
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by viju9162 » Mon Dec 14, 2009 11:09 pm
Hi Testluv,

I realized my mistake. Thank you for pointing it out :).

Regards,
Viju
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by Ian Stewart » Tue Dec 15, 2009 5:15 am
biker317 wrote:Given the fraction p/q, if p and q are positive integers and p<=6, what is the value of q?

1) The lowest common denominator of p/q and 1/8 is 24.
2) The sum of p/q and 1/8 is p/6 .
Statement 2 is almost sufficient alone; because we have a stringent restriction on our unknowns (p must be an integer between 1 and 6 inclusive, and q must be a positive integer) one cannot count equations and unknowns here, since in many cases, there will only be one set of integers that satisfy a given equation. We can find a common denominator, then cross multiply and solve for q:

p/q + 1/8 = p/6
(8p + q)/8q = p/6
48p + 6q = 8pq
24p = 4pq - 3q
24p = q(4p - 3)
q = 24p/(4p - 3)

So 4p-3 must be a factor of 24p, and we know that 1 < p < 6. The only possible values of p are 1 and 3, and the only possible values of q are thus 24 and 8 respectively. So to decide which value q has, we need Statement 1; the answer is C.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by biker317 » Tue Dec 15, 2009 6:06 am
as some of u guys pointed out q=3,6,12,24 so according to testluv he gets

"If q = 3, then, subbing into 2:

p/3 + 1/8 = p/6

or

2p/6 + 1/8 = p/6"


Im not sure how he gets 2p/6 shouldnt it be p/6?
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by biker317 » Tue Dec 15, 2009 6:09 am
Also whats a great stratergy on guessing b/w C and E...is therea rule of thumb? ....After a minute and a half of working this I probably know that neither A or B is suff...any tips? Im a real slow coach, the last thing I need is to spend 4 minutes on this :)
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by Testluv » Tue Dec 15, 2009 12:51 pm
biker317 wrote:as some of u guys pointed out q=3,6,12,24 so according to testluv he gets

"If q = 3, then, subbing into 2:

p/3 + 1/8 = p/6

or

2p/6 + 1/8 = p/6"


Im not sure how he gets 2p/6 shouldnt it be p/6?
p/3 = 2p/6

just as 5/10 = 10/20
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by biker317 » Tue Dec 15, 2009 11:29 pm
ok, i mis read I thought u were substituting for q=3,6,..and so on
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