tricky DS question

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tricky DS question

by fibbonnaci » Sun May 16, 2010 9:40 pm
If pqrst= 4, then is p= (1/q) ??

1) r=s=t

2) three of p,q,r,s,t are integers

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by kstv » Mon May 17, 2010 12:10 am
Combining 1 and 2
r=s=t or rst = r^3
3 of pqrst are integers , has to be r,s and as they are =
pqr^3 = 4
pq= 4/r^3
r^3 can never be 4
so p cannot be 1/q
IMO E.

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by harshavardhanc » Mon May 17, 2010 12:28 am
kstv wrote:Combining 1 and 2
r=s=t or rst = r^3
3 of pqrst are integers , has to be r,s and as they are =
pqr^3 = 4
pq= 4/r^3
r^3 can never be 4
so p cannot be 1/q
if we're getting a definite NO, won't the answer be C in that case.
Regards,
Harsha

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by fibbonnaci » Mon May 17, 2010 1:00 am
and moreover, what is the reason behind directly combining both the statements?? why is that you negate 1 and 2??

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by sanju09 » Mon May 17, 2010 1:39 am
fibbonnaci wrote:and moreover, what is the reason behind directly combining both the statements?? why is that you negate 1 and 2??
Absolutely!
If pqrst= 4, then is p= (1/q) ??

1) r=s=t

2) three of p,q,r,s,t are integers
(1) If r = s = t = k, which is either positive or negative, then p q = 4/k^3, and for p = 1/q be true, p q = 1.

Two questions now,

First one being:

Is 4/k^3 equal to 1, which we don't know or

The second one being:

Can 4/k^3 be equal to 1? Well, the answer is NO for a rational k, but yes for an irrational k.

Insufficient

(2) When three of p, q, r, s, and t are integers, many possible integers n in the range -2 < n < 2, with permissible repetitions are possible for the three of p, q, r, s, and t, leading to no fixed status of p and q.

Insufficient

Take them one now


If r = s = t = k, which is either positive or negative, and three of p, q, r, s, and t are integers, then at least one of r, s, and t is an integer, and if it's so, then k is an integer, and if it's really so then no integer has its cube equal to 4, hence [spoiler]p is definitely NOT equal to 1/q.

C
[/spoiler]
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by kstv » Mon May 17, 2010 2:03 am
harshavardhanc wrote:
kstv wrote:Combining 1 and 2
r=s=t or rst = r^3
3 of pqrst are integers , has to be r,s and t as they are =
pqr^3 = 4
pq= 4/r^3
r^3 can never be 4
so p cannot be 1/q
if we're getting a definite NO, won't the answer be C in that case.
Spot on.
It's a silly mistake on my part.

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by Haldiram Bhujiawala » Tue May 18, 2010 8:51 am
IMOC

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by im2crazyin » Mon Jan 31, 2011 4:35 pm
it should be C

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by Night reader » Mon Jan 31, 2011 5:44 pm
fibbonnaci wrote:If pqrst= 4, then is p= (1/q) ??

1) r=s=t

2) three of p,q,r,s,t are integers
Is pq=1 ?

given pq * rst = 4; if rst=4 then pq=1

st(1) r=s=t, we don't know if r^3=4 Not Sufficient
st(2) r,s,t integers then pq may and may not be 1, as rst={1*1*1, 1*2*2,...} Not Sufficient

Combining st(1&2) r=s=t, when r=1 --> pq*1=4 False; when r=2 --> pq*8=4 False Sufficient

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by cyrwr1 » Wed Feb 02, 2011 4:39 pm
I am going with C on this on as with
1) you get p=4/qr^3. tells nothing
2)3 of them are integers. tells nothing.

But when you combine them, then you know : rst are those 3 integers that are equal.

Thus,

In order for p=1/q then r^3 =4, as they are integers. No way it is possible. Hence, I select C

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by Ananda » Wed Mar 02, 2011 11:52 pm
Need Statement 2, otherwise we bring Fractions to the party.
Need Statement 1, to restrict all kinds of integers to the party too.

Therefore to answer
p= (1/q)
we would need both.

Hence C.

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by pankajks2010 » Mon Apr 11, 2011 11:16 am
Statement 1: r=s=t
the given statement is pqrst=4; pqr^3=4; pq=4/r^3
Now, as there is no integer or fraction whose cube is equal to 4 our RHS in the equation (pq=4/r^3) can never be 1. thus, pq is not equal to 1. Thus, p is not equal to 1/q. Statement 1 is sufficient

Statement 2: three of p, q, r, s & t are integers. Now, this can be any of the three integers. Thus, insufficient to point out whether pq=1 or not. To illustrate:
a) Let p=32, q=1, r=1/2, s=1/4, t=1 this would satisfy pqrst=4. However, pq is not equal to 1.
b) Let p=1/2, q=2, r=1, s=1, t=4 this would satisfy pqrst=4 and pq=1
Thus, statement 2 is not sufficient.

Thus, the answer is A

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by GMATGuruNY » Mon Apr 11, 2011 1:22 pm
fibbonnaci wrote:If pqrst= 4, then is p= (1/q) ??

1) r=s=t

2) three of p,q,r,s,t are integers
If p = 1/q, then pq = 1 and rst = 4.

Question rephrased: Does rst = 4?

Statement 1: r=s=t.
If r=s=t=4^(1/3), then rst = 4.
If r=s=t=1, then rst ≠ 4.
Insufficient.

Statement 2: Three of p,q,r,s,t are integers.
No way to determine whether rst = 4.
Insufficient.

Statements 1 and 2 combined:
Given statement 1, rst = 4 only if r=s=t=4^(1/3).
Given statement 2, at least one of r, s and t must be an integer.
Thus, it is not possible that r=s=t=4^(1/3), so we know that rst ≠ 4.
Sufficient.

The correct answer is C.
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by GmatFred » Fri Jul 08, 2011 3:25 am
Perfect explanation, thanks Mitch.

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by ronnie1985 » Thu Mar 29, 2012 7:38 am
S1 and S2 are alone not sufficient here as there is absolutely no clue to prove or disprove p = 1/q.
Combining, pq*r^3 = 4 => pq = 4/r^3. No way can this be 1. As if r,s and t are integers, pq cannot be 1.
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